MHT CET 2023 12th May Morning Shift | Fluid Mechanics Question 16 | Physics

Consider a soap film on a rectangular frame of wire of area $$3 \times 3 \mathrm{~cm}^2$$. If the area of the soap film is increased to $$5 \times 5 \mathrm{~cm}^2$$, the work done in the process will be (surface tension of soap solution is $$\left.2.5 \times 10^{-2} \mathrm{~N} / \mathrm{m}\right)$$

$$9 \times 10^{-6} \mathrm{~J}$$

$$16 \times 10^{-6} \mathrm{~J}$$

$$40 \times 10^{-6} \mathrm{~J}$$

$$80 \times 10^{-6} \mathrm{~J}$$

MHT CET 2023 12th May Morning Shift

MCQ (Single Correct Answer)

-0

A spherical drop of liquid splits into 1000 identical spherical drops. If '$$\mathrm{E}_1$$' is the surface energy of the original drop and '$$\mathrm{E}_2$$' is the total surface energy of the resulting drops, then $$\frac{E_1}{E_2}=\frac{x}{10}$$. Then value of '$$x$$' is

MHT CET 2023 12th May Morning Shift

MCQ (Single Correct Answer)

-0

The excess pressure inside a first spherical drop of water is three times that of second spherical drop of water. Then the ratio of mass of first spherical drop to that of second spherical drop is

1 : 3

1 : 6

1 : 9

1 : 27

MHT CET 2023 11th May Evening Shift

MCQ (Single Correct Answer)

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A liquid drop of radius '$$R$$' is broken into '$$n$$' identical small droplets. The work done is [T = surface tension of the liquid]

$$4 \pi \mathrm{R}^2\left(\mathrm{n}^{\frac{2}{3}}-1\right) \mathrm{T}$$

$$4 \pi R^2\left(n^{\frac{1}{3}}-1\right) T$$

$$4 \pi R^2\left(1-n^{\frac{1}{3}}\right) T$$

$$4 \pi \mathrm{R}^2\left(1-\mathrm{n}^{\frac{2}{3}}\right) \mathrm{T}$$

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