PYQ NEET- Work Energy And Power L-8

Question: Consider a drop of rain water having mass $1 \mathrm{~g}$ falling from a height of $1 \mathrm{~km}$. It hits the ground with a speed of $50 \mathrm{~m} \mathrm{~s}^{-1}$. Take ‘g’ constant with a value $10 \mathrm{~m} \mathrm{~s}^{-2}$. The work done by the (i) gravitational force and the (ii) resistive force of air is#

A) (i) $1.25 \mathrm{j}$ (ii) $-8.25 \mathrm{~J}$

B) (i) $100 \mathrm{j}$ (ii) $8.75 \mathrm{~J}$

C) (i) $10 \mathrm{j}$ (ii) $-8.75 \mathrm{~J}$

D) (i) $-10 \mathrm{j}$ (ii) $-8.25 \mathrm{~J}$

Answer: (i) $10 \mathrm{j}$ (ii) $-8.75 \mathrm{~J}$#

Solution:#

From work-energy theorem, $$ \mathrm{W}{\mathrm{g}}+\mathrm{W}{\mathrm{a}}=\Delta \mathrm{K} \cdot \mathrm{E} $$ or $\mathrm{mgh}+\mathrm{W}_{\mathrm{a}}=\frac{1}{2} m v^2-0$ $10^{-3} \times 10 \times 10^3+W_a=\frac{1}{2} \times 10^{-3} \times(50)^2$ $$ \Rightarrow W_a=-8.75 J $$ which is the work done due to air resistance

Work done due to gravity = mgh $$ =10^{-3} \times 10 \times 10^3=10 \mathrm{~J} $$