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Standard XII

Physics

Area and Volume Expansion

In the figure...

Question

In the figure shown, the heavy cylinder (radius R) resting on a smooth surface separates two liquids of densities $2 ρ$ and $3 ρ$ The height h for the equilibrium of cylinder must be

3 R / 2

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R \sqrt{\frac{3}{2}}

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R \sqrt{2}

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None of these

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Solution

The correct option is B $R \sqrt{\frac{3}{2}}$
Torque of hydrostatic force about centre of sphere is already zero as hydrostatic force passes through the center.
$∴$ $F_{L H S} = F_{R H S}$
$∴$ $\frac{1}{2} (2 ρ) (g) h^{2} = \frac{1}{2} (3 ρ) (g) R^{2}$
$∴$ $h = \sqrt{\frac{3}{2}} R$

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Similar questions

Q. In the figure shown, the heavy cylinder (radius R) resting on a smooth surface separates two liquids of densities

2 ρ

and

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. The height 'h' for the equilibrium of cylinder must be.

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Q. A cylinder with radius r and height h is closed on the top and bottom. Which of the following expressions represents the total surface area of this cylinder?

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2 πr (r + h)

(b)

πr (r + 2 h)

(c)

πr + (2 r + h)

(d)

2 {πr}^{2} + h

A cylinder with radius r and height h is closed on the top and botton. Which of the following expressions represents the total surface area of this cylinder ?

(a) $2 π$ r(r+h) (b) $π$ r(r+2h)

The figure consists of 2 cylinders, the inner cylinder is a solid cylinder whose radius is r and the outer cylinder is a hollow cylinder whose radius is R and height is h, the volume of fluid it can hold is:

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In the figure shown, the heavy cylinder (radius R) resting on a smooth surface separates two liquids of densities 2ρ and 3ρ The height h for the equilibrium of cylinder must be

The correct option is B R√32Torque of hydrostatic force about centre of sphere is already zero as hydrostatic force passes through the center.∴ FLHS=FRHS∴ 12(2ρ)(g)h2=12(3ρ)(g)R2∴ h=√32R

In the figure shown, the heavy cylinder (radius R) resting on a smooth surface separates two liquids of densities $2 ρ$ and $3 ρ$ The height h for the equilibrium of cylinder must be

The correct option is B $R \sqrt{\frac{3}{2}}$
Torque of hydrostatic force about centre of sphere is already zero as hydrostatic force passes through the center.
$∴$ $F_{L H S} = F_{R H S}$
$∴$ $\frac{1}{2} (2 ρ) (g) h^{2} = \frac{1}{2} (3 ρ) (g) R^{2}$
$∴$ $h = \sqrt{\frac{3}{2}} R$