## 1st PUC Economics Question Bank Chapter 6 Measures of Dispersion

Students can Download Economics Chapter 6 Measures of Dispersion Questions and Answers, Notes Pdf, 1st PUC Economics Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## Karnataka 1st PUC Economics Question Bank Chapter 6 Measures of Dispersion

### 1st PUC Economics Measures of Dispersion TextBook Questions and Answers

Question 1.
A measure of dispersion is a good supplement to the central value in understanding a frequency distribution comment.
A measure of dispersion:
A good supplement to central value: A central value condences the series into a single figure. The measure of central tendencies indicate the central tendency of a frequency distribution in form of an average. These averages tell us something about the general level of the magnitude of the distribution, but they fail to show anything further about the distribution. The averages represent the series as a whole.

One may now be keen to know how for the various values of the series tend to dispense from each other or from their averages. This brings us to yet another important brand of statistical methods, viz measures of dispersion. Only when we study dispersion along with average of series that we can have a comprehensive information about the nature and composition of a statistical series.

In a country, the average income or wealth maybe equal. Yet there maybe great disparity in its distribution. As a residt, there of, a majority of people may be below the poverty line. There is need to measure variation in dispersion and express it as a single figure, it can be further explained with an example.

Below are given the family’s incomes of Ram, Rahim and Maria. Ram, Rahim and Maria have four, six and five members in their families respectively.

From the table use come to know that each family have average income of Rs. 15,000

But there are considerable differences in individual methods. It is quite obvious that averages try to tell only one aspect of a distribution i.e. representative size of values, to understand it better, we need to know the spread of values also. The Ram’s family, differences in incomes are comparatively lower.

In Rahim’s family, differences are higher and Maria’s family differences are the highest. Knowledge of only average is sufficient. A measure of dispersion improves the understanding of the distribution series.

Question 2.
Which measure of dispersion is the best and how?
Standard deviation is the best measures of dispersion, because it Posseses’s most of the characterstics of an ideal measures disperrsion.

The Standard Deviation Calculator is a tool to calculate the standard deviation from the data, the standard error, the range, percentiles.

Question 3.
Some measures of dispersion depend upon the spread of values where as some calculate the variation of values from a control value. Do you agree?
Yes, we agree with the statement. Range and quartile deviation measure the dispersion by calculating the spread within which the values lie ie., they depend on the spread of value. On the other hand, mean deviation & standard deviation calaculate the variation of value from a central value.

Question 4.
In a town 25% of the persons earned more than Rs. 45000, where as 76% earned more than Rs. 18000. Calculate the absolute & relative values of dispersion.
1. Absolute value of dispersion ie.
Quartile deviation = $$\frac{\mathrm{Q}_{3}-\mathrm{Q}_{1}}{2}=\frac{45000-18000}{2}$$ = 13500

2. Relative Value of dispersion ie., coefficient of
Quartile deviation =

Question 5.
The yield of wheat and rice per acre for 10 districts of a state is as under:

Calculate for each crop:
i) Range
ii) Q.D
v) Standard Deviation
vi) Which crop has greater variation
vii) Compare the values of different measures for each crop.
WHEAT
i) Range = L – S
= 25 – 9
=16

ii) Qusirtile Deviation (Q.D)
Data in ascending order
9, 10, 10, 12, 15, 16, 18, 19,21,25

Median (M) $$=\frac{N+1}{2}=\frac{10+1}{2}=\frac{11}{2}$$
= 5.5 th item

v) Standard Deviation

vi) Coefficient of Variation
$$\mathrm{C.V}_{\mathrm{w}}=\frac{\sigma}{\mathrm{X}} \times 100$$
= $$\frac{5.044}{15.5}$$ × 100
= 32.54

Rice:

1. Range = L – S = 34 -12 = 22
2. Quartile Deviation Q.D.

Data in ascending order 12, 12, 12, 15, 18, 18, 22, 23, 29, 34

v) Standard Deviation

vi) Coefficient of Variation
C. Vw = $$\frac{\sigma}{\mathrm{X}}$$ × 100 = $$\frac{7.158}{19.5}$$ × 100 = 636.70

Since coefficient of variation of Rice is more, therefore, Rice crop has greater variation.

vii) Comparison of the values of different measures:

Question 6.
In the previous question, calculate the relative measures of variation and indicate the value, which in your opinion is more reliable.
Coefficient of range is the relative measure of range. Hence we will calculate coefficient of range.
Coefficient of Range of wheat.

Coefficient of Range of Rice.

Question 7.
A batsman is to be selected for a cricket team. The choice is between X and Y on the basis of their five previous scores which are:

 X Y 25 50 85 70 40 65 80 45 120 80

Which batsman should be selected if we want

1. a higher run getter or
2. a more reliable batsman in the team?

Batsman – X

Coefficient of S.D. = $$\frac{\sigma}{\bar{X}}=\frac{33.90}{70}$$ = 0.484
Variance = $$\frac{\sum X^{2}}{N}=\frac{5750}{5}$$ = 1150 Runs
Coefficient of Variance = $$\frac{33.91}{70}$$ × 100 = 48.44%

Batsman – Y

Question 8.
To check he quality of two breands of light bulbs, their lie in burning hours was estimated as under for 100 bulbs of each brand.

1. Which brand gives higher life?
2. Which brand is more dependable?

= $$\sqrt{1.8939 \times 50}$$ = 1.376 × 50 = 68.8 hrs
CVA = $$\frac{\sigma A}{\bar{X} A}$$ × 100 = $$\frac{68.8}{134.5}$$ × 100 CVA = 51.15%

AM = 125
C = 50

= 37.32 hrs

CVB = $$\frac{\sigma \mathrm{B}}{\overline{\mathrm{X}} \mathrm{B}}$$ × 100
= $$\frac{37.62}{136.5}$$ × 100 = 27.34%

1. Since the average life of bulbs of Brand B (136.5) is greater that of Brand A (134.5hrs) therefore the bulbs of Brand B gives a higher life.

2. Since C.V. of bulbs of Brand B (27.34%) is less than that of Brand A (51.55), therefore the bulbs of brand B are more dependable.

Question 9.
Average daily wager of 50 worker of a factory was Rs. 200 with a standard deviation of Rs. 40. Each worker is given a raise of Rs. 20, What is the new average daily wage and standard deviation? Have the wages become more or less uniform?
Increase in each worker wages = Rs. 20
Total Increase in wages = 50 × 20 = Rs. 1000
Total of wages before increase worker in wagers = 50 × 200 = Rs. 1000
Total wages after increase in wages = 10000 + 1000 + Rs. 11000

New Average wage
Hence, mean wages will be affected but standard deviation will not be affected are the stndard deviation is independent of origin. Have the wages become or less uniform? In order to calculate uniformity wages, we will have to calculate co-efficient of variation.

Afterwards

Now more uniformity in wages has taken place.

Question 10.
If in the previous question, each worker is given a hike of 10% in wages, how are the mean and standard deviation values affected?
When each worker is given a hikes of 10% in wages both the mean and standard deviation values will be affected.
New Mean = Rs. 200 + 10 % of Rs. 200
= Rs. 200 + $$\frac{10}{100}$$ × 200 = Rs. 220

New Standard Deviation

= Rs. 40 + 10% of Rs. 40 = Rs. 40 + $$\frac{10}{100}$$ × 40 = Rs. 44.

Question 11.
Calculate the mean deviation about mean and standard deviation for the following distribution:

 Classes Frequencies 20 – 40 3 40 – 80 6 80 – 100 20 100 – 120 12 120 – 140 9 50

Calculation of mean deviation about mean:

Mean Deviation from A.M. = $$\frac{\sum \mathrm{fd}}{\sum \mathrm{f}}=\frac{998.4}{50}$$ = 19.96

Calculation of standard deviation from mean:

Question 12.
The sum of 10 Values is 100 and sum of their squares is 1090. Find the coefficient of Variation
Given N-10, Σ X = 10, Σ X2 = 1090

Coefficient of variation = 30

1st PUC Economics Measures of Dispersion Very Short Answer Type Questions

Question 1.
Define disperation.
Dispersion measures the extent to which different items tend to dispersion away from the central tendency.

Question 2.
How many methods are there to calculate dispersion?
Following are the methods of absolute and relative measures of dispersion:
i) Absolute measure: Range, quartile deviation, mean deviation, standard deviation and Lorenz curve.
ii) Relative measure – Coefficient of range, coefficient of quartile deviation, coefficient of mean deviation, standard deviation, coefficient of variation.

Question 3.
Define range.
Range is the difference between the highest value and lowest value in a series.

Question 4.
Define quartile deviation.
Quartile deviation is half of Inter quartile Range.
Quartile Deviation $$\frac{\mathrm{Q}_{3}-\mathrm{Q}_{1}}{2}$$

Question 5.
How is coefficient of quartile deviation calculated?
Coefficient of quartile deviation is calculated by using the following formula: Coefficient of

Question 6.
Define mean deviation
Mean deviation is the arithmetic average of the deviation of all the values taken from some average value (mean, median, mode) of the series, ignoring sign (+or-) of the deviations.

Question 7.
Define standard deviation
Standard deviation is the square root of the arithmetic mean of the squares of deviation of the items from their mean value.

Question 8.
Give formula to calculate standard deviation of a frequency distribution series by direct method.
$$\overline { X } \quad =\quad \frac { \sum { f\left( x \right) } }{ N }$$ SD or $$\sqrt{\frac{\sum f_{x}^{2}}{N}}$$
Where x = x – $$\overline { X }$$
x = deviation
x = mid value
$$\overline { X }$$ = Mean value

Question 9.
What is Lorenz curve?
Lorenz curve is a measure of deviation of actual distribution from the line of equal distribution.

Question 10.
What do you mean by coefficient of variation?
Coefficient of variation is a percentage expression of standard deviation. It is 100 times the coefficient of dispersion based on standard deviation of a statistical series.
C.V = $$\frac{\sigma}{\overline{\mathrm{x}}}$$ × 100

Question 11.
Given formula of standard deviation for discrete series.

Question 12.
Give formula of mean deviation through mean for individual series. Mean deviation through mean for individual series.

where x = – $$\overline { X }$$
Deviation from the arithmetic
average
N = No of Items

Question 13.
What is standard deviation?
Standard deviation is the positive square root of the mean of squared deviation from mean S.D. is always calculated on the basis of mean only.

Question 14.
What is Variance?
Variance is the square of standard deviation. In Equation Variance = (SD)2

Question 15.
Name the four methods available for the calculation of standard deviation of individual series.

1. Actual mean method
2. Assumeed mean method
3. Direct method & step deviation method.

Question 16.
Write down the formula for calculating standard deviation directly.
$$\sigma=\sqrt{\frac{\sum x^{2}}{n}-(\bar{x})^{2}}$$

Question 17.
Which formula is used for calculating standard deviation of individual series by step deviation method.
$$\sigma=\sqrt{\frac{\sum \mathrm{d}^{2}}{\mathrm{n}} \times \mathrm{C}}$$

Question 18.
What is dispersion?
The degree to which numerical data spread about an average value is called the variation of dispersion. It is an average of second order.

Question 19.
What is measure of dispersion?
The measure of the deviation of the size of items from an average is called a measure of dispersion.

Question 20.
Name the important measure of dispersion?
Range Quartile deviation, mean deviation and standard deviation arc the important measures of dispersion.

Question 21.
Define the range
The range is defined as the difference between the largest and the smallest value of variable in the given set of value R = L – S.

Question 22.
Define quartile deviation.
Quartile deviation is defined as: $$\frac{\mathrm{Q}_{3}-\mathrm{Q}_{1}}{2}$$

Question 23.
What is the Variance?
Variance is the square of standard deviation. In equation, variance = $$(\sigma)^{2}$$

Question 24.
What is the difference between Variance and standard deviation?
The Variance is the average squared deviation from mean and standard deviation is the square of Variance.

Question 25.
Write down the Unique feature of mean deviation.
Mean deviation is the least when taken about median.

Question 26.
What is Coefficient of Variation?
Coefficient of Variation is the percentage Variation in the mean, the standard being treated as the total Variation in the mean.

Question 27.
Write down the formula of calculating Coefficient of Variation
C.V = $$\frac{\sigma}{\overline{\mathrm{x}}}$$ × 100

Question 28.
What is a Lorenz Curve?
Lorenz Curve is a curve which measures the distribtution of wealth and income. Now it is also used tar the study of the distribution of profit wages etc.

Question 29.
What does higher value of range imply?
Higher value of range implies higher dispersion.

Question 30.
How do Range and Quartile deviation measure the dispersion?
Range and quatile deviation measure the dispersion by calculating the spread within which the values lie.

Question 31.
Which aspect of distribution is indicated and which is not indicated by the average.
Average try to tell only one aspect of a distribution ie. a representative size of the Value. It does not tell us the spread of values.

Question 32.
Name the measures based on the spread of values
Range and Quantile deviation based on the speread of values.

Question 33.
Give two limitation of Range.
1. Range is unduly affected based on the spread of values.

Question 34.
Calculate the range of the following observations :
20, 25, 29, 30, 35, 39, 41, 48, 51, 60 & 70
Range = L – S = 70 – 20 = 50

Question 35.
Name the measures of dispersion from average.
Mean deviation and standard deviation are the measure of dispersion from average.

Question 36.
How is standard deviation independent of origin?
Standard deviation is independent of origin as it is not affected by the Value of constant from which deviations are calculated. The value of the constant does not figure in the standard deviation formula.

Question 37.
What are open ended distribution?
Open ended distributions are those distribution in which either the lower limit of the lower class or the upper limit of the highest class or both are not specified.

Question 38.
What is the main limitation of range?
Range is unduely uneffected by extreme range.

Question 39.
Calculate range from the following date:
20, 25, 29, 30, 70,100
Range = 100 – 20 = 80

Question 40.
In a town 25% of person earned more than Rs. 45000 where are 75% earned more than Rs. 18000 calculate absolute Value of dispersion.
Absoluite value ofdispersioa (Quartile value of deviation)= $$\frac{\mathrm{Q}_{3}-\mathrm{Q}_{1}}{2}=\frac{45000-18000}{2}$$ = 13500

1st PUC Economics Measures of Dispersion Short Answer Type Questions

Question 1.
Illustrate the meaning of the term dispersion with example.
Dispersion is a measure of the Variation of the items. According to Prof. C.R. Conn, Dispersion is a measure of the extent to which the individual item vary. The measures of dispersion are required to measure the amount of variation of values about the central values.

Example:
Suppose the monthly incomes in rupees of five households are as Rs. 4500, 6000, 5500, 3750 and 4700
The arithmatic mean of income is Rs. 4700. The amount of Variation in income is shown by deviation from the central values. In this case the deviation from the arithmean. are Rs. 390, 110,610,1140 & 190. Deviation being considered as the total Variation in the mean.

Question 2.
What are the properties of a good measure of dispersion?
Properties of a good measure of disperison.

1. It should be based on all the observation.
2. It should be readily comprechensible.
3. It should be fairly & easily understood.
4. It should be medable to further algebric treatment.
5. It should be affected as little as possible by fluctuation in sampling.

Question 3.
Give the absolute and relative measures dispersion.
Absolute Measure:

1. Range
2. Quartile deviation
3. Mean deviation
4. Standard deviation

Relative Measure:

1. Coefficient of Range
2. Coefficient of quartile deviation
3. Coefficient of mean deviation
4. Coefficient of standard deviation.

Question 4.
Write down the steps involved in the calculation of mean deviation in case of discrete series.
Steps involved in the calculation of mean deviation:
Following steps are involved in the calculation of mean deviation.

1. Find out the mean / median / mode of a series.
2. Find out the deviation of different items from mean /median /mode
3. Add the deviations ignoring positive & negative signs. Treat all deviations as positive.
4. Calculate mean deviation by dividing the sum total of the deviation by the number of items.

Question 5.
Write down the steps involved in the calculation of mean deviation for the discrete series.
Steps

1. Find out central tendency of the series (mean or median) from which deviations are to be taken.
2. Take from central tendency ignoring signs (+, -) express it as | dx | Or (|dm|)
3. Multiply each deviation value by frequency facing it.
4. Add the multiplies & express it as $$\sum \int(\mathrm{d})$$
5. Divide $$\sum \int(\mathrm{d})$$ by sum total of frequency.

The resuttant value will be mean deviation.

Question 6.
Write down the features of mean deviation.
Features of mean deviation.

1. Mean deviation is rigidly defined.
2. It depends on all the values of the variable.
3. It is based on absolute deviation from central values.
4. It is easy to understand.
5. It involves a harder calculation than the range & quantile deviation.
6. It is amendable to algebraic treatment.
7. The units of measurement of the mean deviation are the same as there of the variable.

Question 7.
Differentiate between mean deviation & Standard deviation.
Difference between mean deviation and standard deviation.
Mean Deviation:

1. In calculating mean deviation algebric signs are ignored.
2. Mean or Median is used in calculating the mean deviation.

Standard Deviation:

• In calculating standard deviation, algebraic signs are taken into account.
• Only mean is used in calculating the standard deviation.

Question 8.
What are the uses of co-efficient of Variation?
Coefficient of Variation is used to compose the variability, homogeneity, stability and uniformity of two different statistical series. Higher value of co-efficient variation suggests greater degree of variation & lesser degree of stability. On the other hand, a lower value of coefficient variation suggests lower degree of variability & higher degree of stability, uniformity, homogeneity ‘and consistency.

Question 9.
Explain merits & demerits of Quartile Deviation?
Merits:

1. It is easy understand and to calculate.
2. It is unaffected by the extreme values.
3. It is quite satisfactory when only the middle half of the group is dealt with.

Demerits :

1. It is not capable of algebraic treatment
2. It ignores 50 percent of the extreme items.
3. This is not useful when extreme items are to be given special height.

Question 10.
Write down the merits of mean deviation?
Merits:

1. It is easy to understand mean Deviation.
2. Mean Deviation is less affected by extreme value than the Range.
3. Mean deviation is based on all the items of the series: It is therefore, more representative than the range quartile Deviation.
4. It is very simple & easy measure of dispersion

Demerits of Mean Deviation :

1. Mean deviation is not capable of algebraic treatment, because it ignores plus and minus signs.
2. It is not a well – defined measure since mean deviation from different, average (Mean, median & mode) will not be same.

