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

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

Updated on 20 Jan 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:
$B.\; 4\pi$
Hint:
Here, we will use the formula,
$v=\frac{4}{3}\pi r^{3}$
Given:
Here,
$r=10\; \; and \; \; \frac{dr}{dt}=0.01$
Solution:
$\begin{gathered} v=\frac{4}{3} \pi r^{3} \\ \Rightarrow \frac{d v}{d t}=4 \pi r^{2} \frac{d r}{d t} \\ \end{gathered}$
$\Rightarrow \text { Substituting values of } r=10 \quad \text { and } \frac{d r}{d t}=0.01$
$\begin{aligned} &\text { We get }\\ &\frac{d v}{d t}=4 \pi \times 10^{2} \times 0.01=4 \pi \end{aligned}$

Derivative as a Rate Measurer exercise multiple choise question 2

Answer:
$B. \quad 10\sqrt{3}\; cm^{2}/sec$
Hint:
The area of an equilateral triangle side is
$A=\frac{\sqrt{3}}{4}a^{2} \quad \quad \quad \quad.....(i)$
Given:
$a=10 \text { and } \frac{d a}{d t}=2 \mathrm{~cm} / \mathrm{sec}$
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
$a=10 \text { and } \frac{d a}{d t}=2 \mathrm{~cm} / \mathrm{sec}$
We get
$\begin{aligned} &\frac{d A}{d t}=\frac{\sqrt{3}}{4} \times 2 a\times \frac{d a}{d t} \\ &\quad=\frac{\sqrt{3 }a}{2} \times \frac{d a}{d t} \\ &\text { When } a=10 \\ &\frac{d A}{d t}=\frac{10 \sqrt{3}}{2} \times 2=10 \sqrt{3} \mathrm{~cm}^{2} / \mathrm{sec} \end{aligned}$

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:
$\frac{d r}{d t}=0.1 \mathrm{~cm} / \mathrm{sec} \text { and } r=200 \mathrm{~cm}$
Solution:
The surface area of a sphere of radius r is defined by
$\begin{aligned} &A(r)=4 \pi r^{2} \quad \ldots \ldots . . \text { (i) }\\ &\rightarrow \text { Differentiating (i) with respect to } \mathrm{t}\\ &\frac{d A}{d t}=8 \pi r \cdot \frac{d r}{d t}=8 \pi \times 200 \times 0.1=160 \pi \mathrm{cm}^{2} / \mathrm{sec} \end{aligned}$

Derivative as a Rate Measurer exercise multiple choise question 4

Answer:
$0.002 \mathrm{~cm} / \mathrm{sec} \quad (D)$
Hint:
$V(r, h)=\frac{1}{3} \pi r^{2} h$
Given:
$h=2 r \text { and } \frac{d V}{d t}=40 \mathrm{~cm}^{3} / \mathrm{sec}$
Solution:
$\begin{aligned} &V(r)=\frac{2}{3} \pi r^{3}\\ &\rightarrow \text { Differentiating (i) with respect to } t\\ &\frac{d V}{d t}=2 \pi r^{2} \frac{d r}{d t}\\ &40=2 \pi \times 10000 \times \frac{d r}{d t}\\ &\frac{d r}{d t}=0.002 \mathrm{~cm} / \mathrm{sec} \end{aligned}$

Derivative as a Rate Measurer exercise multiple choise question 5

Answer:
$1m/minute$
Hint:
The volume of a cylinder, with radius r and height h is defined by
Given:
$V(r, h)=\pi r^{2} h$
Solution:
$Substituting\; r=0.5\; \mathrm{~m} \\ we\; get \\ Given\; that\; \frac{d V}{d t}=0.25 \pi \mathrm{m}^{3} / \mathrm{min}, \;we \;have \;to \;calculate \;\frac{d h}{d t} \\ \rightarrow Di\! f\! \! f\! erentiating \;(i)\; with\; respect\; to\; t$
$\begin{aligned} &\frac{d V}{d t}=\pi r^{2} \frac{d h}{d t} \\ &\frac{d h}{d t}=\frac{1}{\pi r^{2}} \times \frac{d V}{d t} \\ &\frac{d h}{d t}=\frac{0.25 \pi}{\pi(0.5)^{2}} \\ &\frac{d h}{d t}=\frac{0.25 \pi}{0.25 \pi} \\ &\frac{d h}{d t}=1 \mathrm{~m} / \mathrm{min} \end{aligned}$

