A raindrop of mass 1.00 g falling from a height of 1 km hits the ground with a speed of 50 ms-1. Calculate
(a) The loss of P.E. of the drop.
(b) The gain in K.E. of the drop.
(c) Is the gain in K.E. equal to loss of P.E.? If not why. Take g = 10 ms-2
A raindrop of mass 1.00 g falling from a height of 1 km hits the ground with a speed of 50 ms-1. Calculate
(a) The loss of P.E. of the drop.
(b) The gain in K.E. of the drop.
(c) Is the gain in K.E. equal to loss of P.E.? If not why. Take g = 10 ms-2
This is a short answer type question as classified in NCERT Exemplar
mass of the rain drop = 1g=1
Height of falling h= 1km = 103m and g = 10m/s2 and sped of drop =50m/s
(a) Loss of PE of the drop =mgh= 1
(b) Gain in KE of the drop = 1/2mv2= ½ =1.250J
(c) No gain in KE is not equal to the loss in
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Using Newton’s formula,
This is a multiple choice answer as classified in NCERT Exemplar
This is a multiple choice answer as classified in NCERT Exemplar
(b) conserving energy between “O” ans ”A”

Ui + Ki = Uf + Kf
0+1/2mv2= mgh + 1/2mv’
(v’)2=v2-2gh = v’= ……….1
Let speed after emerging be v1 then
=1/2mv12=1/2[1/2mv’2]
1/2m(v1)2=1/4m(v’)2=1/4m[v2-2gh]
V1= ………….2
From eqn 1 and 2
So v1 = v’/
This is a multiple choice answer as classified in NCERT Exemplar
(b, d) When a man of mass m climbs up the staircases of height L, work done by the gravitational force on the man is mgl work done by internal muscular forces will be mgL as the change in kinetic is almost zero.
Hence total work done =-
This is a multiple choice answer as classified in NCERT Exemplar
(c) m =150g =3/20kg
Time of contact =0.001s
U=126km/h=
V= -35m/s
Change in momentum of the ball = m (v-u)=
=21/2
F= dp/dt=- = - 1.05
Here – negative sign indicates that force will be opposite to the direction of movement of the ball be
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Physics NCERT Exemplar Solutions Class 11th Chapter Six 2025
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