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Standard XII

Physics

Capillary Action

A ring of ext...

Question

A ring of external and internal radii $r_{1}$ and $r_{2}$ just touches the horizontal surface of a liquid of surface tension $σ$ . The force required to pull the ring away from the surface is

$2 π (r_{1} + r_{2}) σ$

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$2 π (r_{1} - r_{2}) σ$

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$4 π (r_{1} + r_{2}) σ$

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$4 π (r_{1} - r_{2}) σ$

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Solution

The correct option is A
$2 π (r_{1} + r_{2}) σ$

Force required to pull the ring away, $F = 2 π r_{1} σ + 2 π r_{2} σ = 2 π (r_{1} + r_{2}) σ$
Hence, the correct choice is (a).

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A ring of external and internal radii r1 and r2 just touches the horizontal surface of a liquid of surface tension σ. The force required to pull the ring away from the surface is

The correct option is A 2π(r1+r2)σ Force required to pull the ring away, F=2πr1σ+2πr2σ=2π(r1+r2)σ Hence, the correct choice is (a).

A ring of external and internal radii $r_{1}$ and $r_{2}$ just touches the horizontal surface of a liquid of surface tension $σ$ . The force required to pull the ring away from the surface is

The correct option is A
$2 π (r_{1} + r_{2}) σ$

Force required to pull the ring away, $F = 2 π r_{1} σ + 2 π r_{2} σ = 2 π (r_{1} + r_{2}) σ$
Hence, the correct choice is (a).