Question

A cubical block(a×a×a), with a coin of mass 'm' over it is floating over liquid of density d. In this case x1 depth of the block is immersed .Now the coin is removed and it is found that x2 depth is immersed in liquid. Value of(x1-x2) is-​

Answer 1

Answer:

The value of(x1-x2) is $\frac{m}{d a^2}$​

Explanation:

When mass m is placed on the cubical block then total weight is counter balanced by the buoyancy force

so we will have

$d x_1(a \times a) g = (M + m) g$

so we have

$x_1 = \frac{(M + m)}{d a^2}$

now when the mass is removed from the top then we have

$d x_2 (a\times a) g = M g$

now we have

$x_2 = \frac{M}{d a^2}$

now we have

$x_1 - x_2 = \frac{(M + m)}{d a^2} - \frac{M}{d a^2}$

$x_1 - x_2 = \frac{m}{d a^2}$

#Learn

Topic : Buoyancy Force

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