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In the arrangements as shown Mgt2m. Find the extension in the spring of forces constant `K_(1)`. All pulleys are smooth and massless and the blocks do not oscillates. A. `(Mmg)/(K_(1)(M+m))`B. `(4Mmg).((M+m)).K/(K_(1))`C. `(Mmg)/(K_(1)(M+2m))`D. `(4Mmg)/(K_(1)(M+2m))` |
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Answer» Correct Answer - 4 `a=((M-2m)g)/((M+2m))` Now, `Mg-K_(1)x_(1)=Ma` `x_(1)=(M(g-a))/(K_(1))` `=(4Mmg)/(K_(1)(M+2m))` |
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