IIT-JEE 2012 · Paper 1 · Physics Q4

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PhysicsSingle correct+3 / -1Question 4 of 60

In the determination of Young's modulus $\left(Y=\dfrac{4MLg}{\pi ld^2}\right)$ 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 is observed. Quantities $d$ and $l$ are measured using a screw gauge and a micrometer, respectively. They have the same pitch of 0.5 mm. The number of divisions on their circular scale is 100. The contributions to the maximum probable error of the $Y$ measurement

due to the errors in the measurements of

d

and

l

are the same.

due to the error in the measurement of

d

is twice that due to the error in the measurement of

l

due to the error in the measurement of

l

is twice that due to the error in the measurement of

d

due to the error in the measurement of

d

is four times that due to the error in the measurement of

l

Answer and worked solution

Official answer: A

Least counts. Both instruments have pitch 0.5 mm and 100 divisions, so $\Delta d=\Delta l=0.005$ mm.

Contributions to $\Delta Y/Y$ . From $Y\propto\dfrac1{ld^2}$ :

from $d$ : $2\dfrac{\Delta d}d=2\times\dfrac{0.005}{0.5}=0.02$ ;
from $l$ : $\dfrac{\Delta l}l=\dfrac{0.005}{0.25}=0.02$ .

They are equal.

(A)

Topics:Young's modulus by Searle's method Errors in measurement Vernier callipers and screw gauge

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IIT-JEE 2012 · Paper 1, Physics question 4