A null point is found at 200   cm in potentiometer when cell in secondary circuit is shunted by 5 &#160…

Let potential gradient be $x$.

For the first case, we can write

$\frac{\epsilon }{r+5}\times 5=200x$ ......(1)

and after adding resistance, we can write

$\frac{\epsilon }{r+15}\times 15=300x$ .....(2)

From both equations, we get

$\Rightarrow r=5 \mathrm{\Omega }$