JEE Advanced
Physics
Mechanical Properties of Fluids
2024
JEE Advanced 2024 (Paper 2)
JEE Advanced Physics Question (2024) — Solution
Question
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 , viscosity of water =1 10^ -3 ~Pa -s.]
Options
- A. The work done in pushing the ball to the depth d is 0.077 ~J .
- B. If we neglect the viscous force in water, then the speed v=7 ~m / s .
- C. If we neglect the viscous force in water, then the height H=1.4 ~m .
- D. The ratio of the magnitudes of the net force excluding the viscous force to the maximum viscous force in water is 500 / 9.
Answer
D. The ratio of the magnitudes of the net force excluding the viscous force to the maximum viscous force in water is 500 / 9.
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 ) 7 \\ & =18 11 10^ -5 ~N aligned Now, F_ net (F_v )_ = 500 9 [4 is correct]
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Related: Physics — Mechanical Properties of Fluids · All PYQ Banks