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For the arrangement shown in the figure, the reading of spring balance is A. `50 N`B. `100 N`C. `150 N`D. None of the above |
Answer» Correct Answer - D `a = ("Net pulling force")/("Total mass") = (10 xx 10 - 5 xx 10)/(10 + 5) = (10)/(3) m//s ^(2)` `5 kg` `T - 5 xx 10 = 5 xx a = (50)/(3)` :. `T = (200)/(3) N` This is also the reading of spring balance. |
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