Question

A ring of mass $M$, radius $R$, cross-section area $‘a^{\prime }$ and Young's modulus $Y$ is kept on a smooth cone of base radius $2R$ and semi vertex angle ${45}^{\circ }$, as shown in figure. Assume that the extension in the ring is small. Pick the correct alternative $\left(s\right)$.

• A. The tension in the ring will be same throughout.
• B. The tension in the ring will be independent of its radius.
• C. The elongation in the ring is $\frac{MgR}{aY}$
• D. The elastic potential energy stored in the ring will be $\frac{M^2{g}^{2}R}{8\pi Ya}$

Answer 1

Answer: (A), (B), (C)

The elongation in the ring is $\frac{MgR}{aY}$

When some force is applied on ring, then it is displaced by angle $d\theta$ and the elongation in ring will be $dy$. If the tension in ring is $T$ then.
$T.d\theta =\frac{Mg}{2\pi R}.Rd\theta$
$T=\frac{Mg}{2\pi }$
$\because M,g$ are constant therefore $T$ is same throughout the ring. It is independent of radius $R$.

Young's Modulus
$Y=\frac{\text{stress}}{\text{strain}}=-\frac{F/A}{\frac{\Delta L}{L}}$
$\therefore Y=\frac{\frac{T}{a}}{\frac{\Delta y}{l}}$

$\Delta y=\frac{Tl}{aY}=\frac{Mg\left(2\pi R\right)}{2\pi aY}$
$\Delta y=\frac{MgR}{aY}$
Now elastic energy
$U=\frac{1}{2}\times \text{stress}\times \text{strain}\times \text{Volume}$
$U=\frac{1}{2}\times \frac{T}{a}\times \frac{\Delta y}{l}2\pi Ra$
$U=\frac{1}{2}\times \frac{Mg}{2a\pi }\times \frac{MgR}{aY\left(2\pi R\right)}2\pi Ra$
$\Rightarrow U=\frac{M^2{g}^{2}R}{4\pi aY}$