Ncert Exercises & Solutions

Q.38 A raindrop of mass 1.00 g falling from a height of 1 km hits the ground with a speed of 50 m/s. Calculate

(a) the loss of PE of the drop.
(b) the gain in KE of the drop.
(c) Is the gain in KE equal to the loss of PE? If not why? Take, g 10 m/ $s^{2}$

Hint: Use energy conservation.

Step 1: Find the loss in potential energy of the raindrop.

Given, mass of the raindrop (m) = 100 g
$= 1 \times 10^{- 3} kg$
Height of falling (h) $= 1 km = 10^{3} m$
$g = 10 m / s^{2}$
of the raindrop (v)= 50 m/s

(a) Loss in PE of the drop = mgh
$= 1 \times 10^{- 3} \times 10 \times 10^{3} = 10 J$

Step 2: Find gain in KE of the drop.

(b) Gain in KE of the drop

$\begin{array}{l} = \frac{1}{2} {mv}^{2} \\ = \frac{1}{2} \times 1 \times 10^{- 3} \times (50)^{2} \\ = \frac{1}{2} \times 10^{- 3} \times 2500 \\ = 1 .250 J \end{array}$

Step 3: Compare the above two results and interpret the answer.

(c) No, gain in KE is not equal to the loss in its PE, because a part of PE is utilized in doing work against the viscous drag of air.

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions