2nd PUC

Students can Download 2nd PUC Maths Previous Year Question Paper March 2019, Karnataka 2nd PUC Maths 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 Maths Previous Year Question Paper March 2019

Time: 3 Hrs 15 Min
Max. Marks: 100

Part-A

Answer ALL the following questions: (10 × 1 = 10)

Question 1.
Define Binary Operation.
Answer:
A binary operation * on a set ‘A’ is defined as a function.
* : A × A → A
a * b = ab + 1 ∈ Z ∀ a, b, ∈ Z
∴ * is binary operation on Z.

Question 2.
Find the principal value of cos^-1\(\left(-\frac{1}{2}\right)\).
Answer:
\(\frac{2 \pi}{3}\)

Question 3.
Define a Scalar Matrix.
Answer:
In a square matrix the principal diagonal elements are same and all the other elements are zeroes.

Question 4.
Find a value of x for which

Answer:

3 – x² = 3 – 8 ⇒ x² = 8 ⇒ x = ±2\(\sqrt{2}\)

Question 5.
If y = sin(x² + 5) then find \(\frac{d y}{d x}\)
Answer:
y = sin(x² + 5)
\(\frac{d y}{d x}\) = cos(x² + 5)(2x)

Question 6.
Find \(\int(1-x) \sqrt{x} d x\)
Answer:
\(\int(1-x) \sqrt{x} d x\)

Question 7.
Find a value of x for which x(î + ĵ + k̂) is a unit vector.
Answer:
x(î + ĵ + k̂) = 1 ⇒ xî+ xĵ+ xk̂ = 1
⇒\(\sqrt{x^{2}+x^{2}+x^{2}}\) = 1 ⇒ \(\sqrt{3 x^{2}}\) = 1 ⇒ x = \(\frac{1}{\sqrt{3}}\)

Question 8.
If a line has direction ratios 2, -1, -2 then determine its direction cosines.
Answer:
\(\left(\frac{2}{3}, \frac{-1}{3}, \frac{-2}{3}\right)\)

Question 9.
Define objective function in Linear Programming Problem.
Answer:
The linear function Z = ax + by, where ‘a’ and ‘b’ are constants which has to be maximised or minimised is called objective function.

Question 10.
If P(E) = 0.6, P(F) = 0.3 and P(E n f) = 0.2 then find P(F/E).
Answer:
\(P(F / E)=\frac{P(F \cap E)}{P(E)}=\frac{0.2}{0.6}=\frac{2}{6}=\frac{1}{3}\)

Part-B

Answer any TEN questions: (10 × 2 = 20)

Question 11.
Show that the function f: N → N given by f(x) = 2x is one-one but not onto.
Answer:
The function/is one-one, for f(X₁) = f(x₂)
⇒ 2x₁ = 2x₂ ⇒ x₁ = x₂ Further,/is not onto, as for 1 ∈ N, there does not exist any x in N such that f(x) = 2x = 1.

Question 12.
Prove that sin^-1 x + cos^-1 x = \(\frac{\pi}{2}\), x ∈ [-1,1].
Answer:
Let sin^-1 x = θ then.
x = sin θ ⇒ x = cos(\(\frac{\pi}{2}\) – θ)
⇒ Cos^-1x = \(\frac{\pi}{2}\) – θ ⇒ θ + cos^-1 x = \(\frac{\pi}{2}\)
∴ sin^-1 x + cos^-1 x = \(\frac{\pi}{2}\)

Question 13.
Write cot^-1\(\left(\frac{1}{\sqrt{x^{2}-1}}\right)\), x > 1 simplest form.
Answer:
Let x = sec q ⇒ q = sec^-1 x.

Question 14.
Find the area of the triangle with vertices (2, 7), (1, 1) and (10, 8) using determinant method.
Answer:
The area of triangle is given by

= \(\frac { 1 }{ 2 }\) |[-2 (1 – 8) -7 (1 – 10) + 1 (8 – 10)]|
= \(\frac { 1 }{ 2 }\) |-14 + 63 – 2| = \(\frac { 47 }{ 2 }\) sq.units

Question 15.
Find \(\frac{d y}{d x}\) if y = (logx)^cosx.
Answer:
Let y = (log x)^cosx
⇒ log y = cos x log (logx)
⇒ \(\frac{d}{d x}\) log y = \(\frac{d}{d x}\) cos x log(log x)

Question 16.
Find ax + by² = cosy then find \(\frac{dy}{d x}\)
Answer:
ax + by² = cosy
⇒ \(\frac{d}{d x}\)ax + \(\frac{d}{d x}\)by² = \(\frac{d}{d x}\)cosy
⇒ a + 2by \(\frac{dy}{d x}\) = (-sin y)\(\frac{dy}{d x}\) dx dx
(2by + sin y) \(\frac{dy}{d x}\) = -a ⇒ \(\frac{d}{d x}\) = \(-\frac{a}{2 b y+\sin y}\)

Question 17.
Find the approximate change in the volume V of a cube of side x meters caused by increasing side by 2%.
Answer:
Note that V = x³
or dV = \(\left(\frac{d V}{d x}\right)\) ∆x = (3x²) ∆x
= (3x²) (0.02x) = 0.06x³m³
Thus, the approximate change in volume is 0.06 x³m³

Question 18.
Find \(\int \frac{1}{\cos ^{2} x(1-\tan x)^{2}} d x.\)
Answer:
Put (1 – tan x) = t ⇒ – sec² x dx = dt

Question 19.
Find \(\int \sin 2 x \cos 3 x d x\)
Answer:
W.K.T sin x cosy = \(\frac { 1 }{ 2 }\)[sin(x + y) +sin(x – y)]
Then \(\int \sin 2 x \cos 3 x d x=\frac{1}{2}\left[\int \sin 5 x d x-\int \sin x d x\right]\)
= \(\frac { 1 }{ 2 }\) [\(\frac { 1 }{ 5 }\)cos 5x + cos x] + C = \(\frac { -1 }{ 10 }\) cos5x + \(\frac { 1 }{ 2 }\) cos x + C.

Question 20.
Find the order and degree (if defined) of the differential equation
\(\left(\frac{d^{3} y}{d x^{2}}\right)^{2}+\cos \left(\frac{d y}{d x}\right)\)=0
Answer:
Order = 2, Degree = not defined.

Question 21.
If \((\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})\) = 8 and \(|\vec{a}|\) = 8\((\vec{b})\) then find \(|\vec{b}|\).
Answer:

Question 22.
Find the projection of the vector ā = 2î + 3ĵ + 2k̂ and b̄ = î – ĵ + k̂
Answer:
Projection of vector ā on b̄ is given by

Question 23.
Find the distance of the point (3, -2, 1) from the plane 2x – y + 2Z + 3 = 0.
Answer:

Question 24.
Probability of solving specific problem independently by A and B are \(\frac { 1 }{ 2 }\) and \(\frac { 1 }{ 3 }\) respectively. If both try to solve the problem independently, find the probability that the problem is solved.
Answer:
Given P(A) = \(\frac { 1 }{ 2 }\) and P(B) = \(\frac { 1 }{ 23}\)
(i) The problem is solved
= 1 – P(not solved) = 1 – P(A’ ∩ B’)
= 1 – P(A’) P(B’) = 1 – [(1 – \(\frac { 1 }{ 2 }\) – (1 – \(\frac { 1 }{ 3 }\)]
= 1 – (\(\frac { 1 }{ 2 }\) × \(\frac { 2 }{ 3 }\)) = 1 – \(\frac { 1 }{ 3 }\) = \(\frac { 2 }{ 3 }\)

(ii) Exactly one of them solves problem
= P(A ∩ B) + P(A ∩ B’)
= P(A’) + P(B) + P(A) P(B’)
\(=\left(1-\frac{1}{2}\right)\left(\frac{1}{3}\right)+\frac{1}{2}\left(1-\frac{1}{3}\right)\)
\(=\left(\frac{1}{2}\right)\left(\frac{1}{3}\right)+\frac{1}{2}\left(\frac{2}{3}\right)=\frac{1}{2}\)

Part-C

Answer any TEN questions: (10 × 3 = 30)

Question 25.
Check whether the relation R in 1R of real numbers defined by R = {(a, b): a < b³} is reflexive, symmetric or transitive.
Answer:
(i) For (a, a) ∈R, a < a³ which is false.
As for a = -2, a³ = -8
∴ a \(\nleq\) a³ as – 2 \(\nleq\) – 8
∴ R is not reflexive.

(ii) For (a, b), a < b³ and (b, a), b < a³ which is false
As for a = 1 and b = 2
a < b³ as 1 < 8
Now, b \(\nleq\) a³ as 2 \(\nleq\) 1
∴ R is not symmetric.

(iii) For (a, b), (b, c) and (a, c)
a < b³ and b < c³
Now a = 25, b = 3, c = 2
a < b³ and 25 < 27
b < c³ and 3 < 8
but a \(\nleq\) c³ as 25 \(\nleq\) 8
.’. R is not transitive.
Hence R is neither reflexive, nor symmetric and nor transitive.

Question 26.
Prove that
cos^-1 (\(\frac { 4 }{ 5 }\)) + cos^-1 (\(\frac { 12 }{ 13 }\)) = cos^-1 (\(\frac { 33 }{ 65 }\))
Answer:
Let
cos^-1 (\(\frac { 4 }{ 5 }\)) ⇒ cos A = \(\frac { 4 }{ 5 }\) ⇒ Sin A = \(\sqrt{1-\frac{16}{25}}=\frac{3}{5}\)
cos^-1 (\(\frac { 12 }{ 13 }\)) = cos B = \(\frac { 12 }{ 13 }\) ⇒ sin B = \(\sqrt{1-\frac{144}{169}}=\frac{5}{13}\)
cos(A + B) = cosA.cosB – sinA.sinB \(=\frac{4}{5} \cdot \frac{12}{13}-\frac{3}{5} \cdot \frac{5}{13}=\frac{48}{65}-\frac{15}{65}=\frac{33}{65}\)
cos(A + B) = \(\frac { 33 }{ 65 }\) ⇒ A + B = cos^-1 (\(\frac { 33 }{ 65 }\))
cos^-1 (\(\frac { 4 }{ 5 }\)) + cos^-1 (\(\frac { 12 }{ 13 }\)) = cos^-1 (\(\frac { 33 }{ 65 }\))

Question 27.
By using elementary transformations, find the inverse of the matrix

Answer:

Question 28.
If x = a(θ + sin θ), y = a(1 – cosθ) then show that \(\frac{d y}{d x}=\tan \left(\frac{\theta}{2}\right)\)
Answer:
We have
\(\frac{d x}{d \theta}\) = a(1 + cos θ), \(\frac{d y}{d \theta}\) = a(sin θ)

Question 29.
Verify Rolle’s theorem for the function f(x) = x² + 2x – 8, x ∈ [-4, 2].
Answer:
f(x) is a polynomial in x. Hence it is continuous over [-4, 2] and differentiable over (-4,2).
f(-4) = (-4)² + 2 (-4) – 8 = 16 – 8 – 8 = 0
f(2) = 4 + 4 – 8 = 0
∴ f(-4) = f(2)
∴ All the conditions of the Rolle’s theorem are satisfied.
∴ there exists a c ∈[-4,2] such that
f'(c) = 0 f'(x) = 2x + 2
f(c) = 2c + 2
f(c) = 0
⇒ 2c + 2 = 0
⇒ 2c = -2
⇒ c = -1 ∈[-4, 2]
∴ Rolle’s theorem is verified.

