2nd PUC Basic Maths Question Bank Chapter 19 Application of Derivatives Ex 19.1

Students can Download Basic Maths Exercise 19.1 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Basic Maths Question Bank Chapter 19 Application of Derivatives Ex 19.1

Part-A

2nd PUC Basic Maths Application of Derivatives Ex 19.1 Two Marks Questions and Answers

Question 1.
The displacement ‘s’ of a particle at time ‘t’ is given by S = 4t3 – 6t2 + t – 7. Find the velocity and acceleration when t = 2 sec.
Answer:
Given S = 4t3 – 6t + t – 7
Velocity = v = \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = 12t2 -12t + 1
At t = 2secs, v = 12(2)2 – 12(2) + 1 = 48 – 24 + 1 = 25 units/sec.
At t = 2 sec, acceleration = 24.2 – 12 = 48 – 12 = 36 units/sec2.

Question 2.
If S = 5t2 + 4t – 8. Find the initial velocity and acceleration, (s = displacement, t = time).
Answer:
Given s = 5t2 + 4t – 8
V = \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = 10t + 4 dt
Initial velocity is velocity when t = 0
i.e., = 10.0 + 4 = 4 units/sec.
Acceleration = \(\frac{\mathrm{dv}}{\mathrm{dt}}\) = 10 units /sec2.

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Question 3.
A stone thrown vertically upward rises ‘s’ ft. in ‘t’ sec. where s = 80t – 16t2. What its velocity after 2 sec.? Find the acceleration?
Answer:
Given S = 80t -16t2
v = \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = 80 – 32t
At t=2 sec, v = 80 – 32 (2)
= 80 – 64 = 16 ft./sec.
Acceleration = \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = -22 ft/sec2 .

Question 4.
A body is thrown vertically upwards its distance S feet is’t’ sec. is given by S = 5 + 12t – t2. Find the greatest highest by the body.
Answer:
Given s = 5 + 12t – t2
v = \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = 12 – 2t dt
Maximum height ⇒ Kinetic energy = 0 ⇒ v = 0 ⇒ 12 – 2t = 0 ⇒ t = 6 sec
∴ Greatest height = s = 5 + 12.6 – 62
= 5 + 72 – 36 = 77 – 36 = 41 feet.

Question 5.
If v = \(\sqrt{s^{2}+1}\) prove that acceleration is ‘S’ (V = velocity, S = displacement).
Answer:
Given v = \(\sqrt{s^{2}+1}\)
Acceleration = \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = \(\frac{1}{2 \sqrt{s^{2}+1}} \cdot 2 s \frac{d s}{d t}=\frac{1}{2 v}\) . 2 . s . v . s units / sec2.

Question 6.
If S = at3 + bt. Find a and b given that when t = 3 velocity is ‘O’ and the acceleration is 14 unit. (S = displacement, t = time).
Answer:
Given s = at3 + bt; v = \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = 3at2+b
Acceleration = \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = 6at
When v = 0 then 3at2 + b = 0 ⇒ 27a + b = 0.
t = 3 sec
When acceleration = 14 then 14 = 6a.3, b = -27a = \(-\frac{27.7}{9}\) = -21
F = 3 sec. a = \(\frac{14}{18}=\frac{7}{9}\)
∴ a = \(\frac { 7 }{ 9 }\) and b = -21.

Question 7.
When the brakes are applied to moving car, the car travels a distance ‘s’ ft. in ‘t’ see given by s = 8t – 6t2 when does the car stop?
Answer:
Given s = 8t – 6t2
v = \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = 8 – 12t car stops when v = 0
∴ 8 – 12t = 0 ⇒ t = \(\frac{8}{12}=\frac{2}{3}\) sec.

KSEEB Solutions

Part-B

2nd PUC Basic Maths Application of Derivatives Ex 19.1 Three Marks Questions and Answers.

Question 1.
The radius of sphere is increasing at the rate of 0.5 mt/sec. Find the rate of increase of its surface area and volume after 3 sec.
Answer:
Given \(\frac{\mathrm{dr}}{\mathrm{dt}}\) = 0.5 , t = 3 sec, \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = ? \(\frac{\mathrm{dv}}{\mathrm{dt}}\) = ?
surface area = s = 4πr2
\(\frac{d s}{d t}\) = 4π . 2r . \(\frac{d r}{d t}\)

= 4π × 2 × 1.5 × 0.5
= 6π m2/sec
dr = 0.5 × dt
⇒ r = 0.5 t
= 0.5 × 3
= 1.5

Question 2.
The surface area of a spherical bubble is increasing at the rate of a 0.8cm2 / sec. Find at what rate is its volume increasing when r = .25cm [r = radius of the sphere].
Answer:
Given \(\frac{\mathrm{dr}}{\mathrm{dt}}\) = 0.8 cm2 / sec. r = 2.5 cm, \(\frac{\mathrm{dv}}{\mathrm{dt}}\) = ?
s = 4πr2
\(\frac{\mathrm{dr}}{\mathrm{dt}}\) = 4π . 2r × \(\frac{\mathrm{dr}}{\mathrm{dt}}\)
0.8 = 4π × 2 × 2.5. \(\frac{\mathrm{dr}}{\mathrm{dt}}\) ⇒ \(\frac{0.8}{8 \pi \times 2.5}=\frac{0.1}{2.5 \pi}\)
v = \(\frac { 4 }{ 3 }\)πr3
\(\frac{d v}{d t}=\frac{4}{3} \pi 3 r^{2} \cdot \frac{d r}{d t}=4 \pi \times(2.5)^{2} \times \frac{0.1}{2.5 \pi}=1 \mathrm{cc} / \mathrm{sec}\)

Question 3.
A spherical balloon is being inflated at the rate 35cc/sec. Find the rate at which the surface area of the balloon increases when its diameter is 14cm.
Answer:
Given \(\frac{\mathrm{dr}}{\mathrm{dt}}\) = 35cc /sec 2r = 14 ⇒ r = 7, \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = ?, \(\frac{\mathrm{dr}}{\mathrm{dt}}\) = ?
v = \(\frac { 4 }{ 3 }\)πr3 s = 4πr2
\(\frac{d V}{d t}=\frac{4}{3} \pi 3 r^{2} \frac{d r}{d t} \quad \frac{d S}{d t}=4 \pi 2 r \frac{d r}{d t}\)
35 = 4π . 72 \(\frac{\mathrm{dr}}{\mathrm{dt}}\) = 4π × 2 × 7 × \(\frac{5}{28 \pi}\)
\(\frac{d r}{d t}=\frac{35}{196 \pi}=\frac{5}{28 \pi}=10 \mathrm{cm}^{2} / \mathrm{sec}\)

Question 4.
The radius of a circular plate is increasing at the rate of \(\frac{2}{3 \pi}\) cm/sec. Find the rate of change of its area when the radius is 6cm.
Answer:
Given \(\frac{d r}{d t}=\frac{2}{3 \pi} r=6 \mathrm{cm} \frac{d A}{d t}=?\)
A = πr2
2nd PUC Basic Maths Question Bank Chapter 19 Indefinite Integrals Ex 19.1 - 1

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Question 5.
A circular patch of oil spreads on water the area growing at the rate of 16cm2/min. How fast are radius and the circumference increasing when the diameter is 12cm.
Answer:
Given \(\frac{\mathrm{dA}}{\mathrm{dt}}\) = 16cm2/min, d = 2r = 12cm, \(\frac{\mathrm{dr}}{\mathrm{dt}}\) = ? \(\frac{\mathrm{dc}}{\mathrm{dt}}\) = ? r = 6cm
A = πr2 c = 2πr
\(\frac{\mathrm{dA}}{\mathrm{dt}}\) = π . 2r . \(\frac{\mathrm{dr}}{\mathrm{dt}}\) \(\frac{\mathrm{dc}}{\mathrm{dt}}\) = 2π \(\frac{\mathrm{dr}}{\mathrm{dt}}\)
16 = 2π .6 . \(\frac{\mathrm{dr}}{\mathrm{dt}}\) 2π . \(\frac{4}{3 \pi}=\frac{8}{3}\)cm / min
⇒ \(\frac{d r}{d t}=\frac{16}{12 \pi}=\frac{4}{3 \pi} \mathrm{cm} / \mathrm{min}\)

Question 6.
A stone is dropped into a pond waved in the form of circles are generated and the radius of the outer most ripple increases at the rate 2 inches/sec. How fast is the area increasing when the (a) radius is 5 inches (b) after 5 sec.?
Answer:
Given \(\frac{\mathrm{dr}}{\mathrm{dt}}\) = 2inch/sec, r = 5 inch, \(\frac{\mathrm{dA}}{\mathrm{dt}}\) = ?
(a) A = πr2
\(\frac{\mathrm{dA}}{\mathrm{dt}}\) = π2r. \(\frac{\mathrm{dr}}{\mathrm{dt}}\) = π × 2 × 5 × 2 = 20π sq. inches / sec.

(b) After 5 sec, dr = 2dt
r = 2t
⇒ when t = 5, r = 10 inches
\(\frac{\mathrm{dA}}{\mathrm{dt}}\) = π . 2r \(\frac{\mathrm{dr}}{\mathrm{dt}}\) = π × 2 × 20 × 2 = 40π square inches/sec.

Question 7.
The side of an equilateral triangle is increasing at the rate \(\sqrt{3}\) cm./sec. Find the rate at which its area is increasing when its side is 2 meters.
Answer:
Given \(\frac{\mathrm{dx}}{\mathrm{dt}}\) = \(\sqrt{3}\) cm/sec., x = 2 meters, \(\frac{\mathrm{dA}}{\mathrm{dt}}\) = ?
Area of equilateral Δle = A = \(\frac{\sqrt{3}}{4}\) x2
\(\frac{d A}{d t}=\frac{\sqrt{3}}{4} \cdot 2 x \frac{d x}{d t}\)
= \(\frac{\sqrt{3}}{4}\) . 2. 200. \(\sqrt{3}\) = 300cm2 / sec.

Question 8.
Water is being poured at the rate of 30 mt3/min. into a cylindrical vessel whose base is a circle of radius 3 mt. Find the rate at which the level of water is rising?
Answer:
Given r = 3mts, \(\frac{\mathrm{dv}}{\mathrm{dt}}\) = 30 m3/min, \(\frac{\mathrm{dh}}{\mathrm{dt}}\) = ?
V = πr2h, r = constant
\(\frac{\mathrm{dv}}{\mathrm{dt}}\) = π . (3)2 . \(\frac{\mathrm{dh}}{\mathrm{dt}}\)
30 = 9π \(\frac{\mathrm{dh}}{\mathrm{dt}}\) ⇒ \(\frac{\mathrm{dh}}{\mathrm{dt}}\) = \(\frac{10}{3 \pi}\) meter/min.

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Question 9.
Sand is being dropped at the rate of 10 mt3/sec. into a conical pile. If the height of the pile twice the radius of the base, at what rate is the height to the pile is increasing when the sand in the pile is 8mt high.
Given
\(\frac{d v}{d t}\) = 10m3/sec, h = 2r, h = 8, \(\frac{d h}{d t}\) = ?
r = \(\frac{\mathrm{h}}{2}\) \(\mathrm{v}=\frac{1}{3} \pi \mathrm{r}^{2} \mathrm{h}=\frac{1}{3} \pi \cdot\left(\frac{\mathrm{h}}{2}\right)^{2} \cdot \mathrm{h}=\frac{1}{3} \pi \frac{\mathrm{h}^{3}}{4}\)
\(\mathrm{v}=\frac{1}{12} \pi \mathrm{h}^{3} \Rightarrow \frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\pi}{12} \cdot 3 \mathrm{h}^{2} \frac{\mathrm{dh}}{\mathrm{dt}} ; \quad 10=\frac{\pi}{4} \cdot 8^{2} \cdot \frac{\mathrm{dh}}{\mathrm{dt}}\)
\(\frac{d h}{d t}=\frac{40}{64 \pi}=\frac{5}{8 \pi} \mathrm{m} / \mathrm{sec}\)

Question 10.
A ladder of 15ft. long leans against a smooth vertical wall. If the top slides downwards at the rate of 2ft sec. Find how fast the lower and is moving when the lower end is 12ft. from the wall.
Answer:
2nd PUC Basic Maths Question Bank Chapter 19 Indefinite Integrals Ex 19.1 - 2
Given \(\frac{\mathrm{d} y}{\mathrm{dt}}\) = 2ft /sec, x = 12
From fig, x2 + y2 = 152
122 + y2 = 152
y2 = 152 – 122
y = \(\sqrt{225-144}\)
y = \(\sqrt{81}\) = 9
x2 + y2 = 152 ⇒ 2x \(\frac{\mathrm{dx}}{\mathrm{dt}}\) + 2y\(\frac{\mathrm{dy}}{\mathrm{dt}}\) = 0; 2.12.\(\frac{\mathrm{dx}}{\mathrm{dt}}\) + 2 .9 . 2 = 0
⇒ \(\frac{d x}{d t}=\frac{-36}{24}=\frac{-3}{2} f t / \sec\)

Question 11.
An edge of a variable cube is increasing at the rate of 10cm/sec. How fast the volume and also its surface area is increasing when the edge is 5cm long.
Answer:
Given \(\frac{d x}{d t}\) = 10 cm/sec, x = 5 cm, \(\frac{d v}{d t}\) = ? \(\frac{d s}{d t}\) = ?
(i) V = x3
\(\frac{d v}{d t}\) = 3x2 \(\frac{d x}{d t}\) = 3 × (52) × 10 = 750 cm3/sec.

(ii) S = 6x2 .
\(\frac{d s}{d t}\) = 12 × .\(\frac{d x}{d t}\) = 12.5 .10 = 600 cm2/sec.

