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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RD Sharma Solutions Class 12 Mathematics Chapter 13 MCQ

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RD Sharma Solutions Class 12 Mathematics Chapter 13 MCQ

RD Sharma Solutions Class 12 Mathematics Chapter 13 MCQ

Updated on 27 Jan 2022, 10:01 AM IST

RD Sharma class 12th exercise MCQ is one of the best NCERT solutions available in the market today. Hundreds of students trust RD Sharma solutions because they know how beneficial the answers can be for exam preparations. Students in class 12 especially are requested by teachers to use the RD Sharma class 12 chapter 13 exercise MCQ for home practice. Maths is a tough subject, and students may need a little help to master it. Self-practice and testing at home are pretty important in maths, and the class 12 RD Sharma chapter 13 exercise MCQ solution will definitely be a top choice for a study guide.

This Story also Contains

RD Sharma Class 12 Solutions Chapter 13MCQ Differentials, Errors and Approximations - Other Exercise
Differentials, Errors and Approximations Excercise:MCQ
RD Sharma Chapter wise Solutions

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

RD Sharma Class 12 Solutions Chapter 13MCQ Differentials, Errors and Approximations - Other Exercise

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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