Directions : In the following question more than one of the answers given may be correct. Select the correct answer and mark it according to the code:
A lead ball is dropped into a lake from diving board 5 m above the water. If hits the water with certain velocity and then sinks to the bottom with the same constant velocity. It reaches the bottom 5.0 after it is dropped. If\[g=10\text{ }m/{{s}^{2}}\] (1) the depth of lake in 40 m (2) the depth of lake in 50 m (3) the average velocity of ball is 9 m/s (4) the average velocity of ball is 5 m/sA) 1, 2 and 3 are correct
B) 1 and 2 are correct
C) 2 and 4 are correct
D) 1 and 3 are correct
Correct Answer: D
Solution :
If S is the height of ball from surface of lake and\[{{t}_{1}}\]is the time taken by ball to reach the water surface, then \[{{t}_{1}}=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2\times 5}{10}}=1s\] The velocity of ball, when it strikes the water surface, \[v=\sqrt{2gs}=\sqrt{2\times 10\times 5}\] \[=10\text{ }m/s\] Time taken by ball to travel depth h of the lake with uniform velocity v is \[{{t}_{2}}=\frac{h}{v}=\frac{h}{10}\] Given, \[{{t}_{1}}+{{t}_{2}}=5\]or\[1+\frac{h}{10}=5\]or\[h=40\text{ }m\] Displacement\[=5+40=45\text{ }m\] Total time \[=5\text{ }s\] Average velocity\[=\frac{45}{5}=9m/s\]
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