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← JEE Advanced 2018 · Paper 2
PhysicsNumerical+3Question 11 of 54

JEE Advanced 2018 · Paper 2, Physics question 11

A steel wire of diameter 0.5 mm and Young’s modulus 2×1011 N m−22 \times 10^{11}\ \mathrm{N\,m^{-2}} carries a load of mass MM. 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−2g = 10\ \mathrm{m\,s^{-2}} and π=3.2\pi = 3.2.

Answer and worked solution

Official answer: 3

Extension.
ΔL=FLAY=12×1.0π(0.25×10−3)2×2×1011=123.2×6.25×10−8×2×1011=124×104=0.3 mm.\Delta L=\frac{FL}{AY}=\frac{12\times1.0}{\pi(0.25\times10^{-3})^2\times2\times10^{11}}=\frac{12}{3.2\times6.25\times10^{-8}\times2\times10^{11}}=\frac{12}{4\times10^4}=0.3\text{ mm}.

Vernier. 10 VSD = 9 MSD, so the least count is 1−0.9=0.11-0.9=0.1 mm. A shift of 0.3 mm makes the 3rd vernier division coincide with a main-scale division.

Answer: 3