KSEEB Solutions

Question 12.
A man 6ft. tall is moving directly away from a lamp post of height 10ft. above the ground. If he is moving at the rate 3ft./sec. Find the rate at which the length of his shadow is increasing and also the tip of his shadow is moving?
Answer:
2nd PUC Basic Maths Question Bank Chapter 19 Indefinite Integrals Ex 19.1 - 3
Let the shadow be x & y be the distance the man walks
Given \(\frac{d y}{d t}\) = 3ft/sec From a similar Δles we have
\(\frac{6}{10}=\frac{x}{x+y}\)
6x + 6y = 10x
6y = 4x ⇒ 3y = 2x
⇒ \(3 \frac{d y}{d t}=2 \frac{d x}{d t} \Rightarrow 2 \frac{d x}{d t}=3.3=9\)
∴ the shadow is increasing at the rate.
∴ \(\frac{d x}{d t}=\frac{9}{2}\) ft/sec & the tip of the shadow moves is
\(\frac{d x}{d t}+\frac{d y}{d t}=\frac{9+6}{2}=\frac{15}{2} \mathrm{ft} / \mathrm{sec}\)

Question 13.
The height of circular cone is 30 cm. and it is constant. The radius of the base is increasing at the rate of 0.25cm/sec. Find the rate of increase of volume of the cone when the radius of base is 10cm.
Answer:
dr
Given h = 30 cm, \(\frac{d r}{d t}\) = 0.25cm/sec. r = 10cm. dt
V = \(\frac { 1 }{ 3 }\) πr2h
\(\frac{d v}{d t}\) = \(\frac{\pi}{3}\)h.2r. \(\frac{d r}{d t}\) = π. \(\frac { 30 }{ 3 }\) . 20.(0.25) = 50π cm2 / sec

Question 14.
The volume of a spherical ball in increasing at the rate 4πcc/sec. Find the rate of increase of the radius of the ball when the volume is 288πCC.
Answer:
Given V = 288π C.C., \(\frac{d v}{d t}\) = 4πcc/ sec \(\frac{d r}{d t}\) = ?
V = \(\frac { 4 }{ 3 }\) πr3
2nd PUC Basic Maths Question Bank Chapter 19 Indefinite Integrals Ex 19.1 - 4
r = 6cm
\(\frac{d v}{d t}=\frac{4}{3} \pi \cdot 3 r^{2} \frac{d r}{d t} 4 \pi=\frac{4}{3} \pi \times 3 \times 36 \times \frac{d r}{d t} \Rightarrow \frac{d r}{d t}=\frac{1}{36} \mathrm{cm} / \mathrm{sec}\)

Question 15.
A drop of ink spreads over a blotting paper so that the circumferences of the blot is 4πcm and it changes 3cm/sec. Find the rate of increase of its radius and also find the rate of increase of its area?
Answer:
Given c = 4π, \(\frac{d C}{d t}\) = 3cm / sec \(\frac{d A}{d t}\) = ? \(\frac{d r}{d t}\) = ?
Circumference = c = 2πr
4π = 2πr ⇒ r = 2
Again
C = 2πr & A = πr2
\(\frac{d c}{d t}\) = 2π . \(\frac{d r}{d t}\) \(\frac{d A}{d t}\) = π . 2r. \(\frac{d r}{d t}\)
3 = 2π . \(\frac{d r}{d t}\) = π . 2. 2. \(\frac{3}{2 \pi}\)
⇒ \(\frac{d r}{d t}\) = \(\frac{3}{2 \pi}\) cm/ sec \(\frac{d A}{d t}\) = 6cm2 / sec

Question 16.
A circular plate of metal is heated so that its radius increase at the rate of O.lmm/min. At what rate is the [plate’s area increasing when the radius is 25cm [1cm = 10mm].
Answer:
Given \(\frac{d r}{d t}\) = 0.1 mm/min, r = 25 cm, \(\frac{d A}{d t}\) A = πr2
\(\frac{d A}{d t}\) = π. 2r . \(\frac{d r}{d t}\) = π . 2. /250 (0.1) = 50πmm2 /min.

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Question 17.
The surface area of a spherical soap bubble increasing at the rate of 0.6cm2/sec. Find the rate at which its volume is increasing when its radius is 3cm.
Answer:
Given \(\frac{ds}{d t}\) = 0.6 cm2 / sec, r = 3cm, \(\frac{d v}{d t}\) = ?
s = 4 πr2 &
\(\frac{d s}{d t}\) = 4π. 2r. \(\frac{d r}{d t}\)
0.6 = 4π × 3 × 3 × \(\frac{d r}{d t}\)
∴ \(\frac{d r}{d t}=\frac{0.6}{6 \times 4 \pi}=\frac{0.1}{4 \pi} \mathrm{cm} / \mathrm{sec}\)
\(\frac { 1 }{ 40 }\) πcm/sec.
v = \(\frac { 4 }{ 3 }\) πr3
\(\frac{d v}{d t}=\frac{4}{3} \pi \cdot 3 r^{2} \frac{d r}{d t}\)
= 4πr2 \(\frac{d r}{d t}\)
= 4π(3)2 . \(\frac{0.1}{4 \pi}\)
= 0.9 cm3 / sec.

Question 18.
A rod 13 feet long slides with it end A and B as two straight lines at right angles which meet at ‘O’. If A is moved away from O with a uniform speed at 4ft./sec., find the speed of the end B move when A is 5 feet from O.
Answer:
2nd PUC Basic Maths Question Bank Chapter 19 Indefinite Integrals Ex 19.1 - 5
From fig we have
x2 + y2 = 132
y2 = 132 – 152 = 144
y = \(\sqrt{144}\) = 12
Als0
x2 + y2 = 132 ⇒ 2x \(\frac{d x}{d t}\) + 2y\(\frac{d y}{d t}\) = 0
\(5 \times 4=-12 \frac{\mathrm{dy}}{\mathrm{dt}} \Rightarrow \frac{\mathrm{dy}}{\mathrm{dt}}=-\frac{20}{12}=\frac{-5}{3}=\frac{\mathrm{dy}}{\mathrm{dt}}=\frac{-5}{3} \mathrm{ft.} / \mathrm{sec}\)

Question 19.
A street lamp is hung 12 feet above a straight horizontal floor on which a man of 5 feet is walking how fast his shadow lengthening when he is walking away from the lamp post at the rate of 175ft./min.
Answer:
2nd PUC Basic Maths Question Bank Chapter 19 Indefinite Integrals Ex 19.1 - 6
Let the shadow be y & the distance from the man walks is x.
Given \(\frac{d x}{d t}\) = 175 ft/min.
From figure we have
\(\frac{12}{5}=\frac{x+y}{y}\) ⇒ 12y = 5x + 5y ⇒ 7y = 5x
⇒ \(\frac{7 \mathrm{dy}}{\mathrm{dt}}=5 \frac{\mathrm{dx}}{\mathrm{dt}}\)
∴ the shadow is lengthening ⇒ \(\frac{d y}{d t}=\frac{5}{7} \times 175\) = 125 ft/ min.

Question 20.
Find a point on the parabola y2 = 4x at which the ordinate increases at twice the rate of the abscissa [Ordinate = y, abscissa = x].
Answer:
Given y2 = 4x diff. w.r.t. x
2y \(\frac{d y}{d x}\) = 4 \(\frac{d x}{d t}\)
Also given \(\frac{d y}{d x}\) = 2. \(\frac{d x}{d t}\) ⇒ 2y .2 \(\frac{d x}{d t}\) = 4. \(\frac{d x}{d t}\)
⇒ y = 1 ⇒ 12 = 4x
⇒ x = \(\frac { 1 }{ 4 }\)
∴ the point on the parabola is (\(\frac { 1 }{ 4 }\), 1 ).

KSEEB Solutions

2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.2

Students can Download Maths Chapter 6 Application of Derivatives Ex 6.2 Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.2

2nd PUC Maths Application of Derivatives NCERT Text Book Questions and Answers Ex 6.2

Question 1.
Find the intervals in which the function f given by f (x) = 3x + 17 is strictly increasing in R
Answer:
f (x) = 3x + 17
f’ (x) = 3 > 0 , ∀ x ∈ R
hence f(x) is strictly increasing on R.

Question 2.
Show that the function given by f (x) = e2x is strightly increasing on R
Answer:
f’ (x) = 2 e2x > 0 ∀ x ∈ R
hence f (x) is strictly increasing on R.

KSEEB Solutions

Question 3.
Show that the function given by f(x) = sin x is
(a) strictly increasing in \((0, \pi / 2)\)
(b) strictly decreasing in \((\pi / 2, \pi)\)
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.2.1

Question 4.
Find the intervals in which the function f given by f (x) = 2x2 – 3x is
(a) strictly increasing
(b) strictly decreasing
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.2.2

Question 5.
Find the intervals in which the function f given by f (x) = 2x2 – 3x2 – 36x +7 is
(a) strictly increasing
(b) strictly decreasing
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.2.3
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.2.4

Question 6.
Find the intervals in which the following functions are strictly increasing or decreasing:
(a) x2 + 2x – 5
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.2.5

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(b) 10 – 6x – 2x2
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.2.6

(c) -2x3 – 9x2 – 12x + 1
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.2.7

(d) 6 – 9x – x2
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.2.8

(e) (x + 1)3 (x -3)3
Answer:
f (x) = (x + 1)3 (x – 3)3
f’ (x) = (x + 1)3 x 3(x – 3)2 + (x – 3)3 x 3 (x + 1)2
= (x + 1)2 (x – 3)2 [3x + 3 + 3x – 9]
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.2.9

KSEEB Solutions

Question 7.
Show that \(y=\log (1+x)-\frac{2 x}{2+x}, x>-1\) an increasing function of x throughout its domain.
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.2.10

Question 8.
Find the values of x for which y = [x(x – 2)]2 is an increasing function.
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.2.11
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.2.12

KSEEB Solutions

Question 9.
Prove that \(y=\frac{4 \sin \theta}{(2+\cos \theta)}-\theta\) is an a increasing function of θ in \(\left[0, \frac{\pi}{2}\right]\)
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.2.13

Question 10.
Prove that the logarithmic function is strictly increasing on (0, ∞).
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.2.14

Question 11.
Prove that the function f given by f (x) = x2 – x + 1 is neither strictly increasing nor strictly decreasing on (- 1, 1).
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.2.15

KSEEB Solutions

Question 12.
Which of the following function are strictly decreases on \((0, \pi / 2)\)
(a) cos x
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.2.16

(b) f(x) = cos 2x
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.2.17
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.2.18

(c) cos 3x
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.2.19

(d) tan x
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.2.20

Question 13.
On which of the following intervals is the function f given by f (x) = x100 + sin x – 1 strictly decreasing ?
(A) (0,1)
(B) \(\left(\frac{\pi}{2}, \pi\right)\)
(C) \(\left(0, \frac{\pi}{2}\right)\)
(D) None of these
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.2.21

Question 14.
Find the least value of a such that the function f given by f (x) = x2 + ax + 1 is strictly increasing on (1, 2).
Answer:
f (x) = x2 + ax + 1
f’ (x) = 2x + a
f (x) is increasing if f’ (x) > 0
2x + a > 0 is x > – a/2
2x > -a – a < 2x ⇒ a > – 2x
since x ∈ (1,2) a > -2
The least value is -2.

KSEEB Solutions

Question 15.
Let I be any interval disjoint from (-1, 1). Prove that the function f given by \(f(x)=x+\frac{1}{x}\) is strictly increasing on I.
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.2.22

Question 16.
Prove that the function f given by f (x) = log sin x is strictly increasing on
\(\left(0, \frac{\pi}{2}\right)\) and strictly increasing on \(\left(0, \frac{\pi}{2}\right)\)and strictly decreasing on \(\left(\frac{\pi}{2}, \pi\right)\)
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.2.23

Question 17.
Prove that the function f given by f (x) = log cos x is strictly decreasing on
\(\left(0, \frac{\pi}{2}\right)\)and strictly increasing on \(\left(\frac{\pi}{2}, \pi\right)\)
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.2.24

Question 18.
Prove that the function given by f (x) = x3 – 3x2 + 3x – 100 is increasing in R.
Answer:
f (x) = x3 – 3x2 + 3x
f'(x) = 3x2 – 6x + 3
= 3 (x – 2)2
f'(x)>0 ∀ x ∈ R
∴ function is continuous on R

KSEEB Solutions

Question 19.
The interval in which y = x2 e-x is increasing is
(A) (- ∞ , ∞)
(B) ( – 2, 0)
(C) (2, ∞)
(D) (0,2).
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.2.25

2nd PUC Basic Maths Question Bank Chapter 5 Partial Fractions Ex 5.1

Students can Download Basic Maths Exercise 5.1 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Basic Maths Question Bank Chapter 5 Partial Fractions Ex 5.1

Part – A

2nd PUC Basic Maths Partial Fractions Ex 5.1 Two or Three Marks Questions and Answers

Question 1.
Express the following as a sum of polunomial and proper rational fraction ; \(\frac{x^{2}+x+1}{x^{2}-1}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 5 Partial Fractions Ex 5.1 - 1
Question 2.
\(\frac{3 x^{2}-4 x+7}{x+7}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 5 Partial Fractions Ex 5.1 - 2

KSEEB Solutions

Question 3.
\(\frac{x^{2}-1}{x^{2}+1}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 5 Partial Fractions Ex 5.1 - 3

Question 4.
\(\frac{5 x^{2}}{x^{2}+4 x+3}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 5 Partial Fractions Ex 5.1 - 4

KSEEB Solutions

Question 5.
\(\frac{x^{4}}{(x-1)\left(x^{2}+1\right)}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 5 Partial Fractions Ex 5.1 - 5

Question 6.
\(\frac{4 x^{3}-2 x^{2}+3 x+1}{2 x^{2}+4 x-1}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 5 Partial Fractions Ex 5.1 - 6

Question 7.
\(\frac{4 x^{2}-4 x-1}{2 x-1}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 5 Partial Fractions Ex 5.1 - 7

KSEEB Solutions

Question 8.
\(\frac{x^{3}+7}{x^{2}-2 x+1}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 5 Partial Fractions Ex 5.1 - 8

2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2

Students can Download Basic Maths Exercise 4.2 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2

Part – A

2nd PUC Basic Maths Binomial Theorem Ex 4.2 Two Marks Questions and Answers

1. Find
Question (i).
The 5th term in \(\left(\frac{4 x}{5}-\frac{5}{2 x}\right)^{8}\)
Answer:
\(\left(\frac{4 x}{5}-\frac{5}{2 x}\right)^{8}\) compare with (x + a)n
⇒ x → \(\frac{4 x}{5}\), a → \(\frac{4 x}{5}\) n → 8,
To find th term put r = 4
Tr+ = nCr.xn – r.ar
2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2 - 1
T5 = 8C4.(22).2-4
8C4.28-4 = 8C4.24 = 1120

