Properties Of Matter Ques 47
- A pendulum made of a uniform wire of cross-sectional area $A$ has time period $T$. When an additional mass $M$ is added to its bob, the time period changes to $T _M$. If the Young’s modulus of the material of the wire is $Y$, then $\frac{1}{Y}$ is equal to $(g$ = gravitational acceleration $)$
(2015 Main)
(a) $\left[1-\left(\frac{T}{T _M}\right)^{2}\right] \frac{A}{M g}$
(b) $\left[\left(\frac{T _M}{T}\right)^{2}-1\right] \frac{M g}{A}$
(c) $\left[1-\left(\frac{T _M}{T}\right)^{2}\right] \frac{A}{M g}$
(d) $\left[\left(\frac{T _M}{T}\right)^{2}-1\right] \frac{A}{M g}$
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Answer:
Correct Answer: 47.(d)
Solution:
$$ \begin{aligned} T & =2 \pi \sqrt{\frac{L}{g}} \\ T _M & =2 \pi \sqrt{\frac{L+\Delta L}{g}} \end{aligned} $$
Here, $\quad \Delta L=\frac{F L}{A Y}=\frac{M g L}{A Y}$
$$ \therefore \quad T _M=2 \pi \sqrt{\frac{L+\frac{M g L}{A Y}}{g}} $$
Solving Eqs. (i) and (ii), we get
$$ \begin{aligned} & \frac{1}{Y}=\frac{A}{M g}\left[\left(\frac{T _M}{T}\right)^{2}-1\right] \\ & \Delta l=\frac{F L}{A Y}=\frac{F L}{\left(\pi r^{2}\right) Y} \Rightarrow \Delta l \propto \frac{L}{r^{2}} \\ & \therefore \quad \frac{\Delta l _1}{\Delta l _2}=\frac{L / R^{2}}{2 L /(2 R)^{2}}=2 \end{aligned} $$
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