Question

A ring of mass ‘M’ and radius ‘R’ slipped into cone kept on a horizontal surface. The cone is accelerated towards right with acceleration ‘a’. There is no friction between ring and the cone. Tension in ring at point A is-

• A. $\frac { \mathrm { Ma } } { 2 \pi }$
• B. $\frac { M g \cot \theta } { 2 \pi }$
• C. $\frac { M ( a + g \cot \theta ) } { 2 \pi }$
• D. $\frac { \mathrm { M } ( \mathrm { a } + \mathrm { g } \tan \theta ) } { 2 \pi }$

Answer 1

Answer: (C)

$\frac { M ( a + g \cot \theta ) } { 2 \pi }$

Answer 2

Tension in string = F.R

where, F = Net force in radially outward direction (excluding tension) per unit length

\ F = (l g cot q + la)R

(l = mass/length)

= $\frac { M } { 2 \pi } ( g \cot \theta + a )$