Question

A small metallic ball is placed in a liquid filled in a closed container. Initially, the ball is at rest with respect to the container. Now the container is accelerated with an acceleration of $2m/s^2$ towards the right. The density of the ball is thrice that of the liquid. Find the time (in seconds) after which the ball will collide with the wall of the container at a distance of $6m$. (neglect viscous forces and assume the floor of the container to be smooth)

• A. $2$
• B. $3$
• C. $1$
• D. $5$

Answer 1

The correct option is B $3$

Effective acceleration in the frame of container $=-2\hat{i}-10\hat{j}$
Therefore upthrust acting on the ball$=v{\rho }_1(2\hat{i}+10\hat{j})$
Also the pseudo force acting on the ball$=v{\rho }_s(-2\hat{i})$
Therefore Net force acting on the ball in horizontal direction $=v{\rho }_{1}2\hat{i}+v{\rho }_s(-2\hat{i})$
Therefore Acceleration of the ball$=[v{\rho }_{1}2\hat{i}+v{\rho }_s(-2\hat{i}\left)\right]/v{\rho }_s=-2\hat{i}(1-{\rho }_1/{\rho }_s)=4/3\left(\widehat{-i}\right)$.
Now using $s=ut+1/2a{t}^2$, we have $6=\left(1/2\right)\left(4/3\right)t^2,t=3sec$