Question

An object weighing 302 N in air is immersed in water after being tied to a string connected to a balance. The scale now reads 268 N . Immersed in oil, the object appears to weigh 276 N . Find the density of the object. Answer in units of kg/m 3 .

Answer 1

The density of the object
`\rho =` 8883
`(Kg)/(m^(3) )`

Step-by-step explanation:

Given data

`F_(air)` = 302 N

`F_(water)` = 268 N

`F_(oil)` = 276 N

Buoyant force in water on the body is given by

`F_B =`
`F_(air)` -
`F_(water)`

`F_B =` 302 - 268

`F_B =` 34 N

Volume of the object

`V = (F_B)/(\rho g)`

`V = (34)/((1000)(9.81))`

V = 3.465 ×
`10^(-3)`

Mass of the object

`m = (F_(air) )/(g)`

`m = (302)/(9.81)`

m = 30.78 N

Now density of the body

`\rho = (m)/(V)`

`\rho = (30.78)/(3.465 (10^(-3) ))`

`\rho =` 8883
`(Kg)/(m^(3) )`

Therefore the density of the object
`\rho =` 8883
`(Kg)/(m^(3) )`