M.C.Q (1 Marks)

MCQ 11 Mark

While measuring the side of an equilateral triangle an error of k % is made, the percentage error in its area is:

A
k%
✓
2k%
C
$\frac{\text{K}}{2}\%$
D
3k%

Answer

Correct option: B.

2k%

Area of equilateral triangle is,
$\text{A}=\frac{\sqrt{3}}{4}\text{a}^2$
Given that $\frac{\text{da}}{\text{a}}\times100=\text{k}$
and
$\frac{\text{dA}}{\text{da}}=\frac{\sqrt{3}}{2}\text{a}$
$\Rightarrow\frac{\triangle\text{A}}{\text{a}}=\frac{\frac{\sqrt{3}}{2}\text{da}}{\frac{\sqrt{3}}{4}\text{a}^2}$
$\Rightarrow\frac{\triangle\text{A}}{\text{A}}=\frac{2}{\text{a}}\times\frac{\text{Ka}}{100}=2\text{k}$
The error in the area of the triangle is 2K%

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MCQ 21 Mark

If the ratio of base radius and height of a cone is 1 : 2 and percentage error in radius is $\lambda\%,$ then the error in its volume is:

A
$\lambda\%$
B
$2\lambda\%$
✓
$3\lambda\%$
D
$\text{None of these}$

Answer

Correct option: C.

$3\lambda\%$

Let x be the radius of the cone and V be the volume.
Given that $\frac{\triangle\text{r}}{\text{r}}\times100=\lambda$
$\text{V}=\frac{1}{3}\pi\text{r}^2\text{h}$
$\Rightarrow\text{V}=\frac{2}{3}\pi\text{r}^3$ $\Big(\because\frac{\text{r}}{\text{h}}=\frac{1}{2}\Rightarrow\text{h}=2\text{r}\Big)$
$\Rightarrow\frac{\text{dV}}{\text{dr}}=2\pi\text{r}^2$
$\Rightarrow\frac{\text{dV}}{\text{V}}=\frac{2\pi\text{r}^2}{\frac{2}{3}\pi\text{r}^3}\times\frac{\lambda\text{r}}{100}$
$\Rightarrow\frac{\text{dv}}{\text{V}}=3\lambda\%$

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MCQ 31 Mark

The circumference of a circle is measured as $28\ cm$ with an error of $0.01\ cm$. The percentage error in the area is:

✓
$\frac{1}{14}$
B
$0.01$
C
$\frac{1}{7}$
D
$\text{None of these}$

Answer

Correct option: A.

$\frac{1}{14}$

Let $x$ be the radius of the circle and $y$ be its circumference.
$x = 28\ cm$
$\triangle\text{x}=0.01\text{cm}$
$\text{x}=2\pi\text{r}$
$\text{y}=\pi\text{r}^2=\pi\times\frac{\text{x}^2}{4\pi^2}=\frac{\text{x}^2}{4\pi}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{2\pi}$
$\Rightarrow\frac{\triangle\text{y}}{\text{y}}=\frac{\text{x}}{2\pi\text{y}}\text{dx}=\frac{2}{\text{x}}\times0.01$
$\Rightarrow\frac{\triangle\text{y}}{\text{y}}\times100=\frac{2}{\text{x}}=\frac{1}{14}$
Hence, the percentage error in the area is $\frac{1}{14}$

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MCQ 41 Mark

The function $\text{f}(\text{x})=\log_\text{e}\Big(\text{x}^3+\sqrt{\text{x}^6+1}\Big)$ is of the following type:

A
Even and increasing.
✓
Odd and increasing.
C
Even and decreasing.
D
Odd and decreasing.

Answer

Correct option: B.

Odd and increasing.

