NCERT Solutions for Maths Class 12 Exercise 6.1

# NCERT Solutions for Maths Class 12 Exercise 6.1

Hello Students! In this post, you will find the complete NCERT Solutions for Maths Class 12 Exercise 6.1.

You can download the PDF of NCERT Books Maths Chapter 6 for your easy reference while studying NCERT Solutions for Maths Class 12 Exercise 6.1.

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NCERT Solutions for Maths Class 12 Exercise 6.2

NCERT Solutions for Maths Class 12 Exercise 6.3

## NCERT Solutions for Maths Class 12 Exercise 6.1

### Maths Class 12 Ex 6.1 Question 1.

Find the rate of change of the area of a circle with respect to its radius r when

(a) r = 3 cm                    (b) r = 4 cm

Solution:
Let A be the area of the circle. Then, Ï€r2

Therefore, the rate of change of the area with respect to its radius is given by

a. When r = 3 cm, then

dA/dr = 2Ï€(3) = 6Ï€

Hence, the rate of change of the area of the circle is 6Ï€.

b. When r = 4 cm, then

dA/dr = 2Ï€(4) = 8Ï€

Hence, the rate of change of the area of the circle is 8Ï€.

### Maths Class 12 Ex 6.1 Question 2.

The volume of a cube is increasing at the rate of 8 cm³/s. How fast is the surface area increasing when the length of an edge is 12 cm?

Solution:
Let x be the side length of the cube, then volume V = x³, and surface area S = 6x2

Hence, the surface area is increasing at the rate of 8/3 cm3/s.

### Maths Class 12 Ex 6.1 Question 3.

The radius of a circle is increasing uniformly at the rate of 3 cm/s. Find the rate at which the area of the circle is increasing when the radius is 10 cm.

Solution:
Let r be the radius of the circle.
Area of the circle, A = Ï€r², and also dr/dt = 3 cm/s.

Hence, the area of the circle is increasing at the rate of 60Ï€ cm2/s.

### Maths Class 12 Ex 6.1 Question 4.

An edge of a variable cube is increasing at the rate of 3 cm/s. How fast is the volume of the cube increasing when the edge is 10 cm long?

Solution:
Let the edge of the cube be x cm. Then, dx/dt = 3 cm/s

Now, volume of the cube, V = x3 and dV/dt = 3x2

Hence, the volume of the cube is increasing at the rate of 900 cm3/s.

### Maths Class 12 Ex 6.1 Question 5.

A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/s. At the instant when the radius of the circular wave is 8 cm, how fast is the enclosed area increasing?

Solution:
Let r be the radius of a wave circle. Given, dr/dt = 5 cm/s.

We have, the area of the circle, A = Ï€r² and dA/dr = 2Ï€r

Hence, the enclosed area is increasing at the rate of 80Ï€ cm2/s.

### Maths Class 12 Ex 6.1 Question 6.

The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate of increase of its circumference?

Solution:
The rate of change of the radius of a circle is given by 0.7 cm/s, i.e., if the radius is r, then dr/dt = 0.7 cm/s.
Now, circumference of the circle, C = 2Ï€r

Hence, the rate of increase of its circumference is 1.4Ï€ cm/s.

### Maths Class 12 Ex 6.1 Question 7.

The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minute. When x = 8 cm and y = 6 cm, find the rates of change of
(a) the perimeter, and
(b) the area of the rectangle.

Solution:
(a) The length x of a rectangle is decreasing at the rate of 5 cm/min.

Therefore, dx/dt = –5 cm/min        … (i)
The width y is increasing at the rate of 4 cm/min.

### Maths Class 12 Ex 6.1 Question 8.

A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm.

Solution:
Volume of the spherical balloon, V = (4/3)Ï€r3

### Maths Class 12 Ex 6.1 Question 9.

A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the later is 10 cm.

Solution:
Let r be the variable radius of the balloon which is in the form of sphere.

Let the volume of the balloon be V. Then,

Hence, the volume is increasing at the rate of 400Ï€ cm3/s.

### Maths Class 12 Ex 6.1 Question 10.

A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2 cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall?

Solution:
Let AB be the ladder and OB be the wall. At an instant, let OA = x, OB = y,
x² + y² = 25         … (i)
On differentiating,

### Maths Class 12 Ex 6.1 Question 11.

A particle moves along the curve 6y = x3 + 2. Find the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate.

Solution:
We have, 6y = x3 + 2       … (i)

### Maths Class 12 Ex 6.1 Question 12.

The radius of an air bubble is increasing at the rate of 1/2 cm/s. At what rate is the volume of the bubble increasing when the radius is 1 cm?

Solution:
Let r be the radius of the air bubble, then the volume, V = (4/3)Ï€r3

Hence, the rate of increase of volume is 2Ï€ cm3/s.

### Maths Class 12 Ex 6.1 Question 13.

A balloon, which always remains spherical, has a variable diameter (3/2)[2x + 1]. Find the rate of change of its volume with respect to x.

Solution:
Diameter of the sphere = (3/2)[2x + 1]

### Maths Class 12 Ex 6.1 Question 14.

Sand is pouring from a pipe at the rate of 12 cm3/s. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4 cm?

Solution:
Let r and h be the radius and height of the sand cone at time t, respectively, h = r/6  … (i)

### Maths Class 12 Ex 6.1 Question 15.

The total cost C(x) in Rupees associated with the production of x units of an item is given by C(x) = 0.007 x3 – 0.003 x2 + 15x + 4000. Find the marginal cost when 17 units are produced.

Solution:
Marginal cost(MC) = Instantaneous rate of change of total cost at any level of output = dC/dx

So, when 17 units are produced, the marginal cost is Rs 20.967.

### Maths Class 12 Ex 6.1 Question 16.

The total revenue in Rupees received from the sale of x units of a product is given by

R(x) = 13x² + 26x + 15. Find the marginal revenue when x = 7.

Solution:

Marginal Revenue (MR) = Rate of change of total revenue w.r.t. the number of items sold at an instant = dR/dx
We know that, R(x) = 13x² + 26x + 15,
MR = dR/dx = 26x + 26 = 26(x + 1)
Now, when x = 7, then MR = 26(x + 1) = 26(7 + 1) = 208
Hence, the marginal revenue is Rs 208.

### Maths Class 12 Ex 6.1 Question 17.

The rate of change of the area of a circle with respect to its radius r at r = 6 cm is

(A) 10Ï€                        (B) 12Ï€                         (C) 8Ï€                         (D) 11Ï€

Solution:

(B)    A = Ï€r²

Therefore, dA/dr = 2Ï€r = 2Ï€ × 6 = 12Ï€ cm²/s

Hence, option (B) is the correct answer.

### Maths Class 12 Ex 6.1 Question 18.

The total revenue in Rupees received from the sale of x units of a product is given by R(x) = 3x² + 36x + 5. The marginal revenue, when x = 15 is

(A) 116                        (B) 96                          (C) 90                           (D) 126

Solution:

(D) We have, R(x) = 3x² + 36x + 5
MR = dR/dx = 6x + 36
At x = 15; dR/dx = 6 × 15 + 36 = 90 + 36 = 126

Hence, option (D) is the correct answer.