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

Question

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 (A)  5× 10-3N/m  (B)  5× 10-2N/m  (C) 5 N/m (D) 50 N/m

Tardigrade Admin · Accepted Answer

Given, r=2 × 10-3 m. F=W=V ρ g=6.28 × 10-4 m Surface tension (T)=(F/2 π r)=(6.28 × 10-4/2 × 3.14 × 2 × 10-3) =0.05 N / m =5 × 10-2 N / m

Q. Radius of a capillary is $2 \times 1 0^{- 3} m . A$ liquid of weight $6.28 \times 1 0^{- 4} N$ may remain in the capillary, then the surface tension of liquid will be

$5 \times 1 0^{- 3} N / m$

$5 \times 1 0^{- 2} N / m$

$5 N / m$

$50 N / m$

Given, $r = 2 \times 1 0^{- 3} m .$
$F = W = V ρ g = 6.28 \times 1 0^{- 4} m$
Surface tension $(T) = \frac{F}{2 π r} = \frac{6.28 \times 1 0 ^{- 4}}{2 \times 3.14 \times 2 \times 1 0 ^{- 3}}$
$= 0.05 N / m = 5 \times 1 0^{- 2} N / m$

Q. 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

5×10−3N/m

5×10−2N/m

5N/m

50N/m