Question (ii).
The 8th term in \(\left(\frac{a}{2}-\frac{3}{b}\right)^{10}\)
Answer:
\(\left(\frac{a}{2}-\frac{3}{b}\right)^{10}\) compare with (x+a)n
⇒ x → \(\frac{a}{5}\), a → \(\frac{2}{b}\) n = 10,
To find 8th term put r = 7
Tr+ = nCr.xn – r.ar
2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2 - 2

KSEEB Solutions

Question (iii).
The 6th term in (√x – √y)17
Answer:
Compare (√x – √y)17 with (x + a)n
x → √x a → -√y and n = 17
To find 6th term put r = 5
2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2 - 3

Question (iv).
The 7th term in (3x2 – \( \frac{y}{3}\) )9
Answer:
Here x → 3x2 a → \(-\frac{y}{3}\) nn = 9
Put r = 6
T6+1 =9C6 (3x2)9-6. ( \(-\frac{y}{3}\) )6

2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2 - 4

Question (v).
the 10th term in \(\left(\frac{a}{b}-\frac{2 b}{a^{2}}\right)^{12}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2 - 5

Question (vi).
the 11th term in \(\left(x+\frac{1}{\sqrt{x}}\right)^{14}\)
Answer:
Here x = x a = \(\frac{1}{\sqrt{x}}\) ,n = 14 and
put r = 10
T10+1 = 14C10 .x14 – 10 . \(\left(\frac{1}{\sqrt{x}}\right)^{10}\) 14C4.x4 \(\frac{1}{x^{5}}\) = \(\frac{1001}{x}\)

KSEEB Solutions

Part – B

2nd PUC Basic Maths Binomial Theorem Ex 4.2 Three Marks Questions and Answers

Question 2.
Find the middele term in the expansion of

Question (i).
\(\left(x-\frac{1}{2 y}\right)^{10}\)
Answer:
\(\left(x-\frac{1}{2 y}\right)^{10}\)
Here n = even i.e 10 ∴ we have only one middle term  \(\frac{\mathrm{n}}{2}+1=\frac{10}{2}+1\) = 6th term
2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2 - 6

Question (ii).
\(\left(\frac{a}{x}+b x\right)^{12}\)
Answer:
\(\left(\frac{a}{x}+b x\right)^{12}\)
Here
n = 12 (even)
∴ We have only one middle term
2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2 - 7

Question (iii).
\(\left(\frac{2 a}{3}-\frac{3}{3 a}\right)^{6}\)
Answer:
Here n = 6 (even)
2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2 - 8

Question (iv).
\(\left(3 x-\frac{1}{6} x^{3}\right)^{8}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2 - 9

Question (v).
\(\left(\frac{a}{3}+\frac{b}{3}\right)^{8}\)
Answer:
Here n = 8(even)
∴ middle term \frac{\mathrm{n}}{2} + 1 = 4 + 1 = 5th
2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2 - 10

KSEEB Solutions

3. Find the middle terms in the expansion of

Question (i).
Find the middle term in the expansion of
\(\left(3 x-\frac{2}{x^{2}}\right)^{15}\)
Answer:
Here n = 15 and odd, so we have two middle terms i.e, \(\frac{n+1}{2}=\frac{15+1}{2}=8^{t h}\) and 8 + 1 = 9th terms to find 8 thterm to find 8 th term pur r = 7
2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2 - 11

Question (ii)
\(\left(\frac{x}{2}+\frac{3}{x^{2}}\right)^{19}\)
Answer:
Here n = 19 (odd)∴ We have two middle terms \(\frac{n+1}{2}=\frac{20}{2}\) 10th and 10 + 1 =11th term To find 10th term put r = 9
2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2 - 12
Question (iii)
\(\left(2 x^{2}+\frac{1}{\sqrt{x}}\right)^{11}\)
Answer:
Here n = 11 (odd)
∴ the two middle terms are \(\frac{13+1}{2}\) = 7th and 6 + 1 = 7th terms
To find 6th term put r = 5
2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2 - 13

Question (iv).
\(\left(\sqrt{x}-\frac{4}{x^{2}}\right)^{11}\)
Answer:
Here n = 11(odd) we have two middle term
i.e,  \( \frac{1+1}{2}\) = 6th and 6 + 1 = 7th terms
To find 6th term put r = 5.
2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2 - 14
2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2 - 15

Question (v).
\(\left(\sqrt{x}-\frac{3}{x^{2}}\right)^{13}\)
Answer:
\(\left(\sqrt{x}-\frac{3}{x^{2}}\right)^{13}\)
Here n = 13 (odd)
∴ the two middle terms are \(\frac{13+1}{2}\) = 7th and 7 + 1 = 8th terms
To find 7th term put r = 6
2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2 - 16

Part – C

2nd PUC Basic Maths Binomial Theorem Ex 4.2 Five Marks Questions and Answers

4.
Question (i).
Find the coefficent of xn in \(\left(x+\frac{2}{x^{2}}\right)^{17}\)
Answer:
Here x = x  a  = \(\frac{2}{x^{2}}\) and  n = 17
Tr+1 = nCr . x n-r.ar
= 17Cr .x 17-r.( \(\frac{2}{x^{2}}\) )r = 17Cr . 2r.x 17-r-2r
= 17Cr 2 r.x 17-3r
To find coefficient of x11 equate the power of x to 11
⇒ 17 – 3r = 11 ⇒ 17 – 11 = 3r
⇒ 3r = 6 ⇒ r = 2
T2+1 = 17C2 . 22.x11
∴ Coefficient of x” is 17C2 .22 = \(\frac{17 \times 16 \times 2}{2 \times 1}\) = 544

Question (ii).
Y3 in \(\left(7 y^{2}-\frac{2}{y}\right)^{12}\)
Answer:
Here x =7y2, a = \(-\frac{2}{y}\) and n = 12
Tr+1 = 12Cr(7Y2)12-r.\(\left(\frac{-2}{y}\right)^{r}\)
= 12Cr.712-r .y24-2r. y-r(-2)r
Tr+1 = 12Cr.712-r.(-2)r.y24-3r
To find the coefficient of y3 equate the power of y ro 3
i.e., 24 – 3r = 3 ⇒ 21 = 3r ⇒ r= 7
∴T7+1 = 12C7 .712-7. (-2)7 y3
= -12C7 .75 27 . y3
∴Coefficient of y3 is -12C7.75.27

Question (iii).
x11 in \(\left(\sqrt{x}-\frac{2}{x}\right)^{17}\)
Answer:
lere, x → √x, a →\(\frac{-2}{x}\) and n = 17
2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2 - 17
To find the coefficient of x2 equate the power of x to -11
∴ \(\frac{17-3 r}{2}\) = -11 ⇒ 17 – 3r = – 22
⇒ 17+22 = 3r ⇒ 39 = 3.r ⇒ r = 13
T13+1 = 17C13(-2)13.x-11
Coefficient of x-11 is -17C13.213

Question (iv).
X18 in \(\left(x^{2}-\frac{6}{x}\right)^{15}\)
lere, x → x2, a →\(\frac{-6}{x}\) and r = 15
Tr+1 = 15Cr.(x2)15-r\(\left(\frac{-6}{x}\right)^{r}\)
Tr+1 = 15Cr.x30-2r.(-6)r.x-r
= 15Cr.(-6)r.x30-2r-r
= 15Cr(-6)r.x30-3r.
To find the coefficient of x18,equate the power of x to 18
∴ 30 – 3r = 18 ⇒ 30 – 18 = 3r ⇒ 3r = 12 ⇒ r = 4
T4+1 = 15C4(-6) 4x18
∴ Coefficient of x18 is 15C4. (6)4

KSEEB Solutions

Question (v).
X-2 in \(\left(x+\frac{1}{x^{2}}\right)^{17}\)
Answer:
Here, x → x, a → \(\frac{1}{x^{2}}\) and n = 17
∴ Tr+1 = 17Cr.x17-r\(\left(\frac{1}{x^{2}}\right)^{r}\)
Tr+1 = 17Cr.x17-r-2r
= 17Crx17-3r
To find the coefficient of x-2,equate the power of x to -2
17 – 3r = -2 ⇒ 17 + 2 = 3r ⇒ 3r ⇒ r = \(\frac{19}{3}\)
Since r is a fraction the coefficient of x-2 is 0.

Question (vi).
X5 in \(\left(x+\frac{1}{x^{2}}\right)^{17}\)
Answer:
Here, x → x, a → \(\frac{1}{x^{2}}\) and n = 17
Tr+1 = 17Cr.x17-r\(\left(\frac{1}{x^{2}}\right)^{r}\)
= 17Crx17-3r
To find the coefficient of x5, equate the power of x to 5
∴ 17 – 3r = 5
12 = 23 ⇒ r = 4
T4+1 = T5 = 17C4.x5
∴ the coefficient of x5 is 17C4

Question (vii).
X18 in \(\left(x^{2}+\frac{3 a}{x}\right)^{15}\)
Answer:
Here x → x2 a → \(\frac{3 \mathrm{a}}{\mathrm{x}}\) , n=15
∴ Tr+1 = 15Cr.(x2)15-r \(\left(\frac{3 a}{x}\right)^{r}\)
= 15Crx30-2r.(3a)rxr
=15Cr .3r.ar.x30-3r
To find the coefficient of x18, we get 30 – 3r = 18
12 = 3r ⇒ r =4
∴ T4+1 = 15C4.34.a4.x18
∴ coefficient of x18 is 15C4.(3a)4

5. Find the term independent of x in

Question (i).
\(\left(\frac{4 x^{2}}{3}+\frac{3}{2 x}\right)^{9}\)
Here x → \(\frac{4 x^{2}}{3}\) , a = \(\frac{3}{2 x}\) and n = 9
2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2 - 18
To find the term independent of x, equate the power of x to zero.
e., 18 – 3r = 0 ⇒ r = 6
2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2 - 19

Question (ii).
\(\left(x^{3}-\frac{3}{x^{2}}\right)^{15}\)
Answer:
Here x → x3, a = \(\frac{-3}{x^{2}}\) and n = 15
Tr+1 = 15Cr.(x3)15-r.\(\left(\frac{-3}{x^{2}}\right)^{r}\)
= 15Cr.x45-r.x-2r (-3)r
= 15Cr.(-3)r.x45-5r
To find the term independent of x we have 45 – 5r = O
:. 45 = 5r ⇒ r = 9
T9+1 = 15C9.(-3)9.x0
T10 = -15C9.(3)9 is the term independent of x.

KSEEB Solutions

Question (iii).
\(\left(\sqrt{x}+\frac{1}{3 x^{2}}\right)^{10}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2 - 20

Question (iv).
\(\left(\frac{3 x^{2}}{2}-\frac{1}{3 x}\right)^{9}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2 - 21

Question (v).
\(\left(3 x-\frac{2}{x^{2}}\right)^{15}\)
Answer:
Here x → 3x a → \(\frac{-2}{x^{2}}\) and n =15
Tr+1 = 15Cr.(3x)15-r.\(\left(\frac{-2}{x^{2}}\right)^{r}\)
= 15Cr.315-r.(-2)r . x15-r
We have x15-3r. = x0 ⇒ 15 = 3r ⇒ r = 5
T = 15C5.315-5.(-2)5 = -15C5 .310 25
∴ The term independent of x is -15C5.310.25

Question (vi).
\(\left(x^{2}-\frac{2}{x^{3}}\right)^{5}\)
Answer:
Here x → x2 a → \(\frac{-2}{x^{3}}\) = 5r(x2)5 – r . \(\left(\frac{-2}{x^{3}}\right)^{r}\)= 5r.x10-2r.(-2)r
= 5Cr(-2)r.x10-5r
We have 10 – 5r = 0 ⇒ r = 2
T2+1 = T3 = 5C2(-2)2.x0 = 4. \(\frac { 5.4}{ 2.1 }\) = 40
∴  The term independent of x is 40

Question (vii).
\(\left(x-\frac{1}{x^{2}}\right)^{21}\)
Answer:
Here x →x and a → \(\frac{-1}{x^{2}}\) and n= 21
Tr+1 = 21Cr.(x)21-r. \(\left(\frac{-1}{x^{2}}\right)^{r}\)
= 21Cr.x21-3r.(-1)r
We have x0= x21-3r ⇒21 = 3r ⇒ r = 7
∴ T7+1 = T8 = 21C7 (-1)7.x0 = -21C7
∴ The term independent of x is -21C7

Question (viii).
\(\left( \sqrt { 2 } \frac { 2 }{ { x }^{ 2 } } \right)\)<sup>20</sup>
Answer:
2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2 - 22

Part – D

2nd PUC Basic Maths Binomial Theorem Ex 4.2 Four Marks Questions and Answers

6. Use binomial theorem to evaluate upt 4 decimals place

Questoin (i)
(102)6
(102)6 = (100 + 2)6
= (100)6 + 6C1(1 0O)5. 2 + 6C2(100)4.22 + 6C3(100)3.236C4(100)2 24 + 6C5.100.25 + 6C626
= 1000000000000 + 120000000000 + 6000000000 + 160000000 + 2400000 + 19200 + 64
= 1,126,162,419,264

Question (ii).
(98)4
Answer:
(98)4 = (100 – 2 )4
= (100) 44C1 (100)3.2 + 4C2(100)2.22 4C3(100).23 + 4C4.24
= 100000000 – 8000000 + 240000 – 3200 + 16
= 92236816

KSEEB Solutions

Question (iii).
(1.0005)4
Answer:
(1.0005)4 = (1 + 0.0005)4
= 14 + 4C1(O.0005) + 4C2(0.0005)2 + 4C3(0.0005)3 + 4C4(0.0005)4
1 + 0.002 + 0.0000015 + …………….
= 1.00200150 ≈ 1.0020

Question (iv).
(0.99)4
Answer:
(0.99)4 = (1- 0.01)4 = 4C0(0.01) – 4C1 (0.01) + 4C2(0.O1)24C3(0.01)3 + 4C4(0.01)4
= 1 – 0.04 + 0.0006 – 0.000004 + 0.00000001
= 0.96059601 ≈ 0.9606

Question 7.
The first three terms in (1 + ax)n where n is a positive integer are 1, 6x, 16x2. Find the vaIue
Answer:
Given . T1 = lin (1 + ax)n; T2 = 6x
nC1 .ax = 6x
nax = 6x
⇒ na = 6 ⇒ a = \(\frac{6}{n}\)
and T3 = 16x2
\(\frac{n(n-1)}{2}\) a2x2 = 16x2

