RD Sharma Solutions Class 12 Mathematics Chapter 13 MCQ

# RD Sharma Solutions Class 12 Mathematics Chapter 13 MCQ

Edited By Kuldeep Maurya | Updated on Jan 27, 2022 10:01 AM IST

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Also Read - RD Sharma Solutions For Class 9 to 12 Maths

## Differentials, Errors and Approximations Excercise:MCQ

Differentials Errors and Approximations exercise multiple choice question 1

Answer: A) $1%$
Hint: Here, we all know the formula of period of a pendulum is
$T=2 \pi \sqrt{\frac{L}{g}}$
Given:
Given, $\left(\frac{\Delta L}{L}\right) \times 100=2$
[If we let the length of pendulum is $L$]
Solution:
Given $\left(\frac{\Delta L}{L}\right) \times 100=2$......equation 1 (if we let the length of pendulum is L)
we all know the formula of period of a pendulum is $\mathrm{T}=2 \pi \times \mathrm{V}(\mathrm{l} / \mathrm{g})$
By the formula of approximation in derivation, we get
Time period,
$T=2 \pi \sqrt{\frac{L}{g}}$
Taking Log on both sides, we get
$\log T=\log 2 \pi+\frac{1}{2} \log \mathrm{L}-\frac{1}{2} \log g$
Differentiating both sides w.r.t x, we get
So,
$\begin{gathered} \frac{1}{T} \frac{d T}{d L}=\frac{1}{2 L} \\\\ \frac{d T}{d L}=\frac{T}{2 L} \end{gathered}$
$\left(\frac{\Delta T}{T}\right) \times 100=\frac{1}{2} \times\left(\frac{\Delta L}{L}\right) \times 100$
$\left(\frac{\Delta T}{T}\right) \times 100=\frac{1}{2} \times 2$ (from equation 1 $\left(\frac{\Delta L}{L}\right) \times 100=2$ )
\begin{aligned} &\left(\frac{\Delta T}{T}\right)=\frac{1}{100} \\\\ &\left(\frac{\Delta T}{T}\right)=1 \% \end{aligned}

Differentials Errors and Approximations exercise multiple choice question 2

Answer: A) $2a%$
Hint: Here we all know the formula about surface area$(A)$ of cube
$A=6 a^{2}$
Given: Percentage error in measuring side$=a%$
Solution:
Here $x$ be the side of cube\begin{aligned} &\frac{\Delta x}{x} \times 100=a \\\\ &\text { So } \quad \Delta x=\frac{a x}{100} \end{aligned}

and $y$ be the surface area

\begin{aligned} &\text { So, }\\ &\begin{aligned} &y=6 x^{2} \\\\ &\frac{d y}{d x}=\Delta y=12 x \times \Delta x \end{aligned}\\\\ &=12 x \times \frac{a x}{100} \end{aligned}

So,$%$ error in surface area \begin{aligned} &=\frac{\Delta y}{\mathrm{y}}=\frac{\frac{12 a}{100} \times x^{2}}{6 x^{2}} \end{aligned}

$=2 a \%$

So, option (A) $2a%$ is correct.

Differentials Errors and Approximations exercise multiple choice question 3

Answer: $3k%$
Hint: Here, we all know the formula of volume of sphere
$V=\frac{4}{3} \pi r^{3}$
Given: Percentage error in measuring radius$=k%$
Solution:
Here, volume of sphere is
$V=\frac{4}{3} \pi r^{3}$
Let’s differentiate,
$\frac{d V}{d r}=4 \pi r^{2}$
Given percentage error in measuring radius$=k%$
So,
\begin{aligned} &\frac{\Delta r}{r}=\frac{k}{100} \\\\ &\Delta r=\frac{r k}{100} \end{aligned}
Now, approximate error in measuring
\begin{aligned} V &=d V=\left(\frac{d V}{d r}\right) \Delta r \\\\ &=\frac{k}{100} 4 \pi r^{3} \\\\ &=\frac{3 k}{100} V \\\\ &=3 k \% \text { of } V \end{aligned}
So, percentage error in measuring $V=3 k \%$

Differentials Errors and Approximations exercise multiple choice question 4 maths

