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 Multiple Choice QuestionsMultiple Choice Questions

11.

On the heating water, bubbles being formed at the bottom of the vessel detach and rise. Take the bubbles to be spheres of radius R and making a circular contact of radius r with the bottom of the vessel. If r < < R and the surface tension of water is T, value of r just before bubbles detach is (density of water is ρw)

  • straight R squared square root of fraction numerator 2 straight rho subscript straight w straight g over denominator 3 straight T end fraction end root
  • straight R squared square root of fraction numerator straight rho subscript straight w straight g over denominator 6 straight T end fraction end root
  • straight R squared square root of fraction numerator straight rho subscript straight w straight g over denominator straight T end fraction end root
  • straight R squared space square root of fraction numerator 3 straight rho subscript straight w straight g over denominator straight T end fraction end root

A.

straight R squared square root of fraction numerator 2 straight rho subscript straight w straight g over denominator 3 straight T end fraction end root

The bubble will detach if, 
Buoyant force ≥ Surface tension force

4 over 3 πR cubed straight rho subscript straight w straight g space greater or equal than integral straight T space straight x space dl space sin space straight theta space


left parenthesis straight rho subscript straight w right parenthesis open parentheses 4 over 3 πR cubed close parentheses straight g greater or equal than left parenthesis straight T right parenthesis left parenthesis 2 πpr right parenthesis space sin space straight theta
sin space straight theta space equals space straight r over straight R
solving space straight r space equals space square root of fraction numerator 2 straight p subscript straight w straight R to the power of 4 straight g over denominator 3 straight T end fraction end root space equals space straight R squared square root of fraction numerator 2 straight rho subscript straight w straight g over denominator 3 straight T end fraction end root

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12.

The following observations were taken for determining surface tension T of water by the capillary method :
Diameter of capillary, D = 1.25 × 10–2 m rise of water, h = 1.45 × 10–2 m
Using g = 9.80 m/s2 and the simplified relation T = (rhg/2) x103 n/m, the possible error in surface tension is closest to:

  • 2.4%

  • 10%

  • 0.15%

  • 1.5%


D.

1.5%

straight T space equals space rhg over 2 space straight x space 10 cubed
fraction numerator increment straight T over denominator straight T end fraction space equals space fraction numerator increment straight r over denominator straight r end fraction space plus space fraction numerator increment straight h over denominator straight h end fraction space plus 0
100 space straight x space fraction numerator increment straight T over denominator straight T end fraction space equals space open parentheses fraction numerator 10 to the power of negative 2 end exponent space straight x space.01 over denominator 1.25 space straight x space 10 to the power of negative 2 end exponent end fraction space plus fraction numerator 10 to the power of negative 2 end exponent space straight x space 0.1 over denominator 1.45 space straight x space 10 to the power of negative 2 end exponent end fraction close parentheses 100
space equals space 0.8 space plus 0.689
equals space 1.489
space 100 space straight x space fraction numerator increment straight T over denominator straight T end fraction space equals space 1.489 percent sign
space approximately equal to space 1.5 space percent sign
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13.

An open glass tube is immersed in mercury in such a way that a length of 8 cm extends above the mercury level. The open end of the tube is then closed and sealed and the tube is raised vertically up by additional 46 cm. What will be the length of the air column above mercury in the tube now? (Atmospheric pressure = 76 cm of Hg)

  • 16 cm

  • 22 cm

  • 38 cm

  • 6 cm


A.

16 cm

Since the system is accelerating horizontally such that no component of acceleration in the vertical direction. Hence, the pressure in the vertical direction will remain unaffected.

straight p subscript 1 space equals straight p subscript 0 space plus ρgh

.
For air trapped in tube, p1V1 = p2V2
p1 = patm = pg76
V1 = A.8 [ A = area of cross section]
p2 = patm - ρg(54-x) =  ρg(22+x)
V2 = A.x
ρg76 x 8A = ρg (22+x) Ax
x2+22x-78x 8
by solving, x = 16

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