Two resistances R_1=X \Omega and R_2=1 \Omega 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 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 \underline\hspace2cm .

Current Electricity PYQ — JEE Advanced 2022 · Shift 2 · Q39 with Solution

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Correct answer: 5

Condition for zero deflection in galvanometer

Since the galvanometer connected between the mid-point of wire $AB$ and the junction of resistors shows zero deflection, the potentials of these two points are equal.

So the arrangement works like a balanced Wheatstone bridge.

Hence,

\frac{R_{AM}}{R_{MB}}=\frac{R_1}{R_2}

where $M$ is the midpoint of the wire along its axis.

Given:

$R_1=X\,\Omega$
$R_2=1\,\Omega$

Therefore,

\frac{R_{AM}}{R_{MB}}=X

So we only need to find the ratio of resistances of the two halves of the tapered wire.

Radius variation along the wire

The wire length is $100\text{ cm}$ , and the midpoint is at $50\text{ cm}$ from each end.

Radius varies linearly from:

at $A$ : $r_A=0.2\text{ mm}$
at $B$ : $r_B=1.0\text{ mm}$

Thus at the midpoint $M$ ,

r_M=\frac{0.2+1.0}{2}=0.6\text{ mm}

Resistance of a tapered wire segment

For a small element $dx$ ,

dR=\frac{\rho\,dx}{A}=\frac{\rho\,dx}{\pi r^2}

Since $r$ varies linearly with $x$ , let

r=r(x)

Then for such a conical/frustum-like wire,

R\propto \int \frac{dx}{r^2}

Using linear variation of $r$ , this gives

R \propto \frac{1}{r_1r_2}

for a segment whose end radii are $r_1$ and $r_2$ .

More explicitly, for a segment of length $l$ with radii varying linearly from $r_1$ to $r_2$ ,

R=\frac{\rho l}{\pi r_1 r_2}

Resistance of the two halves

Each half has length $50\text{ cm}$ .

So,

R_{AM}=\frac{\rho l}{\pi r_A r_M}

R_{MB}=\frac{\rho l}{\pi r_M r_B}

Therefore,

\frac{R_{AM}}{R_{MB}}=\frac{\rho l}{\pi r_A r_M}\cdot \frac{\pi r_M r_B}{\rho l}=\frac{r_B}{r_A}

Substitute values:

\frac{R_{AM}}{R_{MB}}=\frac{1.0}{0.2}=5

Hence,

X=5

Comparison with stored answer

Derived answer: $5$

Stored correct answer: $5$

They agree.

More from Current Electricity

View written solutionFree

Correct answer: 5

Condition for zero deflection in galvanometer

Since the galvanometer connected between the mid-point of wire $AB$ and the junction of resistors shows zero deflection, the potentials of these two points are equal.

So the arrangement works like a balanced Wheatstone bridge.

Hence,

\frac{R_{AM}}{R_{MB}}=\frac{R_1}{R_2}

where $M$ is the midpoint of the wire along its axis.

Given:

$R_1=X\,\Omega$
$R_2=1\,\Omega$

Therefore,

\frac{R_{AM}}{R_{MB}}=X

So we only need to find the ratio of resistances of the two halves of the tapered wire.

Radius variation along the wire

The wire length is $100\text{ cm}$ , and the midpoint is at $50\text{ cm}$ from each end.

Radius varies linearly from:

at $A$ : $r_A=0.2\text{ mm}$
at $B$ : $r_B=1.0\text{ mm}$

Thus at the midpoint $M$ ,

r_M=\frac{0.2+1.0}{2}=0.6\text{ mm}

Resistance of a tapered wire segment

For a small element $dx$ ,

dR=\frac{\rho\,dx}{A}=\frac{\rho\,dx}{\pi r^2}

Since $r$ varies linearly with $x$ , let

r=r(x)

Then for such a conical/frustum-like wire,

R\propto \int \frac{dx}{r^2}

Using linear variation of $r$ , this gives

R \propto \frac{1}{r_1r_2}

for a segment whose end radii are $r_1$ and $r_2$ .

More explicitly, for a segment of length $l$ with radii varying linearly from $r_1$ to $r_2$ ,

R=\frac{\rho l}{\pi r_1 r_2}

Resistance of the two halves

Each half has length $50\text{ cm}$ .

So,

R_{AM}=\frac{\rho l}{\pi r_A r_M}

R_{MB}=\frac{\rho l}{\pi r_M r_B}

Therefore,

\frac{R_{AM}}{R_{MB}}=\frac{\rho l}{\pi r_A r_M}\cdot \frac{\pi r_M r_B}{\rho l}=\frac{r_B}{r_A}

Substitute values:

\frac{R_{AM}}{R_{MB}}=\frac{1.0}{0.2}=5

Hence,

X=5

Comparison with stored answer

Derived answer: $5$

Stored correct answer: $5$

They agree.

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