A) \[{{t}_{1}}={{t}_{2}}\]
B) \[{{t}_{1}}>{{t}_{2}}\]
C) \[{{t}_{1}}<{{t}_{2}}\]
D) \[{{t}_{1}}<{{t}_{2}}\] or \[{{t}_{1}}>{{t}_{2}}\] depending on whether the left is going up or down.
Correct Answer: A
Solution :
If the lift is stationary, the time \[{{t}_{1}}\] to cover a height h falling freely with acceleration \[g\]can be calculated by \[h=\frac{1}{2}g{{t}_{1}}^{2}\] or \[{{t}_{1}}=\sqrt{\frac{2h}{g}}\] ? (i) If it is considered that the base of the lift is moving upwards with say, uniform speed\[u\], the distance covered by the base of the lift in the upwards direction in time \[{{t}_{2}}\] taken by the coin to reach the base is\[u{{t}_{1}}\]. Thus, the distance travelled by the coin in this situation before it touches the base of the lift is\[h-u{{t}_{2}}\]. If the lift is going upwards with speed\[u\], the speed of the coin at the moment it leaves the hands is also \[u\] in upwards direction. Since, the displacement in the coin is \[h-u{{t}_{2}}\] in downwards direction, the initial velocity of the coin in the upwards direction should be taken as negative. Thus, the equation of displacement of the coin can also be written as\[,\]\[-u{{t}_{2}}+\frac{1}{2}g{{t}_{2}}^{2}\]. \[i.e.,\] \[h-u{{t}_{2}}=-u{{t}_{2}}+\frac{1}{2}g{{t}_{2}}^{2}\] or, \[h=\frac{1}{2}g{{t}_{2}}^{2}\] or, \[{{t}_{2}}=\sqrt{\frac{2h}{g}}\] ? (ii) Comparing equations (i) and (ii), we get\[,\]\[{{t}_{1}}={{t}_{2}}\]. Actually, it is so because in the second case, the average speed of the coin in upwards direction is same as that of the base of the lift, i.e., the average relative velocity between the coin and the base of the lift is throughout zero.
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