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

Derivative as a Rate Measurer exercise multiple choise question 7

Answer:
$\frac{d r}{d t}=\frac{160}{3} \mathrm{~cm} / \mathrm{sec}$
Hint:
The relation between height h, radius r and semi vertical angle $a$ is defined by
$tan\: \alpha =\frac{r}{h}$
Given:
$\alpha=30^{\circ} \text { and } \frac{d \alpha}{d t}=2^{0} / \mathrm{sec}$
Solution:

Let, r be the radius, h be the height & α be the semi-vertical angle of cone
$tan\: \alpha =\frac{r}{h} \quad \quad \quad \quad......(i)$
$\Rightarrow We\; have\; to\; find\; \frac{d r}{d t} \\ \Rightarrow Di\! f\! \! ferentiating \; (i) \; with \; respect \; to \; \mathrm{t}, we \;get \\ \sec ^{2} \alpha \frac{d \alpha}{d t}=\frac{1}{20} \frac{d r}{d t}$
$\begin{aligned} &\Rightarrow \text { Substituting values, We get }\\ &\begin{gathered} \sec ^{2} 30^{\circ} \times 2^{0}=\frac{1}{20} \frac{d r}{d t} \quad\quad\quad\quad\quad\left[\mathrm{~h}=20 \mathrm{~cm}, \alpha=30^{\circ} \text { and } \frac{d \alpha}{d t}=2^{0} / \mathrm{sec}\right] \\ \left(\sec ^{2} 30^{\circ}\right) \times 2^{\circ}=\frac{1}{20} \frac{d r}{d t} \\ \frac{4}{3} \times 2=\frac{1}{20} \frac{d r}{d t} \\ \frac{d r}{d t}=\frac{160}{3} \mathrm{~cm} / \mathrm{sec} \end{gathered} \end{aligned}$

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:
$x^{3}-5x^{2}+5x+8$
Solution:
$p(x)=x^{3}-5x^{2}+5x+8 \quad\quad\quad\quad......(i)$
→ Differentiating (i) with respect to t,we get
$\begin{aligned} &\frac{d p(x)}{d t}=\left(x^{3}-5 x^{2}+5 x+8\right) \frac{d x}{d t}=2 \frac{d x}{d t}\\ &2 \frac{d x}{d t}=\left(3 x^{2}-10 x+5\right) \frac{d x}{d t}\\ &=3 x^{2}-10 x+5=2\\ &=3 x^{2}-10 x+3=0\\ &\rightarrow \text { Factorizing the above quadratic equation, }\\ &\text { we get }(3 x-1)(x-3)=0\\ &\Rightarrow x=\frac{1}{3} \text { and } x=3 \end{aligned}$

Derivative as a Rate Measurer exercise multiple choise question 9

Answer:
$\left(3, \frac{16}{3}\right) \text { and }\left(3,-\frac{16}{3}\right)$
Hint:
Here, we use the basic concept of algebra
Given:
$16x^{2}+9y^{2}=400$
Solution: Let
$E(x,y)=16x^{2}+9y^{2}=400$
Solving for y we get
$y=\pm \frac{\sqrt{400-16 x^{2}}}{3} \quad\quad \quad \quad .....(i)$
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
$\begin{aligned} 16 x=\pm 3 \sqrt{400-16 x^{2}}\\ \end{aligned}$
Squaring the equation, we get
$\begin{aligned} &256 x^{2}=9\left(400-16 x^{2}\right) \end{aligned}$
Solving the equation we get
$x^{2}=9 \\ x=\pm 3$
→ Substituting in (i), we get
$\begin{aligned} &y=\pm \frac{\sqrt{400-16 \times 9}}{3}=\pm \frac{16}{3}\\ &\text { So, }\\ &\left(3, \frac{16}{3}\right) \text { and }\left(3,-\frac{16}{3}\right) \end{aligned}$

