Question

An empty plastic box of mass m is found to accelerate up at the rate of g/6 when placed deep inside water. How much sand should be put inside the box so that it may accelerate down at the rate of g/6,
A. $\dfrac{m}{5}$
B. $\dfrac{2m}{5}$
C. $\dfrac{3m}{5}$
D. $\dfrac{4m}{5}$

Answer 1

Hint: The acceleration of the plastic bottle upwards in the first scenario is due to the fact that the magnitude of upthrust due to the water is more than the weight of the plastic bottle. As we add sand, the weight of the bottle will increase and a resultant downward acceleration will be observed.

Complete answer:
If the plastic bottle is placed in water, it will experience two forces:
1. mg downwards due to its weight when mass is m,
2. Upthrust due to water U, which pushes the bottle upwards.

We are given two situations;
(1) In the first scenario, a resultant force on the bottle is acting upwards.
$F_{res} = U - mg$
Given that the bottle experiences acceleration upwards of magnitude g/6, we may write:
$\dfrac{mg}{6} = U - mg$
Which gives us the magnitude of upthrust on the bottle as:
$\dfrac{7mg}{6} = U$

(2) In the second scenario, we add sand to the bottle, so its mass is increased by M (mass of sand) and it starts to accelerate downwards with a magnitude g/6.
$ - \dfrac{(m+M)g}{6} = U - (m+M)g$
Here the negative sign on the LHS containing the resultant force tells us that the acceleration is in the opposite direction to the previous acceleration.
Substituting the upthrust that we obtained, we get
$- \dfrac{(m+M)g}{6} = \dfrac{7mg}{6} - (m+M)g$ .
$- \dfrac{(m+M)g}{6} = \dfrac{mg}{6} - Mg$
which on simplification gives us:
$\dfrac{5M}{6} = \dfrac{2m}{6}$
$M = \dfrac{2m}{5}$

Therefore the mass of the sand is obtained as $\dfrac{2m}{5}$ so the correct answer is option (B).

Note:
For the solution we assume that the upward direction of acceleration to be a positive direction and downward direction to be negative direction. It is totally fine to do it another way but keeping track of the negative sign is something of importance. Otherwise, one might continue to put a positive sign in case of the resultant force in the second scenario also. Remember force is a vector quantity so direction always matters.