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JEE Advance - Physics (2018 - Paper 2 Offline - No. 10)

A steel wire of diameter $$0.5$$ $$mm$$ and Young's modulus $$2 \times {10^{11}}\,\,N{m^{ - 2}}$$ carries a load of mass $$M.$$ The length of the wire with the load is $$1.0$$ $$m.A$$ vernier scale with $$10$$ divisions is attached to the end of this wire. Next to the steel wire is a reference wire to which a main scale, of least count $$1.0$$ $$mm$$ , is attached. The $$10$$ divisions of the vernier scale correspond to $$9$$ divisions of the main scale. Initially, the zero of vernier scale coincides with the zero of main scale. If the load on the steel wire is increased by $$1.2$$ $$kg,$$ the vernier scale division which coincides with a main scale division is _____________. Take $$g = 10\,m\,{s^{ - 2}}.$$ and $$\pi = 3.2.$$
Answer
3

Explanation

We know that $$\Delta L = {W \over {(YA/L)}}$$

where W is weight or load = mg = 1.2 $$\times$$ 10 = 12 kg m s$$-$$2, Y is Young's modulus = 2 $$\times$$ 1011 N m$$-$$2, L is length of wire with load = 1.0 m, A is area of steel wire = $$ = \pi {r^2} = {\pi \over 4}{d^2} = {\pi \over 4} \times {(0.5 \times {10^{ - 3}})^2}$$

Therefore,

$$\Delta L = {{1.2 \times 10} \over {2 \times {{10}^{11}} \times {\pi \over 4}{{(0.5 \times {{10}^{ - 3}})}^2} \times {1 \over {1.0\,m}}}}$$

$$ = {{1.2 \times 10 \times 4} \over {2 \times {{10}^{11}} \times \pi \times {{(0.5)}^2} \times {{10}^{ - 6}}}}$$

$$ \Rightarrow \Delta L = 0.3 \times {10^{ - 3}}$$ m = 0.3 mm

Now, least count of vernier scale $$ = \left( {1 - {9 \over {10}}} \right)$$ mm = 0.1 mm

Therefore, Vernier reading $$ = {{\Delta L} \over {least\,count}}$$

Vernier reading $$ = {{0.3\,mm} \over {0.1\,mm}} = 3$$

Therefore, 3rd vernier scale division coincides with the main scale division.

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