Question

A chain of length l is placed on a smooth spherical surface of radius r with one of its ends fixed at the top of the surface. Length of chain is assumed to be l<πr/2. Acceleration of each element of chain when upper end is released is

• A. $\frac { \lg } { \mathrm { r } } \left( 1 - \cos \frac { \mathrm { r } } { 1 } \right)$
• B. $\frac { \mathrm { rg } } { 1 } \left( 1 - \cos \frac { 1 } { \mathrm { r } } \right)$
• C. $\frac { \lg } { \mathrm { r } } \left( 1 - \sin \frac { 1 } { \mathrm { r } } \right)$
• D. $\frac { \mathrm { rg } } { 1 } \left( 1 - \sin \frac { 1 } { \mathrm { r } } \right)$

Answer 1

Answer: (B)

$\frac { \mathrm { rg } } { 1 } \left( 1 - \cos \frac { 1 } { \mathrm { r } } \right)$

Answer 2

Let mass per unit length of chain = $\frac { \mathrm { m } } { \mathrm { l } }$

Consider an element of chain of length dl subtending an angle of dθ at the centre of spherical surface.

Mass of element dm = $\frac { \mathrm { m } } { \mathrm { l } } \mathrm { dl } = \frac { \mathrm { m } } { \mathrm { l } } \mathrm { rd } \theta$

Force acting on element, dF = dmg sin θ

= $\frac { \mathrm { m } } { \mathrm { l } }$ rg sin θ dθ

∴ F = $\frac { \mathrm { m } } { 1 } \operatorname { rg } \int _ { 0 } ^ { \mathrm { a } } \sin \theta \mathrm { d } \theta$

= $\frac { \mathrm { m } } { 1 } \operatorname { rg } ( 1 - \cos \alpha )$

= $\frac { m } { 1 } \operatorname { rg } \left( 1 - \cos \frac { 1 } { r } \right)$

Then a = $\frac { \mathrm { F } } { \mathrm { m } } = \frac { \mathrm { rg } } { 1 } \left( 1 - \cos \frac { 1 } { \mathrm { r } } \right)$