Question 11.
The height of 11 men were 61, 64, 68, 69, 67, 68, 66, 70, 65, 67 and 72 inches. Calculate the range If the shortest man is omitted, what is the percentage change in the range?
1. Range = L = S = 72 – 61 = 11 inches.
2. New Range (Shortest man is omitted)
= L – S = 72 – 64 = 8
change in range =11 – 8 = 3 inches
Percentage change in range = 3/11 × 100 = 27.2%

Question 12.
What will be the effect of change of origin and change of scale on the standard deviation, mean & variance?
Change of origin & change of scale: Following are the effects of change of origin and change of scale on the mean, standard deviation and variance.
1. Any constant added or substracted (Change of origin) than the standard deviation of original data and of change data after addition as substraction will not change but the mean of new data will change.

2. Any constant multiplied are divided (change of scale) then mean, standard deviation and variation will change of the new changed data.

Question 13.
How is dispersion of the series different from average of the series?
Average of the series refers to the central tendency of the series. It represents behaviour of all the items in the series. But different item tends to different from each other and from and from the averages. Dispersion measures the extent to which different items tend to disperse away from the central tendency.

Question 14.
Why should we measure dispersion about some particular value?
We should measure dispersion about some particular value because in that case.
1. We can assess how precise is the central tendency observations in the series. Greater value of dispersion implies lesser representativeness of the central tendency and vice versa.

2. We can precisely assess how scattered are the actual observation from their representative value.

Question 15.
The following table gives you the height of 100 person. Calculate dispersion by range method.

 Height in centimeters No of person Below 162 2 Below 163 8 Below 164 19 Below 165 32 Below 166 45 Below 167 58 Below 168 85 Below 169 93 Below 170 100

Calculation of dispersion by range method:

 Height in centimeters No of person 161 – 162 2 162 – 163 6 163 – 164 11 164 -165 13 165 – 166 13 166 – 167 13 167 – 168 27 168 – 169 8 169 – 170 7 Total 100

Range = L – S = 170 – 161 =9

1st PUC Economics Measures of Dispersion Long Answer Type Questions

Question 1.
Give the comparison of alternative measure of dispersion.
Comparison of alternative measure of the dispersion has been discussed below.
1. Rigidly defined:
All the four measures the range, quartile deviation, mean deviation & standard deviation are rigidy defined. There is no Vagueness in their definitions.

2. Ease of calculation:
The range is the easiest one to calculate. Quartile deviation requires calculation of the upper & lower quartiles but that is also easy enough. However, the mean deviation & standard deviation require a little more systematic calculation they, too are easy.

3. Simple interpretaion:
All measures of dispersion are easy to interpret. While the range and quartile deviation measure disperrsion in a general way the mean deviation and standard deviation measure dispersion in terms of deviation from a central value. Thus the mean deviation and standard deviation give a better idea about the dispersion of values within the range.

4. Based on all values:
The range and Quantile deviation do not depend on all values where as the mean deviation and standard deviation use all values of the variable the range is affected the most by extreme values.

5. Amendable to algebraic treatment:
The standard deviation is perhaps the earliest for analytical work. Other measure can be also dealt with analytically but derivation are harder.

Question 2.
A variable ‘x’ takes the following values: 7, 9, 18, 11, 10, 8, 17, 13, 11, 16
a. Calculate
i. The arithmetic mean of ‘x’
ii. The standard deviation of the value of ‘x‘
iii. The mean deviation of the value ‘x’ about

Question 3.
Prove by an example that the Variance is uneffected by the choice of the assumed mean.
We take the following example for providing that the variance is uneffected by the choice of the assumed mean.
Example: Calculate variance of 25, 60, 45, 30, 70, 42, 36, 48, 34 and 60 by actual mean assumed mean method.
a. Calculation of variance by Actual mean method

Σ x = 440, $$\overline{\mathrm{X}}=\frac{440}{10}$$ = 44
Variance $$\frac{\sum x^{2}}{N}=\frac{1710}{10}$$ = 171

b. Calculation of Variance by Assumed Mean method.

Assumed mean = 45

= 172 = (-1)2 = 172 – (+1) = 172 – 1 = 171

Question 4.
The distribution of the cost of production (in rupees) of a quaintal of wheat in 50 farms is as follows:

 cost (in rs.) Number off aims 40 – 50 3 50 – 60 6 60 – 70 12 70 – 80 18 80 – 90 9 90 – 100 2 Total 50

1. Calculate the variance

1. By direct method
2. By step deviation method & compare your results with the mean deviation about the arithmetic mean.

2. Calculate the coefficient of variations by using

1. The standard deviation of costs &
2. The mean deviation of cost about the arithmetic mean & compare the two. What is your conclusion about variation of cost.

Solution:

a. Calculation of Variance by step Deviation

A. M. = 75 i = 10

= $$\sqrt{160-0.16 \times 10}$$ = $$\sqrt{1.44 \times 10}$$
= 1.2 × 10 = 12
Variance = σ² = (12)² = 144

b. Calculation of co – efficeient of variation:
1. Variance co – efficent (from S.D.)
= $$\frac{\sigma}{x}=\frac{12}{71}$$ = 0.169 or 0.17 approx.

2. Variance co-coefficient (from mean deviation) = $$\frac{\mathrm{MD}}{\mathrm{x}}$$
M.D. = $$\frac{492}{50}$$ = 9.84
x = 71
Hence V.C. = $$\frac{9.84}{71}$$ = 0.138 = 0.14 approx.

Question 5.
Briefly explain the various measures calculated from standard deviation?
Measures calculated from standard deviation
Mainly following meassures are calculated from standard deviation.
1. Co – efficient of standard deviation:
It is a relative measure of standard deviation. It is calculated to compare the variability in two or more than two series. It is calculated by dividing the standard deviation by arithmetic mean of data symbolically.
Co – efficient of S.D.
Here Standard deviation = $$\frac{\sigma}{\overline{\mathrm{X}}}$$ × 100

2. Arithmetic mean.
Co – efficient of variance. It is most populartly used to measure relative variation of two or more than two series. It should the relationship between the S.D and the arithmetic mean expressed in terms of percentage. It is used to compare unifromly, consistency and variability in two different series.
Symbolically C.V = $$\frac{\sigma}{\overline{\mathrm{X}}}$$ × 100

3. Variance:
It is the square of standard deviation. It is closely Related to standard deviation. It is the average squared deviation from mean whereas standard deviation is the square root of variance. Symbolically variance = σ²

## 1st PUC Economics Question Bank Chapter 5 Measures of Central Tendency

Students can Download Economics Chapter 5 Measures of Central Tendency Questions and Answers, Notes Pdf, 1st PUC Economics Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## Karnataka 1st PUC Economics Question Bank Chapter 5 Measures of Central Tendency

### 1st PUC Economics Measures of Central Tendency TextBook Questions and Answers

Question 1.
Which average would be suitable in the following cases?

2. Average intelligence of students in a class.
3. Average production in a factory per shift.
4. Average wages in an industrial concern.
5. When the sum of absolute deviations from average is least.
6. When quantities of the variable are in ratios.
7. In case of open ended frequency distribution.

1. Mode
2. Mode
3. Mean
4. Median
5. Mean
6. Weighted mean
7. Median.

Use this Median Calculator to easily calculate the arithmetic mean, median and mode of figures.

Question 2.
Indicate the most appropriate alternative from the multiple choices provided against each question:
i) The most suitable average for qualitative measurement is:

1. Arithmetic mean
2. Median
3. Mode
4. Geometric mean
5. None of the above.

ii) Which average is affect most by the presence of extreme items?

1. Median
2. Mode
3. Arithmetic Mean
4. Harmonic Mean

iii) The algebraic sum of deviation of a set of n values from A.M. is

1. N
2. 0
3. None of the above.

 QUESTIONS APPROPRIATE ANSWER i) The most suitable average for qualitative measurement is 3. Mode ii) Which average is affected most by the Presence of extreme items 3. Arithmetic mean iii) The algebraic sum of deviation of a set of n value from A. M. is 2. 0 (zero)

Question 3.
Comment on whether the following statements are True or False

1. The sum of deviation of items from median is zero.
2. An average alone is not enough to compare series
3. Arithmetic mean is a positional value
4. The upper quartile is the lowest value of top 25% of items.
5. Median is unduly affected by extreme observations.

1. False
2. True
3. False
4. True
5. False.

Question 4.
If the arithmetic mean of the data given below is 28, find

1. The missing frequency and
2. The median of the series.
 Profit per retail shop (in Rs.) Number of retail shops 0- 10 12 10-20 18 20-30 27 30-40 – 40-50 17 50-60 6

A.M. = 35

or x + 80 = 100 or X = 100 – 80 = 20
Hence, figure is = 20
Calculation of Median:

Median=Value of the item= 50th items
Value of 50th item lies in 20 – 30 class interval

= 20 + 7.41 = 27.41

Question 5.
The following table gives the daily income of ten workers in a factory. Find the arithmetic mean.

 Workers Daily Income (Rs.) A 120 B 150 C 180 D 200 E 250 F 300 G 220 H 350 I 370 J 260

Question 6.
Following information pertains to the daily income of 150 families Calculate the arithmetic means.

 Income is (Rs) No. of Families More than 75 150 More than 85 140 More than 95 115 More than 105 95 More than 115 70 More than 125 60 More than 135 40 More than 145 25

AM = 120

= 120 – 3.7 = 116.3

Question 7.
The size of land holdings of 380 families in a village is given below. Find the median size of land holdings.

 Size of Land Holdings (in acres) Number of families less than 100 40 100 – 200 89 200 – 300 148 300 – 400 64 400 and above 39

Calculation of median size of land holding.

Median = Value of $$\frac{\mathrm{N}}{2}$$th item = $$\frac{380}{2}$$

= Value of 190th item
Value of 190th item lies in Class interval of 200 – 300.

Hence, Median

= 200 + 4.21
= 241.21 acres

Question 8.
The following series relates to the daily income of workers employed in a firm. Compute the

1. Highest income of lowest 50% workers,
2. Minimum income earned by the top 25% and
3. Maximum income earned by loest 25% workers.
 Daily Income (in Rs.) Number of workers 10- 14 5 15 – 19 10 20-24 15 25-29 20 30-34 10 35 – 39 5

Median = Value of $$\frac{\mathrm{N}}{2}$$th item = $$\frac{65}{2}$$
= $$\frac{65}{2}$$ = Value of 32.5 item
Value of 32.5 item lies in 24.5 – 29.5 class interval

= 24.5 + $$\frac{2.5}{4}$$
= 24.5 + 0.56
= 25.11 Ans.

b) Q1 = Value of $$\left(\frac{\mathrm{N}}{4}\right)$$ th item
$$\frac{65}{2}$$ = Value of 16.25 item
Value of 16.25 item lies in 19.5 – 24.5 class interval

= 19.5 + 0.42
= 19.92 Ans.

c) Q3 = Value of $$\frac{3 \mathrm{N}}{4}$$ th item.
$$=\frac{65 \times 3}{4}=\frac{195}{4}$$
= Value of 48.75 item
Value of 47.75 lies in 24.5 – 29.5 class interval.

= 24.5 + 4.69
= 29.19 Ans.

Question 9.
The following table gives production yield in kg. per hectare of wheat of 150 farms in a village. Calculate the mean, median and mode production yield.

Calculation of Mean, Median and Mode

Mean A. M. = 63.5
C = 3

= 63.5 + 0.32
Median = 63.82 Ans.
Medain = Value of $$\frac{N}{2}$$ th value
Value of 75th item which lies in 62 – 65 class interval.
Hence the modal class is 62 – 65

= 62 + 1.67 – 63.37 Ans.
Model Class = 62 – 65 because it has the maximum frequency i.e., 36.

= 62 + $$\frac{6}{72-58}$$ × 3
= 62 + $$\frac{6}{14}$$ × 3
= 62 + 1.29 = 62.29

### 1st PUC Economics Measures of Central Tendency Additional Questions and Answers

1st PUC Economics Measures of Central Tendency Very Short Answer Type Questions

Question 1.
What do you mean by ‘Average’?
Average value or central tendency is the critical value which represents all the items in a series.

Question 2.
Pocket expenditures of 6 students are (Rs) 6, 12, 18, 24, 30 and 36 respectively. Find out arithmetic mean?

Question 3.
Name various kind, of stastistical averages.
i) Mathematical average:

• Arithmetic mean
• Geometric mean
• Harmonic means

ii) Positional Averages:

• Median
• Partition value
• Mode

Question 4.
Give formula of weighted averages

Question 5.
Give formula of calculating arithmetic averages of a continuous series?

Question 6.
What is arithmetic mean?
Arithmetic mean is a simple average of all items in a series it is simply mean value which is obtained by adding the values of all the items and dividing the total by the number of etc.

Question 7.
Define Median?
Median is a centrally located value of series such that half of the values of the serives are above it and other half below it.

Question 8.
Define mode?
Mode is the value which occurs mode frequently in the series.

Question 9.
Give formula for finding out median.
M = size of $$\left[\frac{N+1}{2}\right]$$ the item
M = Median N = No.of items

Question 10.
Give formula for estimating mode in frequency distribution series?
1. First we find out the model class
2.

Z = Value of the mode
I1 = Lower limit of the model classes
i = Size of the model group.

Question 11.
Give formula for finding out median of a continuous series?

I1 = Lower limit of the median class
C.f = Cumulative frequency of the classes
f = Preceding the median classes
i = size of the median class interval

Question 12.
Define Quartile?
If a statistical series is divided into four equal parts then the end value of each part is called a quartile.

Question 13.
What does central Tendency refer to?
Central Tendency refers to a central value or a representive value of statistical date. Measures of central tendency- arithmetic mean.

Question 14.
Enumerate any three essential, of a good average?

1. It should be simple
2. It must be certain on character.
3. It should be capable of further mathematical or algebraic treatment.

Question 15.
Name the three statistical measures of central Tendency?
Three statistical measure of central tendency are:

1. Arithmetic mean
2. Median and
3. Mode

Question 16.
Write down the formuala for calculating mean for ungrouped data?
$$\bar{x}=\frac{\sum x}{N}$$
$$\overline{\mathrm{x}}$$ Stands for arithmetic mean
Σ x stands for summation of all the observations in a series
N = Stands for number of observations

Question 17.
Write down an interesting property of AM
The sums of deviations of items about arithmetic mean is always equal to zero symbolically.

Question 18.
Write down the formula for calculating weighted average (mean)?
$$\overline{\mathrm{X}}_{\mathrm{w}}=\frac{\sum_{\mathrm{x}} \mathrm{W}_{\mathrm{x}}}{\sum \mathrm{W}}$$
$$\overline{\mathbf{X}}_{\mathrm{w}}$$ = Weighted average (mean)
W = Weights
X = variables

Question 19.
Give a formula for calculating arithmetic mean of continuous series by step deviation method?
$$\bar{x}=A+\frac{f d^{\prime}}{N} \times C$$

Question 20.
Give the formula for calculating corrected Arithmetic mean?
Corrected arithmetic mean
Σ x (Wrong) + (correct value)
= $$-\left(\frac{\text { Uncorrected value }}{\mathrm{N}}\right)$$

Question 21.
What is the arithmetic mean of 5, 6, 7, 8 and 9?
Arithmetic mean of 5,6,7, 8 and 9 is $$\frac{35}{5}$$ = 7

Question 22.
When can the direct method of computation of arithmetic mean be used?
The direct method of computation of arithmetic mean can be used when the items in the serial are less.

Question 23.
Write down the formula of calculating the weighted arithmetic mean of three related groups?
$$\bar{X}_{1,2,3}=\frac{N_{1} \bar{X}+N_{2} \bar{X}_{2}+N_{3} \bar{X}_{3}}{N_{1}+N_{2}+N_{3}}$$

Question 24.
When will the weighted mean be greater than the simple arithmetic mean?
The weighted arithmetic mean will be greater than the simple arithmetic mean when items of small values are given less weights and items of big values are given more weights.

Question 25.
Name any two positional values?

1. Median
2. Mode.

Question 26.
What is median?
Median is that positional value of the variable which divides the distribution into two equal parts. One part comprises all values greater than or equal to median values and the other comprises all values less than or equal to it.

Question 27.
What is the first and most important rule for calculating the median?
The first and most important rule for calculating median is that data should be arranged in ascending or descending order.

Question 28.
Calculate the median of 5.7.6.6.1.8.10.12.4 and 3
By arranging the date in ascending order we get 1.3.4.5.6.7.8.10.12
Median = value of $$\frac{N+1}{2}$$ item = $$\frac{9+1}{2}$$ = 5th item i.e. 6.

Question 29.
Calculate the median of 3.5.7 and 4
Medain = size of $$\frac{N+1}{2}$$ item
= $$\frac{4+1}{2}$$ = 2.5 item = $$\frac{5+1}{2}$$ = 6

Question 30.
Write down the formula for locating the median from median group?

I1 = Lower limit of median group.
c. f. = Cumulative frequency of the class Preceding the median class.
f = Frequency of the median goup.
i = The class interval of the median group.

Question 31.
What is partition value?
The value that divides the series into more than two parts is called partition value Q1 Q2 Q3 etc are partition value.

Question 32.
Write down the difference between averages and partition values.
An average is representitive of shole series while quartiles are average of parts of series.

Question 33.
What are Percentiles?
Percentiles are measures which divide the distribution to hundred equal parts there are 99 percentiles denoted by P1 P2P3 P99.

Question 34.
Suppose a person has secured 82 percentiles in a management entrance examination. What does it mean?
It means that his position is below 18 percent of total candidates appeared in the examination.

Question 35.
What is mode?
Mode is the most frequenctly observed date value it is denoted by M0.

Question 36.
Calculate mode from the following discrete series?

The value of mode is 30 because it is most frequently observed data.

Question 37.
What do you mean by unimodal data.
Unimodal data is that which has only one model for example there is one modal in the following data.
2.23.4.4 and 5

It is unimodal because it has one mode 30.

1st PUC Economics Measures of Central Tendency Short Answer Type Questions

Question 1.
Write any four merits of mean?
Merits mean are as follows:
1. Certainity:
Arithemetic mean is a certain value it has no scope for estimated values.

2. Based on all items:
Arithmetic mean is based on all items in a series it is therefore based on all the items in series it is therefore representative value of the different items.

3. Simplicity:
From the view point of calculation and the use of arithmetic mean is the simplest of all the measures of central tendency.

4. Stability:
Arithmetic mean is a stable measure of centrel tendency this is because charges in the sample of a series having minimum effect or the arithmetic average series.

Question 2.
Write any four Demerits of mean?
Demerits of arithmetic average are as follows:

1. Effect of extreme value – The main effect of it is that it gets distorted by extreme values
of the series.
2. Unsuitability – Arithmetic mean is a suitable measures in case of percentage or proportionate values.
4. Laughable conclusion – Arithmetic mean sometimes offer laughable conclusion.

