Question

A wire of length 1 m fixed at one end has a sphere attached to it at the other end. The sphere is projected horizontally with a velocity of $\sqrt{9g}$. When it describes a vertical circle, the ratio of elongations of the wire when the sphere is at the top and bottom of the circle is :

• A. $2:5$
• B. $5:2$
• C. $3:5$
• D. $5:3$

Answer 1

The correct option is A $2:5$

When sphere is at top point, it's kinetic energy will be kinetic energy at bottom - raise in potential energy
So, $K.E_{top}=K.E{.}_{bottom}-P.E.$

$=\frac{1}{2}mv{{}_b}^2-mgh$

$=\frac{1}{2}m9g-mg\left(2r\right)$

$=\frac{9}{2}mg-2mg$ $(\because r=1)$

$=\frac{5}{2}mg$

$K.E.=\frac{1}{2}mv{{}_t}^2$

$\frac{5}{2}mg=\frac{1}{2}mv{{}_t}^2\Rightarrow v_t=\sqrt{5g}$

So, tension in top position is: $m\frac{v{{}_t}^2}{r}-mg=5mg-mg=4mg$
And tension in bottom position is: $m\frac{v{{}_b}^2}{r}+mg=9mg+mg=10mg$

So, $Y=\frac{F/A}{\Delta l/l}$
$\Rightarrow \Delta l=\frac{Fl}{AY}$

So, $\Delta l\propto F$
$\therefore \frac{\Delta l_t}{\Delta l_b}=\frac{F_t}{F_b}=\frac{4mg}{10mg}=\frac{2}{5}$