RD Sharma Class 12 Exercise 12 MCQ Derivative as a rate measure Solutions Maths

RD Sharma Class 12 Exercise 12 MCQ Derivative as a rate measure Solutions Maths - Download PDF Free Online

Updated on Jan 20, 2022 07:23 PM IST

RD Sharma Class 12 Solutions Derivative as a Rate Measure Chapter 12 MCQs is an expert-designed coursebook for CBSE Class 12 students. The solutions book will contain answers to all questions in the latest edition of NCERT maths books for the CBSE board. Students should use the RD Sharma Class 12th MCQs Solutions while solving problems from the maths book so that they can check the accuracy of their answers. Practising the book thoroughly before the board exams are highly recommended as it will improve your chances of scoring well and enhance your problem-solving abilities.

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RD Sharma Class 12 Solutions Chapter 12 MCQ Derivative as a Rate Measure - Other Exercise
Derivative as a Rate Measurer Excercise: 12 MCQ
RD Sharma Chapter wise Solutions

Also Read - RD Sharma Solution for Class 9 to 12 Maths

RD Sharma Class 12 Solutions Chapter 12 MCQ Derivative as a Rate Measure - Other Exercise

Derivative as a Rate Measurer Excercise: 12 MCQ

Derivative as a Rate Measurer exercise multiple choise questions question 1

Answer:

B . 4 π

Hint:
Here, we will use the formula,

v = \frac{4}{3} π r^{3}

Given:
Here,

r = 10 a n d \frac{d r}{d t} = 0.01

Solution:

\begin{matrix} v = \frac{4}{3} π r^{3} \\ \Rightarrow \frac{d v}{d t} = 4 π r^{2} \frac{d r}{d t} \end{matrix}

\Rightarrow Substituting values of r = 10 and \frac{d r}{d t} = 0.01

\begin{aligned} We get \\ \frac{d v}{d t} = 4 π \times 10^{2} \times 0.01 = 4 π \end{aligned}

Derivative as a Rate Measurer exercise multiple choise question 2

Answer:

B . 10 \sqrt{3} c m^{2} / s e c

Hint:
The area of an equilateral triangle side is

A = \frac{\sqrt{3}}{4} a^{2} . . . . . (i)

Given:

a = 10 and \frac{d a}{d t} = 2 cm / \sec

Solution:
Differentiating (i) with respect to t
We get

\frac{d A}{d t} = \frac{\sqrt{3}}{4} \times 2 a \frac{d a}{d t}

→ Substituting values of

a = 10 and \frac{d a}{d t} = 2 cm / \sec

We get

\begin{aligned} \frac{d A}{d t} = \frac{\sqrt{3}}{4} \times 2 a \times \frac{d a}{d t} \\ = \frac{\sqrt{3} a}{2} \times \frac{d a}{d t} \\ When a = 10 \\ \frac{d A}{d t} = \frac{10 \sqrt{3}}{2} \times 2 = 10 \sqrt{3} {cm}^{2} / \sec \end{aligned}

Derivative as a Rate Measurer exercise multiple choise question 3

Answer:

(c)

Hint:
Here, we use the formula and concept of volume of sphere
Given:
$\frac{d r}{d t} = 0.1 cm / \sec and r = 200 cm$
Solution:
The surface area of a sphere of radius r is defined by

\begin{aligned} A (r) = 4 π r^{2} \dots \dots . . (i) \\ \to Differentiating (i) with respect to t \\ \frac{d A}{d t} = 8 π r \cdot \frac{d r}{d t} = 8 π \times 200 \times 0.1 = 160 π {cm}^{2} / \sec \end{aligned}

Derivative as a Rate Measurer exercise multiple choise question 4

Answer:

0.002 cm / \sec (D)

Hint:

V (r, h) = \frac{1}{3} π r^{2} h

Given:

h = 2 r and \frac{d V}{d t} = 40 {cm}^{3} / \sec

Solution:

\begin{aligned} V (r) = \frac{2}{3} π r^{3} \\ \to Differentiating (i) with respect to t \\ \frac{d V}{d t} = 2 π r^{2} \frac{d r}{d t} \\ 40 = 2 π \times 10000 \times \frac{d r}{d t} \\ \frac{d r}{d t} = 0.002 cm / \sec \end{aligned}