Question 8.
In the expansion of (3 + kx)9 the x2 and x3 are equal. Find k.
Answer:
Given , Tr+1= 9Cr.39-r.(kx) r
= 9Cr.39-r.krxr
Coefficient of x2 ⇒ x2 ⇒ r = 2
2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2 - 27
36x 37.k2 x x2
Coefficient of x3 ⇒ x3 = xr⇒ r = 3
2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2 - 24
T3+1 = T4 = 93.39-3.k3.x3 = 9C3.36
k3.x3 = 84.36k3.x3
84 x 36k2= 36 x 37 x k2
2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2 - 25

KSEEB Solutions

Question 9.
Find the ratio of the coefficient of x4 in the two expansions (1+x)7 and (1+x)10?
Answer:
2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2 - 26

KSEEB Solutions for Class 8 Maths Chapter 11 Tribhujagala Sarvasamate Ex 11.7

Students can Download Maths Chapter 11 Tribhujagala Sarvasamate Ex 11.7 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths in Kannada helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

KSEEB Solutions for Class 8 Maths Chapter 11 Tribhujagala Sarvasamate Ex 11.7

KSEEB Solutions for Class 8 Maths Chapter 11 Tribhujagala Sarvasamate Ex 11.7 1
KSEEB Solutions for Class 8 Maths Chapter 11 Tribhujagala Sarvasamate Ex 11.7 2

KSEEB Solutions for Class 8 Maths Chapter 11 Tribhujagala Sarvasamate Ex 11.7 3
KSEEB Solutions for Class 8 Maths Chapter 11 Tribhujagala Sarvasamate Ex 11.7 4
KSEEB Solutions for Class 8 Maths Chapter 11 Tribhujagala Sarvasamate Ex 11.7 5

2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.3

Students can Download Maths Chapter 5 Continuity and Differentiability Ex 5.3 Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.3

2nd PUC Maths Continuity and Differentiability NCERT Text Book Questions and Answers Ex 5.3

Find \(\frac{\mathrm{d} y}{\mathrm{d} x}\) in the following

Question 1.
2x + 3y = sin x
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.3.1

KSEEB Solutions

Question 2.
2x + 3y = sin y
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.3.2

Question 3.
ax + by2 = cos y
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.3.3

Question 4.
xy + y2 = tan x + y
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.3.4

Question 5.
x2 + xy + y2 = 100
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.3.5
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.3.6

KSEEB Solutions

Question 6.
x3 + x2y + xy2 + y3 = 81
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.3.7

Question 7.
sin2y + cos xy = π
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.3.8

Question 8.
sin2 x + cos2 y = 1
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.3.9

KSEEB Solutions

Question 9.
\(y=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)\)
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.3.10

Question 10.
\(y=\tan ^{-1}\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right),-\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}}\)
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.3.11

Question 11.
\(y=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right), 0<x<1\)
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.3.12

KSEEB Solutions

Question 12.
\(y=\sin ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right), 0<x<1\)
Answer:

Question 13.
\(y=\cos ^{-1}\left(\frac{2 x}{1+x^{2}}\right),-1<x<1\)
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.3.14
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.3.15

KSEEB Solutions

Question 14.
\(y=\sin ^{-1}(2 x \sqrt{1-x^{2}}),-\frac{1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}\)
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.3.16

Question 15.
\(y=\sec ^{-1}\left(\frac{1}{2 x^{2}-1}\right), 0<x<\frac{1}{\sqrt{2}}\)
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.3.17

1st PUC Chemistry Model Question Paper 2 with Answers

Students can Download 1st PUC Chemistry Model Question Paper 2 with Answers, Karnataka 1st PUC Chemistry Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 1st PUC Chemistry Model Question Paper 2 with Answers

Time: 3.15 Hours
Max Marks: 70

Instruction:

  1. The questions paper has five parts A, B, C, D and E. All parts are compulsory.
  2. Write balanced chemical equations and draw labeled diagram wherever allowed.
  3. Use log tables and simple calculations f necessary (use of scientific calculations is not allowed).

Part – A

I. Answer ALL of the following (each question carries one mark): ( 10 × 1 = 1 )

Question 1.
Express 0.001023 into scientific notation.
Answer:
1.023 x 10-3.

Question 2.
Define critical temperature.
Answer:
The temperature above which gas cannot be liquified.

Question 3.
Give the example which acts as Lewis base as well as Bronsted base.
Answer:
NH3 or Ammonia.

KSEEB Solutions

Question 4.
How does electronegativity related to atomic size?
Answer:
Atomic size ∝ \(\frac{1}{\text { Electronegativity }}\)
Electronegativity

Question 5.
What is the oxidation state of oxygen in peroxide?
Answer:
-1 or minus 1

Question 6.
Which alkali metal is act as strong reducing agent?
Answer:
Lithium.

Question 7.
Write the formula of inorganic benzene.
Answer:
B3N3H3

Question 8.
What is dry ice?
Answer:
Solid CO2.

KSEEB Solutions

Question 9.
Complete the reaction NH4CNO
1st PUC Chemistry Model Question Paper 2 with Answers - 1
Answer:
1st PUC Chemistry Model Question Paper 2 with Answers - 2

Question 10.
Write the name of the chain isomer of n-Butane.
Answer:
CH3-CH2-CH2-CH3

Part – B

II. Answer any FIVE of the following questions carrying TWO marks: ( 5 × 2 = 10 )

Question 11.
Calculate the average atomic mass of chlorine using the following data.
1st PUC Chemistry Model Question Paper 2 with Answers - 3
Answer:
1st PUC Chemistry Model Question Paper 2 with Answers - 4

Question 12.
Draw the graph of pressure versus volume of a gas at a different temperatures to illustrate the Boyle’s law.\
Answer:
1st PUC Chemistry Model Question Paper 2 with Answers - 5

Question 13.
Give any two conditions for hybridization of atomic orbitals.
Answer:
(i) Should contain half filled atomic orbitals.
(ii) Energy of the combining atomic orbitals should be equivalent.

Question 14.
Explain the reactivity of second group elements towards hydrogen.
Answer:
Hydrogen combines with second group elements to form metallic hydrides,
i.e. M + H2 → MH2
OR
Ca + H2 → CaH2
when M = Mg, Ca, Ba, etc.

KSEEB Solutions

Question 15.
How many sigma and pi bonds are in carbon monoxide?
Answer:
2σ bonds and one 7t-bond.

Question 16.
Write the structural isomers of an alkene with molecular formula C2H8.
Answer:
H3C-H2C-HC=CH2 , CH3-CH=CH-CH3

Question 17.
What are the characteristics for any ring system to be called as aromatic compound?
(i) It should be planar.
(ii) Obeys Huckel rule i.e. (4n + 2)π electrons.
(iii) Should contain delocalized π – electrons above and below the plane of the molecule.

Question 18.
(a) Which oxide of nitrogen in higher concentration will retard the rate of photosynthesis in plants?
NO2 or Nitrogen dioxide.
(b) Name the compound formed when carbon monoxide binds to haemoglobin.
Carboxyhaemoglobin (Hb + CO → HbCO).

Part – C

III. Answer any FIVE of the following questions carrying THREE marks: ( 5 × 3 = 15 )

Question 19.
Why Beryllium exhibit anomalous behaviour from the rest of the elements in the group.
Answer:
Due to

  1. Smaller size compared to other elements in the group.
  2. High ionisation enthalpy in the ground state.

Question 20.
With the help of hybridization explain the structure of methane.
Answer:
CH4
EC of C = 1s22s22px12py12pz1
The four half filled orbitals overlaps with each other giving 4sp3 hybrid orbitals. These combines with s-orbital of hydrogen along the axis giving 4σ bonds with bond angle 109°28′ and tetraheral geometry.
1st PUC Chemistry Model Question Paper 2 with Answers - 6

Question 21.
(a) What is bond enthalpy? How it is related to the bond order?
Answer:

  1. The amount of energy required to break one mole of bonds of same type to separate them into gaseous atoms.
  2. Bond enthalpy oc Bond order.

(b) Write the resonance structure of CO2.
Answer:
1st PUC Chemistry Model Question Paper 2 with Answers - 7

Question 22.
(a) Give any two differences between bonding and anti-bonding molecular orbitals.
Answer:
BMO :

  1. Formed by addition overlapping ΨAB = ΨA + ΨB
  2. Less energy than ABMO.
  3. More stable.

ABMO :

  1. Formed by the subtraction overlapping of atomic orbitals. of atomic orbitals i.e. ΨAB = ΨAB
  2. More energy than BMO.
  3. Less stable.

(b) What is the dipole moment of BeF2?
Answer:
Dipole moment is 0 (zero)
F ⇌ Be ⇌ F

KSEEB Solutions

Question 23.
Balance the following redox reaction by half reaction method.
MnO4 (aq) + 1 (aq) → MnO2 (s) + I2(s) : Basic medium.
Answer:
1st PUC Chemistry Model Question Paper 2 with Answers - 8

Question 24.
Give the reactions to show amphoteric nature of water.
Answer:
1st PUC Chemistry Model Question Paper 2 with Answers - 9
(ii) Mention any one method of removal of temporary hardness of water.
Answer:
By boiling.

KSEEB Solutions

Question 25.
How caustic soda is commercially prepared from brine by Castner-Kellner cell.
Answer:
Caustic soda by Castner-Kellner cell: NaOH is manufactured by the electrolysis of aqueous solution of NaCl (Brine).
i.e. 2NaCl → 2Na+ + 2Cl
Sodium ions are discharged at mercury cathode.
Sodium deposited at mercury forms sodium amalgam.
Chlorine liberated at anode removed from the cell.
At cathode : 2Na+ + 2e → Na, Na + Hg → Na – Hg
At anode : 2Cl – 2e → Cl2
Na-Hg is treated with water to form NaOH
i.e. Na/Hg + 2H2O → 2NaOH + H2 + Hg.

Question 26.
Give the example of element of group 14
(i) Shows maximum catenation capacity.
Answer:
Carbon

(ii) Used as semiconductor.
Answer:
Silicon

(iii) Which reacts with water.
Answer:
Tin

Part – D

IV. Answer any FIVE of the following questions carry ing FIVE marks: ( 5 x 5 = 25 )

Question 27.
(a) M atch the following :
1st PUC Chemistry Model Question Paper 2 with Answers - 10
Answer:
(i)-(c),
(ii)-(d),
(iii)-(a)

(b) Define molarity. Write the expression to calculate the molarity of the solution for the given mass and volume.
Answer:
It is the number of moles of solute present in 1000 ml solvent or 1dm3.
M = \(\frac{\mathrm{W}}{\mathrm{GMM}} \times \frac{1000}{\mathrm{V}}\)

Question 28.
(a) Give any three posulates of Bohr’s model for hydrogen atom.
Answer:

  • Electrons are revolving around nucleus in a closed circular path called orbits or main shells or energy levels.
  • When an electron jumped from higher energy level to lower energy level, the difference of energy emitted as radiation, i.e. E2 – E1 = ∆E= hγ
  • The angular momentum of an electron has discrete values. It is given by the equation
    mvr = \(\frac{n h}{2 \pi}\)

(b) Calculate the mass of a photon with wavelength 5,OA°
Answer:
1st PUC Chemistry Model Question Paper 2 with Answers - 11

Question 29.
(a) Sketch the shapes of Px and dz2.
Answer:
1st PUC Chemistry Model Question Paper 2 with Answers - 12

(b) Identify the property exhibited by 19K40 and 19Ca40.
Answer:
Isobars (∵ mass number are equal).

(c) Write the orbital (box) type electronic configuration of p4 and d4 according to Hund’s rule of maximum of multiplicity.
Answer:
1st PUC Chemistry Model Question Paper 2 with Answers - 13

KSEEB Solutions

Question 30.
a) Derive ideal gas equation using gas laws.
Answer:
Ideal gas equation:
According to Boyle’s law V ∝ \(\frac{1}{P}\) T at constant T
According to Charle’s law V ∝ at constant P
According to Avogadro’s law V ∝ n at constant T and P
On coming V ∝ \(\frac{1}{P}\) V ∝ \(\frac{1}{P}\) x T x n
or PV=nRT
For ‘n’ moles, PV = RT for 1 mole.
R = gas constant, T = Kelvin Temp, P = Pressure, V = Volume of gas
n = Number of moles of gas.

(b) At 25°C and 760 mm of Hg pressure a gas occupies 600 mL volume. What will be its pressure at a height where temperature is 10°C and volume of the gas is 640 ml.
Initial conditions
P1 = 760 mm
V1 = 600 mm
T1 = 250C = 298K

Final conditions
P 2= ?
v2 = 640 mL
T2= 10°C + 273k = 283K
1st PUC Chemistry Model Question Paper 2 with Answers - 14

Question 31.
(a) If water vapour is assumed to be a perfect gas, molar enthalpy change for vapourization of 1 mole of water at 1 bar ad 100°C is 41kJ mole-1. Calculate the internal energy change when 1 mole of water is vapourized at 1 bar pressure and 100°C.
H2O (l) → H20 (g)
∆ng = 1 – 0 = 1
W.K.T. LH = ∆U + ∆ngRT
OR ∆U = ∆H – ∆ngRT = 41 x 103 – 1 x 8.314 x 373
= 41000— 3101,122 = 37898.878 J/mol
OR = 37.898878 K/J/mol.

(b) State Hess’s law of constant heat summation.
Answer:
The total amount of heat liberated or absorbed is same whether the reaction takes place in one step or more than one step / (or several steps).

(c) What is the value for standard enthalpy of formation of an element.
Answer:
Zero/0

Question 32.
(a) Calculate ∆G° for conversion of oxygen to ozone, -O2(g) —> O3(g) at 298K. If
Kp for this conversation is 2.47 x 10-29.
Answer:
∆rG° = ? R = 8.314J/k/mol T = 298K Kp = 2.47 x 10-29.
W.K.T. ∆rG° = -2.303RTlogKp = 2.303x 8.314 x 298 x 2.47 x 10-29.
= 163229 J/mol or 163.2 KJ/mol.

(b) What is thermochemical equation? Write the thermochemical equation for the molar combustion of ethanol (Given ∆rH° = -1367 kJ mol-1).

