
Solution
If B is upthrust of air on balloon,and a is downward acceleration, then
Mg–B=Ma
⇒ a=Mg−BM=gVρairGVρCO2 =(1−VρairVρCO2)g=(1−28.844)×9.8m/s^{2}
=3.4m/s^{2}
A couple produces a

Solution
Two forces equal in magnitude but opposite in direction form a couple which tends to rotate the body.

Solution
τ=Iα
Two bodies A and B have masses M and m respectively where M>m and they are at a distance d apart. Equal force is applied to them so that they approach each other. The position where they hit each other is

Solution
As net external force on the system is zero therefore position of their centre of mass remains unaffected i.e.they will hit each other at the point of centre of mass.The centre of mass of the system lies nearer to A because
MA > MB.
A gymnast takes turns with her arms and legs stretched.When shepulls her arms and legs

Solution
Since no external torque act on gymnast, so angular momentum (L = Iw)is conserved. After pulling her arms & legs, the angular velocity increases but moment of inertia of gymnast, decreases in, such a way that angular momentum remains constant.

Solution
The boy does notexert atorque to rotating table by jumping, so angular momentum is conserved i.e.,dL⃗ dt=0
⇒L⃗ =constant.
For inelastic collision between two spherical rigid bodies

Solution
For inelastic collision between two spherical rigid bodies K.E. is not conserved as it is converted into other forms like heat, light sound etc. but total linear momentum is conserved.
In an explosion, a body breaks up into two pieces of unequal masses. In this

Solution
Both part will have numerically equal momentum and lighter part will have more velocity.
A jet plane moves up in air because

Solution
When jet plane flies, it ejects gases in backward directionat very high velocity. From Newton’s third law, these gases provides the momentum to jet plane in forward direction plus compensates the force of gravity.
For a body falling freely under gravity from a height