Answer: $\text { (c) } 3 \alpha \%$

Hint: Here, we all know the formula of volume of cylinder
$V=\pi r^{2} h$
Given:
Percentage error in measuring height$=3 \alpha \%$
Solution:
Here, volume of cylinder, $V=\pi r^{2} h$
But here , $h=r$
So, \begin{aligned} &V=\pi r^{2} \times r \\\\ & \end{aligned}
$V=\pi r^{3}$
Let’s differentiate
$\frac{d V}{d h}=3 \pi h^{2}$
Given percentage error in measuring height$=3 \alpha \%$
So, $\Delta h=\frac{\alpha h}{100}$
Now, approximate error in measuring
$\begin{gathered} V=d V=\left(\frac{d V}{d h}\right) \Delta h \\\\ =\frac{3 \alpha}{100} \pi h^{3} \\\\ =3 \alpha \% \text { of } V \end{gathered}$
So, percentage error in measuring, $3 \alpha \%$

Differentials Errors and Approximations exercise multiple choice question 5

Answer : B) $2k%$
Hint: Here, we know the formula of area of equilateral triangle,
$A=\frac{\sqrt{3}}{4} x^{2}$
Given:
Percentage error in measuring side$=k%$
Solution:
Area of equilateral triangle, $A=\frac{\sqrt{3}}{4} x^{2}$
Let’s differentiate,
$\frac{d A}{d x}=\frac{\sqrt{3}}{2} x$
Here, given percentage error in measuring side$=k%$
So,
\begin{aligned} &\frac{\Delta x}{x}=\frac{k}{100} \\\\ &\Delta x=\frac{k x}{100} \end{aligned}
Now, approximate error in measuring
\begin{aligned} A &=d A=\left(\frac{d A}{d x}\right) \Delta x \\\\ &=\frac{k}{100} \times \frac{\sqrt{3}}{2} x^{2} \end{aligned}
\begin{aligned} &=\frac{2 k}{100} \times \frac{\sqrt{3}}{4} x^{2} \\\\ &=2 k \% \text { of } A \end{aligned}
So, percentage error in measuring $A=2 k \%$

Differentials Errors and Approximations exercise multiple choice question 6

Answer: (C) $1.3893$
Hint: We all know that
$\Delta y=d y=\frac{d y}{d x} \times d x$
Given:
$\log _{e} 4=1.3868$
Solution:
Consider the function,
$y=f(x)=\log_{e} x$
Let $x=4$
So,
\begin{aligned} &x+\Delta x=4.01 \\\\ &\Delta x=0.01 \text { for } x=4 \\\\ &y=\log _{e} 4=1.3868 \\\\ &y=\log _{e} x \end{aligned}
Let’s differentiate,
\begin{aligned} &\frac{d y}{d x}=\frac{1}{x} \\\\ &\left(\frac{d y}{d x}\right)_{x=4}=\frac{1}{4} \end{aligned}
\begin{aligned} \Delta y &=d y=\frac{d y}{d x} \times d x \\\\ &=\frac{1}{4} \times 0.01 \\\\ &=0.0025 \end{aligned}
\begin{aligned} \log _{e} 4.01 &=y+\Delta y \\\\ &=1.3868+0.0025 \\\\ &=1.3893 \end{aligned}

Differentials Errors and Approximations exercise multiple choice question 8

Answer: $\text { (c) } 3 \lambda \%$
Hint: Here, we all know the formula about volume of cone

$V=\frac{1}{3} \pi r^{2} h$

Given:
$\frac{r}{h}=\frac{1}{2}$
So, volume of cone$V=\frac{1}{3} \pi r^{2} h$
Solution:
So,
\begin{aligned} &V=\frac{1}{3} \pi r^{2}(2 r) \\\\ &V=\frac{2}{3} \pi r^{3} \end{aligned}
Let’s differentiate,
\begin{aligned} &V=\frac{2}{3} \pi r^{3} \\\\ &\frac{d V}{d r}=2 \pi r^{2} \end{aligned}
Percentage error in measuring$=\lambda \%$
\begin{aligned} &\frac{\Delta r}{r}=\frac{\lambda}{100} \\\\ &\Delta r=\lambda r(100) \end{aligned}
Approximate error in $V=dV$
\begin{aligned} &=\frac{d V}{d r} \times \Delta r \\\\ &=\frac{\lambda}{100}\left(2 \pi r^{3}\right) \end{aligned}
\begin{aligned} &=\frac{3 \lambda}{100}\left(\frac{2}{3} \pi r^{3}\right) \\\\ &=3 \lambda \% \text { of } V \end{aligned}
So, percentage error in $V=3\lambda \%$

