A piece of iron of density $7.8 g c m^{- 3}$ and volume 100 $c m^{3}$ is totally immersed in water. Calculate its apparent weight in water.

7 N

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7.8 N

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6.8 N

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10 N

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Solution

The correct option is C 6.8 N
Given:
Density = $7.8 g c m^{- 3} = 7800 k g m^{- 3}$
Volume = $100 c m^{3} = 10^{- 4} m^{3}$
Step 1: Calculate ‘ $F_{b}$ ’
$F_{b}$ = $V ⍴_{w a t e r} g$
= $10^{- 4} \times 10^{3} \times 10$
= 1 N
Step 2: Calculating apparent weight
$F_{n e t} = W e i g h t o f i r o n b l o c k (F_{w}) - F_{b}$
= $V ⍴_{i r o n} g - 1$
= $10^{- 4} \times 7800 \times 10 - (1) = 6.8 N$

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A piece of iron of density 7.8 gcm−3 and volume 100 cm3 is totally immersed in water. Calculate its apparent weight in water.

The correct option is C 6.8 NGiven: Density = 7.8 g cm−3 = 7800 kg m−3 Volume = 100 cm3 = 10−4 m3 Step 1: Calculate ‘Fb’ Fb = V⍴waterg = 10−4×103×10 = 1 N Step 2: Calculating apparent weight Fnet = Weight of iron block (Fw) − Fb = V⍴irong − 1 = 10−4×7800 ×10−(1) = 6.8 N

A piece of iron of density $7.8 g c m^{- 3}$ and volume 100 $c m^{3}$ is totally immersed in water. Calculate its apparent weight in water.