JEE Physics/ Lakshmikanta Satapathy/ Laws of Motion QA part 5/ JEE Question on motion of a balloon and the effect of change in mass on its acceleration solved with the related concepts

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L K Satapathy Laws of Motion 5
Upthrust on Balloon
M
a
M–m
a

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L K Satapathy
Question : A balloon of mass M is descending at
constant acceleration a. When a small mass m is
released from the balloon, it starts rising with the
same acceleration. Assuming that there is no change
in volume, the value of m is
Laws of Motion 5
M
a
M–m
a
2 ( ) ( )
( ) ( ) ( ) ( )
2
Ma Ma M g a M g a
a b c d
g a g a a a
 
 
Answer :
. . . (1)Mg U Ma 
Mg a
U
Let the upthrust on the balloon = U (  )
Weight of the balloon = Mg (  )
Acceleration of the balloon = a (  )
Case 1 : Free Body Diagram for mass M
Applying Newton’s 2nd Law

3. Physics Helpline
L K Satapathy Laws of Motion 5
( ) ( ) . . . (2)U M m g M m a   
(1)&(2) (2 ) 2mg M m a Ma ma    
[ ]
2
( ) 2
Ma
m g a Ma m ns
g
A
a
   

Correct option = (b)
(M-m)g
aU
Let the upthrust on the balloon = U (  )
Weight of the balloon = (M – m)g (  )
Acceleration of the balloon = a ( )
Case 2 : Free Body Diagram for mass (M – m )
Applying Newton’s 2nd Law

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L K Satapathy
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