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Standard XII

Physics

Impulsive Forces in Collision

A cube with a...

Question

A cube with a mass m= 20g floats on the surface of water. Each face of the cube is 3cm long. what is the distance between lower surface of the cube and surface of the water? (Surface tension of water = 70 $\times$ $10^{3}$ N/m , and angle of contact is acute)

3.31cm

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4.31cm

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2.31cm

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1.31cm

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Solution

The correct option is C
2.31cm

Down ward force = Buoyant force

mg+4Ta = $a^{2}$ x $ρ$ g

$\Rightarrow$ x= $\frac{m g + 4 T a}{a^{2} ρ g}$

= $\frac{20 \times 10^{- 3} + 4 \times 70 \times 10^{- 3} \times 3 \times 10^{- 2}}{9 \times 10^{- 4} \times 10^{3}} \times 10$

= $\frac{.2 + 0.0084}{9}$ = $\frac{0.2084}{9}$

= 0.02315m

x=2.31 cm

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A cube with a mass m= 20g floats on the surface of water. Each face of the cube is 3cm long. what is the distance between lower surface of the cube and surface of the water? (Surface tension of water = 70 × 103 N/m , and angle of contact is acute)

The correct option is C 2.31cm Down ward force = Buoyant force mg+4Ta = a2xρg ⇒x= mg+4Taa2ρg = 20×10−3+4×70×10−3×3×10−29×10−4×103×10 = .2+0.00849 = 0.20849 = 0.02315m x=2.31 cm

A cube with a mass m= 20g floats on the surface of water. Each face of the cube is 3cm long. what is the distance between lower surface of the cube and surface of the water? (Surface tension of water = 70 $\times$ $10^{3}$ N/m , and angle of contact is acute)