MHT CET 2021 20th September Evening Shift | Fluid Mechanics Question 51 | Physics

A body of density $$V$$ is dropped from (at rest) height '$$h$$' into a lake of density '$$\delta$$' $$(\delta > \rho)$$. The maximum depth to which the body sinks before returning to float on the surface is [Neglect all dissipative forces]

$$\frac{(\delta-\rho)}{2 h \rho}$$

$$\frac{2 h \rho}{(\delta-\rho)}$$

$$\frac{h \rho}{2(\delta-\rho)}$$

$$\frac{h \rho}{(\delta-\rho)}$$

MHT CET 2021 20th September Evening Shift

MCQ (Single Correct Answer)

-0

The surface energy of a liquid drop is 'U'. It splits up into 512 equal droplets. The surface energy becomes

8 U

6 U

4 U

2 U

MHT CET 2021 20th September Morning Shift

MCQ (Single Correct Answer)

-0

Air is pushed in a soap bubble to increase its radius from 'R' to '2R'. In this case, the pressure inside the bubble

does not change

decrease

becomes zero

increases

MHT CET 2021 20th September Morning Shift

MCQ (Single Correct Answer)

-0

Let '$$\mathrm{R_1}$$' and '$$\mathrm{R_2}$$' are radii of two mercury drops. A big mercury drop is formed from them under isothermal conditions. The radius of the resultant drop is

$$\sqrt{\mathrm{R}_1^2+\mathrm{R}_2^2}$$

$$\left(\mathrm{R}_1^3+\mathrm{R}_2^3\right)^{\frac{1}{3}}$$

$$\sqrt{\mathrm{R}_1^2-\mathrm{R}_2^2}$$

$$\frac{\mathrm{R}_1+\mathrm{R}_2}{2}$$

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