A small hollow sphere which has a small hole in it is immersed in water to a depth of 20 cm before any water is penetrated into it. If the surface tension of water is 0.073 Nm -1, what is the radius of the hole?

Question

A small hollow sphere which has a small hole in it is immersed in water to a depth of 20 cm before any water is penetrated into it. If the surface tension of water is 0.073 Nm -1, what is the radius of the hole? (A) 0.07 mm (B) 0.04 mm (C) 0.01 mm (D) 0.05 mm

Tardigrade Admin · Accepted Answer

P =(2 S / R ) ⇒ R =(2 S / P ) =(2 × 0.073/0.2 × 103 × 9.8) =0.0744 × 10-3 R =0.07 mm

Q. A small hollow sphere which has a small hole in it is immersed in water to a depth of $20 c m$ before any water is penetrated into it. If the surface tension of water is $0.073 N m^{- 1}$ , what is the radius of the hole?

$0.07 mm$

$0.04 mm$

$0.01 mm$

$0.05 mm$

$P = \frac{2 S}{R}$
$\Rightarrow R = \frac{2 S}{P}$
$= \frac{2 \times 0.073}{0.2 \times 1 0 ^{3} \times 9.8}$
$= 0.0744 \times 1 0^{- 3}$
$R = 0.07 mm$

Q. A small hollow sphere which has a small hole in it is immersed in water to a depth of 20cm before any water is penetrated into it. If the surface tension of water is 0.073Nm−1, what is the radius of the hole?

0.07mm

0.04mm

0.01mm

0.05mm