MHT CET 2021 22th September Morning Shift | Fluid Mechanics Question 120 | Physics

Water rises in a capillary tube of radius '$$r$$' up to a height '$$\mathrm{h}$$'. The mass of water in a capillary is '$$\mathrm{m}$$'. The mass of water that will rise in a capillary tube of radius $$\frac{'r'}{3}$$ will be

$$3 \mathrm{~m}$$

$$\frac{\mathrm{m}}{3}$$

$$\mathrm{m}$$

$$\frac{2 \mathrm{~m}}{3}$$

MHT CET 2021 22th September Morning Shift

MCQ (Single Correct Answer)

-0

A drop of liquid of density '$$\rho$$' is floating half immersed in a liquid of density '$$d$$'. If '$$T$$' is the surface tension, then the diameter of the drop of the liquid is

$$\sqrt{\frac{6 \mathrm{~T}}{\mathrm{~g}(2 p-\mathrm{d})}}$$

$$\sqrt{\frac{T}{g(2 \rho-d)}}$$

$$\sqrt{\frac{2 T}{g(2 p-d)}}$$

$$\sqrt{\frac{12 T}{g(2 \rho-d)}}$$

MHT CET 2021 21th September Evening Shift

MCQ (Single Correct Answer)

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Under isothermal conditions, two soap bubbles of radii '$$r_1$$' and '$$r_2$$' combine to form a single soap bubble of radius '$$R$$'. The surface tension of soap solution is ( $$P=$$ outside pressure)

$$\frac{P\left(R^3+r_1^3+r_2^3\right)}{4\left(r_1^2-r_2^2+R^2\right)}$$

$$\frac{P^2+r_1^2+r_2^2}{4\left(r_1^2+r_2^2+R^2\right)}$$

$$\frac{P\left(R^3-r_1^3-r_2^3\right)}{4\left(r_1^2+r_2^2-R^2\right)}$$

$$\frac{P\left(R^2-r_1^2-r_2^2\right)}{4\left(r_1^3+r_2^3-R^3\right)}$$

MHT CET 2021 21th September Evening Shift

MCQ (Single Correct Answer)

-0

In a capillary tube having area of cross-section A, water rises to a height 'h'. If cross-sectional area is reduced to $$\frac{A}{9}$$, the rise of water in the capillary tube is

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