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.
You will be redirected in
3 sec
