JEE MAIN - Physics (2024 - 4th April Morning Shift - No. 24)
A soap bubble is blown to a diameter of $$7 \mathrm{~cm}$$. $$36960 \mathrm{~erg}$$ of work is done in blowing it further. If surface tension of soap solution is 40 dyne/$$\mathrm{cm}$$ then the new radius is ________ cm Take $$(\pi=\frac{22}{7})$$.
คำตอบ
7
คำอธิบาย
$$\begin{aligned}
& \Delta W=8 \pi\left(R_2^2-R_1^2\right) T \\
& 36960=8 \times \frac{22}{7} \times 40\left(R_2^2-\frac{49}{4}\right) \\
& R_2=7 \mathrm{~cm}
\end{aligned}$$