A) \[\frac{L({{M}_{1}}+{{M}_{2}})}{2{{M}_{1}}}\]
B) \[\frac{L{{M}_{1}}}{2({{M}_{1}}+{{M}_{2}})}\]
C) \[\frac{2({{M}_{1}}+{{M}_{2}})}{L{{M}_{1}}}\]
D) \[\frac{2L{{M}_{1}}}{({{M}_{1}}+{{M}_{2}})}\]
Correct Answer: B
Solution :
\[x\]co-ordinate of centre of mass \[{{x}_{CM}}=\frac{{{M}_{1}}{{x}_{1}}+{{M}_{2}}{{x}_{2}}}{{{m}_{1}}+{{m}_{2}}}\] We know that, centre of mass lies at the centre. So, \[{{m}_{2}}=0\]and \[{{x}_{1}}=\frac{L}{2}\] So, \[{{x}_{CM}}=\frac{{{m}_{1}}L+0\times {{x}_{2}}}{2({{M}_{1}}+{{M}_{2}})}=\frac{{{M}_{1}}L}{2({{M}_{1}}+{{M}_{2}})}\]You need to login to perform this action.
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