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The 18th division of the Vernier scale exactly coincides with one of the main scale divisions During Searle's experiment, zero of the Vernier scale lies between $2.20 \times 10^{-2} \mathrm{~m}$ and $2.25 \times 10^{-2} \mathrm{~m}$ of the main scale. When an additional load of $4 \mathrm{~kg}$ is applied to the wire, the zero of the Vernier scale still lies between $3.20 \times 10^{-2} \mathrm{~m}$ and $3.25 \times 10^{-2} \mathrm{~m}$ of the main scale, but now the $55 \mathrm{th}$ division of the Vernier scale coincides with one of the main scale divisions. The length of the thin metallic wire is $1.5 \mathrm{~m}$ and its cross-sectional area is $6 \times 10^{-7} \mathrm{~m}^{2}$ . The least count of the Vernier scale is $1.0 \times 10^{-5} \mathrm{~m}$ . The maximum percentage error in Young's modulus of the wire is:
Option: 1
$0.25 \%$

Option: 2
$0.5 \%$

Option: 3
$0.9 \%$

Option: 4
$1.0 \%$

Answers (1)

best_answer

Young's modulus (Y) is given by the formula $Y=\frac{F L}{A}$ , where F is the force applied, L is the original length, and A is the cross-sectional area.

The relative change in Young's modulus can be related to the relative change in length:

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

Given that the Vernier scale has a least count of $1.0 \times 10^{-5} \mathrm{~m}$ , and the change in divisions from 20 th to 45 th is $55-18=37$ divisions, which corresponds to a change in length of $37 \times 1.0 \times 10^{-5} \mathrm{~m}$ .

Therefore, the relative change in Young's modulus:

$\frac{\Delta Y}{Y}=\frac{37 \times 1.0 \times 10^{-5} \mathrm{~m}}{1.5 \mathrm{~m}}=24.67 \times 10^{-5}$

To express this as a percentage:

$\frac{\Delta Y}{Y} \times 100=24.67 \times 10^{-5} \times 100=.25 \%$

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