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Question

A wooden block in the form of a uniform cylinder floats with one-third length above the water surface. A small chip of this block is cut and held at rest at the bottom of a container containing water to a height of 1 m and is then released. The time in which it will rise to the surface of water is (g=10 m/s2)

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Solution

The correct option is **B** 0.63 s

Since two-thirds of the wooden block is immersed in water, its density =2ρ/3, where ρ is the density of water.

Let the volume of the chip be V. Then its weight is (2/3)Vρg.

The upthrust acting on it inside the water is Vρg. Hence, the unbalanced upward force is Vρg(1−2/3)=Vρg/3.

Hence, its acceleration a in the upward direction is

a=Vρg/3(2/3)Vρ=g2

If t is the required time, then

1=0+12at2=12g2t2

Hence, t2=4g=410=0.4

t=0.63 s

Since two-thirds of the wooden block is immersed in water, its density =2ρ/3, where ρ is the density of water.

Let the volume of the chip be V. Then its weight is (2/3)Vρg.

The upthrust acting on it inside the water is Vρg. Hence, the unbalanced upward force is Vρg(1−2/3)=Vρg/3.

Hence, its acceleration a in the upward direction is

a=Vρg/3(2/3)Vρ=g2

If t is the required time, then

1=0+12at2=12g2t2

Hence, t2=4g=410=0.4

t=0.63 s

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