JEE MAIN - Physics (2021 - 17th March Evening Shift - No. 22)
The electric field intensity produced by the radiation coming from a 100 W bulb at a distance of 3 m is E. The electric field intensity produced by the radiation coming from 60W at the same distance is $$\sqrt {{x \over 5}} $$E. Where the value of x = ____________.
উত্তর
3
ব্যাখ্যা
$$I = {1 \over 2}C{ \in _0}{E^2}$$
$${E^2} \propto I$$
$$I = {{Power} \over {Area}}$$
$${E^2} \propto {P \over A}$$
$$E \propto \sqrt P $$
$${{E'} \over E} = \sqrt {{{60} \over {100}}} $$
$$E' = \sqrt {{3 \over 5}} E$$
So the value of x = 3
$${E^2} \propto I$$
$$I = {{Power} \over {Area}}$$
$${E^2} \propto {P \over A}$$
$$E \propto \sqrt P $$
$${{E'} \over E} = \sqrt {{{60} \over {100}}} $$
$$E' = \sqrt {{3 \over 5}} E$$
So the value of x = 3