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NVAEasyJEE 2023Capacitors & Dielectrics

JEE Physics 2023 Question with Solution

64 identical drops each charged up to a potential of 10 mV10 \, \text{mV} are combined to form a bigger drop. The potential of the bigger drop will be _____ mV\text{mV}.

Answer

Correct answer:160

Step-by-step solution

Standard Method

Given: There are 6464 identical drops, each at potential 10 mV=0.01 V10 \, \text{mV} = 0.01 \, \text{V}.

Find: The potential of the bigger drop formed by combining them.

For a conducting spherical drop,

V=kQrV = \frac{kQ}{r}

and its capacitance is

C=4πϵ0rC = 4\pi \epsilon_0 r

When identical drops combine, charge is conserved and volume adds.

If the radius of each small drop is rr and the radius of the bigger drop is RR, then

64(43πr3)=43πR364 \left(\frac{4}{3}\pi r^3\right) = \frac{4}{3}\pi R^3

So,

R3=64r3R^3 = 64r^3 R=4rR = 4r

Let the charge on each small drop be QQ. Then total charge on the bigger drop is

Q′=64QQ' = 64Q

Since each small drop has potential 10 mV10 \, \text{mV}, we have

kQr=10 mV\frac{kQ}{r} = 10 \, \text{mV}

Now the potential of the bigger drop is

V′=kQ′R=k(64Q)4r=16kQrV' = \frac{kQ'}{R} = \frac{k(64Q)}{4r} = 16\frac{kQ}{r}

Therefore,

V′=16×10 mV=160 mVV' = 16 \times 10 \, \text{mV} = 160 \, \text{mV}

Therefore, the potential of the bigger drop is 160 mV160 \, \text{mV}.

Direct Scaling

Given: n=64n = 64 identical drops are combined.

Find: The new potential.

For a spherical drop, potential varies as

V∝QRV \propto \frac{Q}{R}

When nn identical drops merge:

  • total charge becomes nQnQ
  • radius becomes n1/3rn^{1/3}r

So the new potential is

V′∝nQn1/3r=n2/3QrV' \propto \frac{nQ}{n^{1/3}r} = n^{2/3}\frac{Q}{r}

Hence,

V′=n2/3VV' = n^{2/3}V

Substitute n=64n = 64 and V=10 mVV = 10 \, \text{mV}:

V′=642/3×10 mV=16×10 mV=160 mVV' = 64^{2/3} \times 10 \, \text{mV} = 16 \times 10 \, \text{mV} = 160 \, \text{mV}

Therefore, the answer is 160160.

Common mistakes

  • Using radius proportional to 64r64r instead of 641/3r64^{1/3}r is incorrect because volume, not radius, adds when drops combine. Always equate total volumes first to find the new radius.

  • Assuming the potential also becomes 6464 times is wrong because potential depends on both charge and radius. After combining, charge increases but radius also increases, so both effects must be included.

  • Leaving the final answer as 0.16 V0.16 \, \text{V} in the answer field is incorrect for a numerical-value question here because the blank is asked in mV\text{mV}. Convert 0.16 V0.16 \, \text{V} to 160 mV160 \, \text{mV}.

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