Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

Consider a drop of rainwater having a mass of $1\text{ g}$ falling from a height of $1\text{ km}$ . It hits the ground with a speed of $50\text{ m/s}$ . Take $g$ as constant with a value $10\text{ m/s}^2$ . The work done by the (i) gravitational force and the (ii) resistive force of air is:

(i) $1.25\text{ J}$ ; (ii) $-8.25\text{ J}$

(i) $100\text{ J}$ ; (ii) $8.75\text{ J}$

(i) $10\text{ J}$ ; (ii) $-8.75\text{ J}$

(i) $-10\text{ J}$ ; (ii) $-8.75\text{ J}$

Step-by-Step Solution

Calculate Work Done by Gravity ( $W_g$ ): The work done by the gravitational force is $W_g = mgh$ . Given: $m = 1\text{ g} = 10^{-3}\text{ kg}$ , $h = 1\text{ km} = 1000\text{ m}$ , $g = 10\text{ m/s}^2$ . $W_g = 10^{-3} \times 10 \times 1000 = 10\text{ J}$ [Class 11 Physics, Ch 6, Sec 5.2, Example 5.2]
Calculate Change in Kinetic Energy ( $\Delta K$ ): The drop starts from rest ( $v_i=0$ ) and reaches a final speed $v_f=50\text{ m/s}$ . $\Delta K = K_f - K_i = \frac{1}{2}mv_f^2 - 0 = \frac{1}{2} \times 10^{-3} \times (50)^2 = \frac{2500}{2000} = 1.25\text{ J}$
Apply Work-Energy Theorem: The change in kinetic energy is equal to the total work done by all forces (gravity + resistive force $W_r$ ). $\Delta K = W_g + W_r$ $1.25\text{ J} = 10\text{ J} + W_r$ $W_r = 1.25 - 10 = -8.75\text{ J}$ Thus, the work done by gravity is $10\text{ J}$ and by the resistive force is $-8.75\text{ J}$ .

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started