A) 9cm
B) 6cm
C) 4.5cm
D) 2.25cm
Correct Answer: B
Solution :
[b] Velocity of ball just before entering the surface of water \[\sqrt{2gh}\] \[=\sqrt{2\times 10\times 9}cm/s\] This velocity retarded by up thrust acting on the ball due to water Retardation \[=\frac{V\rho g-V\sigma g}{V\rho }=\left[ \frac{\rho -\sigma }{\rho } \right]g=\left[ \frac{0.4-1}{0.4} \right]g=-\frac{3}{2}g\] Depth of penetration of ball \[{{0}^{2}}-2\times 980\times 9=2\left[ -\frac{3}{2} \right]\times 980h\] \[h=6cm\]You need to login to perform this action.
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