Verified Solution • 2025–26 Session moderate

QJEEM23Phaf83: JEE Main 2023 Physics - Class 12 JEEM23Phaf83 Solution: Two isolated metallic solid spheres of radii and are charged…

Two isolated metallic solid spheres of radii R\mathrm{R} and 2R2 \mathrm{R} are charged such that both have same charge density σ\sigma . The spheres are then connected by a thin conducting wire. If the new charge density of the bigger sphere is σ′\sigma^{\prime} . The ratio σ′σ\frac{\sigma^{\prime}}{\sigma} is :

Multiple Choice Options

  • A 53\frac{5}{3}
  • B 56\frac{5}{6}
  • C 94\frac{9}{4}
  • D 43\frac{4}{3}

Step-by-Step Verified Solution

σ=Q14πR2=Q24π(2R)2\sigma = {{{Q_1}} \over {4\pi {R^2}}} = {{{Q_2}} \over {4\pi {{(2R)}^2}}}

Now Q2′=Qtotal[R2R1+R2]Q{'_2} = {Q_{total}}\left[ {{{{R_2}} \over {{R_1} + {R_2}}}} \right]

=(Q1+Q2)[2R3R]= ({Q_1} + {Q_2})\left[ {{{2R} \over {3R}}} \right]

=σ(20πR2)23= \sigma (20\pi {R^2}){2 \over 3}

∴\therefore σ2′=Q2′4π(2R)2=σ(20πR2)2316πR2\sigma {'_2} = {{Q{'_2}} \over {4\pi {{(2R)}^2}}} = {{\sigma (20\pi {R^2}){2 \over 3}} \over {16\pi {R^2}}}

=54×23σ= {5 \over 4} \times {2 \over 3}\sigma

=56σ= {5 \over 6}\sigma

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