A student determined Young's Modulus of elasticity using the formula Y = MgL^3 \over 4bd^3\delta . The value of g is taken to be 9.8 m/s 2 , without any significant error, his observation are as following. .tg border-collapse:collapse;border-spacing:0;width:100% .tg tdborder-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal; .tg thborder-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal; .tg .tg-baqhtext-align:center;vertical-align:top .tg .tg-c3owborder-color:inherit;text-align:center;vertical-align:top .tg .tg-7bttborder-color:inherit;font-weight:bold;text-align:center;vertical-align:top Physical Quantity Least count of the Equipment used for measurement Observed value Mass (M) 1 g 2 kg Length of bar (L) 1 mm 1 m Breadth of bar (b) 0.1 mm 4 cm Thickness of bar (d) 0.01 mm 0.4 cm Depression ( \delta ) 0.01 mm 5 mm Then the fractional error in the measurement of Y is :

Units and Measurements PYQ — JEE Main 2021 · 1 Sep · Shift 2 · Q54 with Solution

A student determined Young's Modulus of elasticity using the formula

Y = {{Mg{L^3}} \over {4b{d^3}\delta }}

. The value of g is taken to be 9.8 m/s², without any significant error, his observation are as following.

Physical Quantity	Least count of the Equipment used for measurement	Observed value
Mass (M)	1 g	2 kg
Length of bar (L)	1 mm	1 m
Breadth of bar (b)	0.1 mm	4 cm
Thickness of bar (d)	0.01 mm	0.4 cm
Depression ( $\delta$ )	0.01 mm	5 mm

Then the fractional error in the measurement of Y is :

A
0.0083
B
0.0155
C
0.155
D
0.083

View written solutionFree

Correct answer: B

Given formula

$Y=\frac{MgL^3}{4bd^3\delta}$

Since $g$ is taken without significant error and $4$ is a constant, they do not contribute to fractional error.

So,

$\frac{\Delta Y}{Y}=\frac{\Delta M}{M}+3\frac{\Delta L}{L}+\frac{\Delta b}{b}+3\frac{\Delta d}{d}+\frac{\Delta \delta}{\delta}$

Compute fractional errors of each measured quantity

(i) Mass $M$

Observed value:

$M=2\,\text{kg}=2000\,\text{g}$

Least count $=1\,\text{g}$

$\frac{\Delta M}{M}=\frac{1}{2000}=0.0005$

(ii) Length $L$

Observed value:

$L=1\,\text{m}=1000\,\text{mm}$

Least count $=1\,\text{mm}$

$\frac{\Delta L}{L}=\frac{1}{1000}=0.001$

Thus,

$3\frac{\Delta L}{L}=3(0.001)=0.003$

(iii) Breadth $b$

Observed value:

$b=4\,\text{cm}=40\,\text{mm}$

Least count $=0.1\,\text{mm}$

$\frac{\Delta b}{b}=\frac{0.1}{40}=0.0025$

(iv) Thickness $d$

Observed value:

$d=0.4\,\text{cm}=4\,\text{mm}$

Least count $=0.01,\text{mm}$$

$\frac{\Delta d}{d}=\frac{0.01}{4}=0.0025$

Thus,

$3\frac{\Delta d}{d}=3(0.0025)=0.0075$

(v) Depression $\delta$

Observed value:

$\delta=5\,\text{mm}$

Least count $=0.01\,\text{mm}$

$\frac{\Delta \delta}{\delta}=\frac{0.01}{5}=0.002$

Add all fractional errors

$\frac{\Delta Y}{Y}=0.0005+0.003+0.0025+0.0075+0.002$

$\frac{\Delta Y}{Y}=0.0155$

Final answer

The fractional error in $Y$ is

$\boxed{0.0155}$

So, the correct option is B.

More from Units and Measurements

A student determined Young's Modulus of elasticity using the formula

Y = {{Mg{L^3}} \over {4b{d^3}\delta }}

. The value of g is taken to be 9.8 m/s², without any significant error, his observation are as following.

Physical Quantity	Least count of the Equipment used for measurement	Observed value
Mass (M)	1 g	2 kg
Length of bar (L)	1 mm	1 m
Breadth of bar (b)	0.1 mm	4 cm
Thickness of bar (d)	0.01 mm	0.4 cm
Depression ( $\delta$ )	0.01 mm	5 mm

Then the fractional error in the measurement of Y is :

A
0.0083
B
0.0155
C
0.155
D
0.083

View written solutionFree

Correct answer: B

Given formula

$Y=\frac{MgL^3}{4bd^3\delta}$

Since $g$ is taken without significant error and $4$ is a constant, they do not contribute to fractional error.

So,

$\frac{\Delta Y}{Y}=\frac{\Delta M}{M}+3\frac{\Delta L}{L}+\frac{\Delta b}{b}+3\frac{\Delta d}{d}+\frac{\Delta \delta}{\delta}$

Compute fractional errors of each measured quantity

(i) Mass $M$

Observed value:

$M=2\,\text{kg}=2000\,\text{g}$

Least count $=1\,\text{g}$

$\frac{\Delta M}{M}=\frac{1}{2000}=0.0005$

(ii) Length $L$

Observed value:

$L=1\,\text{m}=1000\,\text{mm}$

Least count $=1\,\text{mm}$

$\frac{\Delta L}{L}=\frac{1}{1000}=0.001$

Thus,

$3\frac{\Delta L}{L}=3(0.001)=0.003$

(iii) Breadth $b$

Observed value:

$b=4\,\text{cm}=40\,\text{mm}$

Least count $=0.1\,\text{mm}$

$\frac{\Delta b}{b}=\frac{0.1}{40}=0.0025$

(iv) Thickness $d$

Observed value:

$d=0.4\,\text{cm}=4\,\text{mm}$

Least count $=0.01,\text{mm}$$

$\frac{\Delta d}{d}=\frac{0.01}{4}=0.0025$

Thus,

$3\frac{\Delta d}{d}=3(0.0025)=0.0075$

(v) Depression $\delta$

Observed value:

$\delta=5\,\text{mm}$

Least count $=0.01\,\text{mm}$

$\frac{\Delta \delta}{\delta}=\frac{0.01}{5}=0.002$

Add all fractional errors

$\frac{\Delta Y}{Y}=0.0005+0.003+0.0025+0.0075+0.002$

$\frac{\Delta Y}{Y}=0.0155$

Final answer

The fractional error in $Y$ is

$\boxed{0.0155}$

So, the correct option is B.

Units and Measurements question

Units and Measurements question

(i) Mass $M$

(ii) Length $L$

(iii) Breadth $b$

(iv) Thickness $d$

(v) Depression $\delta$

More from Units and Measurements

Units and Measurements question

Units and Measurements question

(i) Mass $M$

(ii) Length $L$

(iii) Breadth $b$

(iv) Thickness $d$

(v) Depression $\delta$

More from Units and Measurements

(i) Mass MMM

(ii) Length LLL

(iii) Breadth bbb

(iv) Thickness ddd

(v) Depression δ\deltaδ

More from Units and Measurements

(i) Mass MMM

(ii) Length LLL

(iii) Breadth bbb

(iv) Thickness ddd

(v) Depression δ\deltaδ

More from Units and Measurements

(i) Mass $M$

(ii) Length $L$

(iii) Breadth $b$

(iv) Thickness $d$

(v) Depression $\delta$

(i) Mass $M$

(ii) Length $L$

(iii) Breadth $b$

(iv) Thickness $d$

(v) Depression $\delta$