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QJEEAdv18Ph61a4: JEE Advanced 2018 Physics - Class 11 JEEAdv18Ph61a4 Solution: A steel wire of diameter and Young's modulus carries a load of…

A steel wire of diameter 0.50.5 mmmm and Young's modulus 2×1011  Nm−22 \times {10^{11}}\,\,N{m^{ - 2}} carries a load of mass M.M. The length of the wire with the load is 1.01.0 m.Am.A vernier scale with 1010 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 1.01.0 mmmm , is attached. The 1010 divisions of the vernier scale correspond to 99 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 1.21.2 kg,kg, the vernier scale division which coincides with a main scale division is _____________. Take g=10 m s−2.g = 10\,m\,{s^{ - 2}}. and π=3.2.\pi = 3.2.

Step-by-Step Verified Solution

We know that ΔL=W(YA/L)\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 = =πr2=π4d2=π4×(0.5×10−3)2= \pi {r^2} = {\pi \over 4}{d^2} = {\pi \over 4} \times {(0.5 \times {10^{ - 3}})^2}

Therefore,

ΔL=1.2×102×1011×π4(0.5×10−3)2×11.0 m\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}}}}

=1.2×10×42×1011×π×(0.5)2×10−6= {{1.2 \times 10 \times 4} \over {2 \times {{10}^{11}} \times \pi \times {{(0.5)}^2} \times {{10}^{ - 6}}}}

⇒ΔL=0.3×10−3\Rightarrow \Delta L = 0.3 \times {10^{ - 3}} m = 0.3 mm

Now, least count of vernier scale =(1−910)= \left( {1 - {9 \over {10}}} \right) mm = 0.1 mm

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

Vernier reading =0.3 mm0.1 mm=3= {{0.3\,mm} \over {0.1\,mm}} = 3

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

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