Question 3.
Compare the arithmetic mean, median and mode as measure of central tendency. Describe situations where one is more suitable than the others?
As compared to mean and median, mode is less suitable. Mean is simple to calculate its value is definite. It can be given algebraic treatment and is not affected by fluctuations of sampling median and is even more simple to calculate but is affected by fluctuations and cannot be given algebraic treatment.

Mode is the most popular item of a series but it is not suitable for most elementary studies because it is not based on all the observations of the series and is unrepresentative.

In case of arithmetic mean it is the numberical magnitude of the deviations that balances In case of median it is the number of values greater than the median which balance against the number or values less than the median the median, is always between the arithmetic mean and the mode.

Question 4.
The arithmetic mean is described as the centre of gravity of the distribution of values of the variable Explain?
This is rightly said that arithmetic mean is the center of gravity of the distribution of values of the variable because it is the number obtained by dividing the total values of different items by their number. We use this method in our everyday life. It is a popular and most widely used measure.

Question 5.
Find mean and median for aft four values of the series and answer the question given at the end of table?

1. is median affected by extreme values?
2. is median a better method than mean

1. No, median is not affected by extreme values
2. Yes, median is a better method then mean.

Question 6.
Prove that the sum of the squares of the and deviation from arithmetic mean is the least ie. less other squares of the deviation of observation taken from any values.
For providing the statement given in the questions we take the following observations

We take another value and we observe that the same of squares of deviations taken from mean (3) less that taken from other value (4) less.

Question 7.
Prove with an example that the weighted arithmetic mean will be greater than the simple arithmetic mean when items of small values are given less weights and items of big values are given more weights?
In order to prove the statements given in the example we take the following table.

Question 8.
With the help of following data prove that the sum of the deviations of items about arithmetic mean is always equal to zero.
X : 4, 6, 8, 10 and 12

From the given data it has been proved that the sum of the deviation of item, arithmetic mean is equal to zero.
Symbolically
Σ cx – $$\frac{2}{x}$$ = 0

Question 9.
Calculate median from the following series.

Median = Size of $$\frac{N+1}{2}$$ th item
= $$\frac{30+1}{2}$$ = value of 15.5 th item
Value of 15.5 the item = 30
Hence median is equal to 30.

Question 10.
Write down the merits of median?
Merits of median:
Following are the merits of median.

1. Median is definite
2. It is easy to calculate and understand.
3. It can also be determined graphically
4. It is not affected by extreme values.
5. It can be calculated in the absence of one of the item.
6. It is helpful in qualitative facts such as ability stability etc.

Question 11.
Write down the demerits of median?
Demerits of median:
Following are the demerits of median:

1. In it all the items of a series are not given equal importance.
2. If the number of items are given the correct value of median cannot be calculated.
3. It is affected by fluctuation of sampling.
4. It cannot be give further algebraic treatment.
5. Data needs to be arranged in ascending of descending order.

Question 12.
Why is arithmetic mean the most popular measure of the central tendency?
Arithmetic mean is the most popular measure of the central tendency due to following reasons:

1. It is easy to calculate and understand arithmetic mean.
2. Arithmetic mean is based on all the values of the serial.
3. Arithmetic mean is stable measure or central tendency it is because change in the sample of a series have minimum effect on the arithmetic averages.

Question 13.
Daily income of eight families are given below. Calculate average daily family income.

 Sl.No. Daily income (Rs) X 1 170 2 500 3 250 4 700 5 300 6 400 7 200 8 350 N = 8 Σ x = 2870

$$\overline{\mathrm{X}}=\frac{\sum \mathrm{x}}{\mathrm{N}}=\frac{2870}{8}$$
= Rs. 358.75

Question 14.
The average height of 50 student’s in a class are 61 inches and that of 70 students of another class is 58 inches Calculate combined average height of both the classes?
Given N1 = 50, $$\overline{\mathbf{X}}_{1}$$ = 61
N2 = 70, $$\overline{\mathbf{X}}_{2}$$ = 58

= 59.25 inches

Question 15.
The following table gives the daily income of ten workers in a factory. Find the arithmetic mean.

 Workers Daily Income (Rs) A 120 B 150 C 180 D 200 E 250 F 300 G 220 H 350 I 370 J 260

$$\overline{\mathbf{X}}$$ = X1 + X2 + X3 + X4…….+ XN
$$=\frac{120+150+180+200+250+300+220+350+370+260}{10}$$
$$=\frac{2400}{10}$$ = Rs. 240

Question 16.
Calculate the median from the following data.

 Marks obtained No. of Students 20 5 40 10 60 20 80 15 100 10

Median = Value of $$\frac{\mathrm{N}+1}{2}$$ th item
= Value of $$\frac{\mathrm{60}+1}{2}$$ th item
i.e value of 30.5th item
Value of 30.5 is 60
Hence Median = 6.

Question 17.
Monthly incomes of five families are given below. Calculate mean by shortcut method. 6550, 7550, 9550, 4550 and 8000

Assumed mean = 5550
$$\overline{\mathbf{X}}$$ = A.M + $$\frac{\sum \mathrm{d}}{\mathrm{N}}$$
= 5550 + $$\frac{8450}{5}$$
= 5550 + 1690
= 7240.

1st PUC Economics Measures of Central Tendency Long Answer Type Questions

Question 1.
The marks obtained by 30 students of PUC first year of a college in economics paper are given in the college in economics paper are given in the following table. Find the mean marks obtained by the students.

Direct Method:

= 59.3 marks

Question 2.
The table below gives the percentage distribution of female teachers in the primary schools of rural area of Karnataka State. Find mean percentage of female teachers by assumed mean method and step deviation Method.

Direct Method:

= 39.71
Step deviation method:

= 39.71 the mean percentage of female teachers is 39.71

Question 3.
Consider the following distribution of daily wages of 50 workers of a factory. Find the mean daily wages of workers.

Question 4.
The following table gives the literarcy rate (in %) of 35 cities. Find the mean literarcy rate. Literacy rate in

Question 5.
The table below shows the daily expenditure on food of 25 households in a locality.

The mean daily expenditure on food is Rs. 211 per Household

Question 6.
The following series relates to the daily income of workers employed in a firm. Compute the highest income of lowest 50% of workers, ie; median.

Since the series of data given is of inclusive type, first we shall convert the class intervals into exclusive type using a correction factor.
C.F – LL of a CI – UL of the previous CI
= $$\frac{15-14}{2}$$ = 0.5
Where LL = Lower Limit
UL = Upper Limit
CI = Class Interval
CF = Cumulative frequency
Deduct 0.5 from lower limit and add 0.5 to the upper limits

Median class = Size of $$\left[\frac{\mathrm{N}}{2}\right]$$th item
= $$\left[\frac{\mathrm{65}}{2}\right]$$th item = 32.5 th item
so, it lies between 24.5 = 29.5
Medain = $$\mathrm{L}+\left[\frac{\frac{\mathrm{N}}{2}-\mathrm{Cf}}{\mathrm{f}}\right] \times \mathrm{i}$$
Where L – Lower limit
C.f = Cumulative frequency of class preceding the median class
L – Length of the median class
N = Total of frequency
f = frequency of the median class
Median class = 24.5 – 29.5
I = 24.5; i = 5; c.f = 30; N = 65; f = 20
Medain = 24.5 + $$\left[\frac{\frac{65}{2}-30}{20}\right]$$ × 5
Medain = Rs. 25.125
Highest income of lowest 50% of workers

## 1st PUC Question Banks with Answers Karnataka

Expert Teachers at KSEEBSolutions.com has created Karnataka 1st PUC Question Banks with Answers Solutions, Notes, Guide Pdf Free Download of 1st PUC Textbook Questions and Answers, Model Question Papers with Answers, Study Material 2020-21 in English Medium and Kannada Medium are part of KSEEB Solutions. Here We have given the Department of Pre University Education (PUE) Karnataka State Board NCERT Syllabus 1st Year PUC Question Banks with Answers Pdf.

Students can also read 1st PUC Model Question Papers with Answers hope will definitely help for your board exams.

## Karnataka 1st PUC Question Banks with Answers

If you have any queries regarding Karnataka State Board NCERT Syllabus 1st Year PUC Class 11 Question Banks with Answers Pdf, drop a comment below and we will get back to you at the earliest.

## 1st PUC Model Question Papers with Answers 2020-21 Science Commerce Karnataka New Syllabus

Expert Teachers at KSEEBSolutions.com has created New Syllabus Karnataka 1st PUC Model Question Papers with Answers 2020-21 Pdf Free Download of 1st PUC Model Question Papers PCMB with Answers, 1st PUC Commerce Model Question Papers with Answers, 1st PUC Model Question Papers Science Arts, Karnataka 1st PUC Previous Year Question Papers with Answers in English Medium and Kannada Medium are part of KSEEB Solutions. Here We have given www.puc.kar.nic.in the Department of Pre University Education (PUE) Karnataka State Board NCERT Syllabus First Year Model Question Papers for 1st PUC Science Commerce Arts with Answers 2020-2021 Pdf.

Students can also read 1st PUC Question Banks with Answers hope will definitely help for your board exams.

## Karnataka 1st PUC Model Question Papers with Answers 2020-2021 Science Commerce New Syllabus

We hope the given New Syllabus Karnataka 1st PUC Class 11 Model Question Papers with Answers 2020-21 Pdf Free Download of 1st PUC Model Question Papers PCMB with Answers, PUC 1st Year Model Question Papers, 1st PUC Commerce Model Question Papers with Answers, 1st PUC Model Question Papers Science Arts, Karnataka 1st PUC Previous Year Question Papers with Answers in English Medium and Kannada Medium will help you.

If you have any queries regarding Karnataka State Board NCERT Syllabus First PUC Class 11 Model Question Papers for 1st PUC Science Commerce Arts with Answers 2020-2021 Pdf, drop a comment below and we will get back to you at the earliest.

Expert Teachers at KSEEBSolutions.com has created Karnataka 1st PUC Hindi Textbook Answers, Notes, Guide, Summary, Solutions Pdf Free Download of Sahitya Vaibhav Hindi Textbook 1st PUC Solutions Pdf, Hindi Workbook 1st PUC Answers, 1st PUC Hindi Lessons Summary Poems Summary, Textbook Questions and Answers, Hindi Model Question Papers With Answers, Hindi Question Bank, Hindi Grammar Notes Pdf, Hindi Study Material 2020-21 are part of 1st PUC Question Bank with Answers. Here KSEEBSolutions.com has given the Department of Pre University Education (PUE) Karnataka State Board Syllabus 1st Year PUC Hindi Textbook Answers Pdf.

Students can also read 1st PUC Hindi Model Question Papers with Answers hope will definitely help for your board exams.

### 1st PUC Hindi Text Book Sahitya Vaibhav Answers

You can download Karnataka State Board साहित्‍य वैभव Hindi Textbook 1st PUC Questions and Answers Pdf, Notes, Lessons Summary, Poem Summary, Textual Exercises.

Sahitya Vaibhav Hindi Textbook Solutions

प्रथम सोपान – गद्य भाग

द्वितीय सोपान – पद्य भाग

(अ) मध्ययुगीन काव्य

(आ) आधुनिक कविता

तृतीय सोपान – अपठित भाग (कहानियाँ)

### 1st PUC Hindi Workbook Answers

चतुर्थ सोपान – व्याकरण तथा रचना

Karnataka 1st PUC Hindi Blue Print of Model Question Paper

We hope the given 1st PUC Hindi Textbook Answers, Notes, Guide, Summary, Solutions Pdf Free Download of Sahitya Vaibhav Hindi Textbook 1st PUC Solutions Pdf, Hindi Workbook 1st PUC Answers, 1st PUC Hindi Lessons Summary Poems Summary, Textbook Questions and Answers, Hindi Model Question Papers With Answers, Hindi Question Bank, Hindi Grammar Notes Pdf, Hindi Study Material 2020-2021 will help you.

If you have any queries regarding Karnataka State Board Syllabus 1st Year PUC Class 11 Hindi Textbook Answers Pdf Download, drop a comment below and we will get back to you at the earliest.

## Karnataka 1st PUC Physics Question Bank Chapter 4 Motion in a Plane

### 1st PUC Physics Motion in a Plane TextBook Questions and Answers

Question 1.
State, for each of the following physical quantities, if it is a scalar or a vector: volume, mass, speed, acceleration, density, number of moles, velocity, angular frequency, displacement, angular velocity.
Scalar: Volume, mass, speed, density, number of moles, angular frequency.
Vector: Acceleration, velocity, displacement, angular velocity.

Question 2.
Pick cut the two scalar quantities in the following list:
force, angular momentum, work, current, linear momentum, electric field, average velocity, magnetic moment, relative velocity.
Scalar quantities: Work, current.

Question 3.
Pick out the only vector quantity in the following list:
Temperature, pressure, impulse, time, power, total, path length, energy, gravitational potential, coefficient of friction, charge.
Vector quantity: Impulse.

question 4.
State with reasons, whether the following algebraic operations with scalar and vector physical quantities are meaningful:

2. adding a scalar to a vector of the same dimensions,
3. multiplying any vector by any scalar,
4. multiplying any two scalars,
6. adding a component of a vector to the same vector.

1. No, scalars must represent same physical quantity.
2. No, vector can be added only to another vector.
3. YesYes
4. The two vectors must represent the same physical quantity.
5. Adding a component of a vector to same vector is meaningless.

Question 5.
Read each statement below carefully false:

1. The magnitude of a vector is always a scalar,
2. each component of a vector is always a scalar,
3. the total path length is always equal to the magnitude of the displacement vector of a particle,
4. the average speed of a particle (defined as total path length divided by the time taken to cover the path) is either greater or equal to the magnitude of the average velocity of the particle over the same interval of time,
5. Three vectors not lying In a plane can never add up to give a null vector.

1. True. The magnitude of a vector gives its length, which is always a scalar.
2. False. Each component of a vector is a vector by itself.
3. False. The total path length is always greater than or equal to the magnitude of the displacement vector of the particle.
4. True. Since the path length is always greater than or equal to the magnitude of the displacement vector of the particle, the average speed is always greater than or equal to the magnitude of the average velocity.
5. True. If the vectors are not coplanar, their addition will always yield some nonzero component in at least one direction.

Question 6.
Establish the following vector inequalities geometrically or otherwise:

When does the equality sign above apply?
a) Geometrical solution:-
Consider the triangle shown in the figure.

which is true by the triangle inequality (Sum of the lengths of any two sides of a triangle is always greater than the length of the third side.)
Equality holds in the case of a degenerate triangle (collinear vectors).
Hence proved.
Analytical solution:-
We must prove that
sqaure both sides,
To prove that

i.e., to prove that
a² + b² + 2ab cos θ ≤ a² + b² + 2ab,
( here $$|\overrightarrow{\mathrm{a}}|$$ = a , $$|\overrightarrow{\mathrm{b}}|$$ = b)
i.e., to prove that
2ab cos θ ≤ 2ab i.e., TPT cosθ ≤ 1
Since the range of cose is [-1, 1], the above inequality is true. The equality holds when θ = 0° (collinear vectors).
Hence Proved.

b) Geometrical solution:
Consider the triangle shown in the figure.

Which is true by the triangle inequality. (Difference in the lengths of any two sides of a triangle is always lesser than the length of the third side.) The equality holds for a degenerate triangle (collinear vectors).
Proved.
Analytical solution:-
We are to prove that

Square both sides we must prove that

i.e., To prove that
a² + b² + 2ab cos θ ≥ a² + b² – 2ab
i.e., to prove that 2ab cos θ ≥ – 2 ab
to prove that cos θ ≥ -1
Since the range of cos θ is [-1, 1], the above inequality is true.
The equality holds when θ = 180° (collinear vectors)
Hence Proved.

c) geometrical solution:-
Consider the triangle shown in the figure.

we must prove that $$|\overrightarrow{\mathrm{c}}| \leq|\overrightarrow{\mathrm{a}}|+|\overrightarrow{\mathrm{b}}|$$
or to prove that $$|\overrightarrow{\mathrm{c}}| \leq|\overrightarrow{\mathrm{a}}|+|\overrightarrow{\mathrm{- b}}|$$
(Since $$|-\vec{b}|=|\vec{b}|$$) which is true by the triangle inequality. (Sum of the lengths of any two sides of a triangle is always greater than the length of the third side.)
The equality holds when the triangle is degenerate (collinear vectors).
Hence proved.
Analytical solution:-
We must prove that $$|\vec{a}-\vec{b}| \leq|\vec{a}|+|\vec{b}|$$
Square both sides.
To prove that $$(|\vec{a}-\vec{b}|)^{2} \leq(|\vec{a}|+|\vec{b}| |)^{2}$$
i.e., To prove that a² + b² – 2ab cos θ ≤ a² + b² + 2ab
i.e., to prove that -cos θ ≤ 1
Since the range of cos θ is [-1, 1], the range of -cos θ is also [-1, 1].
∴ The above inequality is true. Equality holds when θ = 180° (collinear vectors).
Hence proved.

d) Geometrical solution:-
Consider the triangle shown in the figure.

i.e., To prove that $$|\vec{c}| \geq|| \vec{a}|-|-\vec{b}||$$
(Since $$|-\vec{b}|=|\vec{b}|$$)
Which is true by the triangle inequality. (Difference in the lengths of any two sides of a triangle is always lesser than the length of the third side.)
The equality holds in the case of a degenerate triangle (collinear vectors).
Hence proved.
Analytical solution:-
We must prove that $$|\vec{a}-\vec{b}| \geq|| \vec{a}|-| \vec{b}||$$
Square both sides.
To prove that $$(|\vec{a}-\vec{b}|)^{2} \geq(|| \vec{a}|-| \vec{b}||)^{2}$$
i.e., To prove that a² + b² – 2ab cose ≥ a² + b² – 2ab
i.e., to prove that cos θ ≤ 1
Since the range of cos θ is [-1, 1],
∴ The above inequality is true.
Equality holds when θ= 0 (collinear vectors).
Hence proved.