Differentials Errors and Approximations exercise multiple choice question 9

Answer: $\frac{1}{8}%$
Hint: Here, we use the concept of $P=d P=\frac{d P}{d V} \Delta V$
Given:
$P V^{\frac{1}{4}}=$ Constant
Percentage error in $V=\frac{-1}{2} \%$
Solution:
Let $k$ be a constant
\begin{aligned} &\quad P V^{\frac{1}{4}}=k \\\\ &\text { So, } \quad P=\frac{k}{V^{\frac{1}{4}}} \end{aligned}
Let’s differentiate
$\frac{d P}{d V}=\frac{-k}{4} \times V^{\frac{-5}{4}}$
Here, percentage error in $V=\frac{-1}{2} \%$
\begin{aligned} &\frac{\Delta V}{V}=\frac{-1}{200} \\\\ &\Delta V=\frac{-V}{200} \end{aligned}
Approximate change in $P$
$P=d P=\left(\frac{d P}{d V}\right) \Delta V$
$=\frac{1}{800} k V^{\frac{-1}{4}}$ [From equation (i)]
$=\frac{1}{8} \% \text { of } P$
Percentage increase in $V=\frac{1}{8} \%$

Differentials Errors and Approximations exercise multiple choice question 10 maths

Answer: $n:1$
Hint: Here we use the concept of relative error
$\frac{d y}{y}=\left(\frac{d y}{d x}\right) \Delta x$
Given: $y=x^{n}$
Solution:
$y=x^{n}$
Let’s differentiate
$\frac{d y}{d x}=n x^{n-1}$
Approximate error in $y$ is
\begin{aligned} d y &=\left(\frac{d y}{d x}\right) \Delta x \\\\ &=n x^{n-1} \times \Delta x \end{aligned}
Relative error in $y$ is $\frac{d y}{y}=\frac{n}{x} \Delta x$

Approximate error $x$ is $dx$
\begin{aligned} &=\left(\frac{d x}{d y}\right) \Delta y \\\\ &=\frac{1}{n x^{n-1}} \Delta y \end{aligned}
Relative error in $x$ is $\frac{d x}{x}=\frac{1}{n x^{n}} \Delta y$
Required solution,
$\frac{\frac{n}{x} \times \Delta x}{\frac{1}{n x^{n}} \times \Delta y}=n^{2} x^{n-1} \frac{\Delta x}{\Delta y}$
\begin{aligned} &=\frac{x}{y} \times \frac{n \times x^{n-1} \times \Delta x}{\Delta x} \\\\ &=\frac{n \times x^{n}}{x^{n}} \\\\ &=\frac{n}{1} \end{aligned}
So, ratio is $n:1$

Differentials Errors and Approximations exercise multiple choice question 12

Answer: $\frac{1}{14}%$
Hint: Here, we use this below formula circumference,$C=2\pi r$
Given:Circumference circle with error is $0.01\; cm$
Solution:
Circumference,$C=2\pi r$
\begin{aligned} &r=\frac{14}{\pi} \text { as } \mathrm{C}=28 \mathrm{~cm} \\\\ &\text { Also } \quad \frac{d C}{d r}=2 \pi \end{aligned}
Area circle,
$A=\pi r^{2}$
\begin{aligned} &\text { So, } &A=\frac{(14)^{2}}{\pi} \end{aligned}
$\text { Also } \quad \frac{d A}{d r}=2 \pi r$
$\frac{d A}{d C}=\frac{\frac{d A}{d r}}{\frac{d C}{d r}}=r=\frac{14}{\pi}$
Approximate error in $A$ is
\begin{aligned} d A &=\frac{d A}{d C} \times \Delta C \\\\ &=\frac{14}{\pi} \times \frac{1}{100} \\\\ &=\frac{1}{1400} \text { of } A \end{aligned}
So, percentage of error in $A=\frac{1}{14} \%$

Differentials Errors and Approximations exercise multiple choice question 13

Answer: $(a) 0.32$
Hint: Here we use the basic concept of $x=x+\Delta x$
Given: $y=x^{4}-10$
Solution:
Let$x=1.99$
So
\begin{aligned} &x+\Delta x=2 \\ &\Delta x=2-1.99=0.01 \end{aligned}
Let
$\begin{gathered} d x=\Delta x=0.01 \\ y=x^{4}-10 \end{gathered}$
Let’s differentiate,
\begin{aligned} &y=x^{4}-10 \\\\ &\frac{d y}{d x}=4 x^{3} \end{aligned}
$\left(\frac{d y}{d x}\right)_{x=2}=4(2)^{3}=32$
$\therefore \quad \frac{d y}{d x} \times d x=d y$
So, $d y=32(0.01)=0.32(\text { approx })$
So, approximate change in $y$ $=0.32$

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## RD Sharma Chapter wise Solutions

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1. What is the 13ty Chapter of the Class 12 maths book?

The 13th chapter of the class 12 maths book is titled Differentials, Errors and Approximations.

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