A cubical iron block of side 10 cm is floating on mercury in a vessel. What is the height of the block above the mercury level?
Water is poured into vessel until it just covers the iron block . What is the height of the water column?

Asked by Paresh | 1st Dec, 2015, 08:47: PM

Expert Answer:

begin mathsize 14px style straight i. space The space block space will space float space in space mercury space if space its space weight space is space equal space to space the space weight space of space mercury space displaced space by space it. Density space of space iron comma space straight rho subscript straight i equals 7.2 space straight g divided by cm cubed Density space of space mercury comma space straight rho subscript straight m equals 13.6 space straight g divided by cm cubed Length space of space each space side space of space the space iron space block equals 10 space cm Let space straight x space be space the space height space of space the space block space above space the space mercury space level space in space the space vessel. space Then comma space the space volume space of space the space block space inside space the space mercury space is comma straight V subscript straight m equals 10 space cross times space 10 space cross times space left parenthesis 10 minus straight x right parenthesis This space is space also space equal space to space the space volume space of space mercury space displaced space by space the space block. Weight space of space this space volume space of space mercury space is comma space straight W equals straight V subscript straight m straight rho subscript straight m straight g space equals space straight V subscript straight m space cross times space 13.6 space cross times space 981 space equals space upthrust space on space block Total space volume space of space the space iron space block space is comma straight V equals space 10 space cross times space 10 space cross times space 10 space equals space 1000 space cm cubed Hence comma space its space weight space is comma straight W equals Vρ subscript straight i straight g equals straight V space cross times space 7.2 space cross times space 981 The space iron space block space floats comma space therefore comma space Weight space of space mercury space displaced equals Weight space of space the space iron space block 100 space cross times space left parenthesis 10 minus straight x right parenthesis space cross times space 13.6 space cross times space 981 equals 1000 space cross times space 7.2 space cross times space 981 100 space cross times space left parenthesis 10 minus straight x right parenthesis space cross times space 13.6 space equals 1000 space cross times space 7.2 left parenthesis 10 minus straight x right parenthesis space cross times space 1360 space equals space 7200 10 minus straight x equals 7200 over 1360 10 minus straight x equals 5.29 10 minus 5.29 equals straight h straight x equals 4.71 So space the space height space of space the space block space above space mercury space level equals 4.71 space cm end style
ii. Let the height of the water column be 'h'.
In equilibrium condition: Weight of the iron block = Total upthrust
i.e, V dirong = Upthrust of water + Upthrust of mercury
1000 × 7.2 × g = h × 100 × 1 × g + (10 - h) × 100 × 13.6 g
72 = h + 136 - 13.6 h
72 = 136 - 12.6 h
12.6 h = 136 - 72 
12.6 h = 64 
y = 5.08 cm

Answered by Yashvanti Jain | 2nd Dec, 2015, 10:21: AM

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