← 2020 Paper 2

UPSC 2020 Maths Optional Paper 2 Q8c — Step-by-Step Solution

15 marks · Section B

Sources, sinks, doublets · Mechanics & Fluid Dynamics · asked 8× in 13 yrs · Read the full method →

Question

Two sources of strength $\frac{m}{2}$ are placed at the points $(\pm a,0)$. Show that at any point on the circle $x^2+y^2=a^2$, the velocity is parallel to the $y$-axis and is inversely proportional to $y$.

Technique

Superpose the radial velocity fields of two point sources; evaluate the denominators on the circle (they collapse to $2a(a\mp x)$); show $u=0$ and $v=S/y$.

Solution

Step 1 — Velocity field of the two sources

A two-dimensional source of strength $S$ at a point produces a purely radial outflow; in Cartesian form the velocity at $(x,y)$ due to a source $S$ at $(x_s,0)$ is

(up to the usual constant factor, absorbed into $S$). With sources of strength $S=\tfrac{m}{2}$ at $(a,0)$ and $(-a,0)$, superpose:

Step 2 — Restrict to the circle $x^2+y^2=a^2$

On the circle $x^2+y^2=a^2$, the denominators simplify:

Step 3 — The $x$-component vanishes

So $u=0$ on the circle: the velocity has no $x$-component, i.e. it is parallel to the $y$-axis. $\blacksquare$

Step 4 — The $y$-component is inversely proportional to $y$

On the circle $a^2-x^2=y^2$, hence

Answer