Current Electricity Ques 27
27. If a negligibly small current is passed through a wire of length $15 m$ and of resistance $5 \Omega$ having uniform cross-section of $6 \times 10^{-7} m^{2}$, then coefficient of resistivity of material, is [1996]
(a) $1 \times 10^{-7} \Omega-m$
(b) $2 \times 10^{-7} \Omega-m$
(c) $3 \times 10^{-7} \Omega-m$
(d) $4 \times 10^{-7} \Omega-m$
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Answer:
Correct Answer: 27.(b)
Solution:
- (b) Given : Length of wire $(l)=15 m$
Area $(A)=6 \times 10^{-7} m^{2}$
Resistance $(R)=5 \Omega$.
We know that resistance of the wire material
$ \begin{aligned} & R=\rho \frac{l}{A} \\ & \Rightarrow 5=\rho \times \frac{15}{6 \times 10^{-7}}=2.5 \times 10^{7} \rho \\ & \Rightarrow \rho=\frac{5}{2.5 \times 10^{7}}=2 \times 10^{-7} \Omega-m \end{aligned} $
[where $\rho=$ coefficient of resistivity]
BETA
