Answer:
The situation is shown in Fig. When the balloon is held stationary in air, the forces acting on it get balanced. Up thrust \[=\] Weight of balloon \[+\] Tension in string \[U=Mg+T\] ...(1) For the small block of mass M / 2 floating stationary in air, \[T=\frac{M}{2}g\] ? (2) From (1) and (2), \[U=Mg+\frac{M}{2}g=\frac{3}{2}Mg\] When the string is cut, \[T=\text{ }0\]. The small block will begin to fall freely. The balloon will rise up with an acceleration \[a\] such that \[U-Mg=Ma\] or \[\frac{3}{2}Mg-Mg=Ma\] or \[a=\frac{g}{2},\]in the upward direction.
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