Home Physics Fluid Mechanics JEE Main 2023 - ( Mechanical Properties of Fluids ) 64 identical drops each charged upto potenti…
Physics Fluid Mechanics JEE Main 2023 - ( Mechanical Properties of Fluids ) MCQ (Single Correct)

64 identical drops each charged upto potential of 10 mV are combined to form a bigger drop. The potential of the bigger drop will be_____mV.

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(160)

The formula to calculate the potential (V") of each tiny drop is given by

...

When all the drops forms the bigger drop the radius of the new drop (r’) can be calculated as follows: Volume of big drop = 64 x volume of each tiny drop

So, the potential of the new big drop (V’) can be written as

...

Divide equation by equation and simplify to obtain the potential of the new drop.

...

Substitute the value of the known parameter into equation to calculate the required value of the potential for the new drop.

v'=16 x 10 mV

=160 mV

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