| Direction: Question based on the following paragraph. |
| Two rods 1 and 2 are released from rest as shown in figure. |
| Given:\[{{l}_{1}}=4l,{{m}_{1}}=2m,{{l}_{2}}=2l\]and\[{{m}_{2}}=m.\]There is no friction between the two rods. If \[\alpha \]be the angular acceleration of rod 1 just after the rods are released. Then |
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A) \[\left[ \frac{2\sqrt{3}g}{2l}+2\sqrt{3}\alpha \right]\]
B) \[\left[ \frac{3\sqrt{3}g}{l}-\sqrt{3}\alpha \right]\]
C) \[\left[ \frac{6\sqrt{3}g}{8l}+5\sqrt{3}\alpha \right]\]
D) \[\left[ \frac{3\sqrt{3}g}{8l}-\frac{8}{\sqrt{3}}\alpha \right]\]
Correct Answer: D
Solution :
\[{{\alpha }_{2}}=\frac{({{m}_{2}}gl\cos {{30}^{o}})-N(2\sin {{30}^{o}})}{({{m}_{2}}){{(2l)}^{2}}/3}\] \[\frac{(\sqrt{3}mgl/2)-\frac{32m{{l}^{2}}\alpha }{3\sqrt{3}}}{(4m{{l}^{2}}/3)}\] \[=\left( \frac{3\sqrt{3}g}{8l}-\frac{8}{\sqrt{3}}\alpha \right)\]
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