$\text{f}(\text{x})=\log_\text{e}\Big(\text{x}^3+\sqrt{\text{x}^6+1}\Big)$
$\Rightarrow\text{f}(-\text{x})=\log_\text{e}\Big(-\text{x}^3+\sqrt{\text{x}^6+1}\Big)$
$=\log_\text{e}\Bigg\{\frac{\big(-\text{x}^3+\sqrt{\text{x}^6+1}\big)\big(\text{x}^3+\sqrt{\text{x}^6+1}\big)}{\text{x}^3+\sqrt{\text{x}^6+1}}\Bigg\}$
$=\log_\text{e}\Big(\frac{\text{x}^6+1-\text{x}^6}{\text{x}^3+\sqrt{\text{x}^6+1}}\Big)$
$=\log_\text{e}\Big(\frac{1}{\text{x}^3+\sqrt{\text{x}^6+1}}\Big)$
$=-\log_\text{e}\Big(\text{x}^3+\sqrt{\text{x}^6+1}\Big)$
$=-\text{f}(\text{x})$
Hence, f(-x) = -f(x)
Therefore, it is an odd function.
$\text{f}(\text{x})=\log_\text{e}\Big(\text{x}^3+\sqrt{\text{x}^6+1}\Big)$
$\frac{\text{d}}{\text{dx}}\{\text{f}(\text{x})\}=\Big(\frac{1}{\text{x}^3+\sqrt{\text{x}^6+1}}\Big)\times\Big(3\text{x}^2+\frac{1}{2\sqrt{\text{x}^6+1}}\times6\text{x}^5\Big)$
$=\Big(\frac{1}{\text{x}^3+\sqrt{\text{x}^6+1}}\Big)\times\bigg(\frac{6\text{x}^2\sqrt{\text{x}^6+1}+6\text{x}^5}{2\sqrt{\text{x}^6+1}}\bigg)$
$=\Big(\frac{1}{\text{x}^3+\sqrt{\text{x}^6+1}}\Big)\times\Bigg\{\frac{6\text{x}^2\big(\sqrt{\text{x}^6+1+\text{x}^3}\big)}{2\sqrt{\text{x}^6+1}}\Bigg\}$
$=\Big(\frac{6\text{x}^2}{2\sqrt{\text{x}^6+1}}\Big)>0$
Therefore the given function is an increasing function.

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MCQ 51 Mark

If an error of k% is made in measuring the radius of a sphere, then percentage error in its volume is:

A
k%
✓
3k%
C
2k%
D
$\frac{\text{k}}{3}\%$

Answer

Correct option: B.

3k%

Let, x be the radius of the sphere and y be its volume
Then,
$\frac{\triangle\text{x}}{\text{x}}\times100=\text{k}$
Also, $\text{y}=\frac{4}{3}\pi\text{x}^3$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=4\pi\text{x}^2$
$\Rightarrow\frac{\triangle\text{y}}{\text{y}}=\frac{4\pi\text{x}^2}{\text{y}}\text{dx}=\frac{4\pi\text{x}^2}{\frac{4}{3}\pi\text{x}^3}\times\frac{\text{kx}}{100}$
$\Rightarrow\frac{\triangle\text{y}}{\text{y}}\times100=3\text{k}$
Hence, the error in the volume is 3K%

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MCQ 61 Mark

If $y = x^n,$ then the ratio of relative errors in $y$ and $x$ is :

A
$1 : 1$
B
$2 : 1$
C
$1 : n$
✓
$n : 1$

Answer

Correct option: D.

$n : 1$

Let $\frac{\triangle\text{x}}{\text{x}}$ be the relative error in $x$ and $\frac{\triangle\text{y}}{\text{y}}$ be the error in $y.$
Now, $\text{y}=\text{x}^\text{n}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{n x}^{\text{n}-1}$
$\Rightarrow\frac{\triangle\text{y}}{\text{y}}=\frac{\text{n x}^{\text{n}-1}}{\text{y}}\text{dx}$
$\Rightarrow\frac{\triangle\text{y}}{\text{y}}=\frac{\text{n x}^{\text{n}-1}}{\text{y}}\text{dx}=\text{n}\frac{\triangle\text{x}}{\text{x}}$
$\Rightarrow\frac{\triangle\text{y}}{\text{y}}:\frac{\triangle\text{x}}{\text{x}}=\text{n}:1$

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MCQ 71 Mark

If there is an error of a% in measuring the edge of a cube, then percentage error in its surface is:

✓
2a%
B
$\frac{\text{a}}{2}\%$
C
3a%
D
None of these

Answer

Correct option: A.

