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MCQEasyJEE 2026Electric Potential & Potential Energy

JEE Physics 2026 Question with Solution

Three small identical bubbles of water having same charge on each coalesce to form a bigger bubble. Then the ratio of the potentials on one initial bubble and that on the resultant bigger bubble is:

  • A

    1:22/31 : 2^{2/3}

  • B

    1:31/31 : 3^{1/3}

  • C

    1:32/31 : 3^{2/3}

  • D

    32/3:13^{2/3} : 1

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Three identical charged water bubbles, each of charge qq and radius rr, coalesce to form one bigger bubble.

Find: The ratio of potential on one initial bubble to the potential on the resultant bigger bubble.

Electric potential of a charged spherical bubble is

V=kQRV = \frac{kQ}{R}

After coalescence, total charge becomes

Qbig=3qQ_{\text{big}} = 3q

Using conservation of volume,

43πR3=3×43πr3\frac{4}{3}\pi R^3 = 3 \times \frac{4}{3}\pi r^3

So,

R=31/3rR = 3^{1/3}r

Potential of the bigger bubble is

Vbig=k(3q)31/3r=32/3kqrV_{\text{big}} = \frac{k(3q)}{3^{1/3}r} = 3^{2/3}\frac{kq}{r}

Potential of one small bubble is

Vsmall=kqrV_{\text{small}} = \frac{kq}{r}

Therefore,

Vsmall:Vbig=kqr:32/3kqr=1:32/3V_{\text{small}} : V_{\text{big}} = \frac{kq}{r} : 3^{2/3}\frac{kq}{r} = 1 : 3^{2/3}

Therefore, the correct option is C, and the required ratio is 1:32/31 : 3^{2/3}.

Using charge addition and volume scaling

Given: Identical bubbles coalesce, so charges add directly and the new radius is found from volume conservation.

Find: Ratio of initial potential to final potential.

For a spherical conductor or charged bubble,

V∝QRV \propto \frac{Q}{R}

Initially,

Vsmall∝qrV_{\text{small}} \propto \frac{q}{r}

When three identical bubbles merge, the new bubble has charge 3q3q. Since volume is proportional to r3r^3,

R3=3r3R^3 = 3r^3

Hence,

R=31/3rR = 3^{1/3}r

So the final potential is proportional to

Vbig∝3q31/3r=32/3qrV_{\text{big}} \propto \frac{3q}{3^{1/3}r} = 3^{2/3}\frac{q}{r}

Thus,

Vsmall:Vbig=1:32/3V_{\text{small}} : V_{\text{big}} = 1 : 3^{2/3}

Hence the answer is C.

Common mistakes

  • Using radius proportional to the number of bubbles, that is taking R=3rR = 3r. This is wrong because radius follows volume conservation, not direct addition. Use R3=3r3R^3 = 3r^3, so R=31/3rR = 3^{1/3}r instead.

  • Adding potentials directly after coalescence. This is wrong because the question asks for the potential of the new single bubble, whose charge and radius both change. First find the new charge and new radius, then apply V=kQRV = \frac{kQ}{R}.

  • Using the ratio in reverse order. This is wrong because the required ratio is potential on one initial bubble to that on the resultant bigger bubble. Therefore compute Vsmall:VbigV_{\text{small}} : V_{\text{big}}, not Vbig:VsmallV_{\text{big}} : V_{\text{small}}.

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