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Physics Question on Surface tension

A capillary tube of radius rr can support a liquid of weight 6.28×10−4 N6.28 \times 10^{-4} \,N. If the surface tension of the liquid is 5×10−2 N/m5 \times 10^{-2} \,N / m. The radius of capillary must be

A

2.0×10−3m2.0 \times 10^{-3} m

B

20×10−4m20 \times 10^{-4} m

C

1.5×10−3m1.5 \times 10^{-3} m

D

12.5×10−4m12.5 \times 10^{-4} m

Answer

2.0×10−3m2.0 \times 10^{-3} m

Explanation

Solution

W=T.2πrW=T . 2 \pi r
r=W2πT=6.28×10−42×3.14×5×10−2r =\frac{W}{2 \pi T}=\frac{6.28 \times 10^{-4}}{2 \times 3.14 \times 5 \times 10^{-2}}
=2×10−3m=2 \times 10^{-3} m