A balloon of mass M is descending at a constant acceleration a. When a mass m is released from the balloon, it starts rising with the same acceleration a. Assuming that its volume does not change, what is the value of m
A balloon of mass M is descending at a constant acceleration a. When a mass m is released from the balloon, it starts rising with the same acceleration a. Assuming that its volume does not change, what is the value of m
The correct option is B
You know that hot air balloons can float in air go up or come down. And this is possible because of one extra force of buoyancy provided by air In the the upward direction. This you have read in 9th class remember. So if upward force of air is greater than the balloon's weight then it will rise up. (also force of buoyancy depends upon the volume of the object)
Condition 1 :
When the balloon had mass M its weight was Mg downward. F upward and net acceleration a downwards.
Net external force should be in the direction of net acceleration
⇒ Mg - Fb = Ma ---------------(i)
Condition 2 :
Now mass m is thrown off so balloon's weight is (M - m)g, But buoyant force will be same as it depends on volume which hasn't changed.Now net acceleration is a
upwards
⇒Fb−(M−m)g=(M−m)a ------------------(ii)
Solving (I) & (II)
m = (2aa+g)M
You know that hot air balloons can float in air go up or come down. And this is possible because of one extra force of buoyancy provided by air In the the upward direction. This you have read in 9th class remember. So if upward force of air is greater than the balloon's weight then it will rise up. (also force of buoyancy depends upon the volume of the object)
Condition 1 :
When the balloon had mass M its weight was Mg downward. F upward and net acceleration a downwards.
Net external force should be in the direction of net acceleration
⇒ Mg - Fb = Ma ---------------(i)
Condition 2 :
Now mass m is thrown off so balloon's weight is (M - m)g, But buoyant force will be same as it depends on volume which hasn't changed.Now net acceleration is a
upwards
⇒Fb−(M−m)g=(M−m)a ------------------(ii)
Solving (I) & (II)
m = (2aa+g)M
