In the determination of Young's modulus (Y= 4 M L g π l d^ 2 ) by using Searle's method, a wire of length L=2 ~m and diameter d=0.5 ~mm is used. For a load M=2.5 ~kg , an extension l=0.25 ~mm in the length of the wire

Correct answer: A. due to the errors in the measurements of d and l are the same. The maximum possible error in Y due to l and d (Δ Y)/(Y)=(Δ l)/(l)+(2 Δ d)/(d) Least count = Pitch No. of division on circular scale =(0.5)/(100) ~mm=0.005 ~mm Here, Δ d=Δ l=0.005 ~mm Error contribution of l=(Δ l)/(l)= 0.005 ~mm 0.25 ~mm =(1)/(50) Error contribution of d=(2 Δ d)/(d)= 2 × 0.005 ~mm 0.5 ~mm =(1)/(50) Hence contribution to the maximum possible error in the measurement of y due to l and d is the same.

JEE Advanced 2012Paper 1

Exam: JEE Advanced
Year: 2012
Paper: Paper 1
Subject: Physics
Chapter: Mathematics in Physics

JEE Advanced 2012 · Paper 1 · Physics · Mathematics in Physics

In the determination of Young's modulus (Y= 4 M L g π l d^ 2 ) by...

In the determination of Young's modulus

\left(Y=\frac{4 M L g}{\pi l d^{2}}\right)

by using Searle's method, a wire of length

L=2 \mathrm{~m}

and diameter

d=0.5 \mathrm{~mm}

is used. For a load

M=2.5 \mathrm{~kg}

, an extension

l=0.25 \mathrm{~mm}

in the length of the wire is observed. Quantities

d and l

are measured using a screw gauge and a micrometer, respectively. They have the same pitch of

0.5 \mathrm{~mm}

. The number of divisions on their circular scale is 100 . The contributions to the maximum probable error of the

Y

measurement

Options

Adue to the errors in the measurements of $d$ and $l$ are the same.
Bdue to the error in the measurement of $d$ is twice that due to the error in the measurement of $l$ .
Cdue to the error in the measurement of $l$ is twice that due to the error in the measurement of $d$ .
Ddue to the error in the measurement of $d$ is four times that due to the error in the measurement of $l$ .

Correct answer

A. due to the errors in the measurements of $d$ and $l$ are the same.

Step-by-step solution

The maximum possible error in

Y

due to

l

and

d \frac{\Delta Y}{Y}=\frac{\Delta l}{l}+\frac{2 \Delta d}{d}

Least count

=\frac{\text{ Pitch }}{\text{ No. of division on circular scale }} =\frac{0.5}{100} \mathrm{~mm}=0.005 \mathrm{~mm}

Here,

\Delta d=\Delta l=0.005 \mathrm{~mm}

Error contribution of

l=\frac{\Delta l}{l}=\frac{0.005 \mathrm{~mm}}{0.25 \mathrm{~mm}}=\frac{1}{50}

Error contribution of

d=\frac{2 \Delta d}{d}=\frac{2 \times 0.005 \mathrm{~mm}}{0.5 \mathrm{~mm}}=\frac{1}{50}

Hence contribution to the maximum possible error in the measurement of

y

due to

l

and

d

is the same.

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