  1. It is a balanced reaction, gives the information about physical state of reactants and product as well as heat liberated or absorbed.
  2. C2H5OH(l)+ 3O2(g) → 2CO2(g) + 3H20(/) ; ∆H° = -1367 KJ/mol

(c) What is the value of ∆G in a spontaneous process?
Answer:
∆G = -ve

KSEEB Solutions

Question 33.
For the equilibrium 2NOCl (g) ⇌ 2NO(g) + Cl2(g) the value of the equilibrium constant Kc is 3.75 x 10-6 at 1069K. Calculate the Kp for the reaction at this temperature?
Answer:
For the reaction 2NOCl (g) ⇌ 2NO(g) + Cl2(g)
∆n = 3 – 2 = 1. ∴ Kp = Kc (RT)∆n
Kp = (3.75 x 10-6 x 0.0831 x 1069) x 1 = 3.75 x 0.0831 x 1069 x 10-6 = 333.13 x 10-6 OR Kp= 3.33 x 10-4

(b) Write any two general characteristics of equilibria involving physical process.
Answer:

  1. It is dynamic in nature.
  2. It can be achieved in a closed vessel.
  3. It depends only on temperature but not on concentration or pressure.

Question 34.
(a) The pH of the blood is 7.4. Calculate the [H+].
pH = 7.4 [H+] = ?
W.K.T [H+] = Antilog (-pH)
[H+]= Antilog 10 (-7.4)
Add-1 to-7 and +1 to-0.4 ∴ [H+] = Antilog \(\overline{8.6}\)
[H+] = 10-8 x Antilog of (+0.6).

(b) Derive the Henderson-Hasselbalch equation for acid buffer.
Answer:
Henderson-Hasselbalch equation:
Consider a weak acid and its salt BA,
1st PUC Chemistry Model Question Paper 2 with Answers - 15

Part – E

V. Answer any TWO of the following questions carrying FIVE marks: ( 2 x 5 = 10 )

Question 35.
(a) For the following bond cleavage, use curved arrows to show electron flow, mention the type of bond cleavage, and reactive intermediate formed.
CH3CH2O + OCH2CH3 → CH3CH2O + OCH2CH3
1st PUC Chemistry Model Question Paper 2 with Answers - 16

  1. Homolytic cleavage
  2. Ethoxide free radicals.

(b) Give the hybridisation and geometry of carbocation.
Answer:
sp2 and planar.

KSEEB Solutions

Question 36.
(a) 0.2033 g of an organic compound on combustion gave 0.3780 g of CO2 and 0.1288 g H2O. In a separate experiment 0.1877 g of the compound on analysis by Dumas method produced 31.7 ml. of nitrogen collected over water at 14°C and 758 mm pressure. Determine the percentage composition of the compound. (Aqueous tension of water at 14°C = 12 mm pressure).
Answer:
Given mass of organic compound = 0.2033 g |
Volume of nitrogen (V1) = 31.7 mL
T1 = 287K P1 = (P – f) = 758 – 12 = 746 mm;
P2 = 760mm U2 = ? T2 = 273K
1st PUC Chemistry Model Question Paper 2 with Answers - 17
To convert the volume of of N2 at STP into mass
22400 mL of nitrogen at STP weighs = 28 g
1st PUC Chemistry Model Question Paper 2 with Answers - 18
1st PUC Chemistry Model Question Paper 2 with Answers - 19
% age of oxygen = 100 – (18.1985 + 50.71 + 7.03) = 24.06%

(b) What are nucleophiles? Give an example.
Answer:
The negatively charged species reacts at nucleus or +ve centre are called nucleophiles.
Example: Cl, Br, I, \(\mathrm{C} \overline{\mathrm{N}}\), NH3, etc.

Question 37.
(a) Explain the mechanism of addition of HBr to propene in the presence of peroxide catalyst.
Answer:
It is a free radical mechanism.
It has the following steps:
(a) Initiation
1st PUC Chemistry Model Question Paper 2 with Answers - 20

(b) Propagation
1st PUC Chemistry Model Question Paper 2 with Answers - 21

(c) Termination
1st PUC Chemistry Model Question Paper 2 with Answers - 22

KSEEB Solutions

(b) Write the structures of cis and trans isomers of But-2-ene.
Answer:
1st PUC Chemistry Model Question Paper 2 with Answers - 23

KSEEB Solutions for Class 8 Maths Chapter 10 Ghatankagalu Ex 10.7

Students can Download Maths Chapter 10 Ghatankagalu Ex 10.7 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths in Kannada helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

KSEEB Solutions for Class 8 Maths Chapter 10 Ghatankagalu Ex 10.7

KSEEB Solutions for Class 8 Maths Chapter 10 Ghatankagalu Ex 10.7 1
KSEEB Solutions for Class 8 Maths Chapter 10 Ghatankagalu Ex 10.7 2

KSEEB Solutions for Class 8 Maths Chapter 10 Ghatankagalu Ex 10.7 3
KSEEB Solutions for Class 8 Maths Chapter 10 Ghatankagalu Ex 10.7 4
KSEEB Solutions for Class 8 Maths Chapter 10 Ghatankagalu Ex 10.7 5
KSEEB Solutions for Class 8 Maths Chapter 10 Ghatankagalu Ex 10.7 6

2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.3

Students can Download Basic Maths Exercise 20.3 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.3

Part-A

2nd PUC Basic Maths Indefinite Integrals Ex 20.3 Two Marks Questions and Answers

Question 1.
\(\int \frac{3 x^{2}}{1+x^{3}} d x\)
Answer:
\(\int \frac{3 x^{2}}{1+x^{3}} d x\) = log(1 + x3)+c or using \(\int \frac{\mathrm{f}^{\prime}(\mathrm{x})}{\mathrm{f}(\mathrm{x})} \mathrm{d} \mathrm{x} = log (f(x) + c\)
put 1 + x3 = t
∴ 3x2dx = dt
∴ = ∫\(\frac { 1 }{ t }\)dt = log t + c = log(1 + x3) + c

Question 2.
\(\int \frac{4 x+3}{2 x^{2}+3 x+5} d x\)
Answer:
\(\int \frac{4 x+3}{2 x^{2}+3 x+5} d x\) = log(2x2 + 3x +5) + c
∴ \(\int \frac{\mathrm{f}^{\prime}(\mathrm{x})}{\mathrm{f}(\mathrm{x})} \mathrm{d} \mathrm{x}\) = log(f(x)) + C

Question 3.
\(\int \frac{e^{x}-1}{e^{x}-x} d x\)
Answer:
\(\int \frac{e^{x}-1}{e^{x}-x}\) = log(ex – x) + c

KSEEB Solutions

Question 4.
2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.3 - 4
Answer:
2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.3 - 5

Question 5.
\(\int \frac{\cos x}{2+\sin x} d x\)
Answer:
\(\int \frac{\cos x}{2+\sin x} d x\) = log(2 + sin x) + c

Question 6.
\(\int \frac{1}{x(2 \log x+5)} d x\)
Answer:
\(\int \frac{1 / x^{d x}}{2 \log x+5}\)
\(\int \frac{d t / 2}{t}=\frac{1}{2} \int \frac{1}{t} d t\)
= \(\frac { 1 }{ 2 }\)log t + c
= \(\frac { 1 }{ 2 }\)log (2 log x + 5) + c
put 2 log x + 5 = t
2. \(\frac { 1 }{ x }\) dx = dt
\(\frac { 1 }{ x }\) dx = \(\frac { dt }{ 2 }\)

Question 7.
\(\int \frac{3 \sin x}{3+4 \cos x} d x\)
Answer:
\(\int \frac{3 \sin x}{3+4 \cos x} d x\)
\(=\int \frac{3 \cdot \frac{d t}{-4}}{t}=\frac{-3}{4} \int \frac{1}{t} d t\)
= \(\frac { -3 }{ 4 }\) log t + c
= \(\frac { -3 }{ 4 }\) log(3 + 4 cos x) + c
put 3 + 4 cos x = t
– 4 sinx dx = dt
sin x dx = \(\frac { dt }{ -4 }\)

KSEEB Solutions

Part-B

2nd PUC Basic Maths Indefinite Integrals Ex 20.3 Five Marks Questions and Answers

Question 1.
\(\int \frac{1}{\sqrt{x}+x} d x\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.3 - 1

Question 2.
\(\int \frac{\sin 2 x}{1+\cos ^{2} x} d x\)
Answer:
\(\int \frac{\sin 2 x}{1+\cos ^{2} x} d x\)
= \(\int \frac{-d t}{t}\)
= -log t + c
= -log (1 + cos2x) + c
put 1 + cos2x = t
2 cosx (-sinx) dx = dt
-sin 2x dx = dt
sin 2x dx = -dt

Question 3.
\(\int \frac{e^{2 x}+1}{e^{2 x}-1} d x\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.3 - 2

Question 4.
2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.3 - 6
Answer:
2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.3 - 3

KSEEB Solutions

Question 5.
\(\int \frac{\cot x}{3+\log (\sin x)} d x\)
Answer:
\(\int \frac{\cot x}{3+\log (\sin x)} d x\)
\(=\int \frac{d t}{t}\)
= log t + c = log(3 + log(sin x)) + c
put 3 + log(sin x) = t
\(\frac{\cos x}{\sin x} d x=d t\)
cos x dx = dt

Question 6.
\(\int \frac{\csc ^{2} x \cdot \cot x}{4+5 \csc ^{2} x} d x\)
Answer:
\(\int \frac{\csc ^{2} x \cdot \cot x}{4+5 \csc ^{2} x} d x\)
\(=\int \frac{-1 / 10^{\mathrm{dt}}}{t}=\frac{-1}{10} \log t+C\)
= \(\frac { -1 }{ 10 }\)log(4 + 5cosec2x) + c
put 4 + 5 cosec2x = t
– 10 cosec x – cosec x × cot x dx = dt
cosec2x . cot dx = \(\frac { -1 }{ 10 }\)dt

2nd PUC Sociology Model Question Paper 1 with Answers

Students can Download 2nd PUC Sociology Model Question Paper 1 with Answers, Karnataka 2nd PUC Sociology Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Sociology Model Question Paper 1 with Answers

Time: 3 Hrs 15 Min
Max. Marks: 100

I. Answer the following questions in a sentence each. (10 × 1 = 10)

Question 1.
What is demography?
Answer:
Demography is the systematic study of the population.

Question 2.
Name any one ancient name of India.
Answer:
Bharatha Kanda.

Question 3.
State any one reason for diversity in India.
Answer:
Languages.

Question 4.
Name any one Scheduled Tribe of Karnataka.
Answer:
Kadukuruba.

Question 5.
Expand NABARD.
Answer:
National Bank for Agriculture and Rural Development.

KSEEB Solutions

Question 6.
In which year was the programme ‘Stree Shakthi’ launched?
Answer:
2000-2001.

Question 7.
State any one characteristic of a village community.
Answer:
Social homogeneity in language.

Question 8.
Who is Kartha?
Answer:
Head of a Hindu Joint family

Question 9.
What is T.R.P.?
Answer:
Television Rating Point.

Question 10.
Who introduced the concept of Teleshopping?
Answer:
Michael Aldrich.

II. Answer any ten of the following questions in 2-3 sentences each. (10 × 2 = 20)

Question 11.
Define National Integration.
Answer:
National integration refers to national unity and a sense of belonging to the nation.

Question 12.
Define caste.
Answer:
M.N SRINIVAS defines caste as a “Hereditary, endogenous, usually localized group, having a traditional association with an occupation and a particular position in the local hierarchy of castes. Relation between castes is governed, by the concept of pollution and purity, and generally maximum commensality occurs within the caste.

KSEEB Solutions

Question 13.
Mention any two social reform movements.
Answer:
Brahma Samaja and Arya Samaja.

Question 14.
Distinguish between sex and gender.
Answer:
Sex is of biological nature. Gender is identity as male or female as per social and psychological perceptions, learnt through a process of socialisation.

Question 15.
Explain the concept of women empowerment.
Answer:
Women Empowerment is the act of empowering women i.e. to give them the power or authority. The term Empowerment has different meanings to women of different classes.

Question 16.
What is micro finance?
Answer:
Micro Finance is defined as the financial services such as Saving A/c, Insurance Funds and credit ‘ facilities provided to the poor and low-income group so as to help them to improve their income and thereby their standard of living also.

Question 17.
What is a self help group?
Answer:
A Self-help Group comprises a group of micro enterpreneurs having homogenous social and economic backgrounds, all voluntarily coming together to save regularly small sums of money, mutually agreeing to contribute to a common fund and to meet their emergency needs from that fund on the basis of mutual help.

Question 18.
Mention any two advantages of joint family.
Answer:
Protection of members and economic advantage.

Question 19.
Write two major problems of Indian cities.
Answer:
Social problems and Agricultural and Economic problems.

KSEEB Solutions

Question 20.
What is virtual market?
Answer:
The new form of marketing and transactions are taking place through online with the help of Information and Communication Technology. E-commerce, online purchase, online trading of stocks and shares are the latest in the market activities. Such transactions and activities are called as virtual activity of the market.

Question 21.
Mention any two factors facilitating farmers’ movement in India.
Answer:

  • The collection of heavy revenue.
  • Land was made as private property.

Question 22.
What do you mean by Social movement?
Answer:
Social movements have broadly been perceived as ‘organized’ or ‘collective effort’ to bring about changes in the thought, beliefs, values, attitudes, relationships and major- institutions in society or to resist any change in the above societal arrangements.

III. Answer any four of the following questions in 15 sentences each. (4 × 5 = 20)

Question 23.
Explain the nature of diversities in India.
Answer:
The term diversity denoting collective differences so as to find out dissimilarities among groups of people: geographical, religious, linguistic etc. All these differences presuppose collective differences or prevalence of variety of groups and culture. Indian society is characterized by unity as well as diversity.
Primarily there are four major types of diversities in India, which are;

  1. Regional diversities
  2. Linguistic diversities
  3. Religious diversities and
  4. Cultural and Ethnic Diversities

1. Regional Diversities:
India is a vast country. From the Himalayas in the North to the Indian Ocean in the south, there are quite lot of differences in altitude, temperature, Flora and Fauna. India has every conceivable type of climate, temperature and physical configuration. There is the scorching heat of Rajastan and the biting cold of the Himalayas, Rainfall varies from 1200 to 7.5 ems per year.

The result is that India has some of the wettest and driest areas in the world. India also possesses arid desserts and fertile riverine lands, bare and hilly tracts and luxuriant open plain.

2. Linguistic Diversities:
Language is another source of diversity. It contributes to collective identities and even to conflicts. The Indian Constitution has recognized 22 languages in the 8th schedule for its official purposes but HN as many as 1652 languages and dialects are spoken in the country.

