Question

The resistivity of uniform rod of length $l$ = 2.52 m and cross sectional area A changes with distance x from one of the rod as $\rho$ = ($\rho_0 x^2$) $\Omega$/m. The length of OA for which its resistance value is half of the net resistance of rod, is (take $\sqrt[3]{2}$ = 1.26)

Answer 1

2 m

Answer 2

To solve this problem, we need to calculate the total resistance of the rod and the resistance of a segment of the rod (OA) and then apply the given condition.

The resistance of a small differential element of the rod of length $dx$ at a distance $x$ from end O is given by: $dR = \frac{\rho dx}{A}$ where $\rho$ is the resistivity and A is the uniform cross-sectional area.

Given that the resistivity $\rho$ changes with distance $x$ as $\rho = \rho_0 x^2$. Substitute this into the expression for $dR$: $dR = \frac{\rho_0 x^2 dx}{A}$

1. Calculate the total resistance of the rod ($R_{total}$): The rod has a total length $l$. To find the total resistance, we integrate $dR$ from $x=0$ to $x=l$: $R_{total} = \int_{0}^{l} \frac{\rho_0 x^2 dx}{A}$ $R_{total} = \frac{\rho_0}{A} \int_{0}^{l} x^2 dx$ $R_{total} = \frac{\rho_0}{A} \left[ \frac{x^3}{3} \right]_{0}^{l}$ $R_{total} = \frac{\rho_0}{A} \left( \frac{l^3}{3} - \frac{0^3}{3} \right)$ $R_{total} = \frac{\rho_0 l^3}{3A}$

2. Calculate the resistance of the segment OA ($R_{OA}$): Let the length of the segment OA be $x_A$. To find the resistance of this segment, we integrate $dR$ from $x=0$ to $x=x_A$: $R_{OA} = \int_{0}^{x_A} \frac{\rho_0 x^2 dx}{A}$ $R_{OA} = \frac{\rho_0}{A} \int_{0}^{x_A} x^2 dx$ $R_{OA} = \frac{\rho_0}{A} \left[ \frac{x^3}{3} \right]_{0}^{x_A}$ $R_{OA} = \frac{\rho_0}{A} \left( \frac{x_A^3}{3} - \frac{0^3}{3} \right)$ $R_{OA} = \frac{\rho_0 x_A^3}{3A}$

3. Apply the given condition: The problem states that the resistance value of OA is half of the net resistance of the rod: $R_{OA} = \frac{1}{2} R_{total}$

Substitute the expressions for $R_{OA}$ and $R_{total}$: $\frac{\rho_0 x_A^3}{3A} = \frac{1}{2} \left( \frac{\rho_0 l^3}{3A} \right)$

Notice that the common terms $\frac{\rho_0}{3A}$ cancel out from both sides: $x_A^3 = \frac{1}{2} l^3$

Solve for $x_A$: $x_A = \sqrt[3]{\frac{l^3}{2}}$ $x_A = \frac{l}{\sqrt[3]{2}}$

4. Substitute the given values: Given $l = 2.52$ m and $\sqrt[3]{2} = 1.26$. $x_A = \frac{2.52}{1.26}$ $x_A = 2$ m

The length of OA for which its resistance value is half of the net resistance of the rod is 2 m.