Question 30.
Find the intervals in which the function f given by f(x) = 2X³ – 3x² – 36x + 7 is strictly increasing.
Answer:
f(x) = 2x³ – 3x² – 36x + 7
f ‘(x) = 6x² – 6x – 36
f ‘(x) = 0 ⇒ 6x² – 6x – 36 = 0

⇒ 6(x² – x – 6) = 0
⇒ 6(x – 3)(x + 2) = 0
6 ≠ 0, x = 3, x = -2

Question 31.
Find \(\int x \log x d x\)
Answer:

Question 32.
Evaluate \(\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{1+\cos ^{2} x} d x\)
Answer:

Question 33.
Find the area of the region bounded by the curve y² = 4x and the line x = 3.
Answer:
Given curve is y² = 4x and line is x = 3
y² = 4x

Question 34.
Form the differential equation of the family of curves y = ae^3x + be^-2x by eliminating arbitrary constants a and b.
Answer:
Given, y = a e^3x + b e^-2x …. (1)
y’ = (a e^3x) 3 – (be^-2x) 2
y’ = 3 a e^3x – 2 be^-2x …. (2)
From (1) and (2)
y = a e^3x + b e^-2x × 3
y’ = 3a e^3x – 2b e^-2x × 1

Diff
⇒ 3y’ – y” = 5be^-2x
3y’ – y” = -10be^-2x ….(4)
From (3) & (4)
3y’ – y” = (-2) (5be^-2x)
3y’ – y” = -2[3y – y’]
3y’ – y” = -6y + 2y’
y” – y’ – 6y = 0.

Question 35.
Find a unit vector perpendicular to each of the vector \((\vec{a}+\vec{b})\) and \((\vec{a}-\vec{b})\), where ā = î + ĵ + k̂, b̄ = î + 2ĵ + 3k̂.
Answer:

Question 36.
Show that the four points with position vectors: 4î + 8ĵ + 12k̂, 2î + 4ĵ + 6k̂, 3î + 5ĵ + 4k̂ and 5î + 8ĵ + 5k̂ are coplanar.
Answer:
Let

= -2(21 + 0) + 4(7 + 8) – 6(0 + 3)
= -42 + 60 – 18 = -60 + 60 = 0
∴ A, B, C, D are coplanar.

Question 37.
Find the vector equation of the plane passing through the points R(2, 5,-3), S(-2, -3, 5) and T(5, 3, -3).
Answer:

Theorem: Derive intercept form of the equation of a plane
Let the equation of a plane be
Ax + By + Cz + 0 = 0 (1)

Let the plane cuts X – axis at p(a, 0, 0), Y-axis at Q(0, b, 0), Z-axis at R(0, 0, c) substituting P(a, 0, 0) in equation (1)
Aa + D = 0 ⇒ Aa = -D ⇒ A = \(\frac{-D}{a}\)
substituting Q(0, b, 0) in equation (1)
Bb + D = 0 ⇒ Bb = D ⇒ B = \(\frac{-D}{b}\)
R(0, 0, c) in equation (1)
Cc + D = 0 ⇒ Cc = -D ⇒ C = \(\frac{-D}{c}\)
substituting A, B, C in (1)
\(\left(\frac{-D}{a}\right) x+\left(\frac{-D}{b}\right) y+\left(\frac{-D}{c}\right) z=-D\)
Dividing through out by -D
\(\Rightarrow \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\)
is the required equation of the plane in intercept form

Question 38.
An insurance company insured 2000 scooter drivers, 4000 car drivers and, 6000 truck drivers. The probability of an accident is 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?
Answer:
Let S : be Insured person is scooter driver
C : be Insured person is car driver
T : be Insured person is truck driver
Let ‘A’ be the event “meets with an accident”.

Part-D

Answer any SIX questions: (6 × 5 = 30)

Question 39.
Let f: N → Y be a function defined as
f(x) = 4x + 3, where
y = {y ∈ N: y = 4x + 3 for some x ∈ N}. Show that f is invertible. Find the inverse of f.
Answer:
f:N → Y
f(x) = 4x + 3
To prove that / is one – one
Let f(x₁) = f(x₂)
4x₁ + 3 = 4x₂ + 3
4x₁ = 4x₂
∴ x₁ = x₂.
∴ f is one – one
To prove that f is onto
Let y ∈ Y there exits x ∈ N such that
f(x) = y
4x + 3 = y
4x = y – 3, x = \(\frac{y-3}{4}\)
∴ f is one-one and onto
⇒ f is bijective. Hence f is invertible
∴ f^-1(y) = \(\frac{y-3}{4}\)

Question 40.
If A = \(\left[ \begin{matrix} 1 & 2 & 3 \\ 3 & -2 & 1 \\ 4 & 2 & 1 \end{matrix} \right] \) then show that A² – 23A – 40I = 0
Answer:
We have

Question 41.
Solve the following system of linear equations by matrix method:
3x – 2y + 3z = 8
2x + y – z = 1
4x – 3y + 2z = 4
Answer:
The system of equations can be written in the form A X = B where

|A| = 3(2 – 3) + 2(4 + 4) + 3(-6 -4) = -17
Hence, A is non singular matrix and so its inverse exists. Now
A₁₁ = -1 A₁₂ = -8 A₁₃ = -10
A₂₁ = -5 A₂₂ = -6 A₂₃ = +1
A₃₁ = -1 A₃₂ = +9 A₃₃ = +7

Question 42.
If y = (sin^-1 x). show that
\(\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x\left(\frac{d y}{d x}\right)=0\)
Answer:
Given that y = sin^-1 x, we have
y₁ = \(\frac{1}{\sqrt{1-x^{2}}}\) i.e., (1 – x²)y₁² = 1
so (1 – x²). 2y₁y₂ + y₁²(0 – 2x) = 0
Hence (1 – x₂) y₂ – xy₁ = 0

Question 43.
The length x of a rectangle is decreasing at the rate of 3 cm / min and the width y is increasing at the rate of 2 cm / min. When x = 10 cm and y = 6 cm, find the rates of change of
(i) the perimeter and
(ii) the area of the rectangle.
Answer:
Since the length x is decreasing and the width y is increasing with respect to time, we have
\(\frac{d x}{d t}\) = – 3 cm /min and \(\frac{d y}{d t}\) = 2 cm/mm
(a) The perimeter P of a rectangel is given by P = 2 (x + y)
Therefore \(\frac{d P}{d t}\) = \(2\left(\frac{d x}{d t}+\frac{d y}{d t}\right)\) = 2 (-3 + 2)
= – 2 cm / min

(b) The area A of the rectangle is given A = x.y
Therefore \(\frac{d A}{d t}\) = \(\frac{d x}{d t}\) . y + x. \(\frac{d y}{d t}\)
= -3(6)+ 10(2) (as x= 10 cm and y = 6 cm)
= 2 cm²/min.

Question 44.
Find the integral of \(\frac{1}{x^{2}-a^{2}}\) with respect to x and hence evalauate \(\int \frac{1}{x^{2}-16} d x\)
Answer:

Question 45.
Using the method of integration, find the smaller area enclosed by the circle x² + y² = 4 and the line x + y = 2.
Answer:
Given circle x² + y² = 4 and the line is x + y = 2
x² + (2 – x)² = 4
[∴ y = 2 – x]

2x² – 4x = 0 ⇒ 2x(x – 2) = 0 ⇒ x = 0, 2
Now, x² + y² = 4 and x + y = 2
⇒ y² = 4 – x² y = 2 – x,
⇒ y = \(y=\sqrt{2^{2}-x^{2}}\)
Required Area

Question 46.
Find the general solution of the differential equation \(\frac{d y}{d x}\)+ (secx)y = tanx, ( 0 ≤ x ≤ \(\frac{\pi}{2}\)).
Answer:
Given circle x² + y² = 4 and the line is x + y = 2
x² + (2 – x)² = 4 [∴ y = 2 – x]

2x² – 4x = 0 ⇒ 2x(x – 2) = 0 ⇒ x = 0, 2
Now, x² + y² = 4 and x + y = 2
⇒ y² = 4 – x²  y = 2 – x,
⇒ y = \(\sqrt{2^{2}-x^{2}}\)
Required Area

Question 47.
Derive the equation of a line in space passing through a given point and parallel to a given vector in both vector and Cartesian form.
Answer:

Let \(\overrightarrow{\mathrm{r}}\) be any position vector on the line
‘L’ and \(\overrightarrow{\mathrm{a}}\) given point on L.
clearly, \(\overrightarrow{\mathrm{AP}}\) || \(\overrightarrow{\mathrm{b}}\)
\(\text { ie, } \overrightarrow{\mathrm{AP}}=\lambda \overrightarrow{\mathrm{b}}\)
where λ – some Real number

Question 48.
Five cards are drawn successively with replacement from a we!! shuffled deck of 52 cards. What is the probability that
(i) all the five cards are spades?
(ii) only five three cards are spaces?
(iii) none of spades?
Answer:

Part-E

Answer any ONE question: (1 × 10 = 10)

Question 49.
(a) Prove that \(\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\) and hence evaluate \(\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{5} x}{\sin ^{5} x+\cos ^{5} x} d x\)
Answer:
Proof: a – x = t
– dx = dt
dx = -dt
L.L: when x = 0, t = a
U.L: When x = a, t = 0

(b) Evaluate

Answer:
Taking a, b, c commn from R₁, R₂, R₃

Question 50.
Minimise and Maximise
z = 5x + 10y
Subject to the constraints :
x + 2y ≤ 120
x + y ≥ 60
x – 2y ≥ 0
x ≥ 0 and y ≥ 0.
Answer:
Draw the graph of the line

Put (0, 0) in the inequality x + 2y ≤ 120
0 + 2(0) ≤ 120
0 ≤ 120 (True)
∴ The half plane lies towards the origin
P (0, 0) in the inequality x + y ≥ 0
0 + 0 ≥ 0 (False)
∴ The half plane lies away from the origin
Put (0, 1) in the inequality x – 2y ≥ 0
0 – 2(1) ≥ 0
-2 ≥ 0 (False)
∴ The half plane does not contain (0, 1)
Check the point (1, 0) in the inequality
x – 2y ≥ 0
1 – 2 (0) ≥ 0
⇒ 1 ≥ 0 (True)
∴ The half plane contains (1,0)
So the half plane is towards x – axis
Corner points Max & Min z = 5x+ 10y
A (40,20) z = 400
B (60, 30) z = 600 → Maximum
C (60, 0) z = 300 → Minimum
D (120, 0) z = 600 → Maximum
∴ At C (60, 0) z is minimum
At B (60,30), D (120,0) z is maximum

(b) Find the value of K, if
f(x) = \(\left\{\begin{array}{ll}
{K x+1,} & {\text { if } x \leq 5} \\
{3 x-5} & {\text { if } x>3}
\end{array}\right.\)
is continuous at x = 5. (4)
Answer:
Since given f(x) is continous at x = 5 => lim f(x) = lim f(x)=/(5)

⇒ K(5) + 1 = 3(5) – 5
⇒ 5K+ 1 = 10 ⇒ 5K = 9
⇒ K = \(\frac { 9 }{ 5 }\)

Students can Download 2nd PUC Biology Model Question Paper 4 with Answers, Karnataka 2nd PUC Biology 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 Biology Model Question Paper 4 with Answers

Time: 3.15 Hours
Max Marks: 70

General Instructions:

1. This question paper consists of four parts A, B, C and D. Part D consists of two parts, Section – I and Section – II.
2. All the parts are Compulsory.
3. Draw diagrams wherever necessary. Unlabelled diagrams or illustrations do not attract any marks.

Part – A

I. Answer the following questions in One Word or One Sentence each : (10 × 1 = 10 )

Question 1.
What is implantation of embryo?
Answer:
The process of embedding or attachment of blastocyst (embryo) to the wall of uterus (endometrium) is called implantation.

Question 2.
What is convergent evolution?
Answer:
The phenomenon of producing functionally similar structures (analogous organs) by distantly related organisms is called convergent evolution.

Question 3.
Define metastasis.
Answer:
The property of spreading of cancer cells from one part of the body to other parts is called metastasis.

Question 4.
Who is called the father of green revolution in India?
Answer:
Dr. Swaminathan

Question 5.
Name the organism from which cyclosporin is obtained.
Answer:
Trichoderma polysporum

Question 6.
What is gel electrophoresis?
Answer:
Electrophoresis is a technique used to separate macromolecules, especially proteins and nucleic acids according to their in size, charge or conformation under electric field.

Question 7.
Human testes are extra abdominal in position. Give reason?
Answer:
Because they are situated outside the abdominal cavity within a pouch called scrotum as an adaptation to maintain low temperature of the testes, necessary for spermatogenesis.

