Two resistances \(R_{1}=X \Omega\) and \(R_{2}=1 \Omega\) are connected to a wire \(AB\) of uniform resistivity, as shown in the figure. The radius of the wire varies linearly along its axis from \(0.2 mm\) at \(A\) to \(1 mm\) at \(B\). A galvanometer (G) connected to the center of the wire, \(50 cm\) from each end along its axis, shows zero deflection when \(A\) and \(B\) are connected to a battery. The value of \(X\) is \_\_\_\_.

JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · Current Electricity

Two resistances

R_{1} = X Ω

and

R_{2} = 1Ω

are connected to a wire

A B

of uniform resistivity, as shown in the figure. The radius of the wire varies linearly along its axis from

0.2 mm

A

1 mm

B

. A galvanometer (G) connected to the center of the wire,

50 c m

from each end along its axis, shows zero deflection when

A

and

B

are connected to a battery. The value of

X

is ____.

Answer: 5

For a balanced Wheatstone bridge, the galvanometer shows zero deflection when

\frac{R _{1}}{R _{2}} = \frac{R _{A C}}{R _{C B}}

.

The radius varies linearly:

r_{x} = 0.2 + 0.8 x

mm (where

x

is in meters from A).

Resistance of a segment:

d R = \frac{ρ d x}{π r _{x}^{2}}

For segment AC (0 to 0.5 m):

R_{A C} = \int_{0}^{0.5} \frac{ρ d x}{π r _{x}^{2}}

For segment CB (0.5 to 1 m):

R_{C B} = \int_{0.5}^{1} \frac{ρ d x}{π r _{x}^{2}}

Using the linear variation and evaluating the integrals:

\frac{R _{1}}{R _{2}} = \frac{R _{A C}}{R _{C B}} = 5

Therefore,

R_{1} = 5 R_{2} = 5Ω

JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · Integer answer