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 × 10 - 3 Nm ⁻¹. The value of x is _____________. Numerical answer type. Work it out, then open the answer.

Answer: 500. Liquid drop: Δ P = 2T/R ⇒ T = (Δ P · R)/(2). T = 500 × 2 × 10⁻³2 = 0.5 N/m = 500 × 10⁻³ N/m So x = 500.

The excess pressure inside a liquid drop is 500 Nm ⁻². If the radius… | JEE Main 2022 (30 Jun, Shift 1)

Q41JEE Main 2022 · 30 JUN · SHIFT 1Easy

Excess Pressure: Drops & Bubbles

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 _____________.

Numerical answer type. Work it out, then open the answer.

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Answer: 500

Liquid drop: $\Delta P = \frac{2T}{R} \Rightarrow T = \frac{\Delta P \cdot R}{2}$ .
$T = \frac{500 \times 2 \times 10^{-3}}{2} = 0.5$ N/m $= 500 \times 10^{-3}$ N/m
So $x = 500$ .

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