Derivative as a Rate Measurer exercise multiple choise question 10

Answer:
$A.\; \; 54\pi \; \; cm^{2}/min$
Hint:
The lateral surface area of cone, with radius r and height h is defined as
$\begin{aligned} &L(r, h)=\pi r \sqrt{r^{2}+h^{2}} \quad \ldots \ldots \text { (i) }\\ \end{aligned}$
Given:
$\begin{aligned} &r=7 \mathrm{~cm}, h=24 \mathrm{~cm} \text { and } \frac{d r}{d t}=3 \mathrm{~cm} / \mathrm{min}, \frac{d h}{d t}=-4 \mathrm{~cm} / \mathrm{min}\\ \end{aligned}$
Solution:
→ Differentiating (i) with respect to t,we get
$\begin{aligned} &\frac{d l}{d t}=\pi\left(\sqrt{r^{2}+h^{2}} \frac{d r}{d t}+r \times \frac{1}{2 \sqrt{r^{2}+h^{2}}} \times\left(2 r \frac{d r}{d t}+2 h \frac{d h}{d t}\right)\right)\\ &\text { Substituting values }\\ &\frac{d l}{d t}=\pi\left(\sqrt{7^{2}+24^{2}} \times 3+7 \times \frac{1}{2 \sqrt{7^{2}+24^{2}}} \times(2 \times 7 \times 3+2 \times 24 \times-4)\right)\\ &\frac{d l}{d t}=54 \pi \mathrm{cm}^{2} / \mathrm{min} \end{aligned}$

Derivative as a Rate Measurer exercise multiple choise question 11

Answer:
$B.\; \; 180\pi \; \; cm^{3}/min$
Hint:
Here we use formula of a sphere of radius r is defined by
$\begin{aligned} &V(r)=\frac{4}{3} \pi r^{3} \quad \quad \quad \quad....(i) \end{aligned}$
Given:
$\begin{aligned} &r=15 \mathrm{~cm} \; \; \; \frac{d r}{d t}=0.2 \mathrm{~cm} / \mathrm{min} \end{aligned}$
Solution:
→ Differentiating (i) with respect to t, we get
$\begin{aligned} &\frac{d V}{d t}=4 \pi r^{2} \times \frac{d r}{d t} \\ &\frac{d V}{d t}=4 \pi \times 15^{2} \times 0.2=180 \pi \mathrm{cm}^{3} / \mathrm{min} \end{aligned}$

Derivative as a Rate Measurer exercise multiple choise question 12

Answer:
$B.\; \; \frac{3}{16\pi }\; cm/sec$$B.\; \; \frac{3}{16\pi }\; cm/sec$
Hint:
Here we use formula of a sphere of radius r is defined by
$\begin{aligned} &V(r)=\frac{4}{3} \pi r^{3} \quad \quad.....(i) \end{aligned}$
Given:
$\begin{aligned} &r=2 c m\; \; \frac{d V}{d t}=3 \mathrm{~cm}^{2} / \mathrm{sec} \end{aligned}$
Solution:
→ Differentiating (i) with respect to t, we get
$\begin{aligned} &\frac{d V}{d t}=\frac{4}{3} \pi r^{2} \times \frac{d r}{d t} \\ &3=4 \pi \times 2^{2} \times \frac{d r}{d t} \\ &\frac{d r}{d t}=\frac{3}{16 \pi} \mathrm{cm} / \mathrm{sec} \end{aligned}$

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:
$\begin{aligned} s=45 t+11 t^{2}+3 \end{aligned}$
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
$t=\frac{-5}{3}, \quad t \neq 0 \quad \text { so, } t=9$

