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Two blocks, each having a mass M, rest on frictionless surfaces as shown in the figure. If the pulleys are light and frictionless and M on the incline is allowed to move down, then the tension in string will be A. `(2)/(3) Mg sin theta`B. `(3)/(2) Mg sin theta`C. `(Mg sin theta)/(2)`D. `2Mg sin theta` |
Answer» Correct Answer - C ( c) Acceleration of system , `a=(" Net pulling force")/(" Total mass")=(Mg sin theta)/(2M)` `a=(1)/(2)g sin theta` Now, the block on ground is moving due to tension. Hence, `T=Ma=(Mg sin theta)/(2)` |
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