A cylindrical resistor is connected to a batt | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The resistance per unit length is given by $\frac{dR}{dx} = \rho_0 \left(\frac{x}{L}\right)^\alpha$. Integrating this from $x=0$ to $x=L$ gives th

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(C) The resistance per unit length is given by $\frac{dR}{dx} = \rho_0 \left(\frac{x}{L}\right)^\alpha$. Integrating this from $x=0$ to $x=L$ gives the total resistance $R$: $R = \int_0^L \frac{\rho_0}{L^\alpha} x^\alpha dx = \frac{\rho_0}{L^\alpha} \left[ \frac{x^{\alpha+1}}{\alpha+1} \right]_0^L = \frac{\rho_0 L^{\alpha+1}}{L^\alpha(\alpha+1)} = \frac{\rho_0 L}{\alpha+1}$. Given the power $P = 20 \ W$ and voltage $V = 5 \ V$,we use $P = \frac{V^2}{R}$: $20 = \frac{5^2}{R} \Rightarrow R = \frac{25}{20} = 1.25 \ \Omega$. Substituting $R = \frac{\rho_0 L}{\alpha+1}$ and $\rho_0 L = 10 \ \Omega$: $1.25 = \frac{10}{\alpha+1} \Rightarrow \alpha+1 = \frac{10}{1.25} = 8 \Rightarrow \alpha = 7$.

Similar Questions

$A$ metal rod of length $10 \ cm$ and a rectangular cross-section of $1 \ cm \times 0.5 \ cm$ is connected to a battery across opposite faces. The resistance will be . . . . . . .

Dimensions of a block are $1 \, cm \times 1 \, cm \times 100 \, cm$. If the specific resistance of its material is $3 \times 10^{-7} \, \Omega \cdot m$,the resistance between the square faces is:

When the length and area of cross-section both are doubled,then its resistance

If the resistance of a conductor is $5\,\Omega$ at $50\,^{\circ}\text{C}$ and $7\,\Omega$ at $100\,^{\circ}\text{C}$,then the mean temperature coefficient of resistance of the material is ............... $^{\circ}\text{C}^{-1}$.

$n$ conducting wires of same dimensions but having resistivities $1, 2, 3, . . . , n$ are connected in series. The equivalent resistivity of the combination is

Vedclass Test Series

Exam Paper Generator

Online Exam Module