Question

A piece of alloy has a mass 250g in air. When immersed in water,it has an apperant weight of 1.96N and in oil has an apperant weight of 2.16N. Calculate the density of metal and density of oil.

Answer 1

Answer:

when immersed in water it has an apperant weight of 1.96n and in oil has an apperant weight of 1.16n. calculate the density of metal and dencity of oil.

Answer 2

The density of metal is 4980.08$(\frac{kg}{m^{3}})$ and density of oil is 593.9$(\frac{kg}{m^{3}})$.

Explanation:

1. Here mass of object = 0.25kg.

  So weight of object = 0.25×g =0.25×9.81= 2.4525(N).

2. From equation of buoyant force $F_{B}=\rho _{l}\times V_{o}\times g$

 Where $F_{B}$ is buoyant force in Newton.

             $\rho _{l}$ is density of liquid in which body submerged  $(\frac{kg}{m^{3}})$.

             $V_{o}$  is volume of object in $m^{3}$.

3. First case when object is immersed in water($\rho _{w}= 1000 \frac{kg}{m^{3}}$)

    Apparent weight in water = weight in air - buoyant force in water

      1.96 = 2.4525 - 1000×$V_{o}$×g    

      So above can be written as

      1000×$V_{o}$×g = 0.4925 (N)    ...1)

      after solving equation 1) we get volume of metal object

        $V_{o} =5.02\times 10^{-5}(m^{3})$

4. Second case when object is immersed in oil($\rho _{o}$)

   Apparent weight in oil= weight in air - buoyant force in oil

    2.16 = 2.4525 - $\rho _{o}\times V_{o}\times g$

   $\rho _{o}\times V_{o}\times g$ = 0.2925 (N)     ...2)

5. Divide equation 1) from equation 2) we get density of oil

   $\rho _{o}$ = 593.90 ($\frac{kg}{m^{3}}$)

6.  Now density of metal $\rho _{o}= \frac{mass}{volume}$

        So

$\rho _{o}= \frac{0.25}{5.02\times 10^{-5}}$ = 4980.08 ($\frac{kg}{m^{3}}$)