QJEEM25Ph785e: JEE Main 2025 Physics - Class 12 JEEM25Ph785e Solution: Due to presence of an em-wave whose electric component is given …

We know,

Energy density $= {1 \over 2}{\varepsilon _0}{E^2}C$

$\Rightarrow {{Energy} \over {Volume}} = {1 \over 2}{\varepsilon _0}{E^2}C$

$\Rightarrow Energy = {1 \over 2}{\varepsilon _0}E_c^2x\,vol$

Given that, ${\left( {Energy} \right)_1} = {\left( {Energy} \right)_2}$

$\Rightarrow {1 \over 2}{\varepsilon _0}E_1^2cx\pi R_1^2{L_1} = {1 \over 2}{\varepsilon _0}E_2^2cx\pi R_2^2{L_2}$

$\Rightarrow E_1^2R_1^2{L_1} = E_2^2R_2^2{L_1}\,(as\,{L_1} = {L_2})$

$\Rightarrow E_1^2R_1^2 = E_2^2R_2^2$

$\Rightarrow {E_1}{R_1} = {E_2}{R_2}$

$\Rightarrow {E_1}{R_1} = {E_2}\left( {{{{R_1}} \over 2}} \right)\,\left( {As\,{R_2} = {{{R_1}} \over 2}} \right)$

$\Rightarrow {E_2} = 2{E_1} = 2 \times 100$

$\Rightarrow {E_2} = 200\,N/C$

Hence, option (3) is correct.