MHT CET 2021 24th September Morning Shift | Fluid Mechanics Question 91 | Physics

A thin metal disc of radius 'r' floats on water surface and bends the surface downwards along the perimeter making an angle '$$\theta$$' with the vertical edge of the dsic. If the weight of water displaced by the disc is '$$\mathrm{W}$$', the weight of the metal disc is [T = surface tension of water]

$$2 \pi \mathrm{r} \cos \theta+W$$

$$\mathrm{W}-2 \pi \mathrm{T} \cos \theta$$

$$\mathrm{2 \pi r T+W}$$

$$2 \pi \mathrm{T} \cos \theta-\mathrm{W}$$

MHT CET 2021 24th September Morning Shift

MCQ (Single Correct Answer)

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The work done in blowing a soap bubble of volume '$$\mathrm{V}$$' is '$$\mathrm{W}$$'. The work required to blow a soap bubble of volume '$$2 \mathrm{~V}$$' is [$$\mathrm{T}=$$ surface tension of soap solution]

$$2^{2 / 3} \mathrm{~W}$$

$$2 \mathrm{~W}$$

$$\mathrm{W}$$

$$2^{1 / 3} \mathrm{~W}$$

MHT CET 2021 23rd September Evening Shift

MCQ (Single Correct Answer)

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A glass rod of radius '$$r_1$$' is inserted symmetrically into a vertical capillary tube of radius '$$r_2$$' ($$r_1 < \mathrm{r}_2$$) such that their lower ends are at same level. The arrangement is dipped in water. The height to which water will rise into the tube will be ($$\rho=$$ density of water, T = surface tension in water, g = acceleration due to gravity)

$$\frac{2 T}{\left(r_2-r_1\right) \rho g}$$

$$\frac{T}{\left(r_2^2-r_1^2\right) \rho g}$$

$$\frac{T}{\left(r_2-r_1\right) \rho g}$$

$$\frac{2 \mathrm{~T}}{\left(\mathrm{r}_2^2-\mathrm{r}_1^2\right) \rho g}$$

MHT CET 2021 23rd September Evening Shift

MCQ (Single Correct Answer)

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An ice cube of edge $$1 \mathrm{~cm}$$ melts in a gravity free container. The approximate surface area of water formed is (water is in the form of spherical drop)

$$(36 \pi)^{1 / 3} \mathrm{~cm}^2$$

$$(24 \pi)^{1 / 3} \mathrm{~cm}^2$$

$$(28 \pi)^{1 / 3} \mathrm{~cm}^2$$

$$(12 \pi)^{1 / 3} \mathrm{~cm}^2$$

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