JEE Mains · Physics · STD 11 - 9.2 surface tension
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
- A 15 J
- B 5 J
- C 20 J
- D 10 J
Answer & Solution
Correct Answer
(A) 15 J
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & W_1=s 4 \pi\left(\frac{R}{3}\right)^2 \times 27-4 \pi R^2 \cdot s \\ & \Rightarrow W_1=4 \pi R^2 \cdot s(2)=10 \text { Joule } \end{aligned}\) Now, \(W_2=s 4 \pi R^2(4-1)=4 \pi R^2 s \times 3\) \(\Rightarrow \quad W_2=3 \times \frac{10}{2}=15 \text { Joule }\)
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