Question 7.
Given $$\vec{a}+\vec{b}+\vec{c}+\vec{d}=\vec{0}$$. which of the
following statements are correct:
(a) $$\vec{a}, \vec{b}, \vec{c}$$ and $$\vec{d}$$ must each be a null vector,

(b) The magnitude of $$(\vec{a}+\vec{c})$$ equals the magnitude of $$(\vec{b}+\vec{d})$$,

(c) The magnitude of $$\vec{a}$$ can never be greater than the sum of the magnitudes of $$\vec{b}, \vec{c}$$ and $$\vec{d}$$

(d) $$\vec{b} + \vec{c}$$ must lie in the plane of $$\vec{a}$$ and $$\vec{d}$$ if $$\vec{a}$$ and $$\vec{d}$$ if they are not collinear, and in the line of $$\vec{a}$$ and $$\vec{d}$$, if they are collinear?

c) Correct. By an extension of the triangle inequality (sum of the lengths of any (n-1) sides of an n – sided closed polygon is always greater than the length of the remaining side.)

d) Correct. Given $$\vec{a}+\vec{b}+\vec{c}+\vec{d}=\vec{0}$$
⇒ $$(\vec{b}+\vec{c}) = -(\vec{a}+\vec{d})$$
If $$\vec{a}$$ and $$\vec{d}$$ are not collinear, then $$(\vec{a}+\vec{d})$$ defines a plane. Clearly $$(\vec{b}+\vec{c})$$ has to lie in this plane for the given condition to hold good. Now if $$\vec{a}$$ and $$\vec{d}$$ are collinear, $$\vec{a}$$ + $$\vec{d}$$ is a line and clearly $$(\vec{b}+\vec{c})$$ must be collinear with this line for the given condition to hold good.

Question 8.
Three girls skating on a circular ice ground of radius 200 m start from a point P on the edge of the ground and reach a point Q diametrically opposite to P following different paths as shown in Fig. 4.20. What is the magnitude of the displacement vector for each? For which girl is this equal to the actual length of path skate?

The magnitude of the displacement vector for each girl = Shortest distance between points P and Q = Diameter PQ
= 2 × 200 m = 400 m
Since girl B skates along the diameter PQ, for her, the magnitude of the displacement vector is equal to the actual length of path skated.

Question 9.
A cyclist starts from the centre O of a circular park of radius 1 km, reaches the edge P of the park, then cycles along the circumference, and returns to the centre along QO as shown in Fig. 4.21. If the round trip takes 10 min, what is the
(a) net displacement
(b) average velocity, and
(c) average speed of the cyclist?

a) Since the cyclist returns to the starting point, his net displacement is zero.
b) The average velocity = $$\frac{\text { Net displacement }}{\text { Time taken }}$$ = 0
c) Average speed = $$\frac{\text { length of the path }}{\text { Time taken }}$$
= $$\frac{\mathrm{OP}+\mathrm{arc} \mathrm{PQ}+\mathrm{QO}}{10 \mathrm{min}}$$
= \frac{1 \mathrm{km}+1/{4} \times 2 \pi \times 1 \mathrm{km}+1 \mathrm{km}}{1 / 6 \mathrm{hour}}[/latex] = 21.4 km / hour.

Question 10.
On open ground, a motorist follows a track that turns to his left by an angle of 60° after every 500 m. Starting from a given turn, specify the displacement of the motorist at the third, sixth and eighth turn. Compare the magnitude of the displacement with the total path length covered by the motorist in each case.

Let the origin be O. The path of the motions is as shown in the figure.
a) After 3 rd turn:-
In triangle ABC, AC = AB cos (∠ACB) + BC cos (∠BAC)
∠ACB = ∠BAC = 30° (Since AB = BC and ∠ABC = 120°).
∴ AC = 500 cos 30° + 500 cos 30°
AC = 866 m
In triangle OAC,
OC = $$\sqrt{\mathrm{OA}^{2}+\mathrm{AC}^{2}}$$ (Since ∠OAC – 90°)
= $$\sqrt{500^{2}+866^{2}}$$ = 1000 m
∴ Magnitude of the displacement = 1 km
Angle with the original direction
= tan-1 $$\left(\frac{\mathrm{AC}}{\mathrm{OA}}\right)$$ = tan-1 $$\left(\frac{866}{500}\right)$$ = 60°
Path length = 3 × 500 m = 1500 m Clearly magnitude of displacement < path length b) After sixth turn:- The net displacement here = 0 (because the motionist returns to his starting point) Angle with the original direction = 0° Path length = 6 × 500 m = 3 km Clearly, path length > magnitude of the displacement.

c) After the eighth turn:-
In triangle OAB,
OB = $$\sqrt{\mathrm{OA}^{2}+\mathrm{AB}^{2}+2 . \mathrm{OA} . \mathrm{AB} . \cos \theta}$$
= $$\sqrt{500^{2}+500^{2}+2 \times 500 \times 500 \times \cos 60^{\circ}}$$
= 866 m
Net displacement = 866 m
Angle with the original direction
= $$\tan ^{-1}\left(\frac{A B \sin \theta}{O A+A B \cos \theta}\right)$$
= $$\tan ^{-1}\left(\frac{500 \sin 60^{\circ}}{500+500 \cos 60^{\circ}}\right)$$
= $$\tan ^{-1}\left(1/{\sqrt{3}}\right)$$
= 30°
Path length = 8 × 500 m = 4 km
Clearly, path length > Magnitude of net displacement.

Question 11.
A passenger arriving in a new town wishes to go from the station to a hotel located 10 km away on a straight road from the station. A dishonest cab man takes him along a circuitous path 23 km long and reaches the hotel in 28 min. What is

1. the average speed of the taxi,
2. the magnitude of average velocity?
3. Are the two equal?

1. Average speed of the taxi
= $$\frac{\text { Path length }}{\text { Time taken }}$$
= $$\frac{23 \mathrm{km}}{28 \mathrm{min}}$$ = $$\frac{23 \mathrm{km}}{28 \mathrm{28}_{60} \mathrm{h}}$$ = 49.3 km h-1

2. Magnitude of Average Velocity
= $$\frac{\text { Net displacement }}{\text { Time taken }}$$
= $$\frac{10 \mathrm{km}}{28 \mathrm{min}}$$ = $$\frac{10 \mathrm{km}}{28 \mathrm{28}_{60} \mathrm{h}}$$ = 21.4 km h-1

3. Clearly, average speed is greater than magnitude of average velocity. The two are not equal.

Question 12.
Rain is falling vertically with a speed of 30 ms-1. A woman rides a bicycle with a speed of 10 ms-1 in the north to south direction. What is the direction in which she should hold her umbrella?
The direction of θ is given by

= 18° with the vertical.

Question 13.
A man can swim with a speed of 4.0 km/h in still water. How long does he take to cross a river 1.0 km wide if the river flows steadily at 3.0 km/h and he makes his strokes normal to the river current? How far down the river does he go when he reaches the other bank?
Time taken to cross the river
= $$\frac{\text { Width of the river }}{\text { Speed in this direction }}$$
= $$\frac{1 \mathrm{km}}{4 \mathrm{km} \mathrm{h}^{-1}}$$ = $$1 / 4$$ h = 15 min
Distance along the river covered in this time
= Speed of the river x Time taken to cross the river
= 3 km h-1 × 15 min
= 0.75 km.

Question 14.
In a harbour, wind Is blowing at the speed of 72 km/h, and the flag on the mast of a boat anchored in the harbour flutters along the N-E direction. If the boat starts moving at a speed of 51 km/ h to the north, what is the direction of the flag on the mast of the boat?
Answer: When the boat moves, the flag experiences wind blowing at 51 km h-1 in the south direction. Let the flag be at an angle 0 with the East direction.

∴ The flag blows along the east direction.

Question 15.
The ceiling of a long hall is 25 m high. What Is the maximum horizontal distance, that a ball thrown with a speed of 40 m s-1 can go without hitting the ceiling of the hall?
Given hm = 25 m , g = 9.8 ms-2 v0 = 40 ms-1
hm = $$\frac{\left(v_{0} \sin \theta_{0}\right)^{2}}{2 g}$$
25 = $$\frac{\left(40 \sin \theta_{0}\right)^{2}}{2 \times 9.8}$$
i.e. (40 sin θ0)² = 490
⇒ 40 sin θ0 = $$7 \sqrt{10}$$
⇒ sin θ0 = $$\frac{7 \sqrt{10}}{40}$$
⇒ θ0 = 33.6°
Maximum horizontal distance
R = $$\frac{v_{0}^{2}}{g}$$ Sin 2 θ
= $$\frac{40^{2}}{9.8}$$ sin(2 × 33.6°)
= 150.5 m

Question 16.
A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much high above the ground can the cricketer throw the same ball?
Given R = $$\frac{v_{0}^{2}}{g}$$ sin 2 θ = 100 m
Since R is maximum, θ = 45°
R = $$\frac{v_{0}^{2}}{g}$$ = 100 m
Maximum height is achieved when throwing angle is 90°.
hmax = $$\frac{\left(v_{0} \sin 90^{\circ}\right)^{2}}{29}$$
= $$\frac{v_{0}^{2}}{g}$$ = $$\frac{1}{2} \times \frac{v_{0}^{2}}{g}$$
1/2 × 100 m = 50 m.
∴ Maximum height hmax = 50m.

Question 17.
A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone?
Radius of the circle r = 80 cm = 0.8m Angular speed
υ = $$\frac{14 \times 2 \pi r}{t}$$
υ = $$\frac{14 \times 2 \times 22 / 7 \times 0.8}{25}$$
υ = 2.816 ms-1
Acceleration = $$\frac{v^{2}}{r}$$ = $$\frac{2.816^{2}}{0.8}$$
= 9.9 ms-2
Direction of acceleration: Towards the centre of the circle and along the radius.

Question 18.
An aircraft executes a horizontal loop of radius 1.00 km with a steady speed of 900 km/h. Compare its centripetal acceleration with the acceleration due to gravity.
Radius r = 1.00 km = 1000 m
Speed v = 900 km h-1 = 250 ms-1
Centripetal acceleration
a = $$\frac{v^{2}}{r}=\frac{250^{2}}{1000}$$
a = 62.5 ms-2
Acceleration due to gravity g = 9.8 ms-2.
a = $$\frac{62.5}{9.8}$$ = 6.4 g

question 19.
Read each statement below carefully and state, with reasons, if it is true or false:

1. The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre
2. The velocity vector of a particle at a point is always along the tangent to the path of the particle at that point
3. The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector

1. False. The statement is true only if the particle is executing circular motion at a uniform speed (no tangential acceleration).
2. True. The instantaneous velocity is in the direction tangential to the instantaneous path of the particle.
3. True. Consider the two points and on the circle as shown in the figure.

At both these points, acceleration is towards the centre P. The net acceleration is O (equal magnitude but of opposite directions). Similarly, over one cycle, the net acceleration is a null vector.

Question 20.
The position of a particle is given by
$$r=3.0 t \hat{i}-2.0 t^{2} \hat{j}+4.0 \hat{k} m$$
where t is in seconds and the coefficients have the proper units for r to be in meters.

1. Find the v and an of the particle?
2. What is the magnitude and direction of velocity of the particle at t = 2.0 s?

= 8.54 ms-1
Direction θ = $$\tan ^{-1}\left(\frac{v_{y}}{v_{x}}\right)$$
= $$\tan ^{-1}\left(\frac{-8}{3}\right)$$
= – 69.44° with the horizontal (x – axis)

Question 21.
A particle starts from the origin at t = 0 s with a velocity of 10.0 J m/s and moves in the x-y plane with a constant acceleration of (8.0 I + 2.0 J) m s-2.

1. At what time is the x – coordinate of the particle 16 m? What is the y – coordinate of the particle at that time?
2. What is the speed of the particle at the time?

1. Along x – direction,
vx(0) = 0, ax = 8 ms-22
rx(t) = rx(0) + vx(o) t + 1/2 ax
= 0 + 0 + 1/2 × 8 × t² = 4 t²
Given rx(t1) = 16 m
4 t² = 16m
⇒ t = 2 s
∴ At time t = 2.0 s, x – coordinate of the particle = 16 m
Along y – direction,
vy(o) = 10 ms-1, ay = 2 ms-2
ry(t) = ry(0) + vy(0)t +1/2 ay
0 + 10 t + 1/2 × 2 × t²
= 10 t + t²
ry(2) = 10(2) + (2)² = 24 m
∴ y – coordinate at time t = 2.0 s is 24 m

2. Along x – direction,
Vx(t) = vx (0) + axt
= 0 + 8t = 8t
vx(2) = 16 ms-1
Along y – direction,
vy(t) = vy(0) + ayt
= 10 + 2t
vy (2) = 14 ms-1
Speed at time t a 2.0 s is $$\sqrt{\left(v_{x}(2)\right)^{2}+\left(v_{y}(2)\right)^{2}}$$
= $$\sqrt{16^{2}+14^{2}}$$ = 21.26 ms-1

Question 22.
I and J are unit vectors along the x- and y-axis respectively. What is the magnitude and direction of the vectors I+ J, and I – J? What are the components of a vector A = 2I+3J along with the directions of i + j and i – j? [You may use graphical method]

Question 23.
For any arbitrary motion in space, which of the following relations are true:

1. vaverage = (1/2) (v (t1 + v (t2)))
2. vaverage = [r(t2) – r(t1)] /(t2 – t2)
3. v (t) = v (0) + a t
4. r (t) = r (0) + v (0) t + (1/2) a t²
5. aaverage = [v(t2) – v(t1)]/(t2 – t1)

(The ‘average’ stands for an average of the quantity over the time interval t1 to t2)

1. False. Average velocity is not the arithmetic mean of the individual velocities.
2. True. By definition.
3. False. This is true only if the acceleration $$\overrightarrow{\mathrm{a}}$$ is constant.
4. False. This is true only if the acceleration $$\overrightarrow{\mathrm{a}}$$ is constant.
5. True. By definition.

Question 24.
Read each statement below carefully and state, with reasons and examples, if it is true or false:
A scalar quantity is one that

1. is conserved in a process
2. can never take negative values
3. must be dimensionless
4. does not vary from one point to another in space
5. has the same value for observers with different orientations of axes.

1. False. Mass is a scalar quantity and it is not conserved in a nuclear reaction.
2. False. Angle is a scalar quantity and it can take negative values.
3. False. Mass, current, etc. are scalar quantities that have dimensions.
4. False, Temperature, electric potential, etc. are scalar quantities that vary from point to point in a medium.
5. True. Change of axis does not change a scalar quantity. For example, mass is a scalar quantity that is independent of the axis chosen.

Question 25.
An aircraft is flying at a height of 3400 m above the ground if the angle subtended at a ground observation point by the aircraft positions 10.0 s apart is 30°, what is the speed of the aircraft?
Let O be the observation point. A and B are the initial and final positions of the aircraft.
The angle subtended θ = 30°

= $$\frac{\text { Distance coverd }}{\text { Time taken }}=\frac{1960 \mathrm{m}}{10 \mathrm{s}}$$ = 196 ms-1.

### 1st PUC Physics Motion in a Plane Additional Exercises Questions and Answers

Question 26.
A vector has magnitude and direction. Does it have a location in space? Can it vary with time Will two equal vectors a and b at different locations in space necessarily have Identical physical effects? Give examples in support of your answer.
Yes, a vector has a defined location in space. It may or may not vary with time. For example, $$3 \hat{i}+4 \hat{j}$$ is time independent whereas $$4 t \hat{i}-7 \sqrt{t} \hat{k}$$ is time dependent.
Two vectors at different locations need not have the same physical effects. Consider the weight of a person in newtons, ‘k’ newtons of weight have different meanings on earth and on the moon, and so they have different physical effects.

Question 27.
A vector has both magnitude and direction. Does it mean that anything that has magnitude and direction is necessarily a vector? The rotation of a body can be specified by the direction of the axis of rotation, and the angle of rotation about the axis. Does that make any rotation a vector?
A quantity that has magnitude and direction need not necessarily be a vector. Consider finite rotations of a particle about an axis. 2 complete rotations bring back the particle to its initial position but these rotations do not add up as per. Parallelogram law of addition. Hence, rotation is not a vector.

Question 28.
Can you associate vectors with

1. the length of a wire bent into a loop,
2. a plane area,
3. a sphere? Explain.

1. No.
2. Yes. The area $$\overrightarrow{\mathrm{A}}$$ can be defined as $$\overrightarrow{\mathrm{A}}$$ = $$\overrightarrow{I} \times \overrightarrow{\mathbf{b}}$$ where $$\overrightarrow{i}$$ and $$\overrightarrow{b}$$ are the length and breadth vbctors.
3. No.

Question 29.
A bullet fired at an angle of 30° with the horizontal hits the ground 3.0 km away. By adjusting its angle of projection, can one hope to hit a target 5.0 km away? Assume the muzzle speed to the fixed, and neglect air resistance.
Let ‘u’ be the muzzle speed.
0=30°
Range R = $$\frac{\mathfrak{u}^{2} \sin 2 \mathfrak{\theta}}{\mathfrak{g}}$$
3000 m = $$\frac{\mathrm{u}^{2}}{\mathrm{g}}$$ × sin (2 × 30°)
⇒ $$\frac{\mathrm{u}^{2}}{\mathrm{g}}$$ = $$\frac{3000}{\sin 60^{\circ}}$$ = 3464 m.
Maximum range possible Rmax = $$\frac{\mathrm{u}^{2}}{\mathrm{g}}$$
= 3.46 km
∴ It is not possible to hit a target 5 km away.

Question 30.
A fighter plane flying horizontally at an altitude of 1.5 km with speed 720 km/h passes directly overhead an antiaircraft gun. At what angle from the vertical should the gun be fired for the shell with muzzle speed 600 m s-1 to hit the plane? At what minimum altitude should the pilot fly the plane to avoid being hit?
(Take g = 10 m s-2).
Plane’s altitude
h = 1.5 km = 1500 m
Plane’s speed
Vp = 720 km/h = 200 ms-1
Shell’s speed
Vs = 600 ms-1
Let the shell be fired at an angle θ with the vertical.
Horizontal distance covered by the shell in time t = Horizontal distance covered by the plane in time t
⇒ Vs sin θ .t = Vpt
⇒ 600 sin θ = 200
⇒ θ = sin-1(1/3) = 19.5°
Minimum altitude to ensure no hit = Maximum height achieved by the shell
= $$\frac{\mathrm{u}^{2} \cos ^{2} \theta}{2 \mathrm{g}}$$
= $$\frac{600^{2} \cos ^{2}\left(19.5^{\circ}\right)}{2 \times 10}$$
= 1600 m
= 16 km.

Question 31.
A cyclist is riding with a speed of 27 km/h. As he approaches a circular turn on the road of radius 80 m, he applies brakes and reduces his speed at the constant rate of 0.50 m/s every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn?
Speed of the cyclist v = 27 km/h
= 7.5 ms-1.
Radius of the turn r = 80 m
Centripetal acceleration ac = $$\frac{v^{2}}{r}=\frac{7.5^{2}}{80}$$
= 0.70 ms -2
Tangential retardation ar = $$\frac{0.5}{1}$$ = 0.5 ms-2
Net acceleration a = $$\sqrt{a_{c}^{2}+a_{r}^{2}}$$
= $$\sqrt{0.70^{2}+0.5^{2}}$$
= 0.86 ms-2
Angle θ with the tangent is given by
θ = tan-1 $$\left(\frac{\mathrm{a}_{\mathrm{r}}}{\mathrm{a}_{\mathrm{t}}}\right)$$
θ = tan-1$$\left(\frac{0.70}{0.50}\right)$$
θ = 54.5°.