Derivative as a Rate Measurer exercise multiple choise question 14

Answer:
$\frac{d r}{d t}=\frac{1}{36} \mathrm{~cm} / \mathrm{sec}$
Hint:
The volume of sphere, the radius r is defined by
$\begin{aligned} &V(r)=\frac{4}{3} \pi r^{3} \quad \quad \quad .....(i) \end{aligned}$
Given:
$\begin{aligned} V=288 \pi \mathrm{cm}^{3} \end{aligned}$
Solution:
$\begin{aligned} &288 \pi=\frac{4}{3} \pi r^{3}\\ &\text { Solving for } r^{r}, \text { we get } r=6 \mathrm{~cm} \end{aligned}$
$\begin{aligned} &\rightarrow \text { Given that } \frac{d V}{d t}=4 \pi \mathrm{cm}^{3} / \mathrm{sec}\\ &\rightarrow \text { Differentiating (i) with respect to t, we get } \end{aligned}$
$\begin{aligned} &\frac{d V}{d t}=\frac{4}{3} \pi r^{2} \times \frac{d r}{d t}\\ &\rightarrow \text { Substituting values, we get }\\ &4 \pi=4 \pi \times 6^{2} \times \frac{d r}{d t}\\ &\frac{d r}{d t}=\frac{1}{36} \mathrm{~cm} / \mathrm{sec} \end{aligned}$

Derivative as a Rate Measurer exercise multiple choise question 15

Answer:
$\begin{aligned} &\text { D. } \quad r=\frac{1}{2 \sqrt{n}} \end{aligned}$
Hint:
The volume of sphere of radius r is defined by
$V(r)=\frac{4}{3}\pi r^{3}$
Given:
$\frac{dV}{dt}=\frac{dr}{dt}$
Solution:
$\begin{aligned} &4 \pi r^{2} \times \frac{d r^{\prime}}{d t}=\frac{d r^{\prime}}{d t} \\ &4 \pi r^{2}=1 \end{aligned}$
$\begin{aligned} &r=\frac{1}{2\sqrt{n}}\: units \end{aligned}$

Derivative as a Rate Measurer exercise multiple choise question 16

Answer:
$B.\; r=\frac{1}{\pi }units$
Hint:
The area of circle of radius r is defined by
$A(r)=\pi r^{2}$
Given:
$\frac{d A}{d t}=\frac{2 d r}{d t}$
Solution:
We get
$2 \pi r \frac{d r}{d t}=\frac{2 d r}{d t}\\ \\ \pi r=1 \\ \\ r=\frac{1}{\pi} unite$

Derivative as a Rate Measurer exercise multiple choise question 17

Answer:
$8\sqrt{3}\: cm^{2}/hr$
Hint:
The area of an equilateral triangle with side a, is defined as,
$A(a)=\frac{\sqrt{3}}{4} a^{2}$
Given:
$\frac{d a}{d t}=8 \mathrm{~cm} / \mathrm{hr} \\ \\ a=2 \mathrm{~cm} \\ W\! e\; have\; to\; calculate\; \frac{d A}{d t}$
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
$\begin{aligned} &\frac{d A}{d t}=\frac{\sqrt{3}}{2} \times 2 \times 8=8 \sqrt{3} \mathrm{~cm}^{2} / h r \end{aligned}$

Derivative as a Rate Measurer exercise multiple choise question 18

Answer:
$D. \: \: -\frac{16}{3}\: unit/sec$
Hint:
We are given the distance travelled as a function of time, We can calculate velocity and acceleration by
$\begin{aligned} &V t=\frac{d s}{d t} \text { and } a=\frac{d^{2} s}{d t^{2}} \\ \end{aligned}$
Given:
$\begin{aligned} s t=t^{3}-4 t^{2}+5 \end{aligned}$
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
$\begin{aligned} &a(t)=\frac{d^{2} s}{d t^{2}}=6 t=8 \end{aligned}$
$\begin{aligned} &\rightarrow \text { Given that } a=0 \Rightarrow 6 t-8=0 \\ &\text { Or }=\frac{4}{3} \text { unit } / \mathrm{sec} \\ &V\left(\frac{4}{3}\right)=3 \times \frac{4^{2}}{3^{2}}-8 \times \frac{4}{3}=-\frac{16}{3} \text { unit } / \mathrm{sec} \end{aligned}$

