Question

During Searle’s experiment, zero of Vernier scale lies between $3.20 \times {10^{ - 2}}\;{\text{m}}$ and $3.25 \times {10^{ - 2}}\;{\text{m}}$of the main scale. The 20th division of Vernier scale exactly coincides with one of the main scale divisions. When an additional load of 2 kg is applied to the wire, the zero of the vernier scale still lies between $3.20 \times {10^{ - 2}}\;{\text{m}}$ and $3.25 \times {10^{ - 2}}\;{\text{m}}$of the main scale but now the 45th division of Vernier scale coincides with one of the main scale divisions. The length of the thin metallic wire is 2 m and its cross-sectional area is $8 \times {10^7}\;{{\text{m}}^2}$. The least count of the Vernier scale is $1 \times {10^{ - 5}}\;{\text{m}}$. The maximum percentage error in the Young’s modulus of the wire is:
A. 8%
B. 3%
C. 7%
D. 4%

Answer 1

The least count is the minimum measurable quantity by any instrument. It is also the same as the ratio of one part of the main scale to the number of parts on the vernier scale.
The percentage error is evaluated by taking the proportion of the mean value of the absolute errors to mean value of the quantity.

Complete step-by-step solution :
Given:
The number of divisions of the vernier scale that coincide with the main scale without any load is, ${n_1} = 20$.
The number of divisions of the vernier scale coinciding with the main scale with any load is, ${n_2} = 45$.
The equation to determine the length of the wire without any load is,
${L_1} = M + {n_1}l$
Here, $M$ is the main scale reading and $l$ is the least count.
The equation to determine the length of the wire with any load is,
${L_2} = M + {n_2}l$
The equation to determine the change in the length of the wire is,
$\Delta L = {L_2} - {L_1}$
Substitute all the values in the above equation.
$\Delta L = \left( {M + {n_2}l} \right) - \left( {M + {n_1}l} \right)$
$= \left( {{n_2} - {n_1}} \right)l$
$= \left( {45 - 20} \right)l$
$= 25l$
The equation to determine the Young’s modulus is,
$Y = \dfrac{{FL}}{{lA}}$

The equation to calculate the maximum percentage error in the Young’s modulus of the wire is,
$\dfrac{{\Delta Y}}{Y} = \left( {\dfrac{l}{{\Delta L}}} \right) \times 100$
Substitute all the values in the above equation.
$\dfrac{{\Delta Y}}{Y} = \left( {\dfrac{l}{{25l}}} \right) \times 100$
$= 4\%$
Therefore, the maximum percentage error in the Young’s modulus of the wire is 4% and the option (D) is correct.

Note:- Take the reading carefully from the vernier scale because the percentage error depends on the number of divisions on the vernier scale.