JEE Main 2024 — 6 April, Shift 1 — Question 32

$\quad \mathrm{K}_{\mathrm{i}}=\frac{1}{2} \mathrm{~m}(100)^{2}$

$\mathrm{K}_{\mathrm{f}}=\frac{1}{2} \mathrm{~m}(40)^{2}$

$\% \text { loss }=\frac{\left|\mathrm{K}_{\mathrm{f}}-\mathrm{K}_{\mathrm{i}}\right|}{\mathrm{K}_{\mathrm{i}}} \times 100$

$=\frac{\left|\frac{1}{2} m(40)^{2}-\frac{1}{2} m(100)^{2}\right|}{\frac{1}{2} m(100)^{2}} \times 100$

$=\frac{|1600-100 \times 100|}{100}=84 \%$