Question

A metallic sphere weighs 210g in air, 180 g in water and 120 g in an unknown liquid. Find the density of metal and of liquid.

Answer 1

•A METALLIC SPHERE WEIGHS 210g IN AIR, 180g IN WATER, AND 120g IN AN UNKNOWN LIQUID. THE DENSITY OF METAL would be 7 and that of LIQUID would be 3.

•To find the DENSITY OF METAL, We use the formula DENSITY= weight in air ÷ Change in weight in water.

• DENSITY = 210/(210-180)

=7g/cm²

•To find the DENSITY OF LIQUID, We must use the formula,

Change in LIQUID = Upthrust in LIQUID.

=(V solid)(ρlliquid)g

or

Δwαρliquid

Therefore, ∆w/ϵw = ρl/ρw

=(210−180)/(210−120)(1)gm/cm³

=3g/cm³.

Answer 2

Density of metal = 7 gm/cm³    

Density of liquid = 3 gm/cm³

Given:

A metallic sphere weighs 210 g in air, 180 g in water and 120 g in an unknown liquid.

To find:

Density of metal and of liquid.

Formula used:

Relative density of metal = $\frac{Weight \ in \ air }{Change \ of weight \ in \ water}$

Change in weight = Upthrust force.

density = Relative density × Density of water

Explanation:

Relative density of metal = $\frac{Weight \ in \ air }{Change \ of weight \ in \ water}$

Relative density of metal = $\frac{210 }{210-180}$

Relative density of metal = 7

Density of metal = Relative density × Density of water

                            = 7 × 1 = 7 gm/cm³

Change in weight = Upthrust force

Change in weight = $\left(V_{\text {solid }}\right)\left(\rho_{\text {liquid }}\right) g$

$\left V_{\text {solid }}\right g$ are constant so that

Change in weight ∝ $\rho_ {liquid}$

      $\begin{aligned}&\therefore \frac{\Delta w_{l}}{\Delta w_{w}}=\frac{\rho_{l}}{\rho_{w}}\\&\rho_{l}=\frac{\Delta w_{l}}{\Delta w_{w}} \rho_{w}\end{aligned}$

Density of liquid = $\left(\frac{210-120}{210-180}\right)(1) g m / c m^{3}$

                            = 3 gm/cm³

To learn more....

1)What is upthrustWhat Is unit of upthrust and what formula is used for it.

https://brainly.in/question/1157627

2)Differentiate between thrust and upthrust

https://brainly.in/question/7489896