Question

$P\to \text{Solid sphere},Q\to \text{Solid cube},R\to \text{Solid cone},S\to \text{Solid cylinder}$

Assuming potential energy ${}^{\prime }{U}^{\prime }$ at the ground level to be zero.
All objects are made of the same material.
$U_P=\text{Potential energy of solid sphere}$
$U_Q=\text{Potential energy of solid cube}$
$U_R=\text{Potential energy of solid cone}$
$U_S=\text{Potential energy of solid cylinder}$
Identify the correct option(s).

• A. $U_S>U_P$
• B. $U_Q>U_S$
• C. $U_P>U_Q$
• D. $U_P>U_S$

Answer 1

Answer: (A), (B)

The correct options are
A $U_S>U_P$
B $U_Q>U_S$

Since all are made of same material, density $\left(\rho \right)$ is same for all. Potential energy of all bodies is given by,
$U=Mg{Y}_{\text{CM}}$
where $Y_{\text{CM}}=$ height of centre of mass

For sphere $P$ :
$M_P=\rho \times V_P$
$M_P=\rho \times \frac{4}{3}\pi \times {\left(\frac{D}{2}\right)}^3$
$Y_{\text{CM}}=\frac{D}{2}$
$\Rightarrow U_P=\rho \times \frac{4}{3}\pi \times {\left(\frac{D}{2}\right)}^3\times g\times \frac{D}{2}$
$\therefore U_P=\frac{\pi }{12}\rho g{D}^4$

Similarly for cube $Q$:
$M_Q=\rho \times V_Q=\rho \times D^3$
$Y_{\text{CM}}=\frac{D}{2}$
$\Rightarrow U_Q=\rho \times D^3\times g\times \frac{D}{2}$
$\therefore U_Q=\frac{1}{2}\rho g{D}^4$

For cone $R$:
$M_R=\rho \times \frac{1}{3}\pi \times {\left(\frac{D}{2}\right)}^2\times D$
$Y_{\text{CM}}=\frac{D}{4}$
$\Rightarrow U_R=\rho \times \frac{1}{3}\pi \times {\left(\frac{D}{2}\right)}^2\times D\times g\times \frac{D}{4}$
$\therefore U_R=\frac{\pi }{48}\rho g{D}^4$

For cylinder $S$:
$M_S=\rho \times \pi \times {\left(\frac{D}{2}\right)}^2\times D$
$Y_{\text{CM}}=\frac{D}{2}$
$\Rightarrow U_S=\rho \times \pi \times {\left(\frac{D}{2}\right)}^2\times D\times g\times \frac{D}{2}$
$\therefore U_S=\frac{\pi }{8}\rho g{D}^4$

Comparing the potential energy of all bodies,
$\therefore U_S>U_P\&U_Q>U_S$