A) \[\frac{2M\left( g \right)}{a}\]
B) \[\frac{M\left( g+a \right)}{g}\]
C) \[\frac{Mg}{g+a}\]
D) \[\frac{2Ma}{g+a}\]
Correct Answer: D
Solution :
[d]Up thrust F will remain same in both the cases. | |
Equations of motion be: | |
\[MgF=Ma\] | ...(1) |
and \[F-(M-m)g=(M-m)a\] | ...(2) |
solving equations (1) and (2) we get, mass to be removed \[m=\frac{2Ma}{g+a}\] |
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