Two resistances R _ 1 = XΩ and R _ 2 = 1Ω are connected to a wire...

Resistance of frustum shaped conductor shown is

$R = ρ\dfrac{I}{\piab}$

For the shown conductor in the diagram.

$r = \dfrac{a + b}{2} = \dfrac{0 . 2 + 1}{2} = 0 . 6mm$

Thus, the resistance of left half is $P = \dfrac{ρ \times 0 . 5 \times {10}^{6}}{\pi \times 0 . 2 \times 0 . 6}$

And the resistance of right half is $Q = \dfrac{ρ \times 0 . 5 \times {10}^{6}}{\pi \times 0 . 6 \times 1}$

For Wheatstone to be balanced

$\dfrac{{R}_{1}}{P} = \dfrac{{R}_{2}}{Q}$

$⇒ \dfrac{X\pi \times 0 . 2 \times 0 . 6}{ρ \times 0 . 5 \times {10}^{6}} = \dfrac{1\pi \times 0 . 6 \times 1}{ρ \times 0 . 5 \times {10}^{6}}$

$⇒ X = 5$