Derivative as a Rate Measurer exercise multiple choise question 5

Answer:

1 m / m i n u t e

Hint:
The volume of a cylinder, with radius r and height h is defined by
Given:

V (r, h) = π r^{2} h

Solution:

S u b s t i t u t i n g r = 0.5 m w e g e t G i v e n t h a t \frac{d V}{d t} = 0.25 π m^{3} / \min, w e h a v e t o c a l c u l a t e \frac{d h}{d t} \to D i f f e r e n t i a t i n g (i) w i t h r e s p e c t t o t

\begin{aligned} \frac{d V}{d t} = π r^{2} \frac{d h}{d t} \\ \frac{d h}{d t} = \frac{1}{π r^{2}} \times \frac{d V}{d t} \\ \frac{d h}{d t} = \frac{0.25 π}{π (0.5)^{2}} \\ \frac{d h}{d t} = \frac{0.25 π}{0.25 π} \\ \frac{d h}{d t} = 1 m / \min \end{aligned}

Derivative as a Rate Measurer exercise multiple choise question 6

Answer:

B . - 42

Hint:
We are given the distance travelled x is a function of time, we can calculate velocity V and acceleration

a

V t = \frac{d x}{d t} and a (t) = \frac{d^{2} x}{d t^{2}}

Given:
Here, we use the formula,

x = t^{3} - 12 t^{2} + 6 t + 8

Solution:
Differentiating w.r.t time we get

V (t) = \frac{d x}{d t} = 3 t^{2} - 24 t + 6

Again Differentiating w.r.t time we get

\begin{aligned} a (t) = \frac{d^{2} x}{d t^{2}} = 6 t - 24 \\ a = 0 \Rightarrow 6 t - 24 = 0 or t = 4 units \\ So, Velocity at t = 4 \\ V (t) = \frac{d x}{d t} = 3 t^{2} - 24 t + 6 \\ V (4) = 3 \times 4^{2} - 24 \times 4 + 6 \\ V (4) = (- 42) \end{aligned}

Derivative as a Rate Measurer exercise multiple choise question 7

Answer:

\frac{d r}{d t} = \frac{160}{3} cm / \sec

Hint:
The relation between height h, radius r and semi vertical angle

a

is defined by

t a n α = \frac{r}{h}

Given:

α = 30^{\circ} and \frac{d α}{d t} = 2^{0} / \sec

Solution:

Let, r be the radius, h be the height & α be the semi-vertical angle of cone

t a n α = \frac{r}{h} . . . . . . (i)

\Rightarrow W e h a v e t o f i n d \frac{d r}{d t} \Rightarrow D i f f e r e n t i a t i n g (i) w i t h r e s p e c t t o t, w e g e t \sec^{2} α \frac{d α}{d t} = \frac{1}{20} \frac{d r}{d t}

\begin{aligned} \Rightarrow Substituting values, We get \\ \begin{array}{c} \sec^{2} 30^{\circ} \times 2^{0} = \frac{1}{20} \frac{d r}{d t} [h = 20 cm, α = 30^{\circ} and \frac{d α}{d t} = 2^{0} / \sec] \\ (\sec^{2} 30^{\circ}) \times 2^{\circ} = \frac{1}{20} \frac{d r}{d t} \\ \frac{4}{3} \times 2 = \frac{1}{20} \frac{d r}{d t} \\ \frac{d r}{d t} = \frac{160}{3} cm / \sec \end{array} \end{aligned}

Derivative as a Rate Measurer exercise multiple choise question 8

Answer:
$(d) 3, \frac{1}{3}$
Hint:
Here, we use the basic concept of algebra.
Given:

x^{3} - 5 x^{2} + 5 x + 8

Solution:

p (x) = x^{3} - 5 x^{2} + 5 x + 8 . . . . . . (i)

