Question

A chain of length $l$ and mass $m$ lies on the surface of a smooth sphere of radius $R>l$ with one end tied to the top of the sphere.
Column I Column II

i) Gravitational potential energy w.r.t centre of the sphere	a.$\frac{Rg}{l}[1-\cos \left(\frac{l}{R}\right)]$
ii) The chain is released and slides down, its KE when it has slid by $\theta$	b. $\frac{2Rg}{l}[\sin \left(\frac{l}{R}\right)+\sin \theta -\sin (\theta +\frac{l}{R})]$
iii) The initial tangential acceleration	c. $\frac{M{R}^{2}g}{l}\sin \left(\frac{l}{R}\right)$
iv) The radial acceleration	d. $\frac{M{R}^{2}g}{l}[\sin \left(\frac{l}{R}\right)+\sin \theta -\sin (\theta +\frac{l}{R})]$

• A. i $\to$ c, ii $\to$ d, iii $\to$ a, iv $\to$ b
• B. i $\to$ a, ii $\to$ d, iii $\to$ c, iv $\to$ b
• C. i $\to$ c, ii $\to$ b, iii $\to$ a, iv $\to$ c
• D. i $\to$ b, ii $\to$ c, iii $\to$ d, iv $\to$ a

Answer 1

The correct option is A i $\to$ c, ii $\to$ d, iii $\to$ a, iv $\to$ b


Gravitational P.E. w.r.t the center of the sphere
$$={\int }_0^{l/R}\frac{M}{l}g{R}^2\cos \theta d\theta =\frac{M}{l}g{R}^2\sin \left(\frac{l}{R}\right)$$When the chain has been released and it has slid through angle $\theta$, from mechanical energy theorem, we have

$u_i-u_f=k_f-k_i\Rightarrow u_i-u_f=k_f$


$\frac{M}{l}g{R}^2\sin \left(\frac{l}{R}\right)-{\int }_{\theta }^{\theta +\frac{l}{R}}\frac{M}{l}g{R}^2\cos \theta d\theta =\frac{1}{2}m{v}^2$

If $v$ is the velocity at the moment when angle traversed is $\theta$, then
$$K.E=\frac{m{R}^{2}g}{l}[\sin \left(\frac{l}{R}\right)+\sin \theta -\sin (\theta +\frac{l}{R})]$$From the expression of $K.E,v$ can be obtained and differentiated in order to obtain tangential acceleration which will be $(\theta =0^{\circ })$
$a_t=\frac{Rg}{l}[1-\cos \left(\frac{l}{R}\right)][\text{use}v=R\frac{d\theta }{dt}]$
$\text{[Differentiate w.r.t, time where}\frac{d\theta }{dt}=\omega ]$
Radial acceleration $=\frac{v^2}{R}$
Hence, we have
$$a_r=\frac{2Rg}{l}[\sin \left(\frac{l}{R}\right)+\sin \theta -\sin (\theta +\frac{l}{R})]$$