Question 8.
Give an example for passive immunity.
Answer:
Transfer of antibodies from mother to child during breast feeding or through placenta during pregnancy.

Question 9.
Water hyacinth (Eichornia crassipes) also called as ‘Bengal Terror is World‘s most problematic aquatic weed‘ give reason?
Answer:
It is an alien hydrophyte introduced in India. It has become invasive and cause extinction of native species. It grows faster in eutrophic water bodies and changes the dynamics of ecosystem.

Question 10.
Why do animals like snails enter in to aestivation during summer?
Answer:
To escape from summer related problems like heat and desssication.

Part – B

II. Answer any FIVE of the following questions in 3 – 5 sentences each, wherever applicable : (5 × 2 = 10 )

Question 11.
What is budding? Explain with an example.
Answer:
The method of reproduction through bud formation is known as budding.

Budding is a common method of asexual reproduction in yeast. when yeast cell matures, it develops an outgrowth called bud, the nucleus of the parent divides into two daughter nuclei by mitosis. One of the nuclei is pushed into the bud, later bud gets detached from the parent by the way of constriction and develops into mature individual.

Question 12.
What are biodiversity hot spots? Name the biodiversity hot spots of India.
Answer:
These are regions with very high levels of species richness and endemism. They are the most threatened reservoir of biodiversity on earth.
There are two Biodiversity hotspots in India

1. Western Ghats
2. Eastern Himalayas

Question 13.
What is test cross? Mention the uses of test cross.
Answer:
When offspring with dominant phenotype, whose genotype is not known, is crossed with an individual who is homozygous recessive for the trait is known as test cross

Use of Test Cross:
The test cross is used to find the genotype of an organism.

Question 14.
Distinguish between homologous and analogous organs.
Answer:
Homologous organs

• These are structurally similar functionally dissimilar.
• These are developed by related animals.
• They show similar developmental pattern.
• They show divergent evolution.

Analogous organs

• These are structurally dissimilar but functionally similar.
• These are developed by unrelated animals.
• Their developmental pattern is not similar.
• They show convergent evolution.

Question 15.
What is biofortification? Mention its importance.
Answer:
Breeding crops with higher levels of vitamins and minerals or higher protein and healthier fats is known as biofortification.

The objective of bio fortification is to increase quality and improving nutritional status of the plants by improving

1. Protein content and quality
2. Oil content and quality
3. Vitamin content
4. Micronutrient and mineral content

E.g. Vitamin A enriched carrots, spinach; pumpkin, Vitamin C enriched bitter gourd, calcium enriched spinach, protein enriched beans etc.

Question 16.
What is asexual reproduction? Name the mode of asexual reproduction in the following cases.
a) Yeast b) Amoeba c) Penicillium d) Sponges
Answer:
a) Yeast – Budding
b) Amoeba – Binary fission
c) Penicillium – Sporulation (conidia)
d) Sponges – Gemmules (internal buds)

Question 17.
Show schematic representation of steps of PCR
Answer:

Question 18.
What are polindromic sequences? Give an example.
Answer:
The sequences produced by restriction endonucleases possess the same base sequences in both the strands but in opposite direction i.e. ( 5^l – 3^l and 3^l – 5^l ). Therefore, it reads the same but in opposite direction. Such sequences are called palindrome sequences.

Part – C

III. Answer any FIVE of the following Questions in 40-80 words each, wherever applicable. (5 × 3 = 15 )

Question 19.
What are homologous organs? What is homology? What do the homologous organs explain as an evidence of organic evolution?
Answer:
Organs which are similar in origin but perform varied functions are called homologous organs.
Example: mouth parts of insects.

Homology is phenomenon of having similar origin and basic structures in different species due to their common ancestry for the same phenotype. Homologous organs indicate divergent evolution.

Question 20.
Why should we conserve biodiversity?
Answer:
There are three types of benefits of biodiversity, they are

1. Economic benefits.
2. Ecological benefits.
3. Ethical benefits.

1. Narrow utilitarian reasons for direct or economic benefits
Man derives maximum economic benefits from nature in the form of food (cereals, pulses fruits), fire wood, construction material, industrial products (tannin, resins, gums, latex, perfumes, dyes) and products of medicinal importance (quinine, reserpine, morphine, nimbin, vincristine), more than 25% of the drugs sold in market are derived from plants, more than 85% of the food materials are obtained from just 20 traditional crop species like cereals, millets, pulses etc. Scientists are busy in exploring molecular genetics and species level diversity for products of economic importance this is called biprospecting.

2. Broadly utilitarian reasons for ecological benefits or services
Biodiversity is necessary for maintenance and sustainable utilization of useful products and services from various ecosystems, such as cycling and recycling of nutrients through producers, consumers and decomposers, helps in maintaining climate conditions, soil fertility , pollination of plants, decomposition of waste and services such as purification of air, water mitigating floods and droughts and other environmental disasters.

3. Ethical reasons or benefits
Biodiversity has an ethical role. Every species have an intrinsic right to exist and share the wealth of the planet, therefore, it is our moral duty to ensure the well being of all other organisms and leave the biological legacy to future generations.

Question 21.
Write short notes on Darwin’s finches.
Answer:

• A divergent evolution has occurred in the ground finches of Galapagos Islands situated on the equator, some 900 km west of equator.
• Good example of adaptive radiation illustrated by Charles Darwin.
• He differentiated thirteen species of finches and grouped them into six main types.
• Finches in general possess stout, conical beaks adapted for crushing seeds.
• But they have undergone great diversification in their feeding habits.
• Accordingly shape and size of their beak has changed, ancestral finches on reaching different islands occupied all empty ecological niches in absence of competition and evolved into different species.

Question 22.
Name the causative organism, mode of transmission and symptoms of filariasis.
Answer:
Casual organism
It is caused by two nematode worms Wuchereria bancrofti and Wuchereria malayi.

Mode of transmission
Filarial worms are dimorphic and live in the lymph glands and lymph vessels of the lower limb. A female filarial worm is viviparous, it produces micro filarial larvae, which exhibit nocturnal periodicity. The movement of larvae in the peripheral blood circulation suits the visits of mosquitoes and enters the mosquitoes.

They act as a vector and transmit the micro filarial larvae to the new host, now the larvae penetrate through skin and reach the- lymph vessels of the host, it takes 5-18 months, inside the lymphatic vessels, they increase in the number. This results in blockage and swelling of lymph vessels. The enormous increase in the size of lymph vessels and associated tissue is called filariasis or elephantiasis.

Symptoms
Initial symptoms include inflammation of lymph glands accompanied by filarial fever; in the later stages, it results in blockage of lymph vessels and lymph glands in the legs, arms and other parts.

Question 23.
What is cross pollination. Mention its types. Differentiate geitonogy and xenogamy.
Answer:
When pollen grains are transferred from anthers of one flower to the stigma of another flower present in the same plant or different plant is called cross pollination or allogamy.

Cross pollination is two types namely geitonogamy and xenogamy.

Geitonogamy	Xenogamy
1. It is the cross pollination between the flowers of the same plant.	1. It is the cross pollination between the flowers of different plants.
2. Does not provide opportunity for gametic	2. Provide opportunity for gametic

Question 24.
Enumerate the major steps of isolation of DNA.
Answer:
Major steps for isolation of DNA

1. Cell containing DNA is treated with lysozyme or cellulose or chitinase
2. DNA along with RNA, Protein, lipid are released
3. Treatment with RNAase, protease to remove RNA and Protein
4. Appropriate treatment to remove other impurities
5. Addition of chilled ethanol to get precipitation of purified DNA
6. The fine threads of DNA can be collected by centrifugation

Question 25.
A child has blood group O. If the father has blood B and mother has blood group B. Work out the genotypes of the parents and the possible genotypes of the other offspring.
Answer:

The genotype of the parents of a child having blood group O will be I^AI^O for male and I^BI^O for female in heterozygous condition.

Question 26.
Explain carbon cycle with a suitable diagram.
Answer:
The cyclic movement of carbon between the living and non living systems of the environment is called carbon cycle. It is a perfect gaseous cycle. Its reservoir pool is atmosphere, which is used and repleni shed continuously by living organisms through activities like photosynthesis, respiration and decomposition.

Sources : Carbon is present in the air in the form of carbon dioxide, its concentration is very low (0.0032%), burning of fossils fuels like petrol, diesel, peat, coal adds large amounts of CO₂ into the atmosphere. The decomposition of dead bodies of plants and animals also release CO₂ to the atmosphere.

Atmosphere and hydrosphere are the cycling pool,where as lithosphere is the reservoir pool, therefore, a large amount of fixed carbon is used by marine zooplanktons to make calcium carbonate shells, these are not eaten or decomposed easily, they sink to the bottom and form carboniferous rocks (chalk, limestone), about 99% of carbon is stored in this form, therefore, ocean acts as global sink of CO₂.

Circulation: The CO₂ present in the atmosphere is used by plants for photosynthesis, during this process CO₂ is converted into carbohydrates. Some amount of glucose is used by plants during their respiration and the rest is converted into biomass (starch and fats) and stored as reserved food materials. These organic compounds containing carbon enter animals as food which are digested and absorbed. The absorbed organic nutrients are used to build their body mass.

Some amount of organic matter is also used by animals in respiration and CO₂ is released to the atmosphere after the death of plants and animals, the bodies are decomposed by decomposers like bacteria and helps in releasing back CO₂ to the atmosphere. This cyclic movement of carbon dioxide between plants, animals and back to atmosphere is called carbon cycle.

Part – D

Section – I

IV. Answer any FOUR of the following questions oh 200 – 250 words each, wherever applicable. ( 4 × 5 = 20 )

Question 27.
What are contraceptives? Mention the characters of an ideal contraceptive. Briefly describe the natural methods of birth control.

Question 28.
What is sex determination? Explain XX XY method of sex determination in human beings.
Answer:
The procedure by which the reason for the development of male and female is known in animals is called sex determination.

Sex determination in human beings is of male: heterogametic type, both male and females have the same number of chromosomes but vary in their gamete production. Females are homogametic with XX chromosomes; they produce ova haying X chromosome which are similar, but males are heterogametic with X and Y chromosomes.

They produce two kinds of sperms X and Y, 50% of the sperms carry X chromosome while the rest 50% carry Y chromosome, such a sex which produces two types of gametes is called heterogametic type.

The sex of the offspring largely depends on the sperm that fertilizes the ovum. When sperm with X chromosome fertilizes the egg having X chromosome, the zygote with XX develops into female on the other hand sperm carrying Y chromosome fertilize the egg having X chromosome develops into male.

Question 29.
What is DNA finger printing? Describe the various steps involved in the technique of DNA fingerprinting.
Answer:
DNA finger printing is a technique used for determining the nucleotide sequences of certain regions of DNA which are unique.to each individual and distinguishes him or her more exactly from others. Using DNA for distinguishing and identification of individuals is called DNA finger printing. It works on the principle of polymorphism in DNA sequences.

Various steps involved the technique of DNA fingerprinting

1. The isolation of the desired DNA from small amounts of blood, semen or other cells of body.
2. The digestion of DNA by restriction endonuclease enzyme.
3. The blotting of separate DNA fragments to synthetic membrane e.g., nylon or nitrocellulose membranes.
4. The hybridization by use of labelled VNTR probe (mini satellite).
5. By autoradiography, the detection of hybridised DNA fragments is done.

Question 30.
What is noise pollution? Mention its causes, effects and preventive measures.
Answer:
Noise pollution:
A loud unpleasant or unwanted sound is called noise.

Sources of noise pollution

• The sonic boom produced by air crafts, Jet plane is the extreme cause of noise pollution.
• Textile mills and printing presses, agricultural machines, defense equipments, transport vehicles, public address system, use of crackers on festive occasions. Operations such as blasting, crushing, construction work are other sources of noise pollution.

Effects of Noise Pollution

• It is harmful and causes psychological and physiological disorders in human beings.
• Exposure to extremely loud noise like explosion, sounds of jet plane or rockets damage ear drums; this may cause permanently impairing hearing ability.
• It also causes sleeplessness, increased heartbeat, headache, anxiety, stress, etc.

