1
MHT CET 2023 11th May Morning Shift
+1
-0

A spherical liquid drop of radius $$\mathrm{R}$$ is divided into 8 equal droplets. If surface tension is $$\mathrm{S}$$, then the work done in this process will be

A
$$2 \pi R^2 S$$
B
$$3 \pi \mathrm{R}^2 \mathrm{~S}$$
C
$$4 \pi R^2 S$$
D
$$2 \pi \mathrm{RS}^2$$
2
MHT CET 2023 11th May Morning Shift
+1
-0

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

A
$$\frac{\mathrm{h} \rho}{(\sigma-\rho)}$$
B
$$\frac{h \rho}{(\sigma+\rho)}$$
C
$$\frac{\mathrm{h} \rho}{(\rho-\sigma)}$$
D
$$\frac{2 \mathrm{~h} \rho}{(\sigma-\rho)}$$
3
MHT CET 2023 10th May Evening Shift
+1
-0

'$$n$$' number of liquid drops each of radius '$$r$$' coalesce to form a single drop of radius '$$R$$'. The energy released in the process is converted into the kinetic energy of the big drop so formed. The speed of the big drop is

$$[\mathrm{T}=$$ surface tension of liquid, $$\rho=$$ density of liquid.]

A
$$\sqrt{\frac{T}{\rho}\left[\frac{1}{r}-\frac{1}{R}\right]}$$
B
$$\sqrt{\frac{2 \mathrm{~T}}{\rho}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right]}$$
C
$$\sqrt{\frac{4 \mathrm{~T}}{\rho}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right]}$$
D
$$\sqrt{\frac{6 \mathrm{~T}}{\rho}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right]}$$
4
MHT CET 2023 10th May Evening Shift
+1
-0

At critical temperature, the surface tension of liquid is

A
zero
B
infinity
C
unity
D
same as that at any other temperature
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