Question
In the determination of Young’s modulus $$\left( {Y = \frac{{4MLg}}{{\pi l{d^2}}}} \right)$$ 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-
A.
due to the errors in the measurements of $$d$$ and $$l$$ are the same.
B.
due to the error in the measurement of d is twice that due to the error in the measurement of $$l.$$
C.
due to the error in the measurement of $$l$$ is twice that due to the error in the measurement of $$d.$$
D.
due to the error in the measurement of $$d$$ is four times that due to the error in the measurement of $$l.$$
Answer :
due to the errors in the measurements of $$d$$ and $$l$$ are the same.
Solution :
The maximum possible error in $$Y$$ due to $$l$$ and $$d$$ are $$\frac{{\Delta Y}}{Y} = \frac{{\Delta l}}{l} + \frac{{2\Delta d}}{d}$$
$$\eqalign{
& {\text{Least count}} = \frac{{{\text{Pitch}}}}{{{\text{Number of division on circular scale}}}} \cr
& = \frac{{0.5}}{{100}}\,mm = 0.005\,mm \cr
& {\text{Error contribution of }}l \cr
& = \frac{{\Delta l}}{l} = \frac{{0.005\,mm}}{{0.25\,mm}} = \frac{1}{{50}} \cr
& {\text{Error contribution of }}d \cr
& = \frac{{2\Delta d}}{d} = \frac{{2 \times 0.005\,mm}}{{0.5\,mm}} = \frac{1}{{50}} \cr} $$