An online projectile motion calculator allows you to compute the velocity, maximum height, and flight parameters at a given time in a fraction of a second.

Question 32.
(a) Show that for a projectile the angle between the velocity and the x-axis as a function of time is given by

(b) Shows that the projection angle θ for a projectile launched from the origin is given by

where the symbols have their usual meaning.
(a) Let the horizontal component of the initial velocity be vox and the vertical component be voy.
During motion, the horizontal component is still vox but the vertical component
= Voy – gt

(b) For the projectile

### 1st PUC Physics Motion in a Plane One Mark Questions and Answers

Question 1.
What is the work done by a body in circular motion?
Zero.

Question 2.
What is the trajectory of a projectile?
Parabola.

Question 3.
What is velocity of projection?
The velocity with which the body is projected is called velocity of projection.

Question 4.
What is angle of projection?
The angle between the direction of projection and the horizontal drawn at that point is called the angle of projection.

Question 5.
What is range of projectile?
The maximum horizontal distance covered by the projectile is called the range.

Question 6.
What is time of flight?
The time taken to describe the range is called the time of flight.

Question 7.
When is the range of a projectile is maximum?
When the angle of projection is 45° the range of a projectile is maximum.

Question 8.
Why should an athlete throw the javelin or shot put approximately at an angle of 45°?
An athlete must throw shot – put or a javelin approximately at an angle 45°to achieve the maximum range.

Question 9.
What is uniform circular motion?
Motion of a body along a circular path with constant speed is called uniform circular motion.

Question 10.
What is the direction of motion (or velocity) of a body in uniform circular motion?
The direction of motion of the body at any instant will be along the tangent to the circular path.

Question 11.
Define angular displacement.
The angle through which the radius vector rotates is called angular displacement.

Question 12.
Define angular velocity.
The rate of angular displacement is called angular velocity.

Question 13.
Define period of revolution of a body.
It is the time taken by the body to complete one revolution.

Question 14.
Define frequency of revolution of a body.
Number of revolutions completed by the body in one second is called frequency of revolution.

Question 15.
Define centripetal acceleration.
In uniform circular motion, the acceleration produced on the body acts along the radius, towards the centre of the circular path and is known as centripetal acceleration or radial acceleration.

Question 16.
What is the source of centripetal force for the planets revolving round the sun?
The gravitational force of attraction between the sun and the planets provides the centripetal force for the revolution of the planets round the sun.

Question 17.
Raising the outer edge of the road to get the required centripetal force for a vehicle is known as banking of roads (or track).

Question 18.
What is the work done by centripetal force?

Question 19.
A body moves along a circle of radius 3m. What is the distance travelled. When it completes one circle?
Distance travelled by. the body when it completes one circle = circumference of the circle
= 2 π r = 2 × 3.4 × 3
= 18.84 m

Question 20.
Give an example for three dimen-sional motion.
Flight of Aeroplane or movement of gas molecule in space.

Question 21.
Give an example of centripetal force.
Electrostatic force between an electron and nucleus provides centripetal force for the revolution of the electron in its orbit.

Question 22.
Which vector quantity becomes zero at highest point of motion of projectile?
Vertical component of velocity becomes zero at the highest point.

Question 23.
State the law of parallelogram of two forces.
If two forces acting at a point are represented both in magnitude and direction by the two adjacent sides of a parallelogram then their resultant is represented by the diagonal of the parallelogram drawn from the same point.

Question 24.
What will be the effect on the horizontal range of a projectile when its initial speed is doubled keeping the angle the same?
Since R α v0², doubling the initial speed will increase the range by a factor of 2² = 4 times.

Question 25.
What is the minimum number of forces acting on a object in a plane that can produce a zero resultant force?
Two. (The forces should have equal magnitude but opposite direction).

Question 26.
A unit vector Is represented by $$\mathbf{a} \hat{\mathbf{i}}+\mathbf{b} \hat{\mathbf{j}}+\mathbf{c} \hat{\mathbf{k}}$$. If the values of a and b are
0.6 and 0.8 respectively, find the value of C.

⇒ c = 0

### 1st PUC Physics Motion in a Plane Two Mark Questions and Answers

Question 1.
When is a body said to have two-dimensional motion? Give an example.
Motion of a particle in a plane is known as two-dimensional motion. Example:

• A car moving along a zig-zag path on a road.
• Motion of the planet around the sun in its orbit.

Question 2.
What is a projectile? Give an example.
Anybody was thrown into space, that moves under the effect of gravity is called a projectile.
E.g.:

• A bullet fired from the gun.
• A javelin thrown by an athlete.

Question 3.
Obtain the relation between angular velocity and linear velocity.
Consider a particle moving in a circle of radius ‘r’ with uniform angular velocity ‘ ω ’ and linear speed ‘V’. Let the particle moves from A to B in time ‘t’ seconds through a small distance ‘s’ on the circumference. Let θ be the angular displacement.

Angular velocity ω = $$\frac{\theta}{\mathfrak{t}}$$
Angular displacement θ = $$\frac{s}{\mathfrak{r}}$$
∴ ω = $$\frac{\mathrm{s}}{\mathrm{tr}}$$
But $$\frac{s}{\mathfrak{t}}$$ = v, linear velocity
∴ ω = $$\frac{\mathrm{v}}{\mathrm{r}}$$ or
v = r ω
linear velocity = radius × angular velocity.

Question 4.
What is angle of banking? Give the expression for it.
The angle through which the outer edge of the roads are raised is called the angle of banking. The angle of banking is given by,
θ = tan-1$$\left(\frac{v^{2}}{r g}\right)$$
where m is the mass, v is the velocity and r is the radius.

Question 5.
A body travels one round in a circle of radius ‘R’- What is the

• displacement and.
• distance traveled.

• The displacement of a body = 0
• The distance travelled = 2nR

Question 6.
Write the expressions for the maximum height reached by a projectile, and explain the terms.
The expression for the maximum height reached by the projectile is
$$\mathrm{H}=\frac{u^{2} \sin ^{2} \theta}{2 \mathrm{g}}$$
Where u, is the initial velocity of the projectile, θ is the angle between the horizontal & the initial velocity, g is the acceleration due to gravity.

Question 7.
What is banking of roads? Give an example of it.
Vehicles while taking a turn cannot incline themselves to one side to get the required centripetal force. To provide necessary centripetal force without slipping. The outer edge of the road is raised over the inner edge.

Question 8.
Why a cyclist bends inward when to Is crossing a curve in a road?
He bends inward to get sufficient centripetal force.

Question 9.
Two equal forces have their resultant equal to either. What is the inclination between them?

= A
⇒ 2A² (1 + cos θ) = A²
⇒ 1 + cos θ = 1/2
⇒ cos θ = -1/2
⇒ θ = 120°

Question 10.
A cyclist has to bend a little Inwards from his vertical position while turning. Why?
Bending provides a component of the normal reaction force from the ground to provide the cyclist the necessary centripetal force for turning. Hence, bending is necessary.

Question 11.
Suppose you have two forces $$\overrightarrow{\mathrm{F}}$$ and $$\overrightarrow{\mathrm{F}}$$. How would you combine them in order to have resultant forces of magnitudes

1. zero
2. 2 $$\overrightarrow{\mathrm{F}}$$, and
3. $$\overrightarrow{\mathrm{F}}$$?

1. If the forces act in opposite directions, the resultant is zero.
2. If the forces act in the same direction, resultant $$\overrightarrow{\mathrm{R}}$$ = 2 $$\overrightarrow{\mathrm{F}}$$.
3. For $$\overrightarrow{\mathrm{R}}$$ = 2 $$\overrightarrow{\mathrm{F}}$$
$$\sqrt{\mathrm{F}^{2}+\mathrm{F}^{2}+2 \mathrm{F} \cdot \mathrm{F} \cdot \cos \theta}$$ = F
⇒ 2F² (1 + cos θ) = F²
⇒ 1 + cos θ = 1/2
⇒ θ = 120°

Question 12.
Prove the following statement: “For elevation which exceeds or falls short of 45° by equal amounts, the range Is equal”.
At 45° – θ, the range is

At 45° + θ, the range is

⇒ R2 = R1 Proved.

Question 13.
What is the distance traveled by a point during time ‘t’, if it moves in x – y plane according to the relation x = a sin cot and y = a(1 – cos cot ωt)?
Distance travelled d(t) =

Question 14.
A lady walking due east on a road with velocity 10 ms-1 encounters rain falling vertically with a velocity of 30ms-1. At what angle should she hold her umbrella to protect herself from the rain?

⇒ θ = 18°16′ with the vertical direction.

Question 15.
If the vectors of equal magnitude add to either of them by magnitude, what is the angle between them?

1. If the forces act in opposite directions, the resultant is zero.
2. If the forces act in the same direction, resultant $$\overrightarrow{\mathrm{R}}$$ = 2 $$\overrightarrow{\mathrm{F}}$$.
3. For $$\overrightarrow{\mathrm{R}}$$ = 2 $$\overrightarrow{\mathrm{F}}$$
$$\sqrt{\mathrm{F}^{2}+\mathrm{F}^{2}+2 \mathrm{F} \cdot \mathrm{F} \cdot \cos \theta}$$ = F
⇒ 2F² (1 + cos θ) = F²
⇒ 1 + cos θ = 1/2
⇒ θ = 120°

Question 16.
A swimmer can swim with velocity of 10km h-1 w.r.t. the water flowing in a river with velocity of 5 km h-1. In what direction should he swim to reach the
point on the other bank just opposite to his starting point?
Let the angle of the swimmer be e w.r.t the horizontal.

⇒ cos θ = 1/2
⇒ θ = 60°

Question 17.
The angle between vector $$\overrightarrow{\mathbf{A}}$$ and \overrightarrow{\mathbf{B}} is 60°. What is the ratio of $$\overrightarrow{\mathbf{A}} \cdot \overrightarrow{\mathbf{B}}[latex] and [latex]|\overrightarrow{\mathbf{A}} \times \overrightarrow{\mathbf{B}}|[latex]? Answer: Required ratio [latex]=\frac{\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}}{\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}}$$

= cot θ
= cot (60°)
= $$1 / \sqrt{3}$$

Question 18.
Two forces P=10 N and Q = 15 N are acting at a point making an angle 30° with each other. What is the cross product of P
and Q?
P= 10N, Q = 15N, θ =30°
From the equation P × Q = P.QSine
= 10.15 Sin 30° = 75N.

### 1st PUC Physics Motion in a Plane Three Mark Questions and Answers

Question 1.
Find a unit vector parallel to the vector $$3 \hat{i}+7 \hat{j}+4 \hat{k}$$

Unit vector parallel to $$\overrightarrow{\mathrm{a}}=\frac{\overrightarrow{\mathrm{a}}}{|\overrightarrow{\mathrm{a}}|}$$
$$=\frac{3 \hat{i}+7 \hat{j}+4 \hat{k}}{\sqrt{74}}$$

Question 2.
What is a projectile? Show that the path followed by an oblique projectile Is a parabola.
Anybody was thrown into space, that moves under the effect of gravity is called a projectile.
E.g.:
1. A bullet fired from the gun.
2. A javelin thrown by an athlete., Consider a particle thrown up at an angle θ to the horizontal with a velocity u. The velocity of projection can be resolved into two component u cos θ and usinq along OX and OY respectively.

The horizontal component remains constant while the vertical component is affected by gravity.
Let P(x, y) be the position of the projectile after ‘t’ seconds.
Then distaince travelled ‘x’ along OX is given by,
x = horizontal component of the velocity time
x = u cos θ × t ………… (1)
The distance travelled along the vertical direction in same time‘t’ is,
y = (u sin θ) t – 1/2 gt² ……………… (2)
Substituting, t = $$\frac{x}{u \cos \theta}$$ from (1) in (2),

…………………..(3)
where a = tan θ, $$\mathrm{b}=\frac{\mathrm{y}=\mathrm{gax}-\mathrm{bx}^{2}}{2 \mathrm{u}^{2} \cos ^{2} \mathrm{\theta}}$$ are constant for given values of θ, u and g.
The equation (3) represents a parabola. Therefore trajectory of a projectile is a parabola.

Question 3.
Establish a relation between linear velocity and angular velocity in a uniform circular motion and explain the direction of the velocity.
The distance ‘s’ covered by a body travelling in an arc of radius Y and turning its radial line by ‘θ’ is given by
s = r θ
Differentiating both sides w.r.t. time, we have
$$\frac{\mathrm{ds}}{\mathrm{dt}}=\mathrm{r} \frac{\mathrm{d} \theta}{\mathrm{dt}}$$
i.e., v = rw
or Linear velocity = radius × angular velocity.
At each point the body moves along the tangent.
The presence of centripetal force
F =$$\frac{m v^{2}}{r}$$ makes the body to travel in the circular path.
Thus, the direction of velocity is always along the tangent at any point in the circular path.

Question 4.
Find the magnitude and direction of the resultant of two forces p and Q in terms of their magnitudes and angle Q between them.

P and Q are two forces acting at a point O; making an angle θ. They are represented by OA and OB. the diagonal OC of the parallelogram OBCA represents the resultant R.
Draw CD perpendicular to extended line of OA. In the right angle triangle ODC.
OC² = OD² + CD²
= (OA + AD)² + CD²
= OA² + 2OA. AC cos θ + AC²
R² = P² + 2PQcos θ + Q² v
AC² = AD² + CD² & $$\frac{\mathrm{AD}}{\mathrm{AC}}$$ = cosq
Magnitude of the resultant
R = $$\sqrt{\mathrm{P}^{2}+\mathrm{Q}^{2}+2 \mathrm{PQ} \cdot \cos \theta}$$
To find the direction:
If a is the angle made by R with P, then from the right angled triangle ODC

Question 5.
In long jump, does it matter how high you jump? What factors determine the span of the jump?
In order to increase the span of the jump, it is necessary for the horizontal component of the velocity to be more than the vertical component. The height of the jump does not matter.
The factors that determine the span of the jump are:

1. angle of elevation at the time of jumping.
2. Speed of the runner at the time of jumping.

Question 6.
Determine a unit vector which Is per-pendicular to both $$\vec{A}=2 \hat{i}+\hat{j}+\hat{k}$$ and $$\overrightarrow{\mathbf{B}}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$$
The unit vector $$\hat{n}$$ is given by

Question 7.
An accelerating train is passing over a high bridge. A stone is dropped from the train at an instant when Its speed is 10 ms-1 and acceleration is 1ms-2. Find the horizontal and vertical components of the velocity and acceleration of the stone one second after it is dropped.Take g = 10 ms-2.
Horizontal acceleration = 0 (because there is no force acting on the stone in the horizontal direction to provide any acceleration)
Horizontal velocity = 10 ms-1.
Vertical acceleration = g = 10ms-2.
Vertical velocity = ut + 1/2 gt²
= 0(1) + 1/2 (10)(1)²
= 5 ms-1.

Question 8.
A projectile is projected with velocity ‘u’ making an angle θ with the horizontal direction. Find:

1. Time of flight
2. Horizontal range

1. Let H be the maximum height reached by the projectile in time t1
For vertical motion,
The initial velocity = usin θ
The final velocity = 0
Acceleration = – g
using, v² = u² + 2as
0 = u² sin2 θ – 2gH
2gH =u²sin2 θ
u² sin² θ
H = $$\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{g}}$$

2. Let t, be the time taken by the projectile to reach the maximum height H.
For vertical motion,
initial velocity =usin0
Final velocity at the maximum height = 0
Acceleration a = – g
Using the equation v = u + at1
0 = usin θ – gt1
gt1 = usin θ
t1 = $$\frac{\mathrm{usin} \theta}{\mathrm{g}}$$
Let t2 be the time of descent.
But t1 = t2
i.e. time of ascent = time of descent.
Time of flight T = t1 + t2 = 2t1
∴ T = $$\frac{2 u \sin \theta}{g}$$

Question 9.

1. Show that for two complementary angles of projection of a projectile with the same velocity, the horizontal ranges are equal.
2. For what angle of projectile is the rangle maximum?
3. For what angle of projection of a projectile are the horizontal range and height attained by the projectile equal?

1. For projection angle θ,
R1 = $$\frac{u^{2} \sin 2 \theta}{g}$$
For projection angle 90° – θ,

But sin (180° – 2θ) = sin2θ
∴ R2 = $$\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{g}}$$ = R1

2. Range is maximum when projection angle = 45°.

3. Given R = H
⇒ $$\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{g}}$$ = $$\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{g}}$$
⇒ 2 sin θ cos θ = $$\frac{\sin ^{2} \theta}{2}$$
⇒ cot θ = 1/4
θ = cot-1(1/4)
⇒ θ = 75.96°.

Question 10.
What is meant by centripetal acceleration? Derive the formula for centripetal acceleration.
In uniform circular motion, the acceleration produced on the body acts along the radius, towards the centre of the circular path. This acceleration is known as centripetal acceleration or radial acceleration.
Let a particle of mass m be moving around the circle of radius r with a uniform speed ‘v’.
Let the particle moves from A to B in a time ‘t’ seconds covering a small angle θ.

At the point A the velocity of the particle is along the tangent AC and therefore there is no component along AO (i.e. along radius).
At the point B the velocity of the particle is along BD. This can be resolved into two components v cos θ and v sin θ. Since θ is small sin θ ≅
θ and cos θ ≅ 1.
∴ change in velocity parallel to AC
= v cos θ – v = v -v = 0. ∵ cos θ ≅ 1.
∴ There is no acceleration along the tangent,
Change in velocity parallel to OA is
= vsin θ – 0 = v θ ( sinθ ≅ θ)
∴ Acceleration along AO = $$\frac{\mathrm{v} \theta}{\mathrm{t}}$$
i.e., centripetal acceleration = $$\frac{\mathrm{v} \theta}{\mathrm{t}}$$ = v ω
∴ centripetal force = mass × centripetal acceleration
F=m v ω
F = mw²r = $$\frac{m v^{2}}{r}$$ ∵v = ω r

Question 11.
From the top of a tower of 100m height, a ball is dropped and at the same time another ball is projected vertically upwards from the ground with a velocity of 25ms-1. Find when and where the two balls will meet. Take g = 9.8 ms-2
Let AC = x
Then BC = 100 – x
x = ut + 1/2 g t²
x = 4.9 t² …………… (1)

BC = ut – 1/2 g t²
= 25 t – 1/2 × 9.8 t²
100 – x = 25 t – 4.9 t² ………………. (2)
Substituting 4.9 t² = x from (1) in (2)
100 – 2 = 25 t – x
⇒ t = 4 seconds
x = 4.9 t²
= 4.9 × 4²
= 78.4 m from the top or 21.6 m from the ground.

