Question #9ed66

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Question #9ed66

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Answer 1 · 2016-11-28T14:10:44

Given

$wt of the object in air = 320 N$ $"wt of the object in air"=320N$

$wt of the object in water = 255 N$ $"wt of the object in water"=255N$

$wt of the object in oil = 295 N$ $"wt of the object in oil"=295N$

We know
$the density of water = d_{w} = 1000 k g m^{- 3}$ $"the density of water"= d_w=1000kgm^-3$

$acceleration due to gravity g = 9.8 m s^{- 2}$ $"acceleration due to gravity "g=9.8ms^-2$

Wt of displaced water by the object
$= Its wt in air - wt in water$ $="Its wt in air"-"wt in water"$

$= 320 - 255 = 65 N$ $=320-255=65N$

So mass of displaced water $m_{w} = \frac{65}{9.8} k g = 6.63 k g$ $m_w=65/9.8kg=6.63kg$

Volume of displaced water or volume of the object $v = \frac{m_{w}}{d_{w}} = \frac{6.63}{1000} m^{3} = 6.63 \times 10^{- 3} m^{3}$ $v=m_w/d_w=6.63/1000m^3=6.63xx10^-3m^3$

$mass of the object = m = \frac{its wt}{g} = \frac{320}{9.8} k g = 32.65 k g$ $"mass of the object"=m ="its wt"/g=320/9.8kg=32.65kg$

a) $density of the object = d = \frac{m}{v} = \frac{32.65}{6.63 \times 10^{- 3}} k g m^{- 3} = 4924.6 k g m^{- 3}$ $"density of the object"=d=m/v=32.65/(6.63xx10^-3)kgm^-3=4924.6kgm^-3$

b) The wt of displaced oil by the object
$= Its wt in air - wt in oil$ $="Its wt in air"-"wt in oil"$

$= 320 - 295 = 25 N$ $=320-295=25N$

If the density of oil be $d_{o}$ $d_o$

then

$v \times d_{o} \times g = 25$ $vxxd_o xxg=25$

$\Rightarrow d_{o} = \frac{25}{v \times g} = \frac{25}{6.63 \times 10^{- 3} \times 9.8} k g m^{- 3}$ $=>d_o=25/(vxxg)=25/(6.63xx10^-3xx9.8)kgm^-3$

$= 384.77 k g m^{- 3}$ $=384.77kgm^-3$

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Question #9ed66

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