A container partially filled with water is moved horizontally with acceleration a=g3. A small wooden ball of mass m is tied to the bottom of the container using a string. The ball remains inside water with the string inclined at an angle θ to the horizontal. Assuming that the density of ball is half the density of water, if, then m=310kg find the tension in the string (in N). (Take g = 10 m/s2)

As the container is accelerating the liquid surface will be inclined with horizontal such thattan⁡θ=ag=g/3g=13Hence sin⁡θ=110 and cos⁡θ=310In this case the bouncy force on the ball will be natural to water surface. The magnitude of bouncy forceB=Vdisρwgeff⇒ B=mρρwg2+a2=2mg2+g29which gives B=2103mgThe force acting on the ball are shown in F.B.D. (w.r.t. container)As the ball is in equilibrium with respect to container, thenB=T+mgcos⁡θ+masin⁡θ2103mg=T+mg⋅310+mg3⋅110⇒ T=103mg=103×310×10=100N