Question

A stone of density $2000\;kgm^{-3}$ completely immersed in a lake is allowed to sink from rest. If the effect of friction is neglected, then after 4 seconds, the stone will reach a depth of:
A. 78.4 m
B. 39.2 m
C. 19.6 m
D. 9.8 m
E. 24.6 m

Answer 1

Hint: In order to find out the depth to which the stone will reach, we need to find out the net acceleration acting on the stone because of its weight and upward thrust. Then we need to use the laws of equation to find out the depth.
Since, the density of the water of the lake is not given, let us assume it to be $1000kgm^{-3}$.

Formulae used:
Upward thrust on a body submerged in a fluid $=\rho V g$, where $\rho$ is the density of fluid, V is the volume of solid submerged into it and g is the acceleration due to gravity.

Laws of equation to find the distance travelled, $S=\dfrac{1}{2}at^{2}$, where S is the distance travelled, a is the acceleration and t is the time of travel.

Complete step by step solution:
It has been given the density of stone, ${\rho}_{s}=2000\;kgm^{-3}$ which is completely immersed in the lake. It can be demonstrated in the following figure

Let us consider that the mass of the stone is m and its volume is V, so we can write that
$m={\rho}_{s}V$.

Therefore, the volume of the stone, $V=\dfrac{m}{2.0\times10^{3}}\;m^{3}$. So, the upward thrust on the stone will be $=\rho Vg$, where $\rho$ is the density of water in the lake, V is the volume of stone and g is the acceleration due to gravity.

On putting the values in the equation, we get $\dfrac{m}{2000}\times 1000 \times g=\dfrac{mg}{2}$.

Now, the net force acting on the stone, F = force due to mass of the stone – upward thrust =$mg-\dfrac{mg}{2}=\dfrac{mg}{2}$.

Again, we can write $ma=\dfrac{mg}{2}\implies a=\dfrac{g}{2}=\dfrac{9.8}{2}=4.9\;ms^{-2}$.
Now, let us apply the equations of motion to find out the depth travelled by using the equation $S=\dfrac{1}{2}at^{2}=\dfrac{1}{2}\times 4.9\times 4^{2}=\dfrac{4.9\times 16}{2}=39.2$ m

Hence, the depth covered by the stone is 39.2 m.

Note: We can make mistakes in determining the upward thrust and acceleration of the stone. One may put the density of stone in place of density of water to find out the upward thrust. Then, one may directly consider the acceleration due to gravity as the acceleration of the stone travelling down the lake.