→ Differentiating (i) with respect to t,we get

\begin{aligned} \frac{d p (x)}{d t} = (x^{3} - 5 x^{2} + 5 x + 8) \frac{d x}{d t} = 2 \frac{d x}{d t} \\ 2 \frac{d x}{d t} = (3 x^{2} - 10 x + 5) \frac{d x}{d t} \\ = 3 x^{2} - 10 x + 5 = 2 \\ = 3 x^{2} - 10 x + 3 = 0 \\ \to Factorizing the above quadratic equation, \\ we get (3 x - 1) (x - 3) = 0 \\ \Rightarrow x = \frac{1}{3} and x = 3 \end{aligned}

Derivative as a Rate Measurer exercise multiple choise question 9

Answer:

(3, \frac{16}{3}) and (3, - \frac{16}{3})

Hint:
Here, we use the basic concept of algebra
Given:

16 x^{2} + 9 y^{2} = 400

Solution: Let

E (x, y) = 16 x^{2} + 9 y^{2} = 400

Solving for y we get

y = \pm \frac{\sqrt{400 - 16 x^{2}}}{3} . . . . . (i)

Given that

\frac{d x}{d t} = - \frac{d y}{d t}

we have to calculate
→ Differentiating (i) with respect to t,we get

\begin{aligned} \frac{d y}{d t} & = \pm \frac{1}{3} \times \frac{1}{2 \sqrt{400 - 46 x^{2}}} \times - 32 \times \frac{d x}{d t} \\ = \mp \frac{16 x}{3 \sqrt{400 - 16 x^{2}}} \frac{d x}{d t} \end{aligned}

→ Substituting values
We get

\begin{array}{r} 16 x = \pm 3 \sqrt{400 - 16 x^{2}} \end{array}

Squaring the equation, we get

\begin{aligned} 256 x^{2} = 9 (400 - 16 x^{2}) \end{aligned}

Solving the equation we get

x^{2} = 9 x = \pm 3

→ Substituting in (i), we get

\begin{aligned} y = \pm \frac{\sqrt{400 - 16 \times 9}}{3} = \pm \frac{16}{3} \\ So, \\ (3, \frac{16}{3}) and (3, - \frac{16}{3}) \end{aligned}

Derivative as a Rate Measurer exercise multiple choise question 10

Answer:

A . 54 π c m^{2} / m i n

Hint:
The lateral surface area of cone, with radius r and height h is defined as

\begin{aligned} L (r, h) = π r \sqrt{r^{2} + h^{2}} \dots \dots (i) \end{aligned}

Given:

\begin{aligned} r = 7 cm, h = 24 cm and \frac{d r}{d t} = 3 cm / \min, \frac{d h}{d t} = - 4 cm / \min \end{aligned}

Solution:
→ Differentiating (i) with respect to t,we get

\begin{aligned} \frac{d l}{d t} = π (\sqrt{r^{2} + h^{2}} \frac{d r}{d t} + r \times \frac{1}{2 \sqrt{r^{2} + h^{2}}} \times (2 r \frac{d r}{d t} + 2 h \frac{d h}{d t})) \\ Substituting values \\ \frac{d l}{d t} = π (\sqrt{7^{2} + 24^{2}} \times 3 + 7 \times \frac{1}{2 \sqrt{7^{2} + 24^{2}}} \times (2 \times 7 \times 3 + 2 \times 24 \times - 4)) \\ \frac{d l}{d t} = 54 π {cm}^{2} / \min \end{aligned}

Derivative as a Rate Measurer exercise multiple choise question 11

Answer:

B . 180 π c m^{3} / m i n

Hint:
Here we use formula of a sphere of radius r is defined by

\begin{aligned} V (r) = \frac{4}{3} π r^{3} . . . . (i) \end{aligned}

Given:

\begin{aligned} r = 15 cm \frac{d r}{d t} = 0.2 cm / \min \end{aligned}

Solution:
→ Differentiating (i) with respect to t, we get

\begin{aligned} \frac{d V}{d t} = 4 π r^{2} \times \frac{d r}{d t} \\ \frac{d V}{d t} = 4 π \times 15^{2} \times 0.2 = 180 π {cm}^{3} / \min \end{aligned}

