Question

The drawing shows two cylinders. They are identical in all respects, except that B is hollow. In a setup like that in the figure, identical forces are applied to the right end of each cylinder while the left end is fixed.

• A. The elongation of A is more than that of B.
• B. The elongation of B is more than that of A.
• C. The energy stored in B is more than that in A.
• D. The energy stored in A is more than that in B.

Answer 1

Answer: (B), (C)

The energy stored in B is more than that in A.

As we know that,
$Y=\frac{\frac{F}{A}}{\frac{\Delta l}{l}}$
$\Rightarrow \frac{F}{A}=Y\frac{\Delta l}{l}$
$\Rightarrow \Delta l=\left(\frac{Fl}{Y}\right)\frac{1}{A}$
$\Rightarrow \Delta l\propto \frac{1}{A}....\left(1\right)$
[since $F,l$ & $Y$ are the same for both]

Cross sectional area of A is greater than that of B because B is a hollow cylinder.

$\therefore$ From equation (1), we can say that elongation of B is more than that of A.

$\because$ Potential energy
$U=\frac{1}{2}\times Y\times {\left(\frac{\Delta l}{l}\right)}^2=\frac{1}{2}\times Y\times {\left(\frac{F}{Y}\right)}^2\times \frac{1}{A^2}$
Again $U\propto \frac{1}{A^2}$
$\therefore$ Stored energy in B is more than in A.

$\therefore$ Options (b) and (c) are correct.