Prevention of noise pollution

• Use of loud speakers and amplifiers should be restricted to a fixed intensity and fixed hours of the day.
• Delimitation of horn free zones around hospitals, schools, etc.
• Noise producing industries, railway stations, aerodromes should be located away from human settlements.
• Noisy machines should be installed in sound proof chambers.
• Motor vehicles noise can be reduced by planting many rows of trees.
• Occupational exposure can be reduced by using ear muffs or cotton plugs.

Question 31.
Explain the structure of T.S of mature anther with a neat labeled diagram.
Answer:
Stamens are the male reproductive units of a flower. Each stamen is composed of an anther arid filament; Anther consists of microsporangium, pollen grains which contribute male gametes are formed within microsporangium present in anther.

A typical anther consists of column of sterile tissue called connective and anther lobes on either side. Each anther lobe bears two pollen chambers also called microsporangia. Thus, dithecous anther consists of 4 pollen sacs or micro sporanga.

Anther at maturity consists of sporogenous tissue covered by anther wall which is made up of epidermis, endothelium, middle layer and tapetum.

Epidermis: It is the outmost covering of the anther wall, single layered and protective in function.

Endothelium: It lies below the epidermis. The cells of the endothelium are radially elongated with fibrous thickenings. These cells are hygroscopic and when they lose water they contract as a result anther dehisces which helps in the release of pollen grains.

Middle layers: Two to three layered cells present below the endothelium constitute middle layers. The cells of the middle layer store food materials.

Tapetum: It is the inner most layer of the anther wall and surrounds the sporogenous tissue. The cells have dense cytoplasm and prominent nuclei. Tapetum is nutritive in function which supplies nutritive materials to developing sporogenous tissue.

Question 32.
What is apiculture? Mention the types of honey bees in a honey bee colony. Add a note on the economic importance of apiculture.
Answer:
The practice of rearing honey bees for honey and wax is called apiculture. It is also called Bee keeping.

There are three castes in honey bee colony.

1. Queen (fertile female)
2. Drones (fertile males)
3. Workers (sterile females)

Economic importance

• Honey collected from honey combs is a good nourishing food, it is a sweet syrup contains sugars like levulose (41%), fructose (38%), glucose (31%), maltose, it contains vitamins like ascorbic acid (vitamin c), niacin (B₃), riboflavin (B₂), thiamine (B₁), and minerals (Fe, Cu, mn, mg, Na, k ca p)
• Honey is a good antiseptic, laxative expectorant, blood purifier and chief source of energy.
• Honey is used in the preparation of jellies, Jams, cakes, etc.
• Bees wax secreted from the abdominal wax glands is used in manufacture of polishes, candle, etc.
• Bee venom is used in the treatment of disease like rheumatoid arthritis
• Honey bees are best natural pollinators.

Section – II

Answer any THREE of the following questions in 200 – 250 words each, wherever applicable. (3 × 5 = 15 )

Question 33.
What is biocontrol? Name the principle behind biological method of pest control. Mention examples of bio control agents and their function.
Answer:
It is the use of micro organisms to control or eliminate insect pests. The micro organisms employed in biological control are called bio control agents.

Principle
It is based on prey – predator relationship.

Examples
1. Ladybird and Dragon flies useful to get rid of aphids and mosquitoes.

2. Bad llus thuringiensis (Bt) is used to control butterfly caterpillar. Spores available in sachets are mixed with water and sprayed on plants, eaten by insect larva, toxin released in gut kills larvae.
Example: Bt toxin genes are introduced into cotton plants and made resistant to insect pests such as cotton boll worms, stem borer, aphids and beetles.

3. Nucleo polyhedrovirus ( NPV) is a virus suitable for narrow spectrum insecticide applications. It has no negative impacts on plants, mammals, birds, fish or target insects. It is suitable for overall integrated pest Management programme (IPM) in ecologically sensitive areas.

Question 34.
What is genetic code? Enumerate the characteristics of genetic code.
Answer:
Genetic code is a sequence of three nucleotides on DNA or mRNA, codes for a specific amino acids for protein synthesis.

Features of Genetic code

• Genetic code is triplet: Each codon consists of sequence of three nitrogen bases.
• Genetic code is universal: A particular codon codes for the same amino acid in all organisms.
• Genetic code is non overlapping: The successive triplet codons are read in order without overlapping and they do not share any base.
• Genetic code is degenerate: A single amino acid is coded by more than one codon.
  Example : valine is coded by 4 different codons GUA, GUC, GUU and GUG
• Genetic code is commaless: Codons are without punctuation and written in linear form. There is no signal to indicate the end of one codon or beginning of the next codon.
• Genetic code is non-ambiguous: Each codon specifies a particular amino acid in all organisms.
  Example : AUG codes for methionine .
• Initiator codons: Protein synthesis is always initiated by particular codons called initiator codons.
  Example : AUG in eukaryotas, GUG in prokaryotes
• Terminator codons: Three codons that act as stop signals to terminate protein synthesis are called terminator codons or nonsense codons.
  Example: UAA (Ochre), UGA (Amber) and UGA (Opal).

Question 35.
Explain the role of microbes as biofertlizers.
Answer:
Biofertilizers are organisms which are used to enhance the fertility of the soil and availability of nutrients like nitrogen and phosphorus to the crops.

There are three types of bio fertilizers
Example

1. Bacteria, (Rhizobium Azotobacter).
2. Cyanobacteria or Blue green algae (Anabaena Nostoc).
3. Fungi (Mycorrhizae like glomus, VAM).

1. Free living nitrogen fixing bacteria
They fix atmospheric nitrogen in the soil and made available to plants.
Example: The best example is Azotobacter.

2. Symbiotic nitrogen fixing bacteria
These bacteria show symbiotic association with the root nodules of leguminous plants. They convert atmospheric nitrogen and made available to plants.
Example : Rhizobium the most important symbiotic nitrogen fixing bacteria.
Frankia, mycelia bacterium (actinomycetes) shows symbiotic association with the root nodules of several non leguminous plants like casurina, rubus etc.

3. Free living nitrogen fixing cyanobacteria
Cyanobacteria are group of autotrophic microbes also called Blue green algae (BGA), they help in nitrogen fixation in paddy fields. These are extremely low cost biofertilizers.
Example: Anabaena, Nostoc.

4. Symbiotic nitrogen fixing cyanobacteria.
These nitrogen fixing cyanobacteria lead symbiotic mode of life with several plants like cycas roots, lichens, liver worts, Azolla (fern)
Example : Anabaena and Nostoc in the corolloid roots of cycas.

5. Mycorrhizae
The symbiotic association of fungus with the roots of higher plants is called mycorrhizae.
Example : Glomus species fix phosphorus in the soil and made available to plants.

Question 36.
Explain J shaped or Exponential growth curve.
Answer:
This kind of growth is often observed in bacterial population some bacteria devide once in 22-30 minutes under optimal conditions (2, 4, 8, 16, every 20 minutes.)

Exponential growth is a pattern of growth in which the population density after the initial establish ment phase ( lag phase) increases rapidly in an exponential or logarithmic form, but then stops abruptly as environmental resistance (e.g. seasonality) or some other factor suddenly becomes effective. This type of population growth is termed ‘density-independent’ as the regulation of growth rate is not tied to the population density until the final crash; Population numbers typically show great fluctuation as seen in algal blooms resulting in a J-shaped growth curve when population size is plotted over time.
The equation for exponential population growth is dN/dt = rN
dN/dt = rN (with a definite limit on N)
The intrinsic growth rate (r) is the difference between the birth rates (b) and death rates (d) percapita (r = b – d).
Where
N =is the number of individuals in the population,
t = time,
r = is a constant representing the intrinsic rate of increase (biotic potential) of the organism.

Question 37.
What is oogenesis? Explain oogenesis with a schematic representation.
Answer:
The process of formation of ovum is called oogenesis.
It occurs in the ovary . Each ovary is lined by germinal epithelium. At the time of gamete formation, some cells of the germinal epithelium will become active and detach from the epithelium and then enters into the cortex start dividing mitotically to produce large number of primordial germ cells (2n). The primordial cells pass through three phases before they transform into the ovum.

1. Multiplication phase
The primordial germ cells undergo repeated mitotic divisions to produce large number of cells; each cell is called oogonia or egg mother cell (2n).

2. Growth phase
In this phase egg mother cells grow in size due to synthesis and accumulation of food materials and these enlarged oogonial cells are called primary oocytes (2n).

3. Maturation phase
The primary oocyte undergoes first meiotic division to produce two unequal sized cells, which are haploid (n), a large cell with more amount of cytoplasm is called secondary oocyte and a small cell with negligible amount of cytoplasm is called primary polar body (polocyte).

The secondary oocyte and primary polar body undergo second meiotic division which is also unequal as a result; secondary oocyte produces one large cell and a smaller cell; larger cell is called ootid which develops into ovum. The smaller cell called secondary polar body and primary polar body produces two secondary polar bodies. Thus, in oogenesis, primary oocyte produces one ovum and three polar bodies. The polar bodies disintegrate later.

Students can Download Maths Chapter 3 Matrices Miscellaneous Exercise 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 3 Matrices Miscellaneous Exercise

Question 1.
Let \(A=\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\), Show that (al +bA)ⁿ + na^{n – 1} ba where 1 is the identity matrix of order 2 and n ∈ N.
Answer:

Question 2.
If\(A=\left[\begin{array}{lll}{1} & {1} & {1} \\{1} & {1} & {1} \\{1} & {1} & {1}\end{array}\right]\)
,prove that
Aⁿ = \(\left[\begin{array}{lll}{3^{n-1}} & {3^{n-1}} & {3^{n-1}} \\{3^{n-1}} & {3^{n-1}} & {3^{-1}} \\{3^{n-1}} & {3^{n-1}} & {3^{n-1}}\end{array}\right], n \in N\)
Answer:

Question 3.
\(\mathbf{A}=\left[\begin{array}{ll}{\mathbf{3}} & {-4} \\{\mathbf{1}} & {\mathbf{1}}\end{array}\right]\)
then prove that
Aⁿ = \(\left[\begin{array}{cc}{1+2 n} & {-4 n} \\{n} & {1-2 n}\end{array}\right]\) where n is any positive integer.
Answer:

∴ P (k +1) is true
P(n) is true for all the values of +ve n.

Question 4.
If A and B are symmetric matrices, prove that AB – BA is a skew symmetric matrix.
Answer:
Since A &B are symmetric matrices A = a’ & B = B’
To prove (AB – BA)’ = – (AB – BA)
L.H.S = (AB – BA)’
(AB)’ – (BA)’
= B’A’-A’B’
= BA – AB
= – (AB – BA) = R.H.S
Thus proved.

Question 5.
Show that the matrix g’AB is symmetric or skew symmetric according as A is symmetric or skew symmetric.
Answer:

Question 6.
find the values of x, y, z if the matrix
A = \(\left[\begin{array}{rrr}{0} & {2 y} & {x} \\{x} & {y} & {-z} \\{x} & {-y} & {z}\end{array}\right]\) satisfy the equation A’ A = 1
Answer:

Question 7.
For what values of x:
\(\left[\begin{array}{lll}{1} & {2} & {1}\end{array}\right]\left[\begin{array}{rrr}{1} & {2} & {0} \\{2} & {0} & {1} \\{1} & {0} & {2}\end{array}\right]\left[\begin{array}{l}{0} \\{2} \\{x}\end{array}\right]=0 ?\)
Answer:

Question 8.
\(\mathbf{A}=\left[\begin{array}{rr}{3} & {1} \\{-1} & {2}\end{array}\right]\) show that A² – 5A + 7I = 0
Answer:

Question 9.
Find x, if
\(\left[\begin{array}{ccc}{\mathbf{x}} & {-5} & {-1}\end{array}\right]\left[\begin{array}{ccc}{\mathbf{1}} & {\mathbf{0}} & {\mathbf{2}} \\{\mathbf{0}} & {\mathbf{2}} &{\mathbf{1}} \\{\mathbf{2}} & {\mathbf{0}} &{\mathbf{3}}\end{array}\right]\left[\begin{array}{l}{\mathbf{x}} \\{\mathbf{4}} \\ {\mathbf{1}}\end{array}\right]=\mathbf{0}\)
Answer:

Question 10.
A manufacturer produces three products x, y, z which he sells in two markets. Annual sales are indicated below:

(a) If unit sale prices of, y and z are ₹2. 50, ₹1.50 and ₹1.00, respectively, find the total revenue in each market with the help of matrix algebra.
(b) If the costs of the above three commodities are ₹ 2.00, ₹ 1.00 and 50 paise respectively. Find the gross profit.
Answer:
Matrix representing the sales is

Profit in market I is
₹ 46,000 – ₹ 31,000
= ₹ 15,000

Profit in market 11 is
₹ 53,000 – ₹36,000
= ₹ 17,000

Gross profit
= ₹ 15,000 + ₹17,000
= ₹ 32,000.