Derivative as a Rate Measurer exercise multiple choise question 19

Answer:
$\begin{aligned} &\frac{\pi}{3}+\sqrt{3} \mathrm{~m} / \mathrm{sec}\\ \end{aligned}$
Hint:
Here
$\begin{aligned} &V t=\frac{d s}{d t} \text { and } a=\frac{d^{2} s}{d t^{2}}\\ \end{aligned}$
Given:
$\begin{aligned} &t=2 t^{2}+\sin 2 t \end{aligned}$
Solution:
→ Differentiating with respect to time, we get
$\begin{aligned} &V t=\frac{d s}{d t}=4 t+2 \cos 2 t \quad \quad \quad .....(i) \end{aligned}$
→ Differentiating again with respect to time, we get
$\begin{aligned} &a(t)=\frac{d^{2} s}{d t^{2}}=4-4 \sin 2 t \end{aligned}$
$\begin{aligned} &\text { Given that } a=2 \Rightarrow 4-4 \sin 2 t=2 \\ &\text { Or } \sin 2 t-\frac{1}{2}=\sin \frac{\pi}{6} \\ &\rightarrow t=\frac{\pi}{12} \\ &V\left(\frac{\pi}{12}\right)=4 \times \frac{\pi}{12}+2 \cos \left(\frac{\pi}{6}\right) \Rightarrow \frac{\pi}{3}+\sqrt{3} \mathrm{~m} / \mathrm{sec} \end{aligned}$

Derivative as a Rate Measurer exercise multiple choise question 20

Answer:
$0.24\pi \: cm^{2}/sec$
Hint:
The circumference of a circle
$A(r)=\pi r^{2} \quad \quad \quad \quad \quad....(i)$
Given:
$\begin{aligned} &r=12 \mathrm{~cm} \\ &\frac{d r}{d t}=0.01 \mathrm{~cm} / \mathrm{sec} \end{aligned}$
Solution:
$\begin{aligned} &\rightarrow \text { Differentiating (i) respect to } \mathrm{t} \text { , we get }\\ &\frac{d A}{d t}=2 \pi r \frac{d r}{d t}=2 \pi \times 12 \times 0.01\\ &=0.24 \pi \mathrm{cm}^{2} / \mathrm{sec} \end{aligned}$

Derivative as a Rate Measurer exercise multiple choise question 21

Answer:
$\pi ^{2}\: cm^{2}/sec$
Hint:
The circumference of a circle
$A(r)=\pi r^{2} \quad \quad \quad \quad \quad .....(i)$
Given:
$r=\pi \mathrm{cm} \\ \frac{2 d r}{d t}=1 \mathrm{~cm} / \mathrm{sec}$
Solution:
→ Differentiating (i) respect to t, we get
$\frac{d A}{d t}=2 \pi r \frac{d r}{d t}=\pi \times \pi \times 1 \\ \\ =\pi^{2} \mathrm{~cm}^{2} / \mathrm{sec}$

Derivative as a Rate Measurer exercise multiple choise question 22

Answer:
$Vs=3.2 km/hr$
Hint:
Here, we use basic concept of volume algebra
Given:
$hm=2m \quad Vm=48km/hr \quad h_{1}=5m$
Solution:
$\begin{aligned} &\rightarrow \text { Consider } \Delta A E C \text { and } \Delta B E D\\ &\angle A E D=\angle B E D=\theta\\ &\angle E A C=\angle E B D=90^{\circ}\\ &\text { Therefore, } \Delta A E C \approx \Delta B E D \text { by AA criteria } \end{aligned}$
$\begin{aligned} &\frac{A C}{B D}=\frac{A E}{B E}\\ &\rightarrow \text { Substituting values, we get }\\ &\frac{5}{2}=\frac{4800+x}{x} \end{aligned}$
$\begin{aligned} &\rightarrow \text { Simplifying the equation }\\ &x=3200 \quad \quad......(i) \end{aligned}$
$\begin{aligned} &\rightarrow \text { Differentiating (i) respect to } r^{r}, \text { we get }\\ &V \cdot s=\frac{d x}{d t}=3200 \mathrm{~m} / \mathrm{hr}=3.2 \mathrm{~km} / \mathrm{hr} \end{aligned}$

