JEE MAIN - Physics (2022 - 30th June Morning Shift - No. 21)
The excess pressure inside a liquid drop is 500 Nm$$-$$2. If the radius of the drop is 2 mm, the surface tension of liquid is x $$\times$$ 10$$-$$3 Nm$$-$$1. The value of x is _____________.
Answer
500
Explanation
$\mathrm{P}=\mathrm{P}_{0}+\frac{2 T}{R} $
$\Rightarrow P-P_{0}=\frac{2 T}{R}$
$$ \begin{aligned} &500=\frac{2 \times T}{2 \times 10^{-3}} \\\\ &T=500 \times 10^{-3} \\\\ &\text { So, } x=500 \end{aligned} $$
$\Rightarrow P-P_{0}=\frac{2 T}{R}$
$$ \begin{aligned} &500=\frac{2 \times T}{2 \times 10^{-3}} \\\\ &T=500 \times 10^{-3} \\\\ &\text { So, } x=500 \end{aligned} $$
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