Question

A liquid weighs $50g$ in air and $40g$ in water. How much would it weigh in a liquid of specific gravity $1.5$?

A) 30 grams

B) 35 grams 

C) 65 grams

D) 45 grams

Answer 1

Hint: We will apply the equation for apparent weight of the body when the density of medium is given. Apparent weight of an object varies according to the medium we are taking the object in and calculating its apparent weight.

Complete step by step answer:

Apparent weight is described as a property of objects that refers to how heavy an object is. The apparent weight of a body or an object differs from the actual weight of an object whenever the force of gravity acting on the object is not balanced by an equal and opposite normal force.

The apparent weight of an object immersed in a fluid is given as,

$W'=W-B$

Where,

W' = Apparent weight of object

W = Actual weight of object

B = buoyancy force

$ M_A \times g = M \times g - \rho_{fluid} V \times g$

$ M_A  = M - \rho_{fluid}V$

$ M_A  = M - \rho_{fluid} \dfrac{M}{{\rho }_{\text{object}}}$

$ M_A  = M(1-\dfrac{{\rho_{fluid}}}{{\rho }_{\text{object}}})$

Where,

$M_A$ is the apparent mass of object

When,

${{\rho }_{\text{object}}}>{{\rho }_{\text{fluid}}}$, the apparent weight is positive but less than the actual weight.

${{\rho }_{\text{object}}}={{\rho }_{\text{fluid}}}$, the apparent weight is zero and the object has a neutral buoyancy.

${{\rho }_{\text{object}}}<{{\rho }_{\text{fluid}}}$, the apparent weight is negative and the object is said to be lighter than air.

Specific gravity is another term for relative density. It is the ratio of the density of substance to the density of given reference material. Specific density is a dimensionless quantity, meaning that it is not expressed in units.

We are given that a liquid weighs $50g$ in air and $40g$ in water. And specific gravity of another liquid is given as $1.5$.

Applying the equation for apparent weight,

In first case, we have,

$ 40=50\left( 1-\dfrac{1}{\rho } \right) $

$ \dfrac{1}{\rho }=\dfrac{1}{5} $

$ \rho =5 $

In second case, we have,

$ M_A=50\left( 1-\dfrac{1.5}{\rho } \right) $

$M_A=50\left( 1-\dfrac{1.5}{5} \right)$

$  M_A=50-15 $

$ M_A=35g $

Hence, the correct option is (B).

Note: Apparent weight of a body is observed when the relative density or the specific gravity of a medium does not match the density of air. In that case, the body observes some change in its effective weight. That weight is known as the apparent weight of the body. Apparent weight of a body can be positive or negative depending upon the relative density of the medium.