Radius of a capillary is 2 × 10-3 m. A liquid of weight 6.2 × 10-4 N may remain in the capillary. Then the surface tension of liquid will be :

Question

Radius of a capillary is 2 × 10-3 m. A liquid of weight 6.2 × 10-4 N may remain in the capillary. Then the surface tension of liquid will be : (A) 5 × 10-3 N / m (B) 5 × 10-2 N / m (C) 5 N / m (D) 50 N / m

Tardigrade Admin · Accepted Answer

2 π R T cos θ=mg (θ=0°) T=(m g/2 π R) =(6.2 × 10-4/2 × 3.14 × 2 × 10-3) T=5 × 10-2 N / m

Q. Radius of a capillary is $2 \times 1 0^{- 3} m$ . A liquid of weight $6.2 \times 1 0^{- 4} N$ may remain in the capillary. Then the surface tension of liquid will be :

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

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

$5 N / m$

$50 N / m$

$2 π RT cos θ = m g (θ = 0^{\circ})$
$T = \frac{m g}{2 π R}$
$= \frac{6.2 \times 1 0 ^{- 4}}{2 \times 3.14 \times 2 \times 1 0 ^{- 3}}$
$T = 5 \times 1 0^{- 2} N / m$

Q. Radius of a capillary is 2×10−3m. A liquid of weight 6.2×10−4N 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