Current Electricity Ques 20 — JEE PYQ Chapterwise

Answer:

Correct Answer: 20.$0.2 %$ increase

Solution:

Formula:

Resistance of a Conductor:

20. $R=\rho \frac{l}{A}=\frac{\rho \cdot l}{V / l}=\frac{\rho l^{2}}{V} \quad(V=$ volume of wire $)$

$$ \left.\therefore \quad R \propto l^{2} \quad \text { ( } \rho \text { and } V=\text { constant }\right) $$

For small percentage change

$%$ change $R=2(%$ change in $l)=2(0.1 %)=0.2 %$

Since $R \propto l^{2}$, with increase in the value of $l$, resistance will also increase.