JEE Advanced2024PhysicsMechanical Properties of FluidsActual
A table tennis ball has radius (3 / 2) 10⁻² ~m and mass (22 / 7) 10⁻³ ~kg . It is slowly pushed down into a swimming pool to a depth of d=0.7 ~m below the water surface and then released from rest. It emerges from the water surface at speed v , without getting wet, and rises up to a height H . Which of the following option(s) is(are) correct? [Given: =22 / 7, g=10 ~m ~s ⁻² , density of water =1 10^3 ~kg ~m ⁻³ , visco
Options
- AThe work done in pushing the ball to the depth d is 0.077 ~J .
- BIf we neglect the viscous force in water, then the speed v=7 ~m / s .
- CIf we neglect the viscous force in water, then the height H=1.4 ~m .
- DThe ratio of the magnitudes of the net force excluding the viscous force to the maximum viscous force in water
Correct answer
A. The work done in pushing the ball to the depth d is 0.077 ~J .
Step-by-step solution
Work done in pushing the ball W=(v g) d-(v g) d Where, Density of water Density of ball aligned & W= 4 3 R^3 10 0.7 [1000- 3 4 10⁻³ R^3 ] & W=0.077 ~J aligned [1 is correct] When ball is released at bottom same work (i.e. 0.077 ~J ) is done on ball. aligned & 1 2 m v^2=0.077 & v= 0.077 2 22 7 10⁻³ & =7 ~m / s aligned [2 is correct] also, H= v^2 2 g = 7 7 2 10 =2.45 ~m [3 is incorrect] Net force F_ net =v g-v g=0.11 ~N Also, viscous force is maximum when v=7 ~m / s aligned & (F_v )_ =6 r v & =6 22 7 10⁻³ ( 3 2 10⁻²