Derivative as a Rate Measurer exercise multiple choise question 12

Answer:

B . \frac{3}{16 π} c m / s e c

B . \frac{3}{16 π} c m / s e c

Hint:
Here we use formula of a sphere of radius r is defined by

\begin{aligned} V (r) = \frac{4}{3} π r^{3} . . . . . (i) \end{aligned}

Given:

\begin{aligned} r = 2 c m \frac{d V}{d t} = 3 {cm}^{2} / \sec \end{aligned}

Solution:
→ Differentiating (i) with respect to t, we get

\begin{aligned} \frac{d V}{d t} = \frac{4}{3} π r^{2} \times \frac{d r}{d t} \\ 3 = 4 π \times 2^{2} \times \frac{d r}{d t} \\ \frac{d r}{d t} = \frac{3}{16 π} cm / \sec \end{aligned}

Derivative as a Rate Measurer exercise multiple choise question 13

Answer:

A . 9

Hint:
We are given the distance travelled as a function of time, we can calculate velocity by

\begin{aligned} V (t) = \frac{d s}{d t} \end{aligned}

Given:

\begin{array}{r} s = 45 t + 11 t^{2} + 3 \end{array}

Solution:
→ Differentiating (i) with respect to t, we get

\begin{aligned} V (t) = \frac{d s}{d t} = 45 t + 11 t^{2} + 3 \end{aligned}

→ If the particle is at rest, then it’s velocity will be 0
→ Solving the quadratic equation we get
$t = \frac{- 5}{3}, t \neq 0 so, t = 9$

Derivative as a Rate Measurer exercise multiple choise question 14

Answer:

\frac{d r}{d t} = \frac{1}{36} cm / \sec

Hint:
The volume of sphere, the radius r is defined by

\begin{aligned} V (r) = \frac{4}{3} π r^{3} . . . . . (i) \end{aligned}

Given:

\begin{array}{r} V = 288 π {cm}^{3} \end{array}

Solution:

\begin{aligned} 288 π = \frac{4}{3} π r^{3} \\ Solving for r^{r}, we get r = 6 cm \end{aligned}

\begin{aligned} \to Given that \frac{d V}{d t} = 4 π {cm}^{3} / \sec \\ \to Differentiating (i) with respect to t, we get \end{aligned}

\begin{aligned} \frac{d V}{d t} = \frac{4}{3} π r^{2} \times \frac{d r}{d t} \\ \to Substituting values, we get \\ 4 π = 4 π \times 6^{2} \times \frac{d r}{d t} \\ \frac{d r}{d t} = \frac{1}{36} cm / \sec \end{aligned}

Derivative as a Rate Measurer exercise multiple choise question 15

Answer:

\begin{aligned} D. r = \frac{1}{2 \sqrt{n}} \end{aligned}

Hint:
The volume of sphere of radius r is defined by

V (r) = \frac{4}{3} π r^{3}

Given:

\frac{d V}{d t} = \frac{d r}{d t}

Solution:

\begin{aligned} 4 π r^{2} \times \frac{d r^{'}}{d t} = \frac{d r^{'}}{d t} \\ 4 π r^{2} = 1 \end{aligned}

\begin{aligned} r = \frac{1}{2 \sqrt{n}} u n i t s \end{aligned}

Derivative as a Rate Measurer exercise multiple choise question 16

Answer:

B . r = \frac{1}{π} u n i t s

Hint:
The area of circle of radius r is defined by

A (r) = π r^{2}

Given:

\frac{d A}{d t} = \frac{2 d r}{d t}

Solution:
We get

2 π r \frac{d r}{d t} = \frac{2 d r}{d t} π r = 1 r = \frac{1}{π} u n i t e

Derivative as a Rate Measurer exercise multiple choise question 17

Answer:

8 \sqrt{3} c m^{2} / h r

Hint:
The area of an equilateral triangle with side a, is defined as,

A (a) = \frac{\sqrt{3}}{4} a^{2}

Given:

\frac{d a}{d t} = 8 cm / hr a = 2 cm W e h a v e t o c a l c u l a t e \frac{d A}{d t}

