NEET Practice Questions (MCQs) with Answers & Solutions

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The work done in blowing a soap bubble of radius 0.2 m is (the surface tension of soap solution being 0.06 N/m)

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Explanation

(a) W=8πr2×T=8π×(0.2)2×0.06=192 π×10-4 J

A liquid film is formed in a loop of area 0.05 m2. Increase in its potential energy will be (T = 0.2 N/m)

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Explanation

(b) Increment in Potential energy = T×A

=0.02×2×0.05=2×10-2J

In order to float a ring of area 0.04 m2 in a liquid of surface tension 75 N/m, the required surface energy will be

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Explanation

(a) E=T×A=75×0.04=3J

If two soap bubbles of equal radii r coalesce then the radius of curvature of interface between two bubbles will be

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Explanation

(c) r=r1r2r2-r1=  since r1=r2

When the temperature is increased the angle of contact of a liquid 

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Explanation

(b) Cohesive force decreases so angle of contact decreases.

The angle of contact between glass and mercury is

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Explanation

(d) The angle of contact between glass and mercury is 135°.

A mercury drop does not spread on a glass plate because the angle of contact between glass and mercury is

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Explanation

(b) The mercury does not spread on a glass plate because the angle of contact between glass and mercury is obtuse.

The liquid meniscus in the capillary tube will be convex if the angle of contact is

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Explanation

(a)

For convex meniscus the angle of contact is greater than 90°

The value of contact angle for kerosene with solid surface.

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Explanation

(a) The angle of contact will be equal to zero as it completely wets the solid surface.

Nature of meniscus for liquid of 00 angle of contact

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Explanation

(c) When angle of contact = 0° then, the meniscus for liquid will be semi-spherical.

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