Current Electricity Question 2
Question 2 - 24 January - Shift 1
A hollow cylindrical conductor has length of 3.14 $m$, while its inner and outer diameters are $4 mm$ and $8 mm$ respectively. The resistance of the conductor is $n \times 10^{-3} \Omega$.
If the resistivity of the material is $2.4 \times 10^{-8} \Omega m$. The value of $n$ is
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Answer: (2)
Solution:
Formula: Resistance of a conductor
$R=\rho \frac{\ell}{A}$, the cross-sectional area is $\pi(b^{2}-a^{2})$
$R=\rho \frac{\ell}{\pi(b^{2}-a^{2})}=\frac{2.4 \times 10^{-8} \times 3.14}{3.14 \times(4^{2}-2^{2}) \times 10^{-6}}$
$=2 \times 10^{-3} \Omega$
$\to n=2$
