MCQ
A block is attached to a spring as shown and very-very gradually lowered so that finally spring expands by $"d"$. If same block is attached to spring & released suddenly then maximum expansion in spring will be-


- A$d$
- ✓$2\,d$
- C$3\,d$
- D$4\,d$

therefore $\mathrm{mg}=\mathrm{kx}$
$\therefore \mathrm{x}=\frac{\mathrm{mg}}{\mathrm{k}}=\mathrm{d}(\mathrm{given})$
In second case block variably accelerate from rest
to rest
$0 =-\operatorname{mg} \mathrm{x}+\frac{1}{2} \mathrm{Kx}^{2}$
$\therefore \quad \mathrm{x} =\frac{2 \mathrm{mg}}{\mathrm{K}}=2 \mathrm{d}$
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