Question

In an immersion measurement of an odd-shaped metal object, the weight of the object is found to be 980 N when submerged in water. When it is submerged in a heavier Bromine liquid (density 3100 kg/m3), the object weighs 840 N. What is the volume of this object

Answer 1

Answer:
`0.00680\ m^3`

Step-by-step explanation:

Given

The weight of the object, when submerged in the water is
`980\ N`

When it is submerged in the bromine liquid, it weighs
`840\ N`

Suppose,

`\rho=\text{Density of object}\\\rho_w=\text{Density of water}\\\rho_b=\text{Density of bromine}\\V=\text{Volume of the object}`

for water,

`\Rightarrow V(\rho -\rho_w)g=980\quad \ldots(i)`

For bromine

`\Rightarrow V(\rho-\rho_b)g=840\quad \ldots(ii)`

Divide (i) and (ii)

`\Rightarrow (\rho-1000)/(\rho-3100)=(980)/(840)\\\\\Rightarrow 840\rho -840* 1000=980\rho-980* 3100\\\\\Rightarrow 140\rho=(3038-840)\cdot 1000\\\\\Rightarrow \rho=15,700\ kg/m^3`

Put the density value in equation (i)

`\Rightarrow V(15,700-1000)\cdot 9.8=980\\\\\Rightarrow V=(100)/(14,700)\\\\\Rightarrow V=0.00680\ m^3`