These languages belong to five linguistic families, namely; Indo- Aryan languages, Dravidian languages, Austric languages, Tibeto – Burman languages and European languages. This makes language planning and promotion difficult. But the mother tongue does evoke strong sentiments and reactions.

As a consequence of this multiplicity, there is considerable bilingualism and administration has to use more than one language. Linguistic diversity has posed administrative and political challenges. Apart from that for people with different mother tongues, communication becomes a problem.

3. Religious Diversities:
There are 8 major religious communities in India. Hindus constitute the majority followed by Muslims, Christians and Sikhs. Buddhists, Jains, Zoroastrians and Jews are less than 1% each. Each major religion is further divided along the lines of religious documents, sects and cults. Hindus are broadly divided into Shaivites, Vaishnavaites and Shaktas (worshippers of Shiva, Vishnu and Mother Goddess – Shakthi respectively) and other minor sects.

Even though they took birth in India, both Jainism and Buddhism have lost their hold in India and are confined to a few small pockets. Diganibars and Shw’etambars are the two divisions of Jains. Indian Muslims are broadly divided into Shias and Sunnis. Indian Christians, apart from Roman Catholics and Protestants have other small regional denominational churches.

Sikhism is a synthesizing religion that emphasizes egalitarianism. Parsis even though a small community have played an important role in India’s industrial development. The Jews have a white and black divisions.

4. Cultural and Ethnic Diversities:
Another important source of diversity is the cultural diversity. The people differ considerably in
their social habits. Cultural difference varies from state to state. The conflicting and varying shades of blood, strains, culture and modes of life, the character, conduct, beliefs, morals, food, dress, manners, social norms, Socio-Religious customs, rituals and etc.

causes cultural and ethnic diversities in the country. Dr. R.K. Mukherji rightly said that “India is a museum of cults and customs, creeds and culture, faiths and tongues, racial types and social systems.

KSEEB Solutions

Question 24.
Explain impacts of British rule on caste system.
Answer.
The impact of British rule on caste system in India may be studied under the following heads.

  1. Introduction of Universalistic Legal system.
  2. Impact of English Education.
  3. Impact of Social Reform Movements.
  4. Influence of New Social Formation.
  5. Impact of Freedom Struggle.
  6. Impact of Industrialization and Urbanization.

1. Introduction of Universalistic Legal System:
The establishment of British courts removed authority from the purview of caste panchayats. Under this new principle of justice, all are equal before the law, and the caste panchayats lost their former importance.
Some major legislations were the following:
a. The Caste Disabilities Removal Act of 1850. This act served to remove some of the disabilities associated with castes including the practice of untouchability.

b. The Hindu Widow Remarriage Act 1856, This act made legal, provision for the Hindu widows to remarry.

c. The Special Marriage Act of 1872considered marriage as a civil contract and legalized inter-caste or inter-religious marriages.

d. Other legislative and administrative measures were put into effect, like Government schools to be open to all classes of subjects, stopping of gran s to schools and refusing admission to depressed class, public places to be men to everybody and constitutional provisions for representation in legislative bodies for them.

2. Impact of English Education:
British education was based on scientific, secular and universal principles. It was made accessible to everyone, irrespective of caste or community. It remained liberal in content. It propagated principles such as the liberty equality and fraternity.

As education spread to the lower strata, it kindled libertarian impulses among them. Western education provided an indispensable passport to the new economic opportunities. Members from the lower castes became professionals and embraced the new commercial opportunities offered by western education.

3. Impact of Social Reform Movements:
Social reform movements brought changes in the caste system in British period. They set out to eradicate caste and to establish a casteless and, classless society.

Brahma Samaj by Rajaram Mohan Roy, Prarthana Samaj by Atmaram Pandurang, Arya Samaj by Swamy Dayananda Saraswathi, Ramakrishna Mission by Swami Vivekananda, Theosophical Society by Annie Besant and Divine Life Society by Maharshi Aurobindo Ghosh were leading movements.

All these organizations aimed at the destruction of caste system and social reconstruction of Indian society.

4. Impact of New Social Formations:
The new economic system brought about a new grouping of the population in the economic sphere. The Indians got differentiated into capitalists, workers, peasants, proprietors, merchants, tenants, land lords, doctors, lawyers, teachers and technicians.

Each category being composed of individuals belonging to various castes, but having identical material and political interests. This division weakened the vertical caste lines. Thus there came into existence such, organizations as Mil! Owners Associations, All India Trace Union Congress.

All India Kishan Sabha and etc., these groups struggled for their own interests. In the process of this struggle they developed a new consciousness and outlook and a new solidarity, which slowly weakened the caste consciousness.

5. Impact of Freedom Struggle:
The growth of the nationalist movement played a great role in weakening caste consciousness. In India, the presence of foreign rule was a permanent stimulus to the Indians to unite on a national basis. Thus the growth of the national movement undermined the caste consciousness.

6. Impact of Industrialization and Urbanization:
The growth of Industries destroyed the old craft and provided new ways to earn a livelihood. Occupational mobility and movement from compact ancestral villages started breaking down the caste norms.

New transport facilities, especially crowded trains, and buses, threw together millions of people of all castes and left little room for the necessities of ceremonial purity. Taboos on food and water gradually weakened when industrial workers belonging to various castes started working under one roof and having food at a common canteen.

He demarcation observed by the members of different castes regarding eating food, physical contact with those of other castes, steadily crumbled in cities.

Question 25.
Explain the five causes for changes in joint ‘families.
Answer:
1. Industrialization:
With the establishment of factories in many places of the country, agriculture was pushed to the background and with it changed those social institutions which were it products. The industrial centers pulled persons out of the traditional peasant society comprising of joint families.

This struck at the roots of joint families and the process of change started. Furthermore, the process of change in joint family gained momentum from the rapid development of transport and communication.

2. Urbanization:
The percentage of workers dependent on agriculture has come down and more and more people migrate to cities and towns in search of jobs.

The urban centers also provide people with various amenities of life concerning transport and communication, sanitation and health, education and employment etc., People are tempted by the lure of urban facilities and there is a rural to urban type of migration.

Gradually joint family hold is losing its control Inheritance Act of 1929, and the Hindu women’s Right to Property Act of 1937. Sati Prevention Act 1782, Hindu Widow Remarriage Act 1856, Child Marriage Restraint Act 1902 have brought changes in family relations.

After independence, the process has continued and fundamental changes in the law of inheritance have been brought about by the Hindu Succession Act, 1956. The Hindu Marriage Act, 1955, and the Civil Marriage Act, 1957 gave the freedom to adult males and females to many according to their choice and helped the women to seek divorce on certain grounds.

All these legislations gave enough facility to the members to divide the joint family immediately after the death of the father. The necessity of jointness has also weakened due to various governmental provisions relating to old age pension, widow pension etc.

Question 26.
Explain the characteristics of a village.
Answer:
1. Small in size:
Indian villages are small in size. Due to that the density of population is less in Indian villages.

2. Importance to Primary Relations:
Villages share so many daily requirements and their relationships are close and intimate and face to face interactions.

3. Social Homogeneity:
Village is more homogeneous in language, belief, mores and pattern of behavior. In their occupations, villagers participate together and share common interests.

4. Informal Social Control:
Individual behavior is controlled by family, traditions, customs, religion, etc.

5. Agriculture and its allied occupations:
Agriculture is the main source of livelihood. Along with agriculture, animal husbandry, floriculture, fishing, mining and apiculture and cottage industries are the other occupations.

6. Role of Neighbourhood and simplicity of life:
Neighbourhood relation plays an important role in the social life of village people and a simple way of life is common. There is an interdependent neighbourhood relations.

7. Village Autonomy:
Each village is relatively self-sufficient and independent. Charles Metcalfe called ‘Indian villages as Little Republics’. Recent studies proved that the Indian villages were never self-sufficient and Republic.

KSEEB Solutions

Answer 27.
Explain the major components of Social Movements.
Answer:
M.S.A. Rao in his edited volume on Social Movements in India has highlighted the significance of ideology, collective mobilization, organization and leadership in social movements.
1. Ideology provides a broad frame of action and collective mobilization in the social movement. It also provides legitimacy to the process of interest articulation and organized collective action.

2. Collective Mobilization: The nature and direction of a social movement is widely shaped by the nature of collective mobilization. Collective mobilization may be radical, non-institutionalized, spontaneous, large scale or it may be non-violent, institutionalized, sporadic and restricted.

3. Leadership and Organization are closely linked to the process of collective mobilization.
A leader can be a charismatic figure or a democratically elected one.

IV. Answer any four of the following questions in 15 sentences each. (4 × 5 = 20)

Question 28.
Write a note on demographic dividend.
Answer:
Demographic dividend refers to demographic or population advantage which is obtained due to numerical domination of the young people in the population. It is an advantage due to less dependency ratio.

Dependency ratio means that children less than 14 years and people above 65 years are considered as dependent on the rest of the population. In simple terms, the ratio of the combined age group 0-14 years plus 65 years & above to the 15-65 years age group is referred to as the total dependency ratio.

The younger age groups in the age structure are believed to be an advantage for India. Like the East Asian economies in the past decade and countries like Ireland today, India is supposed to be benefitting from a ‘demographic dividend’.

This dividend arises from the fact that the current generation of working-age people is a relatively large one, and it has only a relatively small preceding generation of old people to support. But there is nothing automatic about this advantage – it needs to be consciously utilised in the following ways.

a. The demographic advantage or ‘dividend’ to be derived from the age structure of the population is due to the fact that India is one of the youngest countries in the world. In 2020, the average Indian will be only 29 years old, compared with an average age of 37 in China and the United States, 45 in Western Europe, and 48 in Japan. This implies a large and growing labour force, which can deliver unexpected benefits in terms of growth and prosperity.

b. But this potential can be converted into actual growth only if the rise in the working age group is accompanied by increasing levels of education and employment.

c. India is indeed facing a window of opportunity created by the demographic dividend. The effect of demographic trends on the dependency ratio defined in terms of age groups is quite visible. The total dependency ratio fell from 79 in 1970 to 64 in 2005.

But the process is likely to extend well into this century with the age-based dependency ratio projected to fall to 48 in 2025 because of continued fall in the proportion of children and then rise to 50 by 2050 because of an increase in the proportion of the aged.

d. This suggests that the advantage offered by a young labour force is not being exploited. Unless a way forward is found, we may miss out on the potential benefits that the country’s changing age structure temporarily offers.

Question 29.
Explain the tribal panchasheela.
Answer:
Jawaharlal Nehru laid down the policy of Integration to five principles (1957) in his foreword note to Verrier Elwin’s book, called “The Philosophy of NEFA” (NEFA – North East Frontier of Assam). The tribal Pancha Sheela as enunciated by him is as follows:
1. People should develop along the lines of their own genius and we should avoid imposing anything on them. We should try to encourage in every way their own traditional – arts and culture.

2. Tribal rights in land and forests should be respected.

3. We should try to train and build up a team of their own people to work and manage administration and development. Some technical personnel from outside will, no doubt be needed especially in the beginning. But we should avoid introducing too many outsiders into tribal territory.

4. We should not over-administer these areas or overwhelm them with a multiplicity of schemes. We should rather work through and not in rivalry’ to their own social and cultural institutions.

5. We should judge the results not by statistics or the amount of money spent but by the quality of human character that is evolved.

KSEEB Solutions

Question 30.
Explain the characteristics of joint family,
Answer:
1. Depth of Generations:
A joint family consists of people of three or more generations including grandparents, parents and children. Sometimes, other kins such as uncles, aunts, cousins and great grandsons also live in a joint family.

2. Common Roof:
Henry Maine called the joint family a ‘Greater Home’. Members of the joint family normally reside together under the same roof. It is a place to uphold the family Heritage. It is a place for Socio, Economic, Religious, Entertainment etc. Due to the scarcity of accommodation members of the joint family may reside separately. Still, they try to retain regular contacts and the feeling of belonging to the same family. They have emotional and economic links with the original family.

3. Common Kitchen:
Members eat the food prepared jointly at the common kitchen. Normally, the eldest female member of the family (the wife of the Karta) supervises the work at the kitchen. Rest of the female members are engaged in different kitchen work. A single kitchen under a common roof is an unique element of joint family.

4. Common Worship:
Joint family derives its strength from religion. Hence, it is associated with various religious rituals and practices. Every family may have its own deity or ‘Kula devata’ and its own religious traditions. Members of the family take part in common worship, rites and ceremonies. At least once a year they join other members to take part in the festivals, feasting, marriage ceremonies and so on.

5. Common Property:
The members hold a common property. As O’ Malley writes: “The joint family is a co-operative institution similar to a joint-stock company in which there is a joint property”. The total earnings of the members are pooled into a common purse of the family and family expenses are met out of that.

6. Exercise of Authority:
In the patriarchal joint family usually the eldest male member known as ‘Karta’ exerscises authority. The super-ordination of the eldest member and the subordination of all the other members to him is a keynote of the joint family.

His commands are normally obeyed by others. Karta ruled his family by love and affection. Similarly, in the matriarchal joint family the eldest female (matriarch) member exercises supreme authority.

7. Arranged Marriages:
In the joint family, the elders consider it as their privilege to arrange the marriages of the members. The individual’s right to select his/her life-partner is undermined. The younger members rarely challenge their decisions and arrangements. But now-a-days selecting a life partner for a family member is more democratic in nature.

8. Identification with Mutual Rights and Obligations towards the Family:
Every member has his own duties and obligations towards the family. The family in turn, protects the interests and promotes the welfare of all. The senior members of the family act as guides for junior members.

9. Self-Sufficiency:
Joint family is relatively self-sufficient. It meets the economic, recreational, medical, educational and other needs of the members. No other type of family is self-reliant that way today.

KSEEB Solutions

Question 31.
Explain the agricultural and Economic problems of Indian villages.
Answer:
Following are the important economic and agricultural problems.
1. Disparities:
Economic growth in contemporary India is marked by considerable disparities of region and class. The Nobel-prize-winning economist Amartya Sen worries that, “As these inequalities intensify, one half of India will come to look and live like California, the other half like sub- Saharan Africa.” Already, prosperity co¬exists with misery, technological sophistication with human degradation.

2. Discriminatory Policies:
Farmers as a group today feel let down by the policies of the State that puts them relatively in a disadvantageous position. This is made abundantly clear by many analysts in the recent past.