### 1st PUC Physics Motion in a Plane FourFive Marks Questions and Answers

Question 1.
Show that the trajectory of a projectile is a parabola.
Consider a particle thrown up at an angle q to the horizontal with a velocity u. The velocity of projection can be resolved into two component u cos θ and u sin θ along OX and OY respectively.

The horizontal component remains constant while the vertical component is affected by gravity.
Let P(x, y) be the position of the projectile after ‘t’ seconds.
Then distance travelled ‘x’ along OX is given by,
x = horizontal component of the velocity × time
x = ucos θ × t …………………. (1)
The distance travelled along the vertical direction in same time ‘t’ is,
y = (usin θ) t – 1/2gt² ………… (2)
Substituting, t = $$\frac{x}{u \cos \theta}$$ from (1) in (2),

y = ax – bx² …………………… (3) where
a = tan θ, $$\mathrm{b}=\frac{\mathrm{y}=\mathrm{gax}-\mathrm{bx}^{2}}{2 \mathrm{u}^{2} \cos ^{2} \mathrm{\theta}}$$, are constants
for given values of θ, u and g.
Equation (3) represents a parabola. Therefore trajectory of a projectile is a parabola.

Question 2.
Derive an expression for the maximum height reached, time of flight and range of a projectile.
(i) Let H be the maximum height reached by the projectile in time t1 For vertical motion,
The initial velocity = usin θ
The final velocity = 0
Acceleration = – g
∴ using, v² = u² + 2as
0 = u²sin² θ – 2gH
2gH = u²sin² θ
H = $$\frac{u^{2} \sin ^{2} \theta}{2 g}$$

(ii) Let t, be the time taken by the projectile to reach the maximum height H.
For vertical motion,
initial velocity = usin θ
Final velocity at the maximum height = 0
Acceleration a = – g
Using the equation v = u + at1
0 = usin θ – gt1
gt1 = usin θ
t1 = $$\frac{\mathrm{usin} \theta}{\mathrm{g}}$$
Let t2 be the time of descent.
But t1 = t2
i.e. time of ascent = time of descent.
∴ Time of flight T = t1 + t2 = 2t1
∴ T = $$\frac{2 u \sin \theta}{\dot{g}}$$

(iii) Let R be the range of the projectile in a time T. This is covered by the projectile with a constant velocity ucos θ.
Range = horizontal component of
velocity × Time of flight
i.e, R = ucos θ. T
R = ucos θ. $$\frac{2 u \sin \theta}{g}$$
R= $$\frac{u^{2} \sin 2 \theta}{g}$$
∵ 2 sin θ.cos θ = sin2 θ.

Question 3.
Derive an expression for the centripetal force.
Let a particle of mass m be moving around the circle of radius r with a uniform speed ‘V’.
Let the particle moves from A to B in a time ‘t’ seconds covering a small angle θ.

At the point A the velocity of the particle is along the tangent AC and therefore there is no component along AO (i.e. along radius).
At the point B the velocity of the particle is along BD. This can be resolved into two components v cos θ and v sin θ. Since θ is small sin θ ≅
θ and cos θ ≅ 1.
∴ change in velocity parallel to AC
= v cos θ – v = v -v = 0. ∵ cos θ ≅ 1.
∴ There is no acceleration along the tangent,
Change in velocity parallel to OA is
= vsin θ – 0 = v θ ( sinθ ≅ θ)
∴ Acceleration along AO = $$\frac{\mathrm{v} \theta}{\mathrm{t}}$$
i.e., centripetal acceleration = $$\frac{\mathrm{v} \theta}{\mathrm{t}}$$ = v ω
∴ centripetal force = mass × centripetal acceleration
F=m v ω
F = mw²r = $$\frac{m v^{2}}{r}$$ ∵v = ω r.

Question 4.
Define centripetal force and give example.
In uniform circular motion the force on the rotating body acts along the radius towards the centre of the circular path and is known as centripetal force.
Eg:

1. In the case of a stone rotated round the circle by means of a string, the centripetal force is provided by the tension in the string.
2. The force of gravitational attraction towards the sun is the centripetal force keeping the planets orbiting round the sun.
3. In an atom the electrons revolve round the nucleus in the circular orbit the centripetal force is provided by the electrostatic force of attraction between the nucleus and the electrons.

Question 5.
What is the angle of banking for a curve road radius 180m suitable for a maximum speed of 30m? Calculate the maximum speed to be maintained when the angle of banking is
θ = 30° (g = 9.8ms1) (Rural 2005)
Angle of banking 0=? Acceleration due to gravity, g =9.8ms-2
From the equation,
tan θ = $$\frac{v^{2}}{r g}$$ = $$\frac{30^{2}}{180 \times 9.8}$$ = $$=\frac{900}{1764}$$
= 0.5102
θ = tan-1(0.5102)
= 27° 1°
When the angle of banking θ = 30°,
From the equation tan θ = $$\frac{v^{2}}{r g}$$
tan θ 30° = $$\frac{v^{2}}{180 \times 9.8}$$
v2 = 180 × 9.8 × tan30°
= 1764 × 0.5774 = 1018.5
v = 32ms-1

Question 6.
A bullet fired from rifle attains a maximum height of 50m and crosses a range of 200m. Find the angle of projection.
Maximum height attained by the bullet, H=50m,
Horizontal range R=200m,
Angle of projection 0 =?
From the equation H = $$\frac{u^{2} \sin ^{2} \theta}{2 g}$$
50 = $$\frac{u^{2} \sin ^{2} \theta}{2 g}$$ ……………….(1)
From the equation R = $$\frac{u^{2} 2 \sin \theta \cos \theta}{g}$$
200 = $$\frac{u^{2} 2 \sin \theta \cos \theta}{g}$$ ………………. (2)
Divide equ (1) by equ (2), we have
$$\frac{50}{200}=\frac{\sin \theta}{4 \cos \theta}$$
i.e $$\frac{1}{4}$$ = $$\frac{1}{4}$$ tan θ Thus tan θ = 1
θ = tan-1(1.0000) = 45°.

Question 7.
A train of mass 10,000 kg moving at 72km ph rounds a curve whose radius of curvature is 200m. What is its acceleration? What is the centripetal force?
m = 10,000kg, v = 72 km/h
= $$\frac{72 \times 1000}{60 \times 60}$$ = 20m/s
r = 200 m.
Acceleration a = v ω
= $$v\left(\frac{v}{r}\right)$$
= $$\frac{v^{2}}{r}=\frac{(20)^{2}}{200}$$ = 2m/s2
Centripetal force, F = $$\frac{m v^{2}}{r}$$
= $$\frac{10000 \times(20)^{2}}{200}$$ = 20000 N.

Question 8.
An object is projected with a velocity of 60ms-1 in a direction making an angle of 60° with the horizontal. Find

1. the maximum height
2. the time taken to reach maximum height
3. the horizontal range

u = 60m/s, θ = 60°.
1. Maximum height
H = $$\frac{u^{2} \sin ^{2}(\theta)}{2 g}$$ = $$\frac{60^{2} \sin ^{2}(60)}{2 \times 9.8}$$ = 137.74 m.

2. Time to reach maximum height
t = $$\frac{u \sin (\theta)}{g}$$ = $$\frac{60 \times \sin (60)}{9.8}$$ = 5.30s.

3. Range
R = $$\frac{u^{2} \sin (2 \theta)}{g}$$ = $$\frac{60^{2} \sin (2 \times 60)}{9.8}$$ = 318.12 m

Question 9.
Drive an expression for magnitude and direction of the resultant of two forces acting at a point.

P and Q are two forces acting at a point O, making an angle θ. They are represented by OA and OB. the diagonal OC of the parallelogram OBCA represents the resultant R.
Draw CD perpendicular to extended line of OA. In the right angle triangle ODC.
OC² = QD² + CD²
= (OA + AD)² + CD²
= OA² + 20A. AC cosq + AC²
R² = P² + 2PQcos θ + Q²
AC² = AD²+ CD² & $$\frac{A D}{A C}$$ = cos θ
Magnitude of the resultant
R = $$\sqrt{\mathrm{P}^{2}+\mathrm{Q}^{2}+2 \mathrm{PQ} \cdot \cos \theta}$$
To find the direction:
If a is the angle made by R with P, then from the right angled triangle ODC

Question 10.
A projectile shot at an angle of 60° above the horizontal ground strikes a vertical wall 30 m away at a point 15m above the ground. Find the speed with which the projectile was launched and the speed with which it strikes the wall. Take g = 10 ms-2
Let the initial velocity of the projectile be u
Horizontal component = ux

Horizontal distance, d = ux t
i.e., 30 = $$\frac{\mathrm{u}}{2}$$ t
⇒ t = $$\frac{60}{\mathrm{u}}$$ ………… (1)
Vertical component uy = usin60°
= $$\frac{\mathrm{u} \sqrt{3}}{2}$$
Vertical displacement s = uyt – 1/2 g t²
15= $$\frac{\mathrm{u} \sqrt{3}}{2}$$ t – 1/2 × 10 × t²
Substitute t = 60/u from (1),

⇒ u = 22.07 ms-1
∴Speed of launch = 22.07 ms-1
At the time of striking the wall,
vx = ux = 22.07 cos 60°
= 11.04 ms-1
vy = uy – gt
= 22.07 sin 60° – 10 × $$\left(\frac{60}{22.07}\right)$$
vy = -8.07 ms-1
Resultant speed v = $$\sqrt{v_{x}^{2}+v_{y}^{2}}$$
= $$\sqrt{11.04^{2}+(-8.07)^{2}}$$
= 13.68 ms-1</sup

Question 11.
Define projectile. Show that the path of projectile is a parabola. Find the angle of projection at which the horizontal range and maximum height of the projectile are equal.
Consider a particle thrown up at an angle q to the horizontal with a velocity u. The velocity of projection can be resolved into two component u cos θ and u sin θ along OX and OY respectively.

The horizontal component remains constant while the vertical component is affected by gravity.
Let P(x, y) be the position of the projectile after ‘t’ seconds.
Then distance travelled ‘x’ along OX is given by,
x = horizontal component of the velocity × time
x = ucos θ × t …………………. (1)
The distance travelled along the vertical direction in same time ‘t’ is,
y = (usin θ) t – 1/2gt² ………… (2)
Substituting, t = $$\frac{x}{u \cos \theta}$$ from (1) in (2),

y = ax – bx² …………………… (3) where
a = tan θ, b = $$\frac{\mathrm{g}}{2 \mathrm{u}^{2} \cos ^{2} \theta}$$, are constants
for given values of θ, u and g.
Equation (3) represents a parabola. Therefore trajectory of a projectile is a parabola.

1st PUC Physics Motion in a Plane Numerical Problems Questions and Answers

Question 1.
A ball is thrown Into air with a speed of 62 ms-1 at an angle of 45° with the horizontal. Calculate

1. the maximum height attained
2. the horizontal range
3. the time of flight and
4. the velocity of the ball after 4 seconds.

Solution:
1. Maximum height, H = $$\frac{u^{2} \sin ^{2} \theta}{2 g}$$ Here,
u = 62 ms-1, θ = 45°, g = 9.8 ms-2

= $$\frac{62 \times 62}{4 \times 9.8}$$ = 98.06 m.

2. Horizontal range. R is given by,
R = $$\frac{u^{2} \sin 2 \theta}{g}$$
= $$\frac{62 \times 62 \times \sin 90}{9.8}$$
= $$\frac{62 \times 62}{9.8}$$ = 392.2m.

3. The time of flight,
T = $$\frac{2 u \sin \theta}{g}$$
= $$\frac{2 \times 62 \times \sin 45}{9.8}$$
= $$\frac{2 \times 62}{\sqrt{2} \times 9.8}$$ = 8.95 s.

4. The horizontal component of the velocity remains constant throughout.
∴ Horizontal component of the velocity after 4 seconds vx = ucos θ
Here u = 62 ms-1 and θ = 45°
∴ vx = 62 cos45° = $$\frac{62}{\sqrt{2}}$$ = 44.3 s.
The vertical component of the velocity changes due to the acceleration due to gravity.
We have vy = uy + at,
Here, uy = usin θ; a = -g; t = 4 s;
From vy = usin θ – 4g
∴ vy = 62 sin45 – 9.8 × 4
= $$\frac{62}{\sqrt{2}}$$ -39.2 = 4.64 ms-1
Magnitude of the resultant velocity after 4 seconds is,
v = $$\sqrt{v_{x}^{2}+v_{y}^{2}}$$
= $$\sqrt{(44.3)^{2}+(4.64)^{2}}$$ = 44.5 ms-1
Direction of the velocity is given by,
tan a = $$\frac{v_{y}}{v_{x}}$$ = $$\frac{4.64}{44.3}$$ = 0.104
∴ a = tan-1(0.104) = 5°5’

Question 2.
A bullet is fired at an angle of 60° with the vertical with certain velocity hits the ground 3 km away is it possible to hit the target 5 km away by adjusting the angle of projection? If not, what must be the velocity of projection for the same angle of projections?
Solution:
Let ‘u’ be the velocity of projection.
Angle of projection ‘θ’ = 90 – 60° = 30°
Horizontal Range,
R = $$\frac{u^{2} \sin 2 \theta}{9}$$ = $$\frac{u^{2} \sin 60}{g}$$
∴ u2 = $$\frac{\mathrm{Rg}}{\sin 60}$$ …………… (1)
For a given velocity, the maximum range attained is given by making sin2 θ =1, and
Rmax $$\frac{u^{2}}{g}$$.
Substituting for u² from (1), we have
Rmax = $$\frac{\mathrm{R}}{\sin 60}$$
= $$\frac{3}{0.866}$$ = 3.46km
Since the maximum range is less than 5 km, it is not possible to hit the target 5km away.
From equation (1), we have,

= 56582
∴ u = 237.9 ms-1
Thus the bullet will hit the target if it is projected with a velocity of 237.9 ms-1

Question 3.
A stone of mass 0.5 kg is tied to one end of a string and whirled along the horizontal circle of 1.2 m radius, If the period of rotation is 5 s, calculate the tension in the string. If breaking tension is 1.5 N, calculate the maximum speed with which it is rotated and also calculate the corresponding period of rotation.
Solution:
When the mass is rotated along a horizontal circle, the centripetal force is equal to the tension of the string.
Hence, T= mw²r,
Here, m = 0.5 kg,
w = $$\frac{2 \pi}{T}$$ = $$\frac{2 \pi}{5}$$
r = 1.2 m.
∴ T = 0.5 × $$\left(\frac{2 \pi}{5}\right)^{2}$$ × 1.2
= $$\frac{0.5 \times 4 \pi^{2} \times 1.2}{25}$$
= 0.9475 N
The maximum tension which the string can withstand without breaking is 1.5 N and the corresponding velocity ‘v’ is given by, 1.5 = $$\frac{m v^{2}}{r}$$
∴ v2 = $$\frac{1.5 \times r}{m}$$ or
v = $$\sqrt{\frac{1.5 \times 1.2}{0.5}}$$
= 1.89 ms-1
The corresponding angular velocity is given by,
w = $$\frac{\mathrm{v}}{\mathrm{r}}$$ = $$\frac{1.89}{1.2}$$
∴ The period of rotation,
T = $$\frac{2 \pi}{1.575}$$ = 3.99 s.

Question 4.
At what angle should a cyclist lean over from the vertical while negotiating a curve of radius 58m with a speed of 15 km/hour?
Solution:
The angle of banking is given by,
tan θ = $$\frac{v^{2}}{r g}$$
Here, v = $$\frac{15000}{60 \times 60}$$ = 4.17 ms-1
∴ tan θ = $$\frac{(4.17)^{2}}{58 \times 9.8}$$ = 0.03059
θ = tan-1(0.03059)
= 1°45’.

Question 5.
A train is moving round a curve of 52m radius and the distance between the rails is 1.2 m by how much should the outer rails be raised above the inner one so that a train running at the rate of 45 kmhr-1 may not skid.
Solution:
The angle of banking is given by,
tan θ = $$\frac{v^{2}}{r g}$$
Here, v = $$\frac{45000}{60 \times 60}$$ = 12.5 ms-1 and
r = 52 m
∴ tan θ = $$\frac{(12.5)^{2}}{52 \times 9.8}$$ = 0.3066
θ = tan-1(0.3066)
= 17°2’
If ‘h’ is the height of the outer rail above the inner one, then
h = 1.2’sin 17.05°
= 0.352 m.

Question 6.
A golf ball leaves the tee with a velocity of 50ms-1 at an angle of 300 with the horizontal. Find its

1. time of flight
2. Range of projectile and
3. The velocity with which it hits the ground at the end of its flight.

u = 50m/s, θ = 30°.
1. Time of flight
T = $$\frac{2 u \sin (\theta)}{g}$$ = $$\frac{2 \times 50 \times \sin (30)}{9.8}$$ = 5.10 s

2. Range R = $$\frac{u^{2} \sin (2 \theta)}{g}$$ = $$\frac{50^{2} \sin \left(2 \times 30^{\circ}\right)}{9.8}$$ = 220.92 m

3. Velocity at the end = velocity of projection = 50 m/s

Question 7.
The greatest and the least resultants of two forces acting at a point are 5 N and1N respectively. Find the forces. What is the resultant of these two forces when they act at an angle of 50°?
Solution:
Let P and Q be the two forces. The greatest resultant is
P + Q = 5 …………… (1)
The minimum resultant is,
P – Q = 1…………….. (2)
2P = 6
P =3N
∴ Q = 2 N
When they are acting at an angle,
θ =50°

= 4.55 N
The direction of the resultant is,

= 0.3575
∴ a = 19.67° = 19° 40′
Therefore the resultant is making an angle of 19°40′ with 3N.