Derivative as a Rate Measurer exercise multiple choise question 23

Answer:
$\frac{dx}{dt}=V\! s=6\: ft/sec$
Hint:
Here, we use basic concept of volume algebra
Given:
$hm=6\, f\! t \quad V\! m=9\, f\! t/sec \quad h_{1}=15\, f\! t$
Solution:
$\begin{aligned} &\rightarrow \text { Consider } \Delta A E C \text { and } \Delta B E D \\ &\angle A E D=\angle B E D=\theta \\ &\angle E A C=\angle E B D=90^{\circ} \end{aligned}$
$\begin{aligned} &\text { Therefore, } \triangle A E C \approx \Delta B E D \text { by AA criteria }\\ &\frac{A C}{B D}=\frac{A E}{B E} \end{aligned}$
$\begin{aligned} &\rightarrow \text { Substituting values, we get }\\ &\frac{15}{6}=\frac{9 t+x}{x} \end{aligned}$
$\begin{aligned} &\rightarrow \text { Simplifying the equation }\\ &\frac{15}{6}=\frac{9 t}{x+x}\\ &x=6 t\\ &V \! s=\frac{d x}{d t}=6 \mathrm{ft} / \mathrm{sec} \end{aligned}$

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:
$V(r)=\frac{4}{3} \pi r^{3} \quad \ldots \ldots \text { (i) } \quad \text { and } \quad A(r)=4 \pi r^{2}$
Solution:
$\begin{aligned} &\rightarrow \text { Differentiating (i) respect to } \mathrm{t}, \text { we get }\\ &V(r)=\frac{4}{3} \pi r^{2} \frac{d r}{d t}=A \times \frac{d r}{d t} \end{aligned}$
(Surface area of sphere) X (rate of change of radius)

Derivative as a Rate Measurer exercise multiple choise question 25

Answer:
D. 8$\pi$times the rate of change of radius.
Hint:
The surface area of a sphere with radius r is defined as
$A(r)=4\pi r^{2} \quad \quad \quad \quad .......(i)$
Given:

Solution:
$\begin{aligned} &\rightarrow \text { Differentiating (i) respect to } \mathrm{t}, \text { we get }\\ &\frac{d A}{d t}=8 \pi r \frac{d r}{d t} \end{aligned}$
$\rightarrow 8 \pi \times \text { current radius } \times \text { rate of change of radius }$

Derivative as a Rate Measurer exercise multiple choise question 26

Answer:
$\begin{aligned} &\text { A. } \frac{d h}{d t}=1 m / h r \end{aligned}$
Hint:
The volume of a cylinder, of radius r and height h is defined as
$\begin{aligned} &V(r, h)=\pi r^{2} h\\ \end{aligned}$
Given:
$\begin{aligned} &r=10 \mathrm{~m} \text { we get }\\ &V(h)=\pi \times 10^{2} \times h=100 \pi h \quad \quad \quad .....(i)\\ &\frac{d v}{d t}=314 m^{3} / h r \end{aligned}$
Solution:
→ Differentiating (i) respect to t, we get
$\begin{aligned} &\frac{d v}{d t}=100 \pi \frac{d h}{d t}\\ &\rightarrow \text { Substituting values and using } \quad \pi =\frac{3}{4} \text { we get } \end{aligned}$
$\begin{aligned} &314=100 \times 3.14 \frac{d h}{d t} \\ &\frac{d h}{d t}=1 \mathrm{~m} / \mathrm{hr} \end{aligned}$

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RD Sharma Class 12th MCQs For Chapter 12 – Derivative as a rate of measure, is a special set of solutions that will clear all your doubts about the 12th chapter of the NCERT book. RD Sharma Solutions are helpful as reference materials as it helps to improve grades and prepares students well for their Mathematics board tests in school.

Rd Sharma class 12 chapter 12 MCQs answers which are simple but detailed which means it will provide assistance to all students. RD Sharma Class 12 Solutions for MCQs has 26 questions. The concept like finding the rate of increase and decrease of the circumference or perimeter for the given shape is present in this chapter.

Benefits of availing the RD Sharma Mathematics Solutions

Free downloadable solutions pdfs which can be found on Career360.
These solutions are easy to follow and understand.
It covers concepts from all chapters and exercises.
Covers all questions from the NCERT exercises.
Helps students to revise at the last minute.
The biggest advantage is that, the RD Sharma Class 12 Solutions Chapter 12 MCQs is available for download from the Career 360 website. Having a copy of it the mobile phones or laptops helps the student access it easily even wen offline.

As Mathematics is a difficult subject for Class 12 students, these solutions will help students get rid of their fears and anxieties.