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JEE Advanced 2018 Paper 2

Section: Physics

Question : 11 of 54

Marks: +1, -0

0.5\text{mm}

2\times10^{11}\,\text{N/m}^2

1.0\text{m}

1.0\text{mm}

1.2\text{kg}

g=10\,\text{m/s}^2

\pi=3.2

Solution:

d=0.5\text{mm}\;\;\;\;Y=2\times10^{11}\;\;\;l=1\text{m}

\Delta l = \frac{Fl}{Ay} = \frac{\frac{mgl}{\pi d^2}}{4y} = \frac{1.2 \times 10 \times 1}{\frac{\pi}{4} \times (5 \times 10^{-4})^2 \times 2 \times 10^{11}} \Delta l = \frac{1.2 \times 10}{\frac{3.2}{4} \times 25 \times 10^{-8} \times 2 \times 10^{11}} =

\frac{12}{0.8\times25\times2\times10^{3}} = \frac{12}{40\times10^{3}} = 0.3\text{mm}

So 3rd division of a vernier scale will coincicle with main scale.

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