Question

A large mass M hangs stationary at the end of a light string that passes through a smooth fixed ring to a small mass m that moves around in a horizontal circular path. If l is the length of the string from m to the top end of the tube and $\theta$ is angle between this part and vertical part of the string as shown in the figure, then time taken by m to complete one circle is equal to?

• A. $2\pi \sqrt{\frac{l}{g\sin \theta }}$
• B. $2\pi \sqrt{\frac{l}{g\cos \theta }}$
• C. $2\pi \sqrt{\frac{ml}{gM\sin \theta }}$
• D. $2\pi \sqrt{\frac{ml}{Mg}}$

Answer 1

The correct option is D $2\pi \sqrt{\frac{ml}{Mg}}$

The body of mass m is revolving in a horizontal circle.

Let the radius of its circular path be r.

r will be perpendicular to the string attached to M.

Now,
$Sin\theta =\frac{r}{l}⟹\left(1\right)$

Since M is fixed its weight will be equal to the tension,T in the string
Therefore, $T=Mg⟹\left(2\right)$

Also,
$Centripetalforce=TSin\theta$

$m\omega ²r=\frac{Mg\times r}{l}$

$\omega ²=\frac{Mg}{ml}$

$\omega =\sqrt{\frac{Mg}{ml}}$

Time taken by m to complete one revolution,

$T=\frac{2\pi }{\omega }$

$T=2\pi \times \sqrt{\frac{ml}{Mg}}$