A small body of density ρ is dropped from rest at a height h into a lake of density σ , where σ > ρ . What would be of the acceleration of body till it moves inside the lake? (Neglect all dissipative effects)

(A) g\left(\frac\sigma\rho-1\right)\;downwards

(B) g\left(\frac\sigma\rho-1\right)\;upwards

(C) g\left(\frac\sigma\rho\right)\;downwards

(D) g\left(\frac\sigma\rho\right)\;upwards

Solution

Let V = Volume of body

\rho = density of body

\sigma = density of lake water

Thus, m=\;\rho\;V

As the density of liquid is more than that of body( σ > ρ), Therefore the body has tendency to float on the liquid because of larger upwards force in comparison to that of weight of body. Hence, net force will be upwards.

W know that F_{up}=V\;\sigma g\;

According to Newton's 2nd law of motion, we can write

F_{net}= F_{up}-mg (upwards)

F_{net}= V\sigma g -V \rho g

ma = V\sigma g -V \rho g

V\rho a = V\sigma g -V \rho g

a=\frac{V(\sigma-\rho)g}{V\rho}

a=\frac{(\sigma-\rho)g}{\rho}

a=g\left(\frac\sigma\rho-1\right)\;upwards

(B) is correct option

A small body of density ρ is dropped from rest at a height h into a lake of density σ , where σ > ρ . What would be of the acceleration of body till it moves inside the lake? (Neglect all dissipative effects)

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