Solution:
→ Differentiating (i) with respect to t, we get

\begin{aligned} \frac{d A}{d t} = \frac{\sqrt{3}}{2} a \frac{d a}{d t} \end{aligned}

→ Substituting values, we get

\begin{aligned} \frac{d A}{d t} = \frac{\sqrt{3}}{2} \times 2 \times 8 = 8 \sqrt{3} {cm}^{2} / h r \end{aligned}

Derivative as a Rate Measurer exercise multiple choise question 18

Answer:

D . - \frac{16}{3} u n i t / s e c

Hint:
We are given the distance travelled as a function of time, We can calculate velocity and acceleration by

\begin{aligned} V t = \frac{d s}{d t} and a = \frac{d^{2} s}{d t^{2}} \end{aligned}

Given:

\begin{array}{r} s t = t^{3} - 4 t^{2} + 5 \end{array}

Solution:
→ Differentiating with respect to time, we get

\begin{aligned} V t = \frac{d s}{d t} = 3 t^{2} - 8 t \end{aligned}

→ Differentiating again with respect to time, we get

\begin{aligned} a (t) = \frac{d^{2} s}{d t^{2}} = 6 t = 8 \end{aligned}

\begin{aligned} \to Given that a = 0 \Rightarrow 6 t - 8 = 0 \\ Or = \frac{4}{3} unit / \sec \\ V (\frac{4}{3}) = 3 \times \frac{4^{2}}{3^{2}} - 8 \times \frac{4}{3} = - \frac{16}{3} unit / \sec \end{aligned}

Derivative as a Rate Measurer exercise multiple choise question 19

Answer:

\begin{aligned} \frac{π}{3} + \sqrt{3} m / \sec \end{aligned}

Hint:
Here

\begin{aligned} V t = \frac{d s}{d t} and a = \frac{d^{2} s}{d t^{2}} \end{aligned}

Given:

\begin{aligned} t = 2 t^{2} + \sin 2 t \end{aligned}

Solution:
→ Differentiating with respect to time, we get

\begin{aligned} V t = \frac{d s}{d t} = 4 t + 2 \cos 2 t . . . . . (i) \end{aligned}

→ Differentiating again with respect to time, we get

\begin{aligned} a (t) = \frac{d^{2} s}{d t^{2}} = 4 - 4 \sin 2 t \end{aligned}

\begin{aligned} Given that a = 2 \Rightarrow 4 - 4 \sin 2 t = 2 \\ Or \sin 2 t - \frac{1}{2} = \sin \frac{π}{6} \\ \to t = \frac{π}{12} \\ V (\frac{π}{12}) = 4 \times \frac{π}{12} + 2 \cos (\frac{π}{6}) \Rightarrow \frac{π}{3} + \sqrt{3} m / \sec \end{aligned}

Derivative as a Rate Measurer exercise multiple choise question 20

Answer:

0.24 π c m^{2} / s e c

Hint:
The circumference of a circle

A (r) = π r^{2} . . . . (i)

Given:

\begin{aligned} r = 12 cm \\ \frac{d r}{d t} = 0.01 cm / \sec \end{aligned}

Solution:

\begin{aligned} \to Differentiating (i) respect to t, we get \\ \frac{d A}{d t} = 2 π r \frac{d r}{d t} = 2 π \times 12 \times 0.01 \\ = 0.24 π {cm}^{2} / \sec \end{aligned}

Derivative as a Rate Measurer exercise multiple choise question 21

Answer:

π^{2} c m^{2} / s e c

Hint:
The circumference of a circle

A (r) = π r^{2} . . . . . (i)

Given:

r = π cm \frac{2 d r}{d t} = 1 cm / \sec

Solution:
→ Differentiating (i) respect to t, we get

\frac{d A}{d t} = 2 π r \frac{d r}{d t} = π \times π \times 1 = π^{2} {cm}^{2} / \sec

Derivative as a Rate Measurer exercise multiple choise question 22

Answer:

