Question

a body weighs 40 gram in air if it's volume is 10 cm cube what will be its weight in water

Answer 1

The object in question  has a mass of 40 gram and a volume of  $10cm^3$  . The density of this object is  $p_o = \frac{40g}{10cm^3}=4g/cm^3$ . Water has a density $p_w=1g/cm^3$. Since the object has a bigger density than water, it will be submerged in water and experience a buoyant force equal to the weight  of the water it displaces. The weigh of an object in water is the difference between the weight of the object outside water and the weight of the water that it displaces. In mathematical form.

$W =mg-B_f$

where $B_f$ is the weight of the water displaced by the object, $m$ is the mass of the object and $g$ is the acceleration due to gravity.

In this problem, the object has a mass of 40 gram or 0.04 kg. The object displaces $10cm^3$ of water.

The mass of water  $m_w$  displaced is the product of  volume of the product of the displaced liquid and the density of water.

$m_w=p_wv=(1g/cm^3)\times(10cm^3)=10g =0.01kg$

The weight of this object in water is calculated using the formula for finding the weight of an object submerged in water.

$W =mg-B_f$

$W=(0.04kg)\times(9.8m/s^2)-(0.01kg)\times(9.8m/s^2)= 0.294N$

Answer 2

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