General Physics Ques 37
Question 16
- A steel wire of diameter $0.5 \mathrm{~mm}$ and Young’s modulus $2 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}$ carries a load of mass $m$. The length of the wire with the load is $1.0 \mathrm{~m}$. 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 $1.0 \mathrm{~mm}$, 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 $1.2 \mathrm{~kg}$, the vernier scale division which coincides with a main scale division is . (Take, $g=10 \mathrm{~ms}^{-2}$ and $\pi=3.2$ ).
(2018 Adv.)
Integer Answer Type Question
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Answer:
Correct Answer: 37.(3)
Solution:
Formula:
- Given, $d=0.5 \mathrm{~mm}$,
$$ \begin{aligned} Y & =2 \times 10^{11} \mathrm{Nm}^{-2} \ l & =1 \mathrm{~m} \ \Delta l & =\frac{F l}{A Y}=\frac{m g l}{\frac{\pi d^{2}}{4} Y}=\frac{1.2 \times 10 \times 1}{\frac{\pi}{4} \times\left(5 \times 10^{-4}\right)^{2} \times 2 \times 10^{11}} \ & =0.3 \mathrm{~mm} \end{aligned} $$
$\mathrm{LC}$ of vernier $=1-\frac{9}{10} \mathrm{~mm}=0.1 \mathrm{~mm}$
So, 3rd division of vernier scale will coincide with main scale.
BETA
