A wooden ball of density 900 kg / m^{3} is immersed in a liquid of density 1200 kg / m^{3} to a depth of 126 cm and the released. The height h above the surface of which the ball rises will be

(A) 40 cm

(B) 42 cm

(C) 45 cm

(D) 50 cm

Solution

Let V = Volume of ball

Given that, \rho = density of wooden ball= 900\frac{kg}{m^3}

\sigma = density of liquid =1200\frac{kg}{m^3}

As, the density of wooden ball = \rho=\frac mV

m=V\rho

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

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

F_{net}= F_{up}-mg

ma= V\sigma g - V\rho g

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

a=\;\frac{V\sigma g-V\rho g}{V\rho}

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

a=\;\frac{(1200-900)g}{900}

a=\;\frac{(300)g}{900}

a=\;\frac{g}{3}

Now, u=0 (because the ball is released from rest)

When ball reaches to surface, S= 126 cm = 1.26 m. Also, let us assume that it attains velocity v at top surface.

According to Kinematics Equation of U.A.M., v^2-u^2=2aS