Question 8.
Four forces of magnitudes 2N, 3N, 4N, and 5N are acting on a body at a point are inclined at 30°, 60°, 90°, and 120° respectively with the horizontal. Find their resultant.
Solution:

Given, P1 =2N, P2 = 3N
P3 = 4N and P4 = 5N
These forces can be resolved into components along the X and Y directions.
Rx= P1 cos θ1 + P2 cos θ2 + P3 cos θ3 + P4 cos θ4
= 2cos 30° + 3cos 60 + 4cos 90 + 5 cos 120
= 2$$\frac{\sqrt{3}}{2}$$ + 3 × $$\frac{1}{2}$$ + 4 × 0 + 5 × $$\frac{1}{2}$$ = 0.732 N

Ry= P1 sin θ1 + P2 sin θ2 + P3 sin θ3 + P4 sin θ4
= 2 sin 30 + 3 sin 60 + 4 sin 90 + 5 cos 120
= 2 × $$\frac{1}{2}$$ + 3 × $$\frac{\sqrt{3}}{2}$$ + 4 × 1 + 5 × $$\frac{\sqrt{3}}{2}$$
= 1 + 1.5$$\sqrt{3}$$ + 4 + 2.5$$\sqrt{3}$$
= 1 + 2.59 + 4 + 4.33
= 11.92 N
∴ Resultant force R = $$\sqrt{\mathrm{R}_{\mathrm{x}}^{2}+\mathrm{R}_{\mathrm{Y}}^{2}}$$
= Vo.7322 +11.92 2
= 11.96 N.

Question 9.
The resultant of two forces acting at 60° is $$\sqrt{13}$$ N. When the same forces act at 90° the resultant is $$\sqrt{10}$$ N. Find the magnitude of the two forces.
Solution:
Let P and Q be the two forces. When the angle between them is 60°, resultant force is $$\sqrt{13}$$ N. Thus
$$\sqrt{\mathrm{P}^{2}+\mathrm{Q}^{2}+2 \mathrm{PQ} \cos 60^{\circ}}$$ = $$\sqrt{13}$$
i.e.$$\sqrt{\mathrm{P}^{2}+\mathrm{Q}^{2}+\mathrm{PQ}}$$ = $$\sqrt{13}$$
P + Q + PQ = 13 ……………(1)
When the angle is 90° resultant force is
$$\sqrt{10}$$ N.
$$\sqrt{\mathrm{P}^{2}+\mathrm{Q}^{2}+2 \mathrm{PQ} \cos 90^{\circ}}$$ = $$\sqrt{10}$$
or $$\sqrt{\mathrm{P}^{2}+\mathrm{Q}^{2}}$$ = $$\sqrt{10}$$
P2+ Q2 =10 ……………….. (2)
Substituting (2) in (1)
10 + P = 13
PQ = 3 ……………………(3)
(P + Q)² = P²+ Q² + 2PQ
= 10 + 2 × 3 = 16
P + Q =4 ………………. (4)
(P – Q)² = P² + Q² – 2PQ
= 10 – 6 = 4
P – Q = 2 ………………. (5)
2P =6; 2Q = 2
P = 3N and Q = 1N.

Question 10.
The sum of the two forces is 8 kg wt and the magnitude of the resultant which is at right angles to the smaller force is 4 kgwt. Find the two forces.
Solution:
Let P be the smaller force.
P + Q = 8 kg wt.,
a = 90°, R = 4 kg wt.,
tan α = $$\frac{\mathrm{Q} \sin \theta}{\mathrm{P}+\mathrm{Q} \cos \theta}$$
tan 90° = $$\frac{\mathrm{Q} \sin \theta}{\mathrm{P}+\mathrm{Q} \cos \theta}$$
∞ = $$\frac{\mathrm{Q} \sin \theta}{\mathrm{P}+\mathrm{Q} \cos \theta}$$
⇒ P + Q cos θ = 0
cos θ = $$-\frac{P}{Q}$$

Q² – P²= 16
(Q-P)(Q+P) = 16
(Q-P) × 8 = 16
Q – P = 2
P + Q = 8
∴ 2Q = 10 ⇒ Q = 5 kg wt. and P = 3 kg wt.

Question 11.
The resultant of two forces acting at an angle of 150° is perpendicular to the smaller force. If the smaller force is 12$$\sqrt{3}$$ N, find the greater force and its resultant.
Solution:
θ = 150°,
Let P be the smaller force P = 12$$\sqrt{3}$$ N
a = 90°, Q = ?, R = ?

Question 12.
Two equal, non parallel forces acting at a point on a body. The square of the resultant is found to be three times the product of the forces, what is the angle between them.
Given, P = Q & R² = 3PQ = 3P²
we have R² = P² + Q² + 2PQ cos θ
i.e. 3P²= P² + P² + 2P² cos θ
i.e. 3P² = 2P² + 2P² cos θ
1P² = 2P²cos θ
$$\frac{1}{2}$$ = cose
⇒ θ = cos-1($$\frac{1}{2}$$) = 60°.

Question 13.
The greatest and least resultant of two forces acting at a point are 17N and 7N respectively. Find the two forces. If these forces act at right angles, find the magnitude of their resultant.
For maximum resultant
P + Q=17N ……………….. (1)
For minimum resultant
P – Q = 7N ……………….. (2)
Eq(1) + (2) gives 2P = 24 Or P = 24/2 = 12N.
From (1), Q = 17 – 12 = 5N
When P and Q are in perpendicular directions,
R = $$\sqrt{\mathrm{P}^{2}+\mathrm{Q}^{2}}$$
= $$\sqrt{12^{5}+5^{2}}$$ = 13N(∵ cos 90° = 0).

Question 14.
Two equal forces are acting at a point and angle between them in 60°. Calculate magnitude of Resultant of these forces.
Q = P and θ = 60°.

1st PUC Physics Motion in a Plane Hard Questions and Answers

Question 1.
The total speed v1 of a projectile at its greatest height is $$\sqrt{\frac{6}{7}}$$ of its speed v2 at half its greatest height. Find the angle of projection.
Velocity at the highest point = ucos θ = v1
Hmax = $$\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{g}}$$
Using v = u – 2gh,
At h = $$\frac{\mathrm{H}_{\mathrm{max}}}{2}$$, vertical component of v2 is
given by,
v2y2 = u2sin2θ – 2g × $$\frac{\mathrm{H}_{\mathrm{max}}}{2}$$
v2y2 = u2 sin2θ – $$\mathrm{g}\left(\frac{\mathrm{u}^{2} \sin 2 \theta}{2 \mathrm{g}}\right)$$
⇒ v = $$\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2}$$
Horizontal component v2 is
v2x = v1 = ucosθ

Squaring both sides,

⇒ 7 cos²θ = 6 cos²θ + 3 sin² θ
⇒ cos²θ = 3 sin²θ
⇒ tan²θ = 1/3
⇒ tan θ = ± $$1 / \sqrt{3}$$
⇒ θ = ± 30°
θ = 30°
(because negative angle is not possible)

Question 2.
Prove that the path of one projectile as seen from another projectile is a straight line.
The coordinates of first projectile as seen from the second projectile are
X = x1 – x1
= u1cos θ1 t – u2cos θ2 t
X = (u1cos θ1 – u2cos θ2)t
Y = y1 – y2
= u1sin θ1 t – 1/2 gt² – (u2sin θ2 t – 1/2 gt²)
Y = (u1 sin θ1 – u2sin θ2)t
∴ $$\frac{Y}{X}=\frac{\left(u_{1} \sin \theta_{1}-u_{2} \sin \theta_{2}\right) t}{\left(u_{1} \cos \theta_{1}-u_{2} \cos \theta_{2}\right) t}$$
= a constant, say m
This equation is of the form Y = mX, which is the equation of a straight line. Thus the path of a projectile as seen from another projectile is a straight line.

Question 3.
A particle is projected with some speed at an angle a to the horizontal from the end B of the horizontal base BC of a triangle ABC. It rises to the vertex A and after just grazing it, falls down to reach point C of the base BC. If the base angles of the triangle are β and γ, show that 4 cot α = cot β + cot γ.

From the figure,
cot β = $$\frac{\mathrm{BD}}{\mathrm{h}_{\max }}$$
and cot γ = $$\frac{\mathrm{DC}}{\mathrm{h}_{\max }}$$
cot β + cot γ = $$\frac{\mathrm{BD}+\mathrm{DC}}{\mathrm{h}_{\max }}$$

$$\frac{\text { Range }}{h_{\max }}$$
⇒ cot β + cot γ = $$\frac{\left(u^{2} \sin 2 \alpha\right) / g}{\left(u^{2} \sin ^{2} \alpha\right) / 2 g}$$
⇒ cot β + cot γ = 2 × $$\frac{2 \sin \alpha \cos \alpha}{\sin ^{2} \alpha}$$
cot β + cot γ = 4 cot α.

## Karnataka 1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation

### 1st PUC Statistics Classification and Tabulation Two Marks Questions and Answers

Question 1.
What is classification of the data?
Solution:
Classification is the process of arranging the data in to groups or classes according to common

Question 2.
What are the objectives of classification?
Solution:

• To reduce the size of the data
• To bring the similarities together.
• To enable further statistical analysis

Question 3.
What are the basis/types of classification?
Solution:

1. Chronological classification
2. Geographical classification
3. Qualitative classification
4. Quantitative classification.

Question 4.
Give the formula of Sturge’s to find the number of classes. Or Give the formula used to determine the number of classes.
Solution:
The numbers of classes are obtained using Sturges’s Rule: k = 1 + 3.22 log N;
N-number of items.

Question 5.
For what purpose is correction factor used in frequency distribution?
Solution:
To get a better continuity between the class interval of a frequency distribution exclusive class intervals are used, so, if the frequency distribution is in inclusive class intervals isconverted into exclusive class intervals using correction factor.

Question 6.
What are the guidelines of classification?
Solution:
following are some guidelines following while classification:

• The number of classes should generally between 4 and 15.
• Exclusive classes should be formed for better continuity between the class intervals.
• The width of the classes should be usually kept constant throughout the distribution.
• Avoid open-end classes.
• The classes should be arranged in ascending or descending order.
• The lower limit of the first class should be either ‘o’ or multiple of 5.

Question 7.
Define the term tabulation
Solution:
Tabulation is a systematic arrangement of the classified data in to columns and rows of a table

Question 8.
Mention the parts of a Table
Solution:
Table No., Title of the table, Headnote/Sub-title, Captions, Stubs, Body of the table, Footnote & Source.

Question 9.
What are the objects of Tabulation?
Solution:
The object/Purpose of tabulation are:-

• It simplifies the complex data
• To facilitate for comparison
• To give an identity to the data
• To reveals trend and patterns of the data

Question 10.
How the number of classes using Prof. H.A.Sturge’s?
Solution:
The number of classes are obtained using Sturges’s Rule: k = 1 + 3.22 Log N

Question 11.
What are inclusive & exclusive class intervals?
Solution:
If in a class, both lower and upper limits are included in the same class are inclusive class intervals, eg. 0-9, 10-19, 20-29…
If in a class, the lower limit is included in the same class but the upper limit is included in the next class are exclusive class intervals, eg. 0-10, 10-20, 20-30…

Question 12.
Define frequency distribution
Solution:
A systematic presentation of the values taken by a variable and the corresponding frequencies is called Frequency distribution.

Question 13.
What are Marginal and Conditional frequency distributions?
Solution:
If in a bivariate frequency distribution, if the distribution of only one variable is considered, the distribution is called marginal frequency distribution.
If in a bivariate frequency distribution, if the distribution of only one variable is formed subject to the condition of the other variable it is called conditional frequency distribution.

Question 14.
What is the tabulation of the data?
Solution:
Tabulation is a process of systematic arrangement of the classified data in rows and columns, in the form of a table.

Question 15.
What are the parts of a table?
Solution:
The parts of a table are: Table No., Title of the table, Headnote/Sub-title, Captions, Stubs, Body of the table, Footnote & Source

Question 16.
What is the purpose of ‘table number’ in tabulation?
Solution:
A number should be given to each and every table, in order to distinguish and also for easy reference.

Question 17.
What are captions and stubs of a table?
Solution:
Captions: Column headings are called captions. They explain what the column represents. Captions are always written in one or two words on the top of each column.
Stubs: Row headings are called Stubs. They explain what the row represents. Stubs are usually written in one or two words at the left extreme sid/of each row.

Question 18.
Solution:
It is a brief explanatory note or statement given just below the heading of the table put in a bracket. The statistical units of measurements, such as in ‘000s, Rs., Million tones, crores, Kgs., etc., are usually put in bracket.

Question 19.
What is indicated by source note of a table?
Solution:
Below foot note or below the table, source of the data may be mentioned for verification to the reader, regarding publications, organizations, pages, Journals etc,

### 1st PUC Statistics Classification and Tabulation Five Marks Questions and Answers

Question 1.
Explain chronological classification and geographical classification of data with examples.
Solution:
Temporal/Chronological classification: when the data enumerated over a different period of time, the type of classification is called chronological classification. The above type of classification is called as time series, e.g. Time series of the population, is listed in chronological order starting with the earliest period.
Table showing the population of India

Geographical classification :
In this type of classification the data are classified on the basis of geographical or locational or area wise differences between various items. Such as cities, districts, states etc. e.g., production of sugar in India may be presented state wise in the following manner:-
Table showing the production of sugar in India

Question 2.
Explain qualitative and quantitative classification with examples.
Solution:
Qualitative classification: Classification of the different units on the basis of qualitative characteristics (Called Attributes). Such as sex, literacy, employment etc.
e. g., the members of a club can be classified on the basis of sex wise distribution as follows:-
Table showing the sex distribution of members of a club

Quantitative classification: classification of the number of units on the basis of quantitative data, such as according to Height, weight, Wages, Age (years), Number of children, etc, Thus the groups of a student may be classified on their heights as follows:-
Table showing the heights of students

Online midpoint calculator and find the midpoint of aline segment joining two points using the midpoint calcultor in just one click.

Question 3.
Define the following terms :
i. Frequency, class frequency:
Solution:
Frequency refers to the number of times an observation repeated (f). The number of observations corresponding to a particular class is known as the class Frequency Class frequency is a positive integer including zero

ii. Class limits:
Solution:
Class limits- Lowest and the highest values that are taken to define the boundaries of the class are called class limits

iii. Range of the class: It is the difference between highest and lowest value in the data, i.e., Range = H.V. – L.V.

iv. Width of the class: The difference between the upper and lower limits of class called width of the class. It is denoted by c or i.
i/c = Upper limit(UL) – Lower limit(LL)

v. Class mid point
Solution:
The central value of a class called mid value/point; $$\mathrm{m} / \mathrm{x}=\frac{\mathrm{LL}+\mathrm{UL}}{2}$$

vi. Define Frequency Density:
Solution:
The frequency per unit of class interval is the frequency density.
i.e. frequency density = $$\frac{\text { Frequency the class }}{\text { width of the class }}=f / w$$

vii. Relative frequency:
Solution:
Relative frequency.. is the ratio of frequency of class to the total frequency of the distribution
Relative Frequency $$\frac{\mathrm{f}}{\mathrm{N}}$$

viii. Class interval-inclusive, exclusive and open-end classes:
Solution:
If in a class, lower as well as upper limits are included in the same class are called Inclusive class
e. g. 30-39,40-49….
If in a class, the lower limit is included in the same class and upper limit is included in the next class, such a class is called Exclusive class, eg. 30-40, 40-50…
If in a class, the lower and upper limits of the class is not specified are called open end classes” e.g. less than/below, or more than/ above 100

ix. Cumulative frequency- less than and more than cumulative frequency:
Solution:
The added up frequencies are called cumulative frequencies.
The number of observations with values less than upper limit is less than cumulative frequency. (l.c.f) i.e. Frequencies added from the top.
The number of observations with values more than lower class limit is more than cumulative frequency (m.c.f)

x. Correction factor:
Solution:
It is half of the difference between lower limit of a class and upper limit of the preceding class. Thus,
Solution:
Correction facctor (C.F) = $$\frac{\text { Lower limit of a class-Upper limit of the precending class }}{2}$$

xi. A Frequency distribution Discrete, Continuous, Bi-variate, Marginal
Solution:
A systematic presentation of the values taken by a variable and the corresponding frequencies is called frequency is called Frequency distribution.
While framing a frequency distribution, if class intervals are not considered, is called discrete frequency distribution.
1. Example:
The number of families according to number of children .

While framing a frequency distribution, if class intervals are considered, is called continuous frequency distribution.

2. Example:
The following table showing the weight (kgs.) of persons

A frequency distribution formed on the basis of two related variables is called bi-variate frequency distribution.
For example, we want to classify data relating to the Height and Weights of a group of individuals, Income and Expenditure of a group of individuals, Ages of Husbands and Wives, Ages of mothers and Number of children, etc .
In a Bivariate frequency distribution, the frequency distribution of only one of the variables is considered, it is marginal frequency distribution.

Question 4.
Mention/what are the rules/principles of formation of Frequency of distribution?
Solution:

1. The lower limit of the first class should be either 0 or a multiple of 5
2. Exclusive classes should be formed for better continuity
3. The number of classes should be generally between 4 & 15
4. The width of the classes should be kept constant throughout the distribution
5. Avoid open-end classes
6. The classes should be arranged in ascending or descending order.

Question 5.
Mention the rules/principles of the tabulation
Solution:

1. The size of the table should be according to the size of the paper
2. The stubs and captions should be arranged logically according to the alphabetical, Chronological, Geographical order and items to be arranged according to the size.
3. If desired to locate the particular quantity in the body of the table that should be showed by thick colored inks or shaded off.
4. The table should not be overloaded with number of characteristics, rather can be prepare another table.
5. The table should be complete in all respects by captions and stubs, titles heading and no cell is left blank if left do not put ‘o’ but give (—) dash marks or write N.A
6. Miscellaneous column can be provided for the data which do not fit to the data, such as Ratio, percentages.
7. Ditto ( “ ) marks should not be used, as they may confuse with the no. 11
9. Sources if provided regarding publications, organizations, pages, Journals etc.

Question 6.
The employees of a college can be classified on the basis of sex wise distribution as follows:-
Solution:
Table showing the sex distribution of employess of a college.

Question 7.
The Employees of a college can be classified according to their occupations as :
Solution:
Table-1
The blank table given below represents the number of employees with different occupation in a college

 Occupations Number of employees Teaching staff – clerks – Attenders – Security men – Total –

Question 8.
The employees of a commercial Bank can be classified according to their occupations and sex is :
Solution:
Table-2
The blank table given below represents the number of employees with different occupation and their sex in a commercial Bank

Question 9.
The employees of a college can be classified according to their occupations , sex and their marital status is :
Solution:
Table-3 : The blank table given below represents the number of employees with different occupation, sex and their marital status in a college

Question 10.
In a survey of 40 families in a certain locality, the number of children per family was recorded and the following data were obtained.

Represent the data in the form of a discrete frequency distribution.
Solution:
Frequency distribution of the number of children.

Question 11.
Prepare a frequency table from the following table regarding the number fatal accidents occurred in a day in Bangalore in June 2010.

Solution:
Frequency distribution of the number of children.