In other words, it is not that the state is discriminatory against the farmers as a group, but the policies are sufficiently provocative in widening the gap between the net incomes of farmers and agricultural laborers on the one hand and the remaining professions on the other.

During the decade of the 1990s the situation became aggravated, both due to policy failure and the successive droughts. At the end, the prices did not pick up even in the event of low production. This was compounded by the economic reforms which took the agricultural sector for granted overlooking their needs.

3. The vulnerability of the Agricultural Sector:
The agricultural sector operates under a large number of constraints. State policies dictate prices of most of the factors of production required for agriculture: electricity, water, fertilizers, pesticides and minimum wages. The credit market operations are largely dictated by the credit policy of the Reserve bank, as well as the difficulties in access to credit.

Difficulties in accessing institutional credit compel the farmers to approach moneylenders and a new emerging institution; namely the input dealer. Weather uncertainties, availability of irrigation water and inputs like fertilizers and pesticides are a cause of concern.

These are compounded by product market imperfections and the price fluctuations that the farmer faces. The process of globalization has intensified some of these concerns, both because of the prominence of trade and the resulting commercialization process in the agricultural sector.

4. Increase in cost cultivation and Environmental degradation:
Increasing cost of cultivation and environmental degradation on one side due to significant increase in the input prices, technology and un-protected farming based on the monsoon on the other makes the farmers hopelessly vulnerable.

Farmers also face high transaction costs and low bargaining power, which leave them with poor returns. The ecological crisis in the rural regions where declining water tables, loss of agricultural biodiversity and the onset of a range of plant diseases and pests have become a challenge to the conduct of agriculture.

5. The deliberate withdrawal of Welfare Programmes from State:
The deliberate withdrawal of the state from its welfare role for the farmers and agriculture labourers has contributed to the accentuation of the agrarian crisis.

The capitalist agriculture in India could thrive because of the proactive role of the state in providing infrastructure, irrigation and credit through institutional agencies. The gradual reduction in the state investment in agriculture was also instrumental in the decline in agricultural productivity and production.

The partial withdrawal of subsidy given to the farmers or to agriculture and free power to agriculture and also the fact that the power tariff has been increased, have added to the woes of farmer drastically.

6. Globalization Resultant Competition and Exploitation by Big Corporates:
The agrarian crisis is due to adoption of World Trade Organization model of agriculture or what is called McKinsey Model of development that created spaces for industry driven agriculture which ultimately resulted in agri-business development including Information Technology.

This model of development has not only exacerbated the crisis leading to an environmental catastrophe but also destroyed millions of rural livelihoods.

7. Peculiar Banking Practices and Non-Availability of Loans from Institutional Sources:
NABARD (National Bank for Agriculture and Rural Development) refinances the cooperative banking institutions and therefore!imposes certain conditions for delivery and recovery of the credit. ‘Eligibility’ is probably the most important concept in dictating the performance of the sector.

A branch of a cooperative bank is categorized as eligible/ non-eligible based on the repayment performance and naturally the Primary Credit Cooperative Societies in the underdeveloped regions have lower repayment performance. As a consequence over the years, these societies, do not get adequate supply of credit and therefore, farmers from these regions have to depend upon the other informal sources of credit.

8. The Failure of the Cooperative Sector:
The Cooperative sector could have helped the farmers in overcoming their debts. The Karnataka government failed to make the cooperative movement a success. For instance, in Karnataka, there are 32,382 cooperative societies at the village level, almost 40 percent of them are running heavy losses while nearly 20 percent of them are either defunct or on the verge of Bankruptcy.

9. Dependence on Ground Water for Irrigation:
Irrigation is another major source of  fof agricultural growth. The actual area under canal and tank irrigation has been declining since the 1990’s. On the other hand, there is a phenomenal increase in the dependencey on the ground water resources through the wells and bore wells.

It is aptly noted that the unstable growth of borewells combined with monsoon failure and increase in surface irrigated area that lias led to drying up of borewells due to inadequate recharge.

10. Rise in Drought prone Areas:
Drought prone Areas in India is rising. Rajasthan, Karnataka, Andhra Pradesh, and Maharastra are considered as the major drought-prone states. Karnataka ranked second in the drought-prone areas.

It has increased from 63% to Tl percent owing to erratic monsoon and lack of drought-proofing methods.In 2011-12, 123 taluks in 23 districts were declared as drought-hit. A total of 157 taluks and 64 taluks were declared drought-hit in 2012-13 and 2013-14 respectively according to NABARD.

Question 32.
Explain the caste-based trade among the Nakarattars of Tamilnadu.
Answer:
Caste-based trade among the Nakarattars of Tamil Nadu banking system resembled an Economist’s model of Western-style banking systems … the Nakarattars loaned and deposited money with one another in caste-defined social relationships based on business, residential location, descent, marriage, and common cult membership.

The Nakarattar banking system was a caste-based banking system. Individual Nakarattars organized their lives around participation in and management of various communal institutions adapted to the task of accumulating and distributing reserves of capital.

The Nattukottai Chettiars (Nakarattars) of Tamil Nadu, provide an interesting illustration of how these indigenous trading networks were organized and worked. A study of this community during the. colonial period shows how its banking and trade activities were deeply embedded in the social organization of the community.

The structures of caste, kinship, and family were oriented towards commercial activity, and business activity was carried out within these social structures. As in most ‘traditional’ merchant communities, Nakarattar banks were basically joint family firms, so that the structure of the business firm was the same as that of the family.

Similarly, trading and banking activities were organised through caste and kinship relationships. For instance, their extensive caste-based social networks allowed Chettiar merchants to expand their activities into Southeast Asia and Ceylon.

Question 33.
Explain the process of modernization in India.
Answer:
Modernization in India is undergoing the following processes:
1. At the economic level, there is a persistent and growing tendency to adopt the rational, mechanized industrial economy in place of older communal — familistic tool economy. This is even responsible for the breakdown of traditional systems like jajmani system.

2. At the political level, the change in the power structure is being introduced through the abolition of semi-feudal group-oriented power structure of the past and by replacing it by a rational parliamentary democratic structure of power.

3. At the cultural level, the change in the realm of values ‘is from sacred value system to secular value system.

4. At the social level, there is a decline in the traditional principle of ascribed status and role to achieve status and role. Yogendra Singh in his work “Modernization of Indian Tradition” is of the opinion that a unique feature of modernization in India is that it is being carried forward through adaptive changes in the traditional structures rather than structural dissociation or breakdown.

V. Answer any two of the following questions in 25-30 sentences each. (2 × 10 = 20)

Question 34.
Define national integration. Explain challenges to National integration.
Answer:
National integration refers to national unity and a sense of belonging to the nation. It is an essential aspect in the making of a nation. Promotion of national integration is regarded as a part and parcel of the policy of any country. According to Benjamin “National integration refers to the assimilation of the entire people of a country to a common identity.

In simple words, National Integration refers to the process wherein a feeling of togetherness, a sense of national unity and above all, a sense of national belongingness is developed among people. It is in this context, the concept of ‘National integration’ has assumed importance.
There are many challenges to National integration.
They are as follows;

  1. Regionalism
  2. Communalism
  3. Linguist and
  4. Extremism and Terrorism

1. Regionalism:
Regionalism is expressed in the desire of people of one region to promote their own regional interest at the expense of the interests of other regions. It has often led to separatism and instigated separatist activities and violent movements. Selfish politicians exploit it. Thus, regionalism has challenged the primacy of the nationalistic interests and undermines national unity. Regionalism is mainly of four forms namely

  • Demand for separation from the Indian Union
  • Demand for a separate statehood
  • Demand for a full-fledged statehood
  • Inter-states disputes-Border disputes.

2. Communalism:
Communalism is the antagonism practiced by the members of one community against the people of other communities and religion. Communalism is the product of a particular society, economy and polity, which creates problems.

Communalism is an ideological tool for the propagation of economic and political interests. It is an instrument in the hands of the upper class to concentrate power by dividing people. The elites strive to maintain a status quo against transformation by dividing people on communal and religious lines.

3. Linguism:
Linguism implies one-sided love and admiration towards one’s language and a prejudice and hatred towards other languages. India is a land of many languages and it has been called as a ‘Museum of languages’.

Diversity of languages has also led to linguism. It has often been manifested into violent movements posing threat to national integration. Linguistic tensions are prevailing in the border areas which are bilingual.

4. Extremism and Terrorism:
Extremism and terrorism have emerged during the recent years as the most formidable challenges to national integration. Extremism refers to the readiness on the part of an individual or group to go to any extreme even to resort to undemocratic, violent and harmful means to fulfil one’s objectives.

In the past India has been facing the problems of terrorism since independence. India has faced this problem in Nagaland (1951), Mizoram (1966), Manipur (1976), Tripura (1980) and West Bengal in (1986).

Terrorism in India is essentially the creation of politics. According to Prof. Rama Ahuja there are four types of terrorism India,

  • Khalistan oriented terrorism in Punjab
  • Militants terrorism in Kashmir.
  • Naxalite terrorism in West Bengal, Bihar, Madhya Pradesh, Orissa, Andhra Pradesh Telangana, Maharastra, Uttar Pradesh, Jharkhand, Chattisgarh.
  • ULFA terrorism in Assam.

The Khalistan oriented Sikh terrorism was based on a dream of theocratic state, Kashmir militants are based on their separate identity. The Naxalite terrorism is based on class enmity. Terrorism in North Eastern India is based on the identity crisis and the grievance situation.

In addition to these factors, corruption, poverty, unemployment/ youth unrest, widening gap between rich and poor, which are also the major challenges for national integration.

KSEEB Solutions

Question 35.
Explain the characteristics of caste.
Answer:
Life of every member of the Indian society is to a large extent influenced by three systems viz., joint family, caste system and village community. They influence one’s occupation, food, dress habits, philosophy and marriage. The study of caste system is important because caste in India is an all pervasive and deep rooted social institution.

Definitions of Caste:
a. Herbert Risley has defined caste as “A collection of families or a group of families bearing a common name, claiming a common descent from a mythical ancestor, human or divine, professing to follow the same hereditary calling and regarding by those who are competent to give an opinion as forming a single homogeneous community”.

b. S. V. Kethkar in his work ‘History of Caste in India’ states that A caste is a group having two characteristics

  • Membership is confined to only those who are born of other members.
  • The members are forbidden by inexorable social law to marry outside the group (Endogamy)”.

1. Caste as a Segmental Division of Society:
The society is divided into various castes with a well developed life of their own. The membership in a caste is determined by birth. Caste has hereditary status, which is determined.by birth. Each caste has a council of its own, known as caste Panchayat.

Caste panchayats imposed certain restrictions on social intercourse between castes like marriages commensal and occupational interactions. By these restrictions each caste had its own way of life. Violation of caste norms attracted punishment from the caste panchayat depending on the seriousness of the violations.

2. Hierarchy:
The whole society is divided into distinct castes with a concept of high and low, or as superior and inferior associated with this gradation or ranking. The Brahmins were placed at the top of the hierarchy and regarded as pure. The degraded castes or untouchables occupied the other end of the hierarchy. They were subjected to manifold disabilities.

3. Restrictions on Feeding and Social Intercourse:
There are minute rules as to what sort of food or drink can be accepted by a person and from what castes, who should accept food or drink at the hands of whom is defined by caste.

4. Civil and Religious Disabilities and Privileges of the Different Sections:
Segregation of individual castes or groups of castes in a village is the most obvious mark of civil privileges and disabilities and it has prevailed in a more or less definite form all over India.

Generally, untouchables were made to live on the outskirts. Certain parts of the town or village are inaccessible to certain castes. Restrictions on using public roads, water facilities, Hotels, etc.

5. Restrictions on occupations:
According to G.S. Ghurye every caste was associated with a traditional occupation. The technical skill of the occupation was made hereditary. Since a distinction was made between occupation being clean and unclean. The hereditary occupations reflected a caste status.

6. Restrictions on Marriages (Endogamy):
Finally, every caste also maintained its rank and status regarding marriages, inter caste marriages were prohibited. Hence they practiced endogamy. Caste is an endogamous group. “Endogamy is the essence of the caste system. Every caste was segmented into sub-casteS, and these sub castes were the units of endogamy.

Question 36.
Define market and explain the characteristics of market.
Answer:
A market is one of the social institutions, whereby parties engage in an exchange of goods and services. Markets rely on sellers offering their goods or services in exchange for money from buyers. It can be said that a market is the proces by which the prices of goods and services are established.

In the field of Sociology, the concept of a market is a structure that allows buyers and sellers to exchange any type of goods, services and information. The exchange of goods or services for money is a Transaction.
Features of Market:
The features of Market are as follow:
1. Market is a place where things are bought and sold:
In common usage, the word ‘market’ may refer to particular markets that we may know of, such as the market next to the railway station, the fruit market, or the wholesale market.

2. Market is not just a physical place, but the gathering of people — buyers and sellers:
Thus, for example, a weekly market may be found in different places on different days of the week in neighboring villages or urban neighborhoods.

3. Market is a type of trade or business:
Market refers to an area or category of trade or business, such as the market for cars or the market for readymade clothes.

4. Market includes the entire spectrum of economic activities and institutions:
In this very broad sense, then, ‘the market’ is almost equivalent to ‘the economy’. We are used to thinking of the market as an economic institution, but the market is also a social institution. In its own way, the market is comparable to more obviously social institutions like caste, religion or family.

KSEEB Solutions

Question 37.
Explain the types of mass media.
Answer:
Mass Media is divided into two major types which are:

  1. Print Media: Newspaper and magazines
  2. Electronic Media: Radio, Television, Internet and Social Networking Sites.

1. Print Media:
The Beginning of Print Media
There are many Kannada language newspapers that has served the media industry significantly and also have earned significant recognition. Some of the leading Kannada language newspapers include. Prajavani, Kannada Prabha, Samyukta Karnataka, Vijaya Karnataka, Hosa Digantha, Sanje Vani, Udaya Vani, Andolan, E-sanje etc.

2. Electronic Media:
a. Radio:
Radio broadcasting which commenced in India through amateur ‘HAM’ Broadcasting Clubs in Kolkata and Chennai in the 1920s, matured into a public broadcasting system in the 1940s during World War II when it became a major instrument of propaganda for Allied forces in South East Asia.

