A 10 cm long wire is placed horizontally on the surface of water and is gently pulled up with a force of 2× 10-2N to keep the wire in equilibrium. The surface tension, in Nm-1 of water is

Question

A 10 cm long wire is placed horizontally on the surface of water and is gently pulled up with a force of 2× 10-2N to keep the wire in equilibrium. The surface tension, in Nm-1 of water is (A) 0.1 (B) 0.2 (C) 0.001 (D) 0.002

Tardigrade Admin · Accepted Answer

: S2l=F (where S is surface tension) Or S=(F/2l)=(2× 10-2/2× 0.1)Nm-1=0.1 NM-1

Q. A 10 cm long wire is placed horizontally on the surface of water and is gently pulled up with a force of $2 \times 10^{- 2} N$ to keep the wire in equilibrium. The surface tension, in $N m^{- 1}$ of water is

0.1

0.2

0.001

0.002

: $S 2 l = F$ (where S is surface tension) Or $S = \frac{F}{2 l} = \frac{2 \times 10 ^{- 2}}{2 \times 0.1} N m^{- 1} = 0.1 N M^{- 1}$

Q. A 10 cm long wire is placed horizontally on the surface of water and is gently pulled up with a force of 2×10−2N to keep the wire in equilibrium. The surface tension, in Nm−1 of water is

0.1

0.2

0.001

0.002

: S2l=F (where S is surface tension) Or S=2lF​=2×0.12×10−2​Nm−1=0.1NM−1

Q. A 10 cm long wire is placed horizontally on the surface of water and is gently pulled up with a force of $2 \times 10^{- 2} N$ to keep the wire in equilibrium. The surface tension, in $N m^{- 1}$ of water is

: $S 2 l = F$ (where S is surface tension) Or $S = \frac{F}{2 l} = \frac{2 \times 10 ^{- 2}}{2 \times 0.1} N m^{- 1} = 0.1 N M^{- 1}$