UPSC 2017 Maths Optional Paper 1 Q8a — Step-by-Step Solution

16 marks · Section B

Work and Potential Energy; Conservation · Dynamics & Statics · Read the full method →

Question

A spherical shot of $W$ gm weight and radius $r$ cm, lies at the bottom of cylindrical bucket of radius $R$ cm. The bucket is filled with water up to a depth of $h$ cm $(h>2r)$ . Show that the minimum amount of work done in lifting the shot just clear of the water must be $\left[W\left(h-\dfrac{4r^3}{3R^2}\right)+W'\left(r-h+\dfrac{2r^3}{3R^2}\right)\right]$ cm gm. $W'$ gm is the weight of water displaced by the shot.

Technique

Minimum work = increase in PE of (shot + water); compute new water depth, shot rise, and water PE change via centroids.

Solution

Let $A=\pi R^2$ be the cross-sectional area, $V_s=\tfrac43\pi r^3$ the shot’s volume, and $\rho g=\dfrac{W'}{V_s}$ the weight per unit volume of water (so $W'=\rho g V_s$ ). The minimum work equals the increase in total potential energy (shot + water) when the shot is raised quasi-statically until it is just clear of the surface.

Step 1 — Final water depth

The amount of water is fixed. Initially the shot is fully submerged at the bottom, so

V_{\text{water}}=A h-V_s.

When the shot is lifted clear, the water occupies a plain cylinder of depth $h'$ with $A h'=V_{\text{water}}$ :

h'=h-\frac{V_s}{A}=h-\frac{\tfrac43\pi r^3}{\pi R^2}=h-\frac{4r^3}{3R^2}.

Step 2 — Rise of the shot (its PE change)

Initially the shot’s centre is at height $r$ . Finally the shot is just clear: its lowest point sits at the new surface $h'$ , so its centre is at $h'+r$ . The rise of the centre of mass is

\Delta_{\text{shot}}=(h'+r)-r=h'=h-\frac{4r^3}{3R^2}.

Hence the change in the shot’s PE is

W\,\Delta_{\text{shot}}=W\Big(h-\frac{4r^3}{3R^2}\Big).

Step 3 — Change in PE of the water

PE of a body of water $=\rho g\displaystyle\int z\,dV$ .

Initially: water = (cylinder $0\le z\le h$ ) minus (sphere centred at $z=r$ ). The full cylinder contributes $\rho g\,A\,\tfrac{h^2}{2}$ ; removing the sphere (centroid at its centre $z=r$ , volume $V_s$ ) subtracts $\rho g\,V_s r$ :

\text{PE}_i=\rho g\Big(A\frac{h^2}{2}-V_s r\Big).

Finally: water = full cylinder $0\le z\le h'$ :

\text{PE}_f=\rho g\,A\frac{h'^2}{2}.

The change:

\Delta\text{PE}_{\text{water}}=\rho g\Big[\frac{A}{2}(h'^2-h^2)+V_s r\Big].

With $h'=h-\dfrac{V_s}{A}$ , $h'^2-h^2=(h'-h)(h'+h)=-\dfrac{V_s}{A}\Big(2h-\dfrac{V_s}{A}\Big)$ , so

\frac{A}{2}(h'^2-h^2)=-\frac{V_s}{2}\Big(2h-\frac{V_s}{A}\Big)=-V_s h+\frac{V_s^2}{2A}.

Therefore

\Delta\text{PE}_{\text{water}}=\rho g\Big[V_s r-V_s h+\frac{V_s^2}{2A}\Big]=W'\Big[r-h+\frac{V_s}{2A}\Big],

using $\rho g V_s=W'$ . Now $\dfrac{V_s}{2A}=\dfrac{\tfrac43\pi r^3}{2\pi R^2}=\dfrac{2r^3}{3R^2}$ , giving

\Delta\text{PE}_{\text{water}}=W'\Big(r-h+\frac{2r^3}{3R^2}\Big).

Step 4 — Total minimum work

Answer

\boxed{\;\text{Work}=W\Big(h-\frac{4r^3}{3R^2}\Big)+W'\Big(r-h+\frac{2r^3}{3R^2}\Big)\ \text{cm·gm}.\;}