Question 12.
From the following paragraph prepare a discrete frequency table with the number of letters present in the words .
“Success in the examination confers no right to appointment unless government is satisfied, after such enquiry as may be deemed necessary that the candidate is suitable for appointment to the public service”
Solution:
The number of digits in the above statement: Highest digit = 11 and Lowest digit = 2
Frequency distribution of the number of letters in the words present in the statement

Question 13.
The following are the marks obtained by 50 college students in a certain test.

Take suitable width of the class interval marks using struge’s rule.
Solution:
Here, N= 50; Range = H.V.-L. V. = 49-12 = 37
The number classes as per Sturge’s rule are obtained as follows:
Number of class intervals (K) = 1 + 3.322 logN= 1 + 3.22 log 50 = 1 + (3.22 × 1.6990) = 6.47=7
classes (Approx.) Size/width of class intervals – e = $$\frac{\text { Range }}{\text { Number of class intervals }}=\frac{37}{7}$$
5.28 = 6 (Approx.)
The size/width of each class is 6 and there are 7 classes. Thus, the required continuous frequency distribution with exclusive class intervals width is prepared as :
Frequency distribution of marks of students

Question 14.
The following data gives ages of 32 individuals in a locality. Using Sturge’s rule form a frequency table with exclusive type of class intervals.

Solution:
Here N= 32; Range = H .V.- L.V. = 59- 01 = 58
The number classes as per Sturge’s rule are:
Number of class intervals (K) =1 + 3.22 logN = 1 + 3.22 log (32) = 1+(3.22 × 1.5051) = 5.85=6
classes (Approx.) Size/width of class intervals – e = $$\frac{\text { Range }}{\text { Number of class intervals }}=\frac{58}{6}$$ = 10
(approx). Teh size / width of each class is 10 and there are 6 classes.
Frequency distribution of marks of students

Question 15.
The following are the marks obtained by 50 students in statistics; prepare a frequency table with class intervals of 10 marks.

Solution:
Range: H.V-L.V = 93 – 23 = 70; take width as 10 marks, and then the number of classes will be: 70/10=7.
Frequency distribution of marks of students in statistics test

Question 16.
Prepare a bivariate frequency distribution of the marks in Accountancy & Statistics:

Solution:
Here the both variables are discrete in nature no need to prepare class interval.
Bi-variate frequency table showing Ages (years) of Mothers and Number of children

Question 17.
Below are the ages of husbands and wives prepare a bivariate frequency distribution with suitable width :

Here both are continuous variables form the class intervals as below :
Solution:
Ages of Husbands: Highest age = 47, Lowest age = 25, Difference = 22/(i)5width =5. classes. Ages of wife: Highest age = 47, Lowest age = 21, Difference = 26/(i)5width = 6 (Approx.) classes.
Let X and Y be the ages of Husbands and ages of Wives.

Question 18.
Below are given the marks obtained by a batch of 20 students in mathematics and statistics:

Solution:
Marks in Mathematics: Highest Marks = 72, Lowest marks = 25, Difference = 47/(i)l Owidth 5 = 5 (Approx.) classes.
Marks in statistics: Highest marks = 85, Lowest marks = 20, Difference = 65/(i)10width = 7 (Approx.) classes.
Let x and y be marks inmathematics and marks in statistics.

TABULATION

Question 19.
Give a general format of a table.
Solution:
General format of a table

Question 20.
What are the requisites of a good table? Or What are the General rules to the tabulation?
Solution:

• Size:- the size of the table should be according to the size of the paper with more rows than columns. Exchange of the data can be done by altering the column and rows. A sufficient space should be provided in a particular to enter any new or to alter any affected figures.
• Logical order:- The stubs and captions should be arranged logically according to the alphabetical, Chronological, Geographical order and items to be arranged according to the size.
• Identity:- If desired to locate the particular quantity in the body of the table that should be showed by thick colored inks or shaded off.
• The table should not be overloaded with number of characteristics, rather can be prepare another table.
• The table should be complete in all respects by captions and stubs, titles heading and no cell is left blank if left do not put ’o’ but give (—) dash marks or write N.A
• Miscellaneous column can be provided for the data which do not fit to the data, such as Radius, percentages.
• Ditto ( “ ) marks should not be used,as they may confuse with the number 11
• Footnote may contain about errors, omissions, remarks
• Sources if provided regarding publications, organizations pages Journals etc.

Question 21.
Elucidate the difference between classification and tabulation.
Solution:
Comparison between Classification and Tabulation:-
The following points may be given as comparison:

• Classification and Tabulation are not two distinct processes. Before tabulation, data are classified and then displayed under different columns and rows of a table.
• Classification is the process of arranging the data in to groups or classes according to common characteristics possessed by the items of the data;
Whereas Tabulation is a process of systematic arrangement of the classified data in rows and columns, in the form of a table.
• Table contains precise and accurate information, where as classification gives only a classified groups of data.
• Classification reduces the size of the data and brings the similarities together and tabulation facilitates comparison, reveal the trend and tendencies of the data.

Question 22.
In the college out of total of 1200 applications received for I puc admission, 450
were applied for science and 580 applied for commerce and remaining applied for
arts faculty. Tabulate the above information.
Solution:
Table 1
Table showing the distribution of applicants for I year puc to different faculties

Question 23.
In the college out of total of 1200 applications received for I puc admission 780 were boys. 450 were applied for science of which 300 were boys and 580 applied for commerce 250 were girls and remaining applied for arts faculty. Tabulate the above information.
Solution
Table 2
Table showing the sex-wise distribution of applicants for I year puc to different faculties

Question 24.
In a Sigma multinational accountant consultants there are 180 were accountants, 210 were article helpers, 300 were practitioner trainees. Of all the members 30% were women among accountants, 20% in Articles helpers and 15% among trainees. Tabulate the data.
Solution:
Table 3
Table represents the members of Sigma accountants according to cadre and sex

Footnote: * 180 × 30%= 54, 210 × 20% = 42

Question 25.
In a trip organized by a college there were 80 persons each of whom paid Rs.15.50 on an average? There were 60 students each of whom paid Rs.16. members of the teaching staff were charged at higher rate. The number of ser ants was 6 (all males) and they were not charged anything. The number of females was 20 percent of the total of which one was a lady staff member. Tabulate the information.
Solution:
Table 4
Table showing the contribution (in Rs.) details made by members of a college trip.

Question 26.
In a state, there are 30 Medical colleges, 10 Dental colleges and 50 Engineering colleges. Among the Medical colleges, 5 are government colleges, 10 are aided private colleges and remaining are the unaided private colleges. Of the unaided colleges, 5 colleges are run by minority institutions.
Among the Engineering colleges, 10 are government colleges, 20 are aided private colleges and the rest are unaided private colleges. Of the unaided colleges, 10 colleges are run by minority institutions.
Among the dental colleges, 2 are aided private colleges and the rest are the unaided private colleges of which one is run by a minority institutions.
Tabulate the above information.
Solution:
Table showing the Medical, Engineering and Dental colleges run by Government, Private Aided and Private Unaided colleges in a State

Question 27.
The number of cases filed, hearing made and disposed by different bench judges in a day at High court of Karnataka are as given:
(i) Criminal cases filed 12, hearings made in 8 cases and disposed 3,
(ii) Land dispute cases filed 18, hearings made in 12cases and disposed 5,
(iii) Government service cases filed 6, hearings made in 4 cases and disposed 3,
(iv) Cheating cases filed 15; hearing made in 12 cases and disposed 8.
Tabulate the above information.
Solution:
Table showing the different types of cases filed, heard and disposed in a day at High court of Karnataka

## 1st PUC Maths Question Bank Chapter 8 Binomial Theorem

Students can Download Maths Chapter 8 Binomial Theorem Questions and Answers, Notes Pdf, 1st PUC Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## Karnataka 1st PUC Maths Question Bank Chapter 8 Binomial Theorem

Question 1.
State and prove Binomial theorem.

Some observations in a binomial theorem:
(1) The expansion of {a + b)n has (n + 1) terms
(2) The coefficients nCr occurring in the binomial theorem are known as binomial coefficients.
(3) The indices of V go on decreasing and that of ‘a’ go on increasing by 1 at each stage.
i.e., for each term: index of a + index of b-n.
(4) Since nCr=nCn_r we have
nC0 =nCn, nCx =nCn_r and so on.
Thus the coefficients of the terms equidistant from the beginning and the end in a binomial theorem are equal.
(5) General term in (a + b)n: Tr+1 – nCr anrbr
(6) Middle terms in (a + b)n

• When ‘n’ is even, the middle term
$$=\left(\frac{n}{2}+1\right)^{t h} \text { term }$$
• When ‘n’ is odd ,the middle term are
$$\frac{1}{2}(n+1)^{n} \text { term an } \frac{1}{2}(n+3)^{n} \text { term }$$

(7) Taking a = x and b = -y in the expansion, we get (x-y)n =[x + (-y)]n

Question 2.
Expand each of the expression
(i)$$\left(x^{2}+\frac{3}{x}\right)^{4}$$
(ii)$$(1-2 x)^{5}$$
(iii)$$\left(\frac{2}{x}-\frac{x}{2}\right)^{5}$$
(iv)$$(2 x-3)^{6}$$
(v)$$\left(\frac{x}{3}+\frac{1}{x}\right)^{5}$$
(vi)$$\left(x+\frac{1}{x}\right)^{6}$$

Question 3.
Using binomial theorem, evaluate each of the following:
(i) (98)5
(ii) (96)3
(iii) (102)5
(iv) (101)4
(v) (99)5

Question 4.
Which is longer $$(1.01)^{1000000} \text { or } 10,000 ?$$

Question 5.
Using Binomial theorem, indicate which number is larger $$(1.1)^{10000} \text { or } 1000 ?$$
Spilitting 1.01 and using Binomial theorem write the first few terms, we have.

Question 6.
Find (a + b)4 – (a -b)4. Hence evaluate
$$(\sqrt{3}+\sqrt{2})^{4}-(\sqrt{3}-\sqrt{2})^{4}$$

Question 7.
Find (x +1)6 +(x -1)6. Hence or otherwise evaluate
$$(\sqrt{2}+1)^{6}+(\sqrt{2}-1)^{6}$$

Question 8.
Evaluate:
$$(\sqrt{3}+\sqrt{2})^{6}+(\sqrt{3}-\sqrt{2})^{6}$$

Question 9.
Find the value of
$$\left(a^{2}+\sqrt{a^{2}-1}\right)^{4}+\left(a^{2}-\sqrt{a^{2}-1}\right)^{4}$$

Question 10.
Show that 9n+1 – 8n – 9 is divisible by 64, whenever ‘n’ is a positive integer.

Question 11.
Using binomial theorem, prove that 6n -5n always leaves remainder 1 when
divided by 25.

Question 12.
Prove that
$$\sum_{n=0}^{n} y \cdot c_{n}=4$$

Question 13.
Find the 4th term in the expansion of (x-2 y)12.

Question 14.
Find the 13th term in the expansion of
$$\left(9 x-\frac{1}{3 \sqrt{x}}\right)^{18}, x \neq 0$$

Question 15.
Write the general term in the expression of
(i) (x2 -y)6
(ii) (x2-yx)12,x≠0

Question 16.
Find the coefficient of
(i) x5 in (x + 3)8
(ii) a5b7 in (a – 2b)12
(iii) x6y3 in (x + 2y)9

Question 17.
Find a, if the 17th and 18th terms of the expansion (2 +a)50 are equal.

Question 18.
In the expansion of (1+ a)m+n, prove that coefficients of am and an are equal.

Question 19.
Prove that the coefficient of xn in the expansion of (1 + x)2n is twice the coefficient of xn in the expansion of (1+x)2n-1
In (1+x)2n we have

From (1) and (2) we get, the coefficient of xn in (1 + x)2n is twice the coefficient of xn in (1 + x)2n-1.

Question 20.
Find a positive value of m for which the coefficient of x2 in the expansion (1 + x)m is 6.

Question 21.
Find the middle terms in the expansions of
(i) $$\left(3-\frac{x^{3}}{6}\right)^{7}$$
(ii)$$\left(\frac{x}{3}+9 y\right)^{10}$$

Question 22.
Show that the middle term in the expansion of $$(1+x)^{2 n} \text { is } \frac{1 \cdot 3 \cdot 5 \ldots(2 n-1)}{\lfloor n} 2^{n} x^{n}$$where ‘n’ is a positive integer.

Binomial Expansion Calculator to make your lengthy solutions a bit easier. Use this and save your time. Binomial Theorem & Series Calculator.

Question 23.
The second, third and fourth terms in the binomial expansion (x + a)n are 240, 720 and 1080, respectively. Find x, a and n

Question 24.
The coefficients of three consecutive terms in the expansion of (1 + a)n are in the ratio 1:7: 42. Find n

Question 25.
The coefficients (r-1)th, rth,and (r + 1)th, terms in the expansion of (x + 1)th, are in the ratio 1:3:5. Find n and r.

Question 26.
Find the term independent of x in the expansion of $$\left(\frac{3}{2} x^{2}-\frac{1}{3 x}\right)^{6}$$

Question 27.
Find the term independent of x in the expansion of $$\left(\sqrt[3]{x}+\frac{1}{2 \sqrt[3]{x}}\right)^{18}, x>0$$

Question 28.
Find a, 6 and n in the expansion of (a + b)n if the first three terms of the expansion are 729,7290 and 30375, respectively.
Given: Tx = 729, T2 = 7290 and T3 = 30375
∴ an=729……………….(1)

Question 29.
Find a if the coefficients of x2 and x3 in the expansion of (3 +ax)9 are equal.

Question 30.
If the coefficients of (r – 5)th and (2r -1)th terms in the expansion of (1 + x)34 are equal, find r.

Question 31.
If the coefficients of ar-1, ar and ar+1 in the expansion of (1 + a)n are in arithmetic progression, prove that n2 – n(4r +1) + 4r2 -2 = 0.

Question 32.
Show that the coefficient of the middle term in the expansion of (1 + x)2n is equal to the sum of the coefficients of two middle terms in the expansion of (1 + x)2n-1.

Question 33.
Find n, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of
$$\left(\sqrt[4]{2}+\frac{1}{\sqrt[4]{3}}\right)^{n} \text { is } \sqrt{6}: 1$$

Question 34.
Find the rth term from the end in the expansion of (x + a)n.
rth term from the end in (x + a)n

Question 35.
If a and b are distinct integers, prove that a-b is a factor of an -bn, whenever n is a positive integer.

Question 36.
The sum of the coefficients of the first three terms in the expansion of
$$\left(x-\frac{3}{x^{2}}\right)^{m}, x \neq 0, m$$being a natural numbers, is 559. Find the term of the expansion containing x3.

Question 37.
Find the coefficient of x5 in the product (1 + 2x)6(1 – x)7 using binomial theorem.

Question 38.
Find the coefficient of a4 in the product (1 + 2a)4(2-a)s using binomial theorem.

Question 39.
Expand using Binomial theorem
$$\left(1+\frac{x}{2}-\frac{2}{x}\right)^{4}, x \neq 0$$

Question 40.
Find the expansion of (3x2 – 1ax + 3a2)3 using binomial theorem.

### Karnataka 1st PUC Business Studies Question Bank with Answers in English

Karnataka 1st PUC Business Studies Blue Print of Model Question Paper

For practical oriented questions, questions are to be selected from each of the 3 groups (Group 1: First 4 chapters, Group 2: Next four chapters and Group 3: Last three chapters. Please follow the chapters list given in this table)
Note: While selecting the Practical Oriented questions in Part E, care should be taken to avoid duplication of questions in Section A, B, C and D).

Total number of questions asked is 43 for 139 marks. Students have to answer 33 for 100 marks.
Section Wise Distribution of Marks

If you have any queries regarding Karnataka State Board NCERT Syllabus 1st Year Class 11 PUC Business Studies Question Bank with Answers Pdf, drop a comment below and we will get back to you at the earliest.

## 1st PUC History Question Bank with Answers Karnataka

Expert Teachers at KSEEBSolutions.com has created Karnataka 1st PUC History Question Bank with Answers Solutions, Notes, Guide Pdf Free Download of 1st PUC History Textbook Questions and Answers, Model Question Papers with Answers, Study Material 2020-21 in English Medium and Kannada Medium are part of 1st PUC Question Bank with Answers. Here KSEEBSolutions.com has given the Department of Pre University Education (PUE) Karnataka State Board NCERT Syllabus 1st Year PUC History Question Bank with Answers Pdf.

Students can also read 1st PUC History Model Question Papers with Answers hope will definitely help for your board exams.

## Karnataka 1st PUC History Question Bank with Answers

### Karnataka 1st PUC History Syllabus

1. Introduction to History
1.1 Meaning and definition of History – Herodotus, Augustine, Karl Marx, J.B-Bury, Nehru and Toynbee
1.2 Importance of the study of History
2. The Story of Human Evolution
3. History of Ancient Civilization
3.1 Introduction
3.2 Egyptian Civilization
3.3 Mesopotamian Civilization
3.4 Chinese Civilization
4. Establishment of Greek and Roman Empires-Contributions
4.1 Greek City – States – Cultural Contributions
4.2. Roman Republics – Legacy of Romans
5. Rise and Spread of Christianity and Islam
5.1. Life and Teachings of Jesus Christ Spread of Christianity
5.2 Life and Teachings of Prophet Mohammed
6. Medieval Period Towards Change
Chruch, Society and State – Feudalism
7. Beginning of Modern Age
7.1 Geographical Discoveries
7.2 Renaissance
7.3 Reformation-Martin Luther-Counter Reformation
8. World Revolutions
8.1 Industrial Revolution
8.2. The American War of Independence
8.3. The French Revolution (1789-1795 CE)
8.4. Russian Revolution
9. Napoleon and Rise of Nationalism
9.1 Napoleon Bonaparte
9.2. Unification of Italy
9.3 Unification of Germany
10. World Wars and International Organizations
10.1 World War I-Treaty of Versailles
10.2 Rise of dictatorships
10.3 World Warn
10.4 UNO-Organs-Achievements
11. Contemporary World
11.1 Cold War
11.2 Disintegration of USSR
11.3 Formation of CIS
12. Non-Aligned Movement-Emergence of the Third world
13. Map Work: Historical Places of World Importance

Karnataka 1st PUC History Blue Print of Model Question Paper

We hope the given Karnataka 1st PUC Class 11 History Question Bank with Answers Solutions, Notes, Guide Pdf Free Download of 1st PUC History Textbook Questions and Answers, Model Question Papers with Answers, Study Material 2020-2021 in English Medium and Kannada Medium will help you.

If you have any queries regarding Karnataka State Board NCERT Syllabus 1st Year PUC Class 11 History Question Bank with Answers Pdf, drop a comment below and we will get back to you at the earliest.

error: Content is protected !!