A wire of length 20 cm is placed horizontall | vedclass

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Question

(B) The wire is in contact with the water surface on both sides. Therefore,the total length of the wire in contact with the surface is $L_{total} = 2

Vedclass · Accepted Answer

$0.0364$

Vedclass · Answer

(B) The wire is in contact with the water surface on both sides. Therefore,the total length of the wire in contact with the surface is $L_{total} = 2 	imes L = 2 	imes 20 \ cm = 40 \ cm = 0.4 \ m$.When the wire is pulled up,the force due to surface tension acts downwards,opposing the pull. For equilibrium,the applied force $F$ must balance the surface tension force $F_s$.The force due to surface tension is given by $F_s = T 	imes L_{total}$,where $T$ is the surface tension.Given $F = 1.456 	imes 10^{-2} \ N$ and $L_{total} = 0.4 \ m$.Equating the forces: $1.456 	imes 10^{-2} = T 	imes 0.4$.Solving for $T$: $T = \frac{1.456 	imes 10^{-2}}{0.4} = 3.64 	imes 10^{-2} \ N \ m^{-1} = 0.0364 \ N \ m^{-1}$.

$A$ wire of length $20 \ cm$ is placed horizontally on the surface of water and is gently pulled up with a force of $1.456 \times 10^{-2} \ N$ to keep the wire in equilibrium. The surface tension of water is (in $N \ m^{-1}$)

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