Question 11.
Find the matrix X so that
\(\mathbf{X}\left[\begin{array}{lll}{1} & {2} & {3} \\{4} & {5} & {6}\end{array}\right]_{2 \times 3}=\left[\begin{array}{rrr}{-7} & {-8} & {-9} \\{2} & {4} & {6}\end{array}\right]_{2 \times 3}\)
Answer:

Question 12.
If A and B are square matrices of the same order such that AB = BA, then prove by induction that ABⁿ = BⁿA. Further, prove that (AB)ⁿ = Aⁿ Bⁿ for all n ∈ N.
Answer:
(i) Let P(n) be the statement
P(n) – ⇒ Ab ⁿ = BⁿA
P(1) ⇒ AB = BA
∴ P( 1) is true
Let P(k) be true
P(k) ⇒ AB^k = B^kA
P(k +1) ⇒ AB^k+1
= A(B^kB) = (AB^k)B
∵ matrix multiplication is associative.
⇒ (B^kA) B ⇒ B^k (AB) = B^k (BA)
= (B^kB) A = B^k+1 A
Hence P(k + 1) is true.
∴ P(n) is true for all the values of n ∈ N

(ii) Let P(n) be the statement
P(n) ⇒ (AB)ⁿ = AⁿBⁿ
P(1) = (AB)¹ = AB
∴ P(1) is true Let P(k) be true
P(k) ⇒ (AB)^k
= A^kB^k P(k+1)
⇒ (AB)^k+1 = (AB)^kAB = (A^kB^k) AB
= A^k (B^kA)B = A^k (AB^k) B
(∵ ABⁿ = BⁿA whenever Ab = BA)
_{= A}k+1 _Bk+1
∴ P (k+1) is true
Hence P(n) is true for all the natures of n ∈ N

Choose correct answer in the following questions:

Question 13.
If \(\mathbf{A}=\left[\begin{array}{ll}{\alpha} & {\beta} \\{\gamma} & {-\alpha}\end{array}\right]\)
is such that A²= I ,then
(A) \(1+\alpha^{2}+\beta \gamma=0\)
(B) \(1-\alpha^{2}+\beta \gamma=0\)
(C) \(1-\alpha^{2}-\beta \gamma=0\)
(D) \(1+\alpha^{2}-\beta \gamma=0\)
Answer:

Question 14.
If the matrix A is both symmetric and skew symmetric, then
(A) A is a diagonal martix
(B) A is a zero matrix
(C) A is a square matrix
(D) None of these
Answer:
Let A be a square matrix such that A is both symmetric & skew-symmetric
⇒ A’ = A – (i) & A’ = – A -(ii)
⇒ A + A = 0  from (i) & (ii)
2A = 0
⇒ A = o
∴ (b) is correct

Question 15.
If A is square matrix such that A² = A, then (I + A)³ – 7A is equal to
(A) A
(B) I – A
(C) I
(D) 3A
Answer:
A² = A
(I + A)³ – 7A = (I + A) (I+A)² – 7A
= (I + A) {(I + A) (I+A)} -7A
= (I + A) (I² + 2A+A²) – 7A
[v IA = AI = A]
= (I + A) (I+2A+A) – 7A (A²= A)
= (I + A) (I + 3A) – 7A
= I² + AI + 3IA + 3A² – 7A
= I + A + 3A + 3A – 7A
[∵ A² = A & IA = AI = A]
=1
∴ (C) is correct.

2nd PUC Maths Matrices Miscellaneous Exercise Additional Questions and Answers

One Mark Questions:

Question 1.
If \(\mathbf{A}=\left(\begin{array}{ll}{\mathbf{0}} & {\mathbf{1}} \\{\mathbf{1}} & {\mathbf{0}}\end{array}\right), \mathbf{F} \text { ind } \mathbf{A}^{4}\)
(UP CET 2010)
Answer:

Question 2.
If
\(\mathbf{A}=\left(\begin{array}{rr}{0} & {2} \\{3} & {-4}\end{array}\right), \mathbf{K A}=\left(\begin{array}{ll}{0} & {3 \mathbf{a}} \\{2 \mathbf{b}} & {24}\end{array}\right)\),find k, a and b.
Answer

Question 3.
Find X and Y
\(2 \mathbf{X}-\mathbf{Y}=\left[\begin{array}{rrr}{6} & {-6} & {0} \\{-4} & {2} & {1} \end{array}\right]\)
\(\mathbf{X}+2 \mathbf{Y}=\left[\begin{array}{rrr}{3} & {2} & {5} \\{-2} & {1} & {-7}\end{array}\right]\)
Answer:

Question 4.
\(\left[\begin{array}{rrr}{1} & {1} & {1} \\{1} & {-2} & {-2} \\{1} & {3} &{2}\end{array}\right]\left[\begin{array}{l}{x} \\{y} \\{z}\end{array}\right]=\left[\begin{array}{l}{0} \\{3} \\{4}\end{array}\right], \text { then }\left[\begin{array}{l}{x} \\{y} \\{z}\end{array}\right]\)(I I T 2006)
Answer:

Question 5.
\(\mathbf{A}=\left[\begin{array}{lll}{6} & {8} & {7} \\{4} & {2} & {3} \\{9} & {7} & {1}\end{array}\right]\)
is the sum of symmentric and skew symmentric matrix, Find B
Answer:

Students can Download 2nd PUC Biology Model Question Paper 3 with Answers, Karnataka 2nd PUC Biology 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 Biology Model Question Paper 3 with Answers

Time: 3.15 Hours
Max Marks: 70

General Instructions:

1. This question paper consists of four parts A, B, C and D. Part D consists of two parts, Section – I and Section – II.
2. All the parts are Compulsory.
3. Draw diagrams wherever necessary. Unlabelled diagrams or illustrations do not attract any marks.

Part – A

I. Answer the following questions in One Word or One Sentence each : ( 10 × 1 = 10 )

Question 1.
Name the site of fertilization in female human beings.
Answer:
Upper limb (Ampullary – isthmic junction) of fallopian tube.

Question 2.
Define adaptive radiation
Answer:
It is the evolution of different animals in different direction from a common ancestor adapting to different ecological niches. It is also called as divergent evolution.

Question 3.
Name the organism that causes typhoid.
Answer:
Salmonella typhi.

Question 4.
What is green revolution?
Answer:
The dramatic increase in food production due to improvement of high yielding and disease resistant crop varieties is called green revolution.

Question 5.
Name the microbe that causes big holes in the Swiss cheese.
Answer:
Propioni bacterium sharmanii

Question 6.
Why does the logistic growth curve becomes S shaped?
Answer:
In sigmoid growth unlimited resources are rare, population growth is, therefore becomes stable due to environmental resistance.

Question 7.
Breast feeding during the initial period of infant growth is recommended. Give reason?
Answer:
The yellowish milk (colostrum) secreted from mammary glands during the initial days contains antibodies (IgA) to boost the development of immune system of the child.

Question 8.
Give an example for chemical carcinogen that causes cancer.
Answer:
Alcohol or tobacco or coal tar or coal gas or asbestos cement etc

Question 9.
What is totipotency?
Answer:
The capacity to generate a whole plant from any cell or explant is called totipotency

Question 10.
How do birds cope with temporary fluctuations in their environmental conditions?
Answer:
They migrate to hospitable areas for food and breeding purposes during unfavourable conditions in their habitat.

Part – B

II. Answer any FIVE of the following questions in 3 – 5 sentences each, wherever applicable : ( 5 × 2 = 10 )

Question 11.
Name the plant that shows unusual flowering phenomenon in hilly tracks of Karnataka, Kerala and Tamilnad.
Answer:
Strobilanthus kuluthiana ( Neelakuranji)

Question 12.
What is Biochemical Oxygen Demand (BOD)? Mention its significance.
Answer:
BOD refers to the amount of oxygen that would be consumed if all the organic matter in one liter of water were oxidized by bacteria.

Significance

• The BOD test measures the rate of uptake of oxygen by micro-organisms in a sample of water.
• Indirectly BOD is a measure of the organic matter present in the water.
• The greater the BOD of waste water more is its polluting potential.

Question 13.
Differentiate test cross and out cross.
Answer:
Test cross
It is a cross between F₁ hybrid and homozygous recessive parent.

Out cross
It is a cross between F₁ hybrid and homozygous dominant parent.

Question 14.
Give two examples for natural selection in organisms.
Answer:

1. Industrial melanism in Biston betularia
2. Antibiotic resistance in bacteria

Question 15.
What are single cell proteins? Mention their importance?
Answer:
Dried biomass of a single species of microbes that can be used as protein source in the diet is known a single cell protein (SCP).

Single cell proteins (SCP) are alternative source of proteins used to feed the ever increasing population with protein rich diet to overcome the problems of protein malnutrition.

Question 16.
Why is India said to have greater ecosystem diversity than Norway?
Answer:
India with diverse ecosystems such as rain forests, coral reefs, wet lands, coastal areas, mangroves, estuaries and deserts has greater ecosystem diversity than Norway.

Question 17.
Draw a diagram of sparged tank bioreactor and label the parts.
Answer:

Question 18.
Explain the mechanism of action of restriction endonucleases.
Answer:
Restriction endonucleases recognize a specific base sequence of a DNA molecule.
They cut double stranded DNA at different sites. This generates protruding (51 or 31) ends. Such ends are called cohesive or sticky ends. These ends can join with single stranded sequences of other DNA having complementary base pairs.
E.g. EcoRI, Bam H1

Cuts symmetrically placed around the line of symmetry

Part – C

III. Answer any FIVE of the following Questions in 40-80 words each, wherever applicable. ( 5 × 3 = 15 )

Question 19.
Describe the different phases of life cycle.
Answer:
Juvenile phase
The period of growth from birth till individual attains sexual maturity is called Juvenile phase. It is known as vegetative phase in plants.

Reproductive phase
The period of growth after Juvenile or vegetative phase during which individual reaches sexual maturity and capable of producing gametes is called reproductive phase. In human beings reproductive phase begins on the onset of puberty usually after 13 or 14 years. In plants flowering marks the beginning of reproductive phase.

Senescent phase
The phase begins after reproductive phase and marks the end of growth cycle in organisms.

