Question

A piece of iron of density 7.8 x 10 power3 kg/m3 and volume 100 cm is totally immersed in water. Calculate :
(a) the weight of the iron piece in air
(b) the upthrust and
(c) apparent weight in water.​

Answer 1

✰ Aиѕωєrs —

• (a) Weight of iron piece in air = 7.65 N
• (b) upthrust = 0.98 N
• (c) apparent weight in water = 6.67 N

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✰ Giνєи —

• Density of piece of iron = $\sf{7.8\times 10^3 kgm^3}$
• Volume = 100 cm

✰ Tσ Fiиd —

• (a) The weight of iron piece in air
• (b) The upthrust
• (c) apparent weight in water

✰ Sσℓυтiσи –

❍ The weight of the iron piece in air :

$\qquad\boxed{\sf{\leadsto weight=m.g}}$

$\qquad\rm{\implies Weight=\big\lgroup( 7.8\times 10^3)\times(100\times 10^{-6})\big\rgroup\times 9.8}$

$\qquad\rm{\implies Weight=0.78\times 9.8}$

$\qquad\rm{\implies Weight=7.65\:N}$

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❍ The upthrust :

$\qquad\boxed{\sf{\leadsto upthrust (F)= pgV}}$

where,

• p = density of liquid (1000kg/m³)
• g = gravitational constant (9.8m/s²)
• V = volume of immensed object${100\times 10^{-6}}$

$\qquad\rm{\implies F = pgV}$

$\qquad\rm{\implies F = 1000\times 9.8\times 100\times 10^{-6} }$

$\qquad\rm{\implies F = 0.98\:N}$

$\qquad\rm{\implies Upthrust = 0.98\:N}$

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❍ Apparent weight when body will be in water :

$\qquad\boxed{\sf{\leadsto Real\: weight-Upthrust}}$

$\qquad\rm{\implies W' = W - F}$

$\qquad\rm{\implies W' = 7.65-0.98}$

$\qquad\rm{\implies W' = 6.67\:N }$

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Answer 2

Answer:

upthrust=7.8*10^3g

Explanation:

upthrust=weight of liquid displaced by the piece of iron

              =mass of liquid displaced*Acceleration due to gravity

               =volume of liquid displaced*density of liquid displaced*g

given

density=7.8*10^3

volume=1m

upthrust=1*7.8*10^3*g=7.8*10^3g