V s = 3.2 k m / h r

Hint:
Here, we use basic concept of volume algebra
Given:

h m = 2 m V m = 48 k m / h r h_{1} = 5 m

Solution:

\begin{aligned} \to Consider Δ A E C and Δ B E D \\ ∠ A E D = ∠ B E D = θ \\ ∠ E A C = ∠ E B D = 90^{\circ} \\ Therefore, Δ A E C \approx Δ B E D by AA criteria \end{aligned}

\begin{aligned} \frac{A C}{B D} = \frac{A E}{B E} \\ \to Substituting values, we get \\ \frac{5}{2} = \frac{4800 + x}{x} \end{aligned}

\begin{aligned} \to Simplifying the equation \\ x = 3200 . . . . . . (i) \end{aligned}

\begin{aligned} \to Differentiating (i) respect to r^{r}, we get \\ V \cdot s = \frac{d x}{d t} = 3200 m / hr = 3.2 km / hr \end{aligned}

Derivative as a Rate Measurer exercise multiple choise question 23

Answer:

\frac{d x}{d t} = V s = 6 f t / s e c

Hint:
Here, we use basic concept of volume algebra
Given:

h m = 6 f t V m = 9 f t / s e c h_{1} = 15 f t

Solution:

\begin{aligned} \to Consider Δ A E C and Δ B E D \\ ∠ A E D = ∠ B E D = θ \\ ∠ E A C = ∠ E B D = 90^{\circ} \end{aligned}

\begin{aligned} Therefore, △ A E C \approx Δ B E D by AA criteria \\ \frac{A C}{B D} = \frac{A E}{B E} \end{aligned}

\begin{aligned} \to Substituting values, we get \\ \frac{15}{6} = \frac{9 t + x}{x} \end{aligned}

\begin{aligned} \to Simplifying the equation \\ \frac{15}{6} = \frac{9 t}{x + x} \\ x = 6 t \\ V s = \frac{d x}{d t} = 6 ft / \sec \end{aligned}

Derivative as a Rate Measurer exercise multiple choise question 24

Answer:
C. Surface area times the rate of change of radius
Hint:
The volume and surface area of a sphere with radius r, is defined as,
Given:
$V (r) = \frac{4}{3} π r^{3} \dots \dots (i) and A (r) = 4 π r^{2}$
Solution:

\begin{aligned} \to Differentiating (i) respect to t, we get \\ V (r) = \frac{4}{3} π r^{2} \frac{d r}{d t} = A \times \frac{d r}{d t} \end{aligned}

(Surface area of sphere) X (rate of change of radius)

Derivative as a Rate Measurer exercise multiple choise question 25

Answer:
D. 8

π

times the rate of change of radius.
Hint:
The surface area of a sphere with radius r is defined as

A (r) = 4 π r^{2} . . . . . . . (i)

Given:

Solution:

\begin{aligned} \to Differentiating (i) respect to t, we get \\ \frac{d A}{d t} = 8 π r \frac{d r}{d t} \end{aligned}

\to 8 π \times current radius \times rate of change of radius

Derivative as a Rate Measurer exercise multiple choise question 26

Answer:

\begin{aligned} A. \frac{d h}{d t} = 1 m / h r \end{aligned}

Hint:
The volume of a cylinder, of radius r and height h is defined as

\begin{aligned} V (r, h) = π r^{2} h \end{aligned}

Given:

\begin{aligned} r = 10 m we get \\ V (h) = π \times 10^{2} \times h = 100 π h . . . . . (i) \\ \frac{d v}{d t} = 314 m^{3} / h r \end{aligned}

Solution:
→ Differentiating (i) respect to t, we get

\begin{aligned} \frac{d v}{d t} = 100 π \frac{d h}{d t} \\ \to Substituting values and using π = \frac{3}{4} we get \end{aligned}

\begin{aligned} 314 = 100 \times 3.14 \frac{d h}{d t} \\ \frac{d h}{d t} = 1 m / hr \end{aligned}

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RD Sharma Chapter wise Solutions

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RD Sharma Class 12 Exercise 12 MCQ Derivative as a rate measure Solutions Maths - Download PDF Free Online

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