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JEE Main 2022 · 30 Jun · Shift 1, Physics question 54

The excess pressure inside a liquid drop is 500 Nm−2^{-2}. If the radius of the drop is 2 mm, the surface tension of liquid is x×10−3x\times10^{-3} Nm−1^{-1}. The value of xx is ______.

Answer and worked solution

Official answer: 500

Drop. ΔP=2Tr\Delta P=\frac{2T}r: T=500×2×10−32=0.5=500×10−3T=\frac{500\times2\times10^{-3}}2=0.5=500\times10^{-3} N m−1^{-1}.

Answer: 500

Topics:Surface tension
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