Question 20.
Give a brief account of Human Origin and evolution.
Answer:

• Human evolution states that humans developed from primate or ape like ancestors.
• Human origin occurred in Central Asia and Africa, China, Java and India (Shivalik hills).
• The first mammals were shrew like terrestrial, insectivores or rat like creatures.
• The insectivorous mammals namely tree shrews gave rise to the primitive primates called prosomians which include tarsiers, lemurs and lories and anthropoids (marmosets, baboons, monkeys, apes and man).
• Modem man Homo sapiens belongs to class mammalia, order primata and suborder anthropoidea.
• Dryopithecus africanus is regarded as common ancestor of man and apes.
• It is lived about 20 – 25 million years ago (Miocene).
• Dryopithecus gave rise to Rama pithecus; he appeared 14-15 million years ago.
• The fossil of Rama pithecus was discovered from Pliocene rocks of Shivalik hills of India by Edward Lewis.
• Rama pithecus gave rise to Australo pithecus after a gap of 9 – 10 million years.
• Rama pithecus the first ape man commonly called southern ape .He appeared about 5 million years ago (early Pleistocene).
• Homo habillus lived in early Pleistocene about 2 million years ago.
• He was the first tool maker or handy man.
• Homo erectus is the direct ancestor of modern man.
• Homo erectus evolved from homo habillus about 1-7 million years ago in middle Pleistocene.
• Homo erectus include Java ape man, pieking man and Heidelberg man.
• The Neanderthal man existed in the late Pleistocene he was roughly equal to the modem man.
• Cro-Magnon is regarded as direct and most recent ancestor of the living modem man.
• He lived about 34.0000 years ago in Holocene epoch.
• The cranial capacity was about 1650cc and considered more intelligent than the man of today.
• The modern man Homo sapiens arose in Africa about 25.0000 years ago in Holocene epoch.
• It is believed that the man of today appeared about 11.000 or 10.000 years ago in the region around Caspian and Mediterranean seas from their members migrated west wards, eastwards and south wards respectively changing into present day white or Caucasoid, Monogoloid and black or Negroid races.
• Modern man underwent cultural evolution. Cave art developed about 18.000 years ago.
• Agriculture came around 10.000 years back and human settlement started.

Question 21.
List the various public health measures as a safeguard against infectious diseases?
Answer:
Following are the public health measures, considered safe against infectious diseases.

1. Maintenance of personal and public hygiene like proper disposal of waste and excreta through proper sanitation.
2. Provision of safe water supply, periodic cleaning and disinfection of water reservoirs, pools, and tanks, etc.
3. Standard practices of hygiene in public health catering.
4. Avoiding stagnation of water in and around residential areas to prevent breeding of mosquitoes and vectors.
5. Spraying of insecticides like DDT in ditches, drainage areas, swamps, etc., to prevent mosquito breeding.
6. Measures against vector borne diseases like dengue and chikungunya, include fixing of wire meshes to doors and windows to prevent the entry of mosquitoes.
7. Regular health camps, immunization (Vaccination) programmes to prevent the spread of infectious diseases.

Question 22.
What is food chain? Describe the types of food chain.
Answer:
The transfer of food energy from producers to decomposers through a series of organisms with repeated eating and being eaten is called food chain.
1. Grazing food chain
The food chain begins with producers (green plants) at the first trophic level. It involves grazing Of grass by herbivores and flow of energy occurs from smaller organisms (herbivores) to larger animals (consumers). It is based on prey – predator relationship and is therefore, called predator food chain.
Grass → Deer → Tiger

2. Parasitic food chain
The food chain begins with producers and flow of energy goes from larger organisms (host) to smaller organisms (parasites) as parasites draw nourishment from the host. This is based on host – parasite relationship.

3. Detritus or saprophytic food chain
The food chain begins with dead and decaying organic matter (detritus). Which is acted upon by decomposers or saprophytes and energy from decomposers goes to detrivores.
Leaf litter → Earth worm → bird → hawk

Question 23.
Define the terms
(a) Emasculation
Answer:
The removal of anthers from the flower bud of a bisexual flower before the anther dehisces is called emasculation.

(b) Bagging
Answer:
The process of covering the emasculated flowers with a bag of suitable size to prevent contamination with unwanted pollen is called bagging.

Question 24.
What is transformation? Explain any two methods of introduction of recombinant DNA in to host organism.
Answer:
Transformation is the process by which plasmid or DNA can be introduced into a cell.

1. Gene gun or particle gun or Biolistics method
It is a popular and widely used direct gene transfer method for delivering foreign genes into any cells and tissues or even intact seedlings.

• The foreign DNA is coated or precipitated onto the surface of minute gold or tungsten particles (1-3 pm).
• It is bombarded or shot onto the target tissue or cells using the gene gun or micro projectile gun or shot gun.with a device much like a particle gun. Hence the term biolistics
• The bombarded cells or tissues are cultured on selection medium to regenerate plants from the transformed cells.

2. Micro injection
In this technique a solution of DNA is directly injected in to the host nucleus using a fine micro capillary pipette or micro syringe, under a phase contrast microscope to aid vision.

Question 25.
Define trophic level. Show schematically the different tropic levels in a food chain.
Answer:
The amount of energy available at each step in food chain is called Trophic level.

Question 26.
What is pedigree? Show diagrammatic representation of pedigree for sickle cell anemia.
Answer:
Pedigree is a chart of graphic representation of record of inheritance of a trait through several generations in a family

Part – D

Section – I

IV. Answer any FOUR of the following questions on 200 – 250 words each, wherever applicable . ( 4 × 5 = 20 )

Question 27.
What are assisted reproductive technologies? Describe the various reproductive technologies to assist an infertile Couple to have children.
Answer:
The couple can be assisted to have children through certain special techniques known as assisted reproductive technologies (ART).
1. In vitro fertilization and embryo transfer (IVF-ET)

• In this technique egg cells are fertilized by sperm (usually 100,000 sperm / ml) outside the body in almost similar conditions as that in the body.
• In this process Ova from the wife or donor female and sperm from husband or donor male are allowed to fuse in shallow containers called Petri dishes. (Made of glass or plastic resins) Under laboratory condition.
• The fertilized egg (zygote) is then transferred to the patient’s uterus this called embryo transfer (E.T.), Where she provides suitable conditions for the development of embryo.
• The baby born by this technique is called test tube baby.

2. ZIFT ( Zygote intra fallopian transfer)
Zygote or early embryo up to eight blastomeres is transferred into the fallopian tube.

3. Gamete intra fallopian transfer (GIFT)
Transfer of an ovum collected from a donor to fallopian tube of another female who cannot produce ova, but provide suitable conditions for fertilization and further development up to parturition.

4. Intra uterine transfer (IUT)
Embryo with more than eight blastomeres is transferred in to the uterus.

5. Intra cytoplasmic sperm injection (ICSI)
The sperm is directly injected into the cytoplasm of the ovum to form an embryo in the laboratory and then embryo transfer is carried out.

6. Artificial insemination (AI)

• This method is used in cases where infertility is due to the inability of the male partner to inseminate the female or due to very low sperm counts in the ejaculates.
• In this method, the semen collected from the husband or a healthy donor is artificially introduced into the vagina or into the uterus (IUI – Intra uterine insemination).

Question 28.
What is dihybrid cross? State and explain Mendel’s second law of independent assortment.
Answer:
When a cross is made between two parents differing in two pairs of contrasting characters like round yellow and wrinkled green is called dihybrid cross.

Based on the results obtained from dihybrid cross, Mendel postulated his second law “The law of independent assortment”. It states that, when two pairs of contrasting characters are brought into the hybrid they behave independently of each other without contamination. The presence of one character is not contaminated with the other character. These characters are assorted independently of one another during the formation of gametes.
Mendel conducted dihybrid cross for which he selected two characters at a time.

Self crossed (F2)

Question 29.
What is meant by semi conservative replication? How did Meselson and Stahl prove it experimentally?
Answer:

It is a process by which DNA produces daughter DNA molecules, which are exact copies of the original DNA. In each new DNA molecule, one strand is old (original) while the other is newly formed. Hence, Watson and Crick described this method as semi-conservative replication.

Meselson and Stahl conducted experiments on E.coli to prove that DNA replication is semi conservative.

1. They grew E.coli cells in a culture medium containing radioactive isotopes N¹⁵ (NH₄cl).
   The ¹⁵N was incorporated into the newly synthesized DNA and other nitrogen containing compounds.
2. Bacteria labeled with ¹⁵N are transferred into a medium containing normal ¹⁴NH₄cl.
3. The analysis was done after every generation at regular intervals.
4. The DNA samples were separated by centrifugation on CSCl₂ gradient.
5. The hybrid DNA separated from fresh generations possesses one heavier strand contributed by parental with ¹⁵N isotope which is radioactive and lighter strand is newly synthesized with ¹⁴N isotope, is non-radioactive.
6. The DNA of the second generation showed that in one DNA molecule one strand was radioactive (¹⁵N) and other was non radioactive (¹⁴N), Whereas both the strands of other DNA molecule were non-radioactive (¹⁴N).
7. It is clear that the DNA of the first generation is intermediate it contains both ¹⁵N and ¹⁴N.
8. In the second generation there were two types of DNA one heavier DNA with ¹⁵N and ¹⁴N isotopes while the lighter DNA with ¹⁴N and ¹⁴N, the ratios of two types of DNA was 50:50.
9. In the third generation the ratio of intermediate (¹⁵N and ¹⁴N) was reduced from 50% to 25% while the lighter fraction (¹⁴N and ¹⁴N) increased from (50% – 75%).
10. In the 4^th generation the percentage of intermediate still reduced to 12.5% while that of lighter fraction increased from 75% to 87.5%.
11. These experiments prove that DNA shows semi conservative replication.

Question 30.
What is water pollution? Write short notes on the following with respect to water pollution.
a) Eutrophication
b) Biomagnifications
Answer:
Any undesirable change in the physical, chemical and biological characteristics of water, which makes water unfit for human use and harmfully affect aquatic organisms is called water pollution.

a) Eutrophication
It the process of nutrient enrichment of water bodies and subsequent loss of species diversity like fishes.
Harmful effects

1. Excess nutrients cause algal bloom, which may cover the whole surface of water body and release toxins.
2. It causes oxygen deficiency in water that leads, to the death of aquatic animals like fishes.

b) Biomagnifications
It refers to increase in concentration of toxic substances at successive trophic levels in a food chain. Example: Biomagnifications of DDT in an aquatic food chain.

Harmful Effect
High concentration of DDT disturbs calcium metabolism in birds, which causes thinning of egg shell and their premature breaking, causing decline in bird’s population.

Question 31.
Explain the structure of an anatropous ovule with a neat labeled diagram.
Answer:

Ovule is the female gametophyte develops inside the ovary from a cushion like structure called placenta. It is attached to placenta by a stalk known as funicle. The nutritive tissue enclosed inside the ovule is the nucellus. Nucellus is surrounded by outer and inner integuments. These integu-ments leave a small pore at the anterior end called micropyle. Anterior end of the ovule is micropyle chalazal end is the posterior part.

Question 32.
Distinguish DNA and RNA with respect to their structure or Chemistry and function.
Answer:

DNA	RNA
1. Present mainly in the chromatin the nucleus.	1. Most of RNA is present in the cell of cytoplasm (90%) and a little (10%) in the nucleolus
2. Normally double stranded and rarely single stranded.	2. Normally single stranded and rarely double stranded.
3. It contains deoxyribose sugar.	3. It has ribose sugar.
4. It contains bases like Adenine, Guanine, cytosine and Thymine.	4. It contains bases like, Adenine, Guanine cytosine and uracil.
5. DNA acts as template for its Synthesis.	5. RNA does not act as a template for its synthesis.

Section – II

Answer any THREE of the following questions in 200 – 250 words each, wherever applicable. ( 3 × 5 = 15 )

Question 33.
What is biocontrol? Name the principle behind biological method of pest control. Mention examples of biocontrol agents and their function.
Answer:
It is the use of micro organisms to control or eliminate insect pests. The micro organisms employed in biological control are called bio control agents.

Principle
It is based on prey – predator relationship.

Examples
1. Ladybird and Dragon flies useful to get rid of aphids and mosquitoes.

2. Bacillus thuringiensis (Bt) is used to control butterfly caterpillar. Spores available in sachets are mixed with water and sprayed on plants, eaten by insect larva, toxin released in gut kills larvae.
Example: Bt toxin genes are introduced into cotton plants and made resistant to insect pests such as cotton boll worms, stem borer, aphids and beetles.

3. Nucleo polyhedrovirus ( NPV) is a virus suitable for narrow spectrum insecticide applications. It has no negative impacts on plants, mammals, birds, fish or target insects. It is suitable for overall integrated pest Management programme (IPM) in ecologically sensitive areas.

Question 34.
What is logistic growth? Explain sigmoid curve with a suitable diagram.
Answer:
The rate of population growth slows as the population size approaches carrying capacity, leveling to a constant level.

