A) \[\frac{2Mm\ell }{(M-m)\,{{t}^{2}}}\]
B) \[\frac{2Mm\ell }{(M+m)\,{{t}^{2}}}\]
C) \[\frac{Mm\ell }{(M-m)\,{{t}^{2}}}\]
D) \[\frac{4Mm\ell }{(M-m)\,{{t}^{2}}}\]
Correct Answer: A
Solution :
Let \[{{a}_{1}}\] and \[{{a}_{2}}\] be the acceleration of M and m respectively. Then \[Mg-F=M{{a}_{1}}\] .........(1) and \[mg-F=m{{a}_{2}}\] .........(2) Now, \[\ell +\frac{1}{2}{{a}_{2}}{{t}^{2}}=\frac{1}{2}{{a}_{1}}{{t}^{2}}\]or \[{{a}_{1}}=\frac{2\ell }{{{t}^{2}}}+{{a}_{2}}\] .........(3) Solving eq. (1), (2) and (3) we get \[F=\frac{2Mm\ell }{(M-m){{t}^{2}}}\]
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