Radius of a capillary is 2×10−3m. A liquid of weight 6.28×10−4N may remain in the capillary then the surface tension of liquid will be

T=F2πr=6.28×10−42×3.14×2×10−3=5×10−2N/m

T=F2πr=6.28×10−42×3.14×2×10−3=5×10−2N/m

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Radius of a capillary is $2 \times 10^{- 3} m .$ A liquid of weight $6 .28 \times 10^{- 4} N$ may remain in the capillary then the surface tension of liquid will be

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$5 \times 10^{- 3} N / m$

$5 \times 10^{- 2} N / m$

$5 N / m$

$50 N / m$

answer is B.

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$T = \frac{F}{2 πr} = \frac{6 .28 \times 10^{- 4}}{2 \times 3 .14 \times 2 \times 10^{- 3}} = 5 \times 10^{- 2} N / m$

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