Logistic growth curve or S-shaped growth curve (sigmoid growth curve) is characteristic of all higher animals including man and microbes like yeast cells growing in a natural environment.

Logistic growth is a pattern of growth in which the population density of few individuals introduced into the new habitat increases slowly initially, in a positive acceleration phase (lag phase) then increases rapidly, approaching an exponential growth rate (log phase) due to abundance of food, favourable environmental conditions such a climate, few predators and low levels of disease. Then declines in a negative acceleration phase.

This decline reflects increasing environmental resistance at higher population densities. This type of population growth is termed density-dependent, since growth rate depends on the numbers present in the population. The point of stabilization, or zero growth rates, is termed the saturation value or carrying capacity (K) of the environment for that organism.

When population growth rate is plotted graphically against time S shaped or sigmoid curve is obtained. This type of population growth is also called Verhulst- pearl logistic growth.
The mathematical equation for sigmoid growth is
dN / dt = rN(N – K) / k
Where
N = Population size at a given time
t = Time,
r = intrinsic rate of natural increase
k = Saturation value or carrying capacity for that organism in that environment.

Question 35.
What is animal breeding? Explain the controlled methods of animal breeding and mention their significance.
Answer:
The production of new varieties of animals which are superior to their parents through selective breeding or mating is called animal breeding.
There are two control breeding methods

1. Artificial insemination
2. Multiple ovulation embryo transfer technology (MOET)

Artificial insemination
Semen of superior male is collected and-injected unto the reproductive tract of selected female.

Significance:

1. It helps to overcome several problems of normal mating.
2. Semen of a desirable bull can be inseminated into cows.
3. Semen can be stored by cryopreservation to use at later stages and can be easily transported to distant places.
4. Semen of single ejaculation can be used to inseminate large number of cows.

Multiple Ovulatuion Embryo transfer technology (MOET)
Technique for herd improvement by successful production of hybrids

1. Hormones (FSH) are administered to the cow for inducing follicular maturation and super ovulation.
2. Cow produces 6-8 eggs instead of one egg and is either mated with, superior bull or artificially inseminated.
3. Fertilized egg at 8-32 cell stage are recovered non-surgically & transferred to surrogate mother.

Significance:

1. It is practiced in cattle, sheep, rabbits, buffaloes, etc.
2. Genetically superior animals can be produced.
3. It solves the problems of infertility.
4. A single female can donate many ovules.
5. Preservation and transport of embryos is convenient.

Question 36.
What is recombinant DNA technology? Draw a schematic representation of various steps involved in recombinant DNA technology.
Answer:
The branch of biotechnology that deals with altering the characters of an organism by introducing genes selected from desired organism is called genetic engineering or gene cloning or gene splicing.

Question 37.
What is gametogenesis? Mention the types. Explain spermatogenesis with a schematic representation.
Answer:
The process of formation of haploid male and female gametes in gonads is called gametogenesis.
It is of two types

1. Spermatogenesis
2. Oogenesis

1. Multiplication phase (mitosis)
Each testis is composed of numerous seminiferous tubules. Epithelial cells lining these tubules are called germinal epithelium. At the time of gametogenesis some cells of the epithelium become active, detach from the layer and start dividing mitotically to produce large number of primordial germ cells (2n), in turn primordial germ cells undergo repeated mitotic divisions to form sperm mother cells or spermatogonial cells which are diploid in nature.

2. Growth phase
Some of the spermatogonial cells (2n) stop dividing and enter into the seminiferous tubule where they grow in size by accumulating cytoplasm and duplication of DNA; sertoli or nurse cells provide necessary nutrients for the growing sperm mother cells, now sperm mother cells are termed as primary spermatocytes.

3. Maturation phase (meiosis)
Each primary spermatocyte (2n) enters into first meiotic division and forms two cells with haploid number of chromosomes called secondary spermatocytes. The secondary spermatocytes undergo second meiotic division to give rise to four haploid spermatids. The spermatids are unspecialized cells and do not undergo further division; all the four spermatids are transformed into four haploid sperms. This process of transformation of spermatids into mature and motile sperms is called spermiogenesis or spermateliosis.

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2nd PUC Maths Matrices NCERT Text Book Questions and Answers
Ex 3.1

Question 1.
In the matrix
\(\mathbf{A}=\left[\begin{array}{rrrr}{2} & {5} & {19} & {-7} \\{35} & {-2} & {\frac{5}{2}} & {12} \\{\sqrt{3}} & {1} & {-5} & {17}\end{array}\right]\)
(i) The order of the matrix,
(ii) The number of elements,
(iii) Write the elements a₁₃, a₂₁, a₃₃, a₂₄ ,a₂₃,
Answer:
(i) The order of the matrix is 3 x 4
(ii) Number of elements = 12
(iii) a₁₃= 19
a₂₁ = 35
a₃₃ = -5
a₂₄ = 12
a₂₃ = \(\frac{5}{2}\)

Question 2.
If a matrix has 24 elements, what are the possible orders it can have? What, if it has 13 elements?
Answer:
1 x 24; 24 x 1; 2 x 12;  12 x 2;
3 x 8; 8 x 3; 4 x 6; 6 x 4
13 elements
1 x 13;   13 x 1

Question 3.
If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements?
Answer:
18 elements
1 x 18;      18 x 1;      2 x 9;      9 x 2;       3 x 6;  6 x 3
5 elements
1 x 5;    5 x 1;

Question 4.
Constract a 2 x 2 matrix A =[a_ij] ,whose element are given by:
\(a_{u}=\frac{(i+j)^{2}}{2}\)
\(a_{i j}=\frac{i}{j}\)
\(a_{i j}=\frac{(i+2 j)^{2}}{2}\)
Answer:

Question 5.
Construct a 3 x 4 matrix, whose elements are given by:
(i) \(a_{i j}=\frac{1}{2}|-3 i+j| \)
(ii) \(a_{i j}=2 i-j\)
Answer:

Question 6.
Find the values of x, y and z from the following equations:
\(\text { (i) }\left[\begin{array}{ll}{4} & {3} \\{x} & {5}\end{array}\right]=\left[\begin{array}{ll}{y} & {z} \\{1} & {5}\end{array}\right]\)
Answer:
x = 1, y = 4, z = 3

(ii) \(\left[\begin{array}{cc}{x+y} & {2} \\{5+z} & {xy}\end{array}\right]=\left[\begin{array}{cc}{6} & {2} \\{5} & {8}\end{array}\right]\)
Answer:
5 + z = 5 = \(\succ \)– z = 0
x + y = 6 = \(\succ \) x = 2
xy = 8 = \(\succ \) x = 4
y = 4
4 = 2

(iii) \(\left[\begin{array}{r}{\mathbf{x}+\mathbf{y}+\mathbf{z}} \\{\mathbf{x}+\mathbf{z}} \\{\mathbf{y}+\mathbf{z}}\end{array}\right]=\left[\begin{array}{l}{\mathbf{9}} \\{\mathbf{5}} \\{\mathbf{7}}\end{array}\right]\)
Answer:
a – b = -1
2a – b = 0
a = 1
∴ 2a – b = 0
⇒ b = 2
2a + c = 5
c = 5 – 2 = 3 ∴ c = 3
d= 13-9 = 4
a = 1, b = 2 ,c = 3, d = 4

Question 8.
A = [a_ij]_{m x n\} is a square matrix, if
(A) m < n
(B) m>n
(C) m = n
(D) None of these
Answer:
(C) m = n as in a square matrix, the number of rows = number of colums.

Question 9.
Which of the given values of x and y make the following pair of matrices equal
\(\left[\begin{array}{cc}{3 x+7} & {5} \\{y+1} & {2-3 x}\end{array}\right],\left[\begin{array}{cc}{0} & {y-2} \\{8} & {4}\end{array}\right]\)
(A) \(x=\frac{-1}{3}, y=7\)
(B) Not possible to find
(C) \(y=7, x=\frac{-2}{3}\)
(D) \(x=\frac{-1}{3}, y=\frac{-2}{3}\)
Answer:

since x has 2 values, matrices are not equal. Hence, it is not possible to find the value. a option (B). is correct.

Question 10.
The number of all possible matrices of order 3 x 3 with each entry 0 or 1 is:
(A) 27
(B) 18
(C) 81
(D) 512
Answer:
There are 9 elements in the matrix. Each elements can be placed in 2 ways is all posible matrices of order 3 x 3 is 2⁹ = 512 a option (D) is correct.

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Karnataka 2nd PUC Physics Model Question Paper 4 with Answers

Time: 3 Hrs 15 Min
Max. Marks: 70

General Instructions:

1. All parts are compulsory.
2. Answers without relevant diagram/figure/circuit wherever necessary will not cany any marks
3. Direct answers to the Numerical problems without detailed solutions will not carry any marks.

Part – A

I. Answer all the following questions ( 10 × 1 = 10 )

Question 1.
State Ohm’s law.
Answer:
Ohm’s Law-At constant temperature the current through a conductor is directly proportional to the potential difference between its ends.

Question 2.
Define current sensitivity of a galvanometer.
Answer:
Deflection per unit current is called current Sensitivity.

Question 3.
Write the expression for force experienced by a straight conductor of length L carrying a steady current I, moving in a uniform external magnetic field B.
Answer:

Question 4.
What is ‘retentivity’ in magnetism?
Answer:
Retaining the magnetism even after the removal of the magnetising field is called retentivity.

Question 5.
Where on the earth’s surface is the magnetic dip zero?
Answer:
At magnetic equator dip is zero. ,

Question 6.
State ‘Lenz’s law in electromagnetic induction.
Answer:
The polarity of induced emf is such that it tends to produce a current which opposes the Change in magnetic flux that produced it.

Question 7.
Write the condition for ‘resonance’ of series LCR circuit.
Answer:
inductive reactance = capacitive reactance X_L = X_c

Question 8.
What is ‘wattless’ current?
Answer:
In purely inductive or capacitive circuit, no power is dissipated even though a current is flowing in the circuit. This current is called wattles current.

Question 9.
Write any one advantage of light emitting diode.
Answer:
Frequency remains constant.

Question 10.
What is attenuation in communication system?
Answer:
Mass number of the daughter nuclide 234

Part – B

II. Answer any Five of the following questions. ( 5 × 2 = 10 )

Question 11.
Represent graphically the variation of resistivity with absolute tempera ture for copper and nichrome metals.
Answer:

Question 12.
Write the expression for cyclotron frequency and explain the terms.
Answer:
V_c = \(\frac{\mathrm{qB}}{2 \pi \mathrm{m}}\)
q- Charge B- magnetic field
m-mass of the charged particle
V_c = cyclotron frequency

Question 13.
State and explain ‘curie’s Law’ in magnetism.
Answer:
The magnetic susceptibility ( χ ) of a para magnetic substance is inversely proportional to the absolute temperature (T).
χ = \(\mathrm{C} \frac{\mu_{0}}{\mathrm{T}}\)
C – Curie constant.

Question 14.
Mention any two factors on which the self inductance of a coil depends.
Answer:
Self-inductance of a coil depends on

1. number of turns of the coil
2. area of cross section and length (geometry)
3. permeability of the medium

Question 15.
Give any two application of ultraviolet radiations.
Answer:

1. used in LASIK eye surgery
2. UV lamps are used to kill germs in water purifiers
3. Disinfection for virus and bacteria
4. To produce photo electric current in burglar alarm.

Question 16.
What is polarisation of light? Name any one method of producing plane polarised light.
Answer:
The phenomenon of confining the vibrations of light in a single plane is called polarisation. Reflection / scattering.

Question 17.
Calculate de Broglie wavelength associated with an electron moving with a speed of 2 × 10⁵ ms^–1 . Given h = 6.625 × 10^–34JS, m_e = 9.11 × 10^–31kg.
Answer:

Question 18.
Write any two advantages of Light Emitting Diode (LED) over conventional in candescent low power lamps.
Answer:
Advantages of LED

1. Low operational voltage and less power consumption
2. Fast action and no warm up time required
3. Long life and ruggedness
4. Fast on-off switching capability

Part – C

III. Answer any five of the following questions. ( 5 × 3 =15)

Question 19.
Give any three properties of electric field lines.
Answer:
Properties electric field lines.

