A balloon of mass M is descending at a consta | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $F$ be the upthrust (buoyant force) acting on the balloon.When the balloon is descending with acceleration $\alpha$,the net force equation is:

Vedclass · Accepted Answer

$\left[ \frac{2\alpha}{\alpha + g} \right] M$

Vedclass · Answer

(B) Let $F$ be the upthrust (buoyant force) acting on the balloon.When the balloon is descending with acceleration $\alpha$,the net force equation is:$Mg - F = M\alpha$  --- $(i)$When a mass $m$ is released,the new mass of the balloon becomes $(M - m)$. It starts rising with the same acceleration $\alpha$. The net force equation is:$F - (M - m)g = (M - m)\alpha$  --- $(ii)$From equation $(i)$,we have $F = Mg - M\alpha = M(g - \alpha)$.Substitute this value of $F$ into equation $(ii)$:$M(g - \alpha) - (M - m)g = (M - m)\alpha$$Mg - M\alpha - Mg + mg = M\alpha - m\alpha$$mg + m\alpha = 2M\alpha$$m(g + \alpha) = 2M\alpha$$m = \left[ \frac{2\alpha}{\alpha + g} \right] M$

$A$ balloon of mass $M$ is descending at a constant acceleration $\alpha$. When a mass $m$ is released from the balloon,it starts rising with the same acceleration $\alpha$. Assuming that its volume does not change,what is the value of $m$?

Similar Questions

$A$ block of mass $M$ is placed on a horizontal surface and it is tied with an inextensible string to a block of mass $m$,as shown in the figure. $A$ block of mass $m_0$ is also placed on $M$. If there is no friction between any two surfaces,then:

The curve between net force and time is shown. Initially,the particle is at rest. Which of the following best represents the resulting velocity-time graph of the particle?

$A$ man is standing on a spring platform. The reading of the spring balance is $60\, kg$ wt. If the man jumps off the platform,what happens to the reading of the spring balance?

Vedclass Test Series

Exam Paper Generator

Online Exam Module