A) \[\left( \frac{\sqrt{3}-1}{4\sqrt{2}} \right)g\]
B) \[(\sqrt{3}-1)g\]
C) \[\frac{g}{2}\]
D) \[\left( \frac{\sqrt{3}-1}{\sqrt{2}} \right)g\]
Correct Answer: A
Solution :
[A]NLM on B \[mg\sin 30{}^\circ -T=m\times a\] ?(1) |
\[T-mg\,\,\sin 60{}^\circ =ma\] ?(2) |
\[a=\frac{mg\,\,\sin 60{}^\circ -mg\,\,\sin 30{}^\circ }{2m}\] |
Where a is Acceleration of system Here, m=mass of each block or \[a=\left( \frac{\sqrt{3}-1}{4} \right)g\] |
Acceleration of center of mass \[\to {{a}_{com}}\] |
Now \[\to {{\overrightarrow{a}}_{\,com}}=\frac{m{{\overrightarrow{\,a}}_{1}}+m\,{{\overrightarrow{a}}_{2}}}{2m}\] |
Here, \[{{\overrightarrow{a}}_{1}}\] and \[{{\overrightarrow{a}}_{2}}\]are at \[\left( \frac{\sqrt{3}-1}{4} \right)g\]right angles. |
Hence, \[\left| \to {{\overrightarrow{a}}_{com}} \right|=\frac{\sqrt{2}}{2}a=\left( \frac{\sqrt{3}-1}{4\sqrt{2}} \right)g\] |
You need to login to perform this action.
You will be redirected in
3 sec