$\begin{aligned}
& W_1=s 4 \pi\left(\frac{R}{3}\right)^2 \times 27-4 \pi R^2 \cdot s \
& \Rightarrow W_1=4 \pi R^2 \cdot s(2)=10 \text { Joule }
\end{aligned}Now,W_2=s 4 \pi R^2(4-1)=4 \pi R^2 s \times 3\Rightarrow \quad W_2=3 \times \frac{10}{2}=15 \text { Joule }$
The amount of work done to break a big water drop of radius ' R ' into 27 small drops of equal radius is 10 J . The work done required to break the…
JEE Main 2025 — Physics Mechanics
The amount of work done to break a big water drop of radius ' R ' into 27 small drops of equal radius is 10 J . The work done required to break the same big drop into 64 small drops of equal radius will be
Held on 24 Jan 2025 · Verified 6 Jul 2026.
Options
- A
15 J
- B
5 J
- C
20 J
- D
10 J
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