Physics · Mechanical Properties of Matter

JEE Advanced 2018 — Paper 2 — Question 10

A steel wire of diameter 0.5 mm and Young's modulus 2×1011Nm−22 \times 10^{11} \mathrm{Nm}^{-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 ms−2\mathrm{g}=10 \mathrm{~ms}^{-2} and is π=3.2\pi=3.2.

Answer: 3

Numerical answer — enter this value.

Step-by-step solution

Δℓ=FℓAY\Delta \ell=\frac{\mathrm{F} \ell}{\mathrm{AY}}

=4 Fℓπ d2Y=\frac{4 \mathrm{~F} \ell}{\pi \mathrm{~d}^{2} \mathrm{Y}}

=4×12×1π×25×10−8×2×1011=0.3 mm=\frac{4 \times 12 \times 1}{\pi \times 25 \times 10^{-8} \times 2 \times 10^{11}}=0.3 \mathrm{~mm}

10VSD=9MSD10 \mathrm{VSD}=9 \mathrm{MSD}

1VSD=910MSD1 \mathrm{VSD}=\frac{9}{10} \mathrm{MSD}

∴\therefore \quad Least count, L.C. =1MSD−1VSD=1 \mathrm{MSD}-1 \mathrm{VSD}

=(1−910)MSD=\left(1-\frac{9}{10}\right) \mathrm{MSD}

=110MSD=0.1 mm=\frac{1}{10} \mathrm{MSD}=0.1 \mathrm{~mm}.

∴3rd \therefore 3^{\text {rd }} vernier scale division coincides with a main scale division.

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2018
Paper
Paper 2
Subject
Physics
Chapter
Mechanical Properties of Matter
Topic
Stress,Strain and Modulus of Elasticity
A steel wire of diameter 0.5 mm and Young's modulus 2 × 10 11 Nm -2… | JEE Advanced 2018 PYQ with Solution · DhiX AI