91Ó°ÊÓ

i) Mass of the object is$\mathrm{m}=5\mathrm{kg}$

ii) Mass of liquid is ${\mathrm{m}}_{\mathrm{liquid}}=3\mathrm{kg}$

iii) Travel time, $\mathrm{t}=0.2\mathrm{s}$ ( to be used to find distance)

Here, we can use the Archimedes Principle to find the net force acting on the object. It states that the buoyant force on the object is equal to the weight of the liquid displaced by the object. We can compare the weight of the object and the weight of the liquid to find the direction of net acceleration. Using Newton’s second law of motion, we can find the net acceleration of the object. Using this acceleration and given time in the kinematic equation, we can find the distance through which the object would move in the given time.

Formula:

Buoyant force exerted by a liquid, ${\mathrm{F}}_{\text{buoyant}}={\mathrm{m}}_{\text{iiquid}}\mathrm{g}$ (i)

The net force on a body at acceleration, ${\mathrm{F}}_{\mathrm{net}}=\mathrm{ma}$(ii)

The second equation of motion, $\mathrm{x}−{\mathrm{v}}_0\mathrm{t}+0.5{\mathrm{at}}^2$(iii)

Weight of a body, $\mathrm{W}=\mathrm{mg}$, where, $\mathrm{g}=$acceleration due to gravity (iv)

Weight of the object using equation (iv) and the given values,

$\left({\mathrm{W}}_1\right)=5\mathrm{kg}×9.8\mathrm{m}/{\mathrm{s}}^2$

$=49.00\mathrm{N}$

Weight of the liquid displaced using equation (i), can be written as:

$\left({\mathrm{W}}_2\right)=3\mathrm{kg}×9.8\mathrm{m}/{\mathrm{s}}^2$

$=29.4\mathrm{N}$

From the above two values, we can clearly see that the weight of the object is greater than the weight of the displaced liquid. As the acceleration is directed downward, we can say that the net acceleration is downwards. Taking downward direction as positive and applying Newton’s second law, we have net force on the body of mass 5 kg as follows:

${\mathrm{F}}_{\mathrm{net}}={\mathrm{W}}_1-{\mathrm{W}}_2$

$ma={\mathrm{W}}_1-{\mathrm{W}}_2$

$\mathrm{a}=\frac{{\mathrm{W}}_1-{\mathrm{W}}_2}{\mathrm{m}}$

Putting the given values in the above equation, we have

$\mathrm{a}=\frac{49.0\mathrm{N}-29.4{\mathrm{N}}_{}}{5\mathrm{kg}}$

$=3.92\raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{s}}^2$}\right.$

Now, substituting the given values and value of acceleration in equation (iii), we get

$\mathrm{x}=\frac{1}{2}×3.92\mathrm{m}/{\mathrm{s}}^2×(0.2\mathrm{s}{)}^2$

$=0.0784\mathrm{m}$

Hence, the object moves $0.0784\mathrm{m}$in $0.2\mathrm{s}$.