a cylinder of conducting material is 2.00 m long and 2.00 cm in diameter. the cylinder is made in such a way…
a cylinder of conducting material is 2.00 m long and 2.00 cm in diameter. the cylinder is made in such a way that its resistivity increases linearly along its length, from zero at one end to 1.0 ω - m at the other end. how does the resistance change if the length of the cylinder is doubled, while keeping its radius constant? assume the resistivity on each end of the cylinder remain the same.\nthe resistance is reduced to half of the original value.\nthe resistance doubles.\nthe resistance quadruples.\nthe resistance does not change.\nhint\nwrite the resistivity as a function of length, z, so you have ρ(z)=ρ0z. then write dr = ρ(z)/a dz and integrate over z.\nsave for later\nsubmit answer
Answer
Explanation:
Step1: Recall resistance formula
The resistance of a small - length element of a conductor is given by $dR=\frac{\rho(z)}{A}dz$, where $\rho(z)$ is the resistivity as a function of length $z$, and $A$ is the cross - sectional area. Given $\rho(z)=\rho_0z$. So $dR = \frac{\rho_0z}{A}dz$.
Step2: Integrate to find total resistance
The total resistance $R$ of the cylinder of length $L$ is obtained by integrating $dR$ from $z = 0$ to $z = L$. Using the integral formula $\int xdx=\frac{x^{2}}{2}+C$, we have $R=\int_{0}^{L}\frac{\rho_0z}{A}dz=\frac{\rho_0}{A}\int_{0}^{L}zdz=\frac{\rho_0}{A}\left[\frac{z^{2}}{2}\right]_{0}^{L}=\frac{\rho_0L^{2}}{2A}$.
Step3: Analyze the effect of doubling the length
Let the initial length be $L_1$ and the new length be $L_2 = 2L_1$. The initial resistance $R_1=\frac{\rho_0L_1^{2}}{2A}$, and the new resistance $R_2=\frac{\rho_0L_2^{2}}{2A}=\frac{\rho_0(2L_1)^{2}}{2A}=\frac{4\rho_0L_1^{2}}{2A}=4R_1$. So the resistance quadruples.