Question #a28fb

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Question #a28fb

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Answer 1 · 2016-11-28T13:09:35

Given
$wt of sample in air = 295 N$ $"wt of sample in air"=295N$

$wt of sample in alcohol = 200 N$ $"wt of sample in alcohol"=200N$

$sp.gravity of alcohol = 0.7$ $"sp.gravity of alcohol"=0.7$

$density of alcohol = 0.7 g c m^{- 3} = \frac{0.7 \cdot 10^{- 3} k g}{10^{- 6} m^{3}} = 700 k g m^{- 3}$ $"density of alcohol"=0.7gcm^-3=(0.7*10^-3kg)/(10^-6m^3)=700kgm^-3$

Now wt of displaced alcohol by the sample

$= Its wt in air - wt in alcohol$ $="Its wt in air - wt in alcohol"$

$= 295 - 200 = 95 N$ $=295-200=95N$

a) Now if $v m^{3}$ $vm^3$ be the volume of the sample,then the weight of displaced alcofol by the sample

$= v \times 700 \times 9.8 N$ $=vxx700xx9.8N$

So $v \times 700 \times 9.8 = 95$ $vxx700xx9.8=95$

$\Rightarrow v = \frac{95}{700 \times 9.8} \approx 0.01385 m^{3}$ $=>v=95/(700xx9.8)~~0.01385m^3$

b) Density of the material

$= \frac{mass}{volume}$ $="mass"/"volume"$

$= \frac{\frac{295}{9.8}}{\frac{95}{700 \times 9.8}} m^{3}$ $=(295/9.8)/(95/(700xx9.8))m^3$

$= \frac{295 \times 700}{95} \approx 2173.7 k g m^{- 3}$ $=(295xx700)/95~~2173.7kgm^-3$

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