JEE WhatsApp Group
Sign in
Please select an account to continue using cracku.in
↓ →
JEE WhatsApp Group
Join Our JEE Preparation Group
Prep with like-minded aspirants; Get access to free daily tests and study material.
A constant voltage is applied between two ends of a metallic wire. If the length is halved and the radius of the wire is doubled, the rate of heat developed in the wire will be:
$$R = \rho \frac{l}{A} = \rho \frac{l}{\pi r^2}$$
$$R' = \rho \frac{l'}{\pi (r')^2} = \rho \frac{\frac{l}{2}}{\pi (2r)^2} = \frac{1}{8}\left(\rho \frac{l}{\pi r^2}\right) = \frac{R}{8}$$
$$H = \frac{V^2}{R}$$
$$H' = \frac{(V')^2}{R'} = \frac{V^2}{\frac{R}{8}} = 8\left(\frac{V^2}{R}\right) = 8H$$
Create a FREE account and get:
Educational materials for JEE preparation
Ask our AI anything
AI can make mistakes. Please verify important information.
AI can make mistakes. Please verify important information.
