Question #43024

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Question #43024

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Answer 1 · 2016-11-28T10:46:47

Given

$the length of edge of the cube = 18.3 c m$ $"the length of edge of the cube"=18.3cm"$
$density of the cube = 651 k g m^{- 3} = \frac{651 \times 10^{3} g}{10^{6} c m^{3}} = 0.651 g c m^{- 3}$ $"density of the cube"=651"kgm^-3=(651xx10^3g)/(10^6cm^3)=0.651gcm^-3$
So volume of the cube $= {18.3}^{3} c m^{3}$ $=18.3^3cm^3$

So mass of the cube $= {18.3}^{3} \times 0.651 g$ $=18.3^3xx0.651g$

Let x cm be the length of the edge of the cube is under water when it is floating in water.

So by the condition of floating the weight of the displaced water by the immersed portion of the cube is equal to the weight of the cube.
Taking density of water $= 1 g c m^{- 3}$ $=1gcm^-3$ we can write

${18.3}^{2} \times x \times 1 \times g = {18.3}^{3} \times 0.651 \times g$ $18.3^2xx x xx1xxg=18.3^3xx0.651xxg$

$\Rightarrow x = 18.3 \times 0.651 = 11.9133 c m$ $=>x=18.3xx0.651=11.9133cm$

a) So the distance from horizontal top surface of cube to water level

$18.3 - x = 18.3 - 11.9133 = 6.3867 c m$ $18.3-x=18.3-11.9133=6.3867cm$

b) Let the mass of lead to be kept on floating cube to just submerge it is $m$ $m$

Then again by condition of floating

$m + {18.3}^{3} \times 0.651 \times g = {18.3}^{3} \times 1 \times g$ $m+18.3^3xx0.651xxg=18.3^3xx1xxg$

$\Rightarrow m = {18.3}^{3} (1 - 0.651) \times 10^{- 3} k g$ $=>m=18.3^3(1-0.651)xx10^-3kg$

$\Rightarrow m = 2.1388 k g$ $=>m=2.1388kg$

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