At the time of independence, there was only 6 radio station located in the major cities catering primarily to an urban audience. A Radio Transmission center called Akashavani was started by Dr. M.V. Gopalaswamy, at Mysore University in 1935 through private effort.

The station was later taken over by the state Government in January 1941 and it was shifted to Bangalore in November 1955. The first AIR station in the North Karnataka Region started functioning at Dharwad, on 8th November 1950. In 1964, Vividh Bharathi (CBS) was added to the Dharwad unit. Auxiliary stations at Bhadravathi and Gulbarga were started in 1965 and 1966 respectively.

Apart from AH India Radio (AIR), there is Vividh Bharati, a channel for entertainment that was primarily broadcasting film songs on listeners’ requests. Vividh Bharati, which soon began to carry sponsored programs and advertisements, grew to become a money-spinning channel for AIR.

Akasha Vani (Kannada version of AIR) headquarters is at Bangalore and there are regional centers at Mysore, Bhadravathi, Dharwad, and Gulbarga covering broadcasting news, entertainment, sponsored programs, and commercial programs, etc.

FM Radio (Frequency Modulator Radio):
The advent of privately owned FM radio stations in 2002, provided a boost to entertainment programs over the radio. In order to attract audiences, these radio stations provide entertainment. They specialize in ‘particular’ kinds of popular music to retain their audiences.

Most of the FM channels which are popular among urbanites and students often belong to media conglomerates. ‘Radio Mirchi’ belong to Times of India group, Red FM is owned by Living Media and Radio City by the Star Network.

b. Television (T.V.):
television programming was introduced experimentally in India to promote rural development in early 1959. ‘Krishi Darshan’ was the first program telecast on Doordarshan. Later, the Satellite Instructional Television Experiment (SITE) was broadcasting directly to community viewers in the rural areas of six states between August 1975 and July 1976.

These instructional broadcasts to 2,400 TV sets directly were for 4 hours daily. Meanwhile, Television stations were set up 1under Doordarshan in 4 cities (Delhi, Mumbai, Srinagar and Amritsar) by 1975. The advent of colour broadcasting during the year 1982 Asian Games in Delhi and the rapid expansion of the national network led to the rapid commercialization of T.V. broadcasting.

Gulbarga was the first center in Karnataka to have a relay center, it was inaugurated on 3-9-1977 and at the outset within a radius of 40 km, 240 villages and towns of Raichur and Vijayapura Districts and Gulbarga were benefitted. Community TV sets were maintained and serviced by the Doordarshan Kendra, Gulbarga, Bangalore city was provided with an interim TV Relay centre on 1-1-1981.

Udaya news was the first private channel to broadcast news in Kannada language. At present, there are 8 Kannada news channels viz., TV-9, Suvarna News, Kasturi 24 × 7, Samaya News, Udaya News, Janashree News and Raj News. ETV News and a few more news channels will also come up short.

c. Internet:
Internet is a global system of interconnected computer networks consisting of millions of private, public, academic, business
networks which are linked with the networking technology. In simple words Internet is a network of networks.

d. Social Networking Sites (SNS):
Social Networking Sites are defined as online platforms that focus on building and reflecting social networks or social relations among people who share interests and activities. Further, social networking sites are a type of virtual community that has grown tremendously in popularly.

VI. Answer any two of the following questions in 15 sentences each. (2 x 5 = 10)

Question 38.
Write a note on the Narasingnavar joint family.
Answer:
The Narasinganavar family is a patriarchal Joint j family of about 206 individuals who are residing together in the village of Lokur in the Dharwad district of Karnataka. All the individuals in the family share a common ancestry and this family is ‘recognized as one of the largest undivided families in the world.

The family spans across five generations. Bhimanna Jinapa Narasinganavar is the part of the family. For India’s largest joint family, balancing the family is a forbidding task. The Narsingnavar family finds that expenditure on its 206 members always seems to be more than its income.

Patriarch Narsingnavar (72), who has been handling money matters of this jumbo family for the past 30 years, says “We believe family finances could be the biggest source of discontent. In their wisdom and sincerity, the elders gave me this job. Whatever I do well be in the interest of the family.

Agriculture is the main occupation for this family. It owns 270 acres of cultivable land, the annual 1 income is Rs 8 lakh to Rs 12 lakh depending on the monsoon and market. Its annual expenditure of around Rs 10 lakh is largely on farm labour and agriculture machinery.

While the family’s requirement of food grains, vegetables and milk are met by its own efforts, it spends a substantial amount on provisions, clothes, medicines, soap and tea. If there’s resource crunch, the earning members contribute to the common kitty and Bheemanna keeps a meticulous record of the transactions.

Weddings are performed every eight or ten years with several marriages being solemnised at the same time. The family’s only source of entertainment is TV.

Question 39.
Write a short note on major problems of urban communities.
Answer:
Problems of Indian cities can be classified in the following ways:
1. Urban Poverty:
Urban poverty is the by-product of industrialization and urbanization. Poverty and overcrowding are the two most visible features of Indian cities. About half of the urbanites are poor and live in substandard life, because of cost of living, lack of regular income, low wages, pro-rich economic policies, inflation, etc.

India has issued its first-ever report on the nature and dynamics of urban poverty in the country undertaken with the support of the United Nations Development Programme (UNDP), India Urban Poverty Report 2009 which identifies the problems faced by the poor and focuses on the systematic changes that are needed to address them.

The report examines various issues related to urban poverty, such as migration, labor, the role of gender, access to basic services and the appalling condition of India’s slums. It also looks at the dynamics of urban land and capital market, urban governance, and the marginalization of the poor to the urban periphery.

2. Slums:
The magnitude of the problem of slums is alarming. The Government of India, in order to implement the various schemes for urban development, has defined a slum area as follows:

A slum area means any area where such dwellings predominate of dilapidation, overcrowding, faulty arrangement of buildings, narrowness and faulty arrangement of streets, lack of ventilation, lack of sanitation facilities, inadequacy of open spaces and community facilities or any combination of these factors, are detrimental to safety, health or morale,” These slum areas are also referred to as the ‘Blighted area’; ‘Renewal area’; ‘deteriorated area’, ‘Gray area’; ‘Lower class neighborhood’; ‘Lower income area’, etc.

3. Problem of Urban Housing:
The bulk of the people in the Indian cities live in one-room or in thatched huts in the sprawling slums or on the pavements. Another sad feature is total lack of essential municipal services like water supply, drainage, sewarage, lighting, -roads, etc.

Further, large proportion of the rural migrants have been bringing with them unskilled persons who take up unskilled jobs in the services, trade, industries, etc.
Generally a single room has to meet all the requirements of the family including cooking, living, sleeping which make confinement.

It is difficult to keep reasonably clean and sanitary washing and bathing facilities. The inconvenience they have to undergo is aggravated during the rainy days. Almost all the above mentioned conditions are found in shawls of Mumbai, that as of Kanpur, bastis of Kolkata, cheris of Chennai as well as in Dhowrahas of the mining centres and barracks of the plantations in India.

These are made of brick walls and iron roof or huts consisting of bamboo walls and thatched roofs. The lanes are too narrow and the huts are built back to back. These lack facilities like bathing, washing and toilets, etc.

4. Sanitation and Pollution:
It is accompanied with corrupt Municipal administration and inefficiency. According to UNICEF, lakhs of urban children in India die or suffer from diarrhea, diphtheria, tetanus and measles etc.

5. Transportation and Traffic:
Transportation and traffic picture in Indian cities is troublesome. Majority of people use buses and other vehicles, while a few use rails as transport system. The increasing number of two wheelers and other types of vehicles make the traffic problem worse.

KSEEB Solutions

Question 40.
Explain the recent trends In Mass Media.
Answer:
In the 21st century, communication technology is such that information can be shared instantaneously by millions of people simultaneously, almost anywhere around the world. Communication the transfer of information from one individual or group to another, whether by speech or through the mass media of modern times is crucial to any society.

According to Marshall McLuhan, society is influenced much more by the type of the media than by the content, or the messages, which the media convey. The electronic media, according to Marshal McLuhan. are creating a Global village people throughout the world see major events unfold and hence participate in them together.

It is the Internet, at the heart of this communications revolution. With the expansion of technologies such as voice recognition, broadband transmission, web casting and cable links, the Internet became the conduit for the delivery of information, entertainment, advertising and commerce to media audiences.

Neil Postman in his book, Amusing Our-selves to Death:
Public Discourse in the Age of Show’ Business, says television presents serious issues as entertainment because the form excludes the content. As Postman states, the medium of print creates a rational population, whereas the medium of television creates an entertained one.

Robert Putnam is referring Media as a social capital to useful social networks, a sense of mutual obligation and trust-worthiness, an understanding of the norms that govern effective behaviour and, in general other social resources that enable people to act effectively.

Putnam in his book Bowling Alone finds significant decline in social capital over the last few decades. TV viewing is strongly and negatively related to social trust and group membership.

Horkheimer and Adorno made an extensive study of what they called the ‘culture industry’ meaning the entertainment industries of film, TV, popular music, radio, newspapers and magazines. They argued that the production of culture had become just as standardized and dominated by the desire for profit as other industries.

Art disappears, swamped by commercialization and culture is replaced by entertainment.
Jurgen Habermas has analysed the media as decay – of the ‘public sphere’. The public sphere is an arena of public debate in which issues of general concern can be discussed and opinions formed.

According to Habermas, the -public sphere developed first in the salons and coffee houses of London, Paris and other European cities. Habermas argues that the salons were vital to the early development of democracy, for they introduced the idea of resolving political problems through public discussion.

The public sphere – at least in principle – involves individuals coming together as equals in a forum for public debate. The spread of mass media and mass entertainment causes the public sphere to become largely a sham. ‘Public opinion’ is not formed through open rational discussion, but through manipulation and control – as, for example, in advertising.

Jean Baudrillard regards impact of modem mass media as Hyper Reality The coming of the mass media, particularly electronic media such as Television, has transformed the very nature of our lives. TV does not just ‘represent’ the world to us it increasingly defines, what the world in which we live actually is Consider as an example the trial of O. J. Simpson, a celebrated court case that unfolded in Los Angeles in 1994-95.

Question 41.
Discuss the factors facilitating globalization.
Answer:
Globalization refers to the growing interdependence of societies across the world, with the spread of the same culture and economic interests across the globe. For example, media and consumer products are often produced for a world market, by the same firms running business all over the world.

Factors Contributing to Globalization:
Anthony Giddens has explained the following factors as contributing to Globalization:
1. The Rise of Information and Communications Technology:
The explosion in global communications has been facilitated by a number of important advances in technology and the world’s telecommunication infrastructure. The spread of communication satellites has also been significant in expanding international communications.

Today a network of more than 200 satellites is in space to facilitate the transfer of information around the globe. The use of Satellites, Internet, Telephones, Computer Networking, known as Information and Communication Technologies – ICT – have revolutionised the way the world communicates.

You could be chatting online, through the internet, with your friend or family, who is thousands of miles away, and feel that you share your everyday travails much more than a person who is closer home like your neighbour. You could be working in India for a company that is located in the United States of America through telecommunication technologies.

2. Information Flows:
It has also facilitated the flow of information about people and events in distant places. Every day, the global media brings news, images and information into homes, linking them directly and continuously to the outside world.

Some of the most gripping events of the past three decades such as the fall of the Berlin Wall, the violent crackdown on democratic protesters in China’s Tiananmen Square and the Terrorist attacks on Mumbai on 11 September 2001, Spring movement in Arabian countries, have unfolded through the media before global audience.

Such events, along with thousands of information, have resulted in a reorientation in people’s thinking from the level of the nation-state to the global stage. In the case of natural disasters, such interventions take the form of humanitarian relief and technical assistance. In recent years, earthquakes in Armenia and Turkey, floods in Mozambique and Bangladesh, famine in Africa and hurricanes in Central America have been rallying points for global assistance.

3. Knowledge Society:
The emergence of the knowledge society has been linked to the development of a broad base of consumers who are technologically literate and eagerly integrate new advances in computing, entertainment and Telecommunications into their everyday lives.

The very operation of the global economy reflects the changes that have occurred in the information age. Many aspects of the economy now work through networks that cross national boundaries, rather than stopping at them.

4. Transnational Corporations: In globalization, the role of trans-national corporations is particularly important. Transnational corporations are companies that produce goods or market services in more than one country.

For example Coca-Cola., Pepsi, Johnson and Johnson, Ford, General Motors, Colgate-Palmolive, Indian corporations like Reliance, TATAs, Birla Groups, Infosys, Mahindras, TVS group, Wipro etc. Even when trans-national corporations have a clear national base, they are oriented towards global markets and global profits. Transnational corporations are at the heart of economic globalization.

5. The Electronic Economy:
Globalization is also being driven forward by the integration of the world economy. In contrast to previous eras, the global economy is no longer primarily agricultural or industrial in its basis. Rather, it is increasingly dominated by activity that is weightless and intangible.

This U weightless economy is one in which products have their base in information, as is the case with computer software, media and entertainment products and Internet-based services.The ‘Electronic Economy’ is another factor ‘that underpins economic globalization.

Banks, corporations, fund managers and individual investors are able to shift funds internationally with the click of a mouse. As the global economy becomes increasingly integrated, a financial collapse in one part of the world can have an enormous effect on distant economies.

6. Political changes:
Another driving force behind contemporary globalization is related to political change.
These are:
a. The collapse of Soviet-style communism in 1991. The collapse of communism has hastened processes of globalization but should also be seen as a result of globalization itself.

b. The important political factor leading to intensifying globalization is the Growth of International and Regional Mechanisms of Government namely The United Nations and the European Union.

SAARC (South Asian Association for Regional Co-operation) and BRICS (Brazil; Russia, India, China, and South Africa) are; the two most prominent examples of international organizations that bring together nation-states into a common political forum.

Finally, globalization is being driven by International Governmental Organizations (IGOs) and International Non-governmental Organizations (INGOs). An IGO is a body that is established by Participating governments and given responsibility for regulating or overseeing a particular domain of activity that is transnational in scope.

The first such body, The International Telegraph Union, was founded in 1865. Since that time, a great number of similar bodies have been created. In 1909, there were only 37 IGOs in existence to regulate transnational affairs; by 1996, there were 260.

Some of the best-known INGOs such as Greenpeace, Medicines Sans Frontiers (Doctors Without Borders), the Red Cross and Amnesty International-are involved in environmental protection and humanitarian efforts. But the activities of thousands of lesser-known groups also link together countries and communities.

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