2a%

Let, x be the sides of a cube and S be the surface area.
Given that $\frac{\triangle\text{S}}{\text{S}}\times100=\text{a}$
$\text{s}=6\text{x}^2$
$\frac{\text{ds}}{\text{dx}}=12\text{x}$
$\frac{\triangle\text{S}}{\text{S}}=\frac{12\text{x}}{\text{S}}\times\text{dx}$
$\frac{\triangle\text{S}}{\text{S}}=\frac{12\text{x}}{6\text{x}^2}\times\frac{\text{ax}}{100}$
$\Rightarrow\frac{\triangle\text{S}}{\text{S}}\times100=2\text{a}$
The error in the surface area is 2a%

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MCQ 81 Mark

The height of a cylinder is equal to the radius. If an error of $\alpha\%$ is made in the height, then percentage error in its volume is:

A
$\alpha\%$
B
$2\alpha\%$
✓
$3\alpha\%$
D
$\text{None of these}$

Answer

Correct option: C.

$3\alpha\%$

Let x be the radius, which is equal to the height of the cylinder.
the cylinder is $3\alpha\%$
$\frac{\triangle\text{x}}{\text{x}}\times100=\alpha$
Also, $\text{y}=\pi\text{x}^2\text{x}=\pi\text{x}^3$ [Radius = Height of the cylinder]
$\Rightarrow\frac{\text{dy}}{\text{dx}}=3\pi\text{x}^2$
$\Rightarrow\frac{\triangle\text{Y}}{\text{y}}=\frac{3\pi\text{x}^2}{\text{y}}\text{dx}=\frac{3}{\text{x}}\times\frac{\text{ax}}{100}$
$\Rightarrow\frac{\triangle\text{y}}{\text{y}}\times100=3\alpha$
Hence, the error in the volume of the cylinder is $3\alpha\%$

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MCQ 91 Mark

If there is an error of 2% in measuring the length of a simple pendulum, then percentage error in its period is:

✓
1%
B
2%
C
3%
D
4%

Answer

Correct option: A.

Also, Let $\triangle\text{l}$ be the error in the length and $\triangle\text{T}$ be the error in the period.
We have
$\frac{\triangle\text{l}}{\text{l}}\times100=2$
$\Rightarrow\frac{\text{dl}}{\text{l}}\times100=2$
Now, $\text{T}=2\pi\sqrt{\frac{\text{l}}{\text{g}}}$
Taking log on both sides, we get
$\log\text{T}=\log2\pi+\frac{1}{2}\log\text{l}-\frac{1}{2}\log\text{ g}$
Differentiating both sides w.r.t.x, we get
$\frac{1}{\text{T}}\frac{\text{dT}}{\text{dl}}=\frac{1}{2\text{l}}$
$\Rightarrow\frac{\text{dT}}{\text{dl}}=\frac{\text{T}}{2\text{l}}$
$\Rightarrow\frac{\text{dl}}{\text{l}}\times100=2\frac{\text{dT}}{\text{T}}\times100$
$\Rightarrow\frac{\triangle\text{T}}{\text{T}}\times100=1$
Hence, there is an error of 1% in calculating the period of the pendulam.

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MCQ 101 Mark

If $\log _e 4=1.3868,$ then $\log _e 4.01=$

A
$1.3968$
B
$1.3898$
✓
$1.3893$
D
None of these

Answer

Correct option: C.

