A) 8.0 cm
B) 10.0cm
C) Any value less than 12.0 cm
D) 4.0cm
Correct Answer: B
Solution :
Let the minimum amplitude of SHM is a. Restoring force on spring \[v_{f}^{2}=v_{i}^{2}+\frac{2G{{m}_{e}}}{{{R}_{e}}}\left( 1+\frac{1}{10} \right)\] Restoring force is balanced by weight mg of block. For mass to execute simple harmonic motion of amplitude a. \[v_{f}^{2}=v_{i}^{2}+\frac{2G{{m}_{e}}}{{{R}_{e}}}\left( 1-\frac{1}{10} \right)\] \[v_{f}^{2}=v_{i}^{2}+\frac{2Gm}{{{R}_{e}}}\left( 1-\frac{1}{10} \right)\] or \[\mathbf{\vec{v}}\] Here, \[\mathbf{\vec{B}}\] \[{{i}_{2}}>{{i}_{3}}>{{i}_{1}}\] \[{{i}_{2}}>{{i}_{1}}>{{i}_{3}}\] \[{{i}_{1}}>{{i}_{2}}>{{i}_{3}}\] Hence, minimum amplitude of the motion should be 10 cm, so that the mass gets detached from the pan.You need to login to perform this action.
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