1. Field lines starts from positive charges and end at negative charges.
2. In a charge free region electric field lines can be taken to be continuous curves without any breaks.
3. Two field lines can never cross each other.
4. Electrostatic field lines do not form any closed loops.

Question 20.
Obtain the expression for effective capacitance of two capacitors connected in series.
Answer:

Let c₁,c₂ -capacitance of 2 capacitors connected in series
Q-charge on each capacitor
V₁ V₂-pd across C₁ and C₂
V-Total voltage drop across the combination
Then V = V₁ + V₂  ……… (1)
Also Q = CV
V₁ = \(\frac{\mathrm{Q}}{\mathrm{C}_{1}}\) and V₂ = \(\frac{\mathrm{Q}}{\mathrm{C}_{2}}\)
If system of capacitors is replaced by a single capacitor of equivalent capacitance cs then
V_s = \(\frac{\mathrm{Q}}{\mathrm{C}_{\mathrm{s}}}\)

Question 21.
Write any three differences between diamagnetic and paramagnetic materials.
Answer:

Diamagnetic	Paramagnetic
1. Weakly magnetised in a direction opposite to the applied magnetic field	1. Weakly magnetised along the direction of the applied magnetic field.
2. Move from stronger to weaker part of the external magnetic field	2. Move from weaker to the stronger part of the external magnetic field
3. Magnetic susceptibility is Low and negative	3. Magnetic susceptibility is low and positive.
4. Example: Bismuth, copper, lead, silicon	4. Example: Aluminium, sodium, calcium, oxygen

Question 22.
Describe the col and barmagnet experiment to demonstrate the phenomenon of electromagnetic induction.
Answer:

Coil and bar magnet experiment In the figure coil C1 is connected to a galvanometer G

1. when N-pole of the bar magnet is pushed towards the coil, there is a momentary deflection in the galvanometer.
2. when the magnet is pulled away from the coil galvanometer shows momentary deflection in the opposite direction.
3. Faster movements result in a larger deflection.
4. But no deflection when the coil and magnet are stationary with respect to each other. Or no deflection when there is no relative motion.

Therefore it shows that the relative motion between the magnet and coil induces electric current.

Question 23.
Derive the expression for effective focal length of two thin lenses kept in contact.
Answer:
Let f₁= focal length of first lens, and
f₂ = focal length of second lens
OP = u= object distance
PI = v = image distance due to the combination
PI₁ = v₁ =  image distance due to first lens
For the image formed by lens A,

Question 24.
Write any three experimental observations of photoelectirc effect.
Answer:

1. The photoelectric emission is an instantaneous process, even when incident radiation is exceedingly dim.
2. Above threshold frequency, the photo current is directly proportional to the intensity of incident radiation.
3. Above the threshold frequency, saturation current is proportional to the intensity
   of the incident radiation and stopping potential is independent of intensity.
4. There exists a certain minimum cut-off frequency called ‘threshold frequency’ below which no photo emission however intense the incident beam.
5. Above threshold frequency the kinetic energy of the photo electrons is directly proportional to the frequency of incident radiation and is independent of intensity.

Question 25.
How zener diode is used as a voltage regulator? Explain.
Answer:
Zener diode as a voltage regulator

The circuit connections are made as shown in the figure. The Zener diode is reverse biased. If the unregulated input voltage increases, the current through Rs and Zener diode also increases. This increases the voltage drop across Rs without any change in the voltage across the Zener diode. This is because in the breakdown region Zener voltage remains constant even though the current through Zener diode changes.

Similarly, if the input voltage decreases, the current through Rs and Zener also decreases. So, any increase or decrease of input results in increase or decrease of voltage drop across Rs without change in voltage across Zener diode. Hence it acts as a voltage regulator.

Question 26.
What is the function of ‘receiver’ in communication system? Draw the block diagram of AM – receiver.
Answer:
A receiver extracts the desired message signals from the received signals at the channel output.

Part – D

IV. Answer any two of the following questions ( 2 × 5 = 10 )

Question 27.
Using Gauss’s law in electrostatics, obtain the expression for electric field due to a uniformly charged thin spherical shell at a point
i. outside the shell and
ii. inside the shell
Answer:

Let σ be the uniform surface charge density of a thin spherical shell of radius R Field outside the shell.
Consider a point P outside the shell at a distance r from the centre of the shell. Imagine a gaussian sphere of radius ‘r’ The electric flux at P due to surface _∆ S is
∆φ= \(\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\Delta \mathrm{S}}\) = E_∆S COSθ=E_∆S {COS θ=1}
Total electric flux due to the sphere is
φ = E4πr²  ……….(1)
From Gauss law the electric flux

where q= total charge enclosed by the surface From (1) & (2)

b) electric field inside shell E = 0

Question 28.
Derive σ = \(\frac{\mathrm{ne}^{2} \tau}{\mathrm{m}}\) Where the symbols have their usual meaning.
Answer:

Consider conductor of length _∆x, area of cross section A.
V_d-drift speed of free electrons, in a time _∆ t, let all the electrons travel a distance _∆ x = V_d ∆ t.
Electric current I = _∆Q /_∆t = neAv_d
The acceleration acquired by the free electrons is given by a = \(\frac{-\mathrm{eE}}{\mathrm{m}}\)
where m= mass of the electron
If τ = relaxation time then velocity

v_d = \(\frac{-e E}{m} \tau\)

Current density J = \(\frac{1}{\mathrm{A}}\), J = σE therefore

σ = \(\frac{n e^{2} \tau}{m}\)

Question 29.
Obtain expression for the force between two infinitely long straight parallel conductros carrying current. Hence define ‘ampere’ the SI unit of electric current.
Answer:

Consider two infinitely long straight conductors a & b carrying currents Ia & Ib respectively are separated by a distance ‘d’. The conductor ‘a’ produces the same magnetic field Ba at all points along the conductor ‘b’.

B_a = \(=\frac{\mu_{0} I_{a}}{2 \pi d}\)   …………( 1 )

The conductor ‘b’ experiences a force Fba

F_ba = B_aLI_b = \(=\frac{\mu_{0} I_{a} I_{b} L}{2 \pi d}\)

where L= length of the conductor

Similarly one can show that Fba = -Fab
(A) ampere- The currents flowing through two infinitely long parallel conductors separated by 1m distance is 1 ampere if they experience a force of 2 X 10^–7N per unit length in air or vacum.

V. Answer any two of the following questions ( 2 × 5 = 10 )

Question 30.
Derive the expression for the fringe width of interference pattern in Young’s double-slip experiment.
Answer:

S₁ & S₂ are two coherent sources. GG| – screen at a distance D from the sources.
Let d = distance between the slits (two sources).
P = a point at a distance from the middle of the screen where a bright fringe is formed.
For constructive interference path difference = nλ
S₂P – S₁P = nλ where n = 0,1,2,3…….
From the diagram

(S₂P )² – (S₁P )²=2xd
(S₂P – S₁P)(S₂P + S₁P) = 2xd
S₂P ≅ S₁P ≅ OP = D
S₂P – S₁P = \(\frac{\mathrm{xd}}{\mathrm{D}}\) = nλ
X_n = \(\frac{\mathrm{n} \lambda \mathrm{D}}{2}\)
Where n = 0, +1, ±2, ± 3 for bright fringes
±1, ±2, ±3 for dark fringes
Since the finges are equally spaced the distance between two consecutive bright or consecutive dark fringes gives finge width.

β = \(\frac{\lambda \mathrm{D}}{\mathrm{d}}\)

Question 31.
Describe with suitable block diagra ms, action of pn-junction diode under forward and reverse bias conditions. Also draw I-V characteristics.
Answer:

PN Junction diode under forward bias.
When the diode is forward biased as shown in the figure the depletion region width decreases and the barrier height is reduced. The electrons from n-side cross the depletion region and reach p-side also holes from p-side cross the junction and reach the n-side. A concentration gradient is developed at the junction boundary. Due to this the motion of charged carriers on either side gives rise to current. The total diode forward current is sum of hole diffusion current and conventional current due to electron diffusion.

PN Junction diode under reverse bias

When the diode is reverse biased the depletion region width increases and the barrier height is increased. This supresses the flow of electrons from n-side to p-side and holes from p-side to n-side. Thus, diffusion current decreases. The conventional current is due to drift of the minority charge carriers which is of the order of micro amperes. The current under reverse bias is voltage independent up to a critical reverse bias voltage known as breakdown voltage.

The I-V characteristics are as shown.
VI Characteristics

Question 32.
Assuming the expression for the radius of electron orbit, obtain the expression for the total energy of the electron in the stationary orbit of hydrogen atom.
Answer:
The radius of the electron orbit is given by r = \(=\frac{\varepsilon 0 n^{2} h^{2}}{\pi m e^{2}}\)
Total energy E of the electron in a hydrogen atom is the sum of kinetic energy K and potential energy U
E=K+U
The electrostatic force of attraction between the revolving electrons and the nucleus is balanced by centripetal force in a dynamically stable orbit of hydrogen atom.

(negative sign signifies that the electrostatic force is in the – r direction)
The total energy of the electron in a hydrogen atom is

VI. Answer any three of the following question ( 3 × 5 = 15 )

Question 33.
The plates of a parallel plate capacitor have an area of 100 cm² each of 100 cm² each and are separated by 3 mm. The capacitor is charged by connecting it to a 400 V supply.
a) Calculate the electrostatic energy stored in the capacitor.
b) If a dielectric of dielectric constant 2.5 is introduced between the plates of the capacitor, then find the electrostatic
Answer:
energy stored and also change in the energy stored.
Given A = 100 cm² = 100 x 10^–4m²,
d = 3 mm = 3 x 10^–3m, V = 400 V, U₁ =?
K = 2.5,U₂ = ? also U₂ – U₁=?
U = \(\frac{1}{2} \mathrm{CV}^{2}\) C = \(\frac{\varepsilon_{0} \mathrm{A}}{\mathrm{d}}\)
Calculating
U₁ = 23.608 x 10^–7J
U₂ = 59.02 x 10^–7J
U₂∼U₁ = 35.412 x 10^–7J

Question 34.

In the given circuit diagram, calculate : (i) The main current through the circuit and (ii) Also current through 9 Ω resistor.
Answer:
RS = R₁ + R₂

Calculating the value of effective resistance R = 2.76 Ω

Current through 9 Ω= 0.308A

Question 35.
A 20 Ω resistor; 1.5 H inductor and 35 μF capacitor are connected in series with a 220 V, 50 Hz ac supply, calculate the impedance of the circuit and also find the current through the circuit.
Answer:
R = 20Ω, L= 1.5 H, C = 35 x 10^–6F,
V = 220V, v=50Hz, Z=?,  I=?

X_L = wL=2πvL = 471 Ω

Calculation of Z = 380.53 Ω
I = \(\frac{v}{z}\) = 0.578A

Question 36.
The radii of curvature of two surfaces of a convex lens is 0.2 m and 0.22 m. find the focal length of the lens of refractive index of the material of lens is 1.5. Also find the change in focal length, if it is immersed in water of refractive index 1.33.
R₁=0.2m, R₂=–022m, n₈=1.5, f_air=?
f_water =?, f_air~f_water =? n_w=1.33

Substituting the values, and calculating f = 0.209 m. Calculating focal length when immersed in water f_water=0.819
change in focal length = 0.61m

Question 37.
The half life of a radioactive sample ₃₈Sr⁹⁰ is 2 years. Calculate the rate of disintegration of 15 mg of this isotope.
Given Avogadro number = 6.023 x 10²³.
Answer:
T_1/2 = 28 years = 28 x 365 x 24 x 600 = 8.83 x 10⁸ seconds
90g \(\mathrm{Sr}_{38}^{90}\) contains 6.023 x 10²³ atoms
λ = \(\frac{0.693}{T_{1 / 2}}\)
Decay constant λ =7.85 x 10¹⁰ x 1.004 x 1.004 × 10¹⁰
Rate of disintegration R= λ N
R = 7.88 x 10¹⁰ Bq