$1.3893$

Consider the function $\text{y}=\text{f(x)}=\log_\text{e}\text{x}.$
Let : $x = 4$
$\text{x}+\triangle\text{x}=4.01$
$\Rightarrow\triangle\text{x}=0.01$
For $x = 4,$
$\text{y}\log4=1.3868$
$\text{y}=\log_\text{e}\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}$
$\Rightarrow\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=4}=\frac{1}{4}$
$\Rightarrow\triangle\text{y}=\text{dy}=\frac{\text{dy}}{\text{dx}}\text{dx}=\frac{1}{4}\times0.01=0.0025$
$\therefore\log_\text{e}4.01=\text{y}+\triangle\text{y}=1.3893$

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MCQ 111 Mark

The pressure P and volume V of a gas are connected by the relation $\text{PV}^{\frac{1}{4}}=$ constant. The percentage increase in the pressure corresponding to a deminition of $\frac{1}{2}\%$ in the volume is:

A
$\frac{1}{2}\%$
B
$\frac{1}{4}\%$
✓
$\frac{1}{8}\%$
D
$\text{None of these}$

Answer

Correct option: C.

$\frac{1}{8}\%$

We have
$\frac{\triangle\text{V}}{\text{V}}=\frac{-1}{2}\%$
$\text{PV}^{\frac{1}{4}}=\text{constant = k}(\text{say})$
Taking log on both sides, we get
$\log\Big(\text{PV}^{\frac{1}{4}}\Big)=\log\text{k}$
$\Rightarrow\log\text{P}+\frac{1}{4}\log\text{V}=\log\text{k}$
Differentiating both sides w.r.t. x we get
$\frac{1}{\text{p}}\frac{\text{dp}}{\text{dv}}+\frac{1}{4\text{V}}=0$
$\Rightarrow\frac{\text{dp}}{\text{P}}=-\frac{\text{dV}}{4\text{V}}=-\frac{1}{4}\times\frac{-1}{2}=\frac{1}{8}$
Hence, the increase in the pressure is $\frac{1}{8}\%$

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MCQ 121 Mark

A sphere of radius 100mm shrinks troximate decreo radius 98mm, then the appase in its volume is:

A
$12000\pi\ \text{mm}^3$
B
$800\pi\ \text{mm}^3$
✓
$80000\pi\ \text{mm}^3$
D
$1200\pi\ \text{mm}^3$

Answer

Correct option: C.

$80000\pi\ \text{mm}^3$

Let x be the radius of the sphere and y be its volume.
x = 100,
$\text{x}+\triangle\text{x}=98$
$\Rightarrow\triangle\text{x}=-2$
$\text{y}=\frac{4}{3}\pi\ \text{x}^3$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=4\pi\ \text{x}^2$
$\Rightarrow\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=100}=40000\pi$
$\therefore\triangle\text{y}=\text{dy}=\frac{\text{dy}}{\text{dx}}\text{dx}=40000\pi\times(-2)=-80000\pi$
Hence, the decrease in the volume of the sphere is $80000\pi\ \text{mm}^3$

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MCQ 131 Mark

The approximate value of $(33)^{\frac{1}{5}}$ is:

✓
2.0125
B
2.1
C
2.01
D
None of these

Answer

Correct option: A.

2.0125

To find $(33)^{\frac{1}{5}}$ we consider $\text{f(x)}=\text{x}^{\frac{1}{5}}$
Here, x = 32,
$\triangle\text{x}=1$
$\text{f}(32)=32^{\frac{1}{5}}=2$
$\Rightarrow\text{f}'(\text{x})=\frac{1}{5}\times\frac{-4}{5}$
$\Rightarrow\text{f}'(32)=\frac{1}{5}(32)^{\frac{-4}{5}}=\frac{1}{80}$
$\triangle\text{y}=\frac{\text{dy}}{\text{dx}}\text{dx}=0.0125$
$\Rightarrow\text{f}(33)=2+0.0123=2.0125$

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Maths M.C.Q (1 Marks)

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