Work Power And Energy Ques 7
- A uniform cable of mass $M$ and length $L$ is placed on a horizontal surface such that its $\frac{1}{n}$ th part is hanging below the edge of the surface. To lift the hanging part of the cable upto the surface, the work done should be
(2019 Main, 9 April I)
(a) $\frac{2 M g L}{n^{2}}$
(b) $n M g L$
(c) $\frac{M g L}{n^{2}}$
(d) $\frac{M g L}{2 n^{2}}$
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Answer:
Correct Answer: 7.(d)
Solution:
Formula:
Modified Form of Work-Energy Theorem:
- Given, mass of the cable is $M$.
So, mass of $\frac{1}{n}$ th part of the cable, i.e. hanged part of the cable is
$$ =M / n \cdots(i) $$
Now, centre of mass of the hanged part will be its middle point. So, its distance from the top of the table will be $L / 2 n$.
$\therefore$ Initial potential energy of the hanged part of cable,
$$ \begin{aligned} U _i & =\frac{M}{n}(-g) \frac{L}{2 n} \\ \Rightarrow \quad U _i & =-\frac{M g L}{2 n^{2}} \cdots(ii) \end{aligned} $$
When whole cable is on the table,
its potential energy will be zero.
$$ \therefore \quad U _f=0 \cdots(iii) $$
Now, using work-energy theorem,
$$ \begin{aligned} & W _{\text {net }}=\Delta U=U _f-U _i \\ & \Rightarrow \quad W _{\text {net }}=0–\frac{M g L}{2 n^{2}} \quad [using \quad Eqs. (ii) and (iii)]\\ & \Rightarrow \quad W _{\text {net }}=\frac{M g L}{2 n^{2}} \end{aligned} $$
BETA
