A) 9 cm
B) 6 cm
C) 4.5 cm
D) 2.25 cm
Correct Answer: B
Solution :
The velocity of ball before entering the water surface \[v=\sqrt{2gh}=\sqrt{2g\times 9}\] When ball enters into water, due to upthrust of water the velocity of ball decreases (or retarded) The retardation, a = \[\frac{\text{apparent weight}}{\text{mass of ball}}\] \[\frac{=V(\rho -\sigma )g}{V\rho }\]\[=\left( \frac{\rho -\sigma }{\rho } \right)g\]\[=\left( \frac{0.4-1}{0.4} \right)\times g\]\[=-\frac{3}{2}g\] If h be the depth upto which ball sink, then, \[0-{{v}^{2}}=2\times \left( -\frac{3}{2}g \right)\times h\]Þ \[2g\times 9=3gh\]\ h = 6 cm.
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