Question

A metallic sphere weighing 3kg in air is held by a string so as to be completely immersed in a liquid of relative density 0.8.the relative density of metallic is 10. The tension in the string is

Answer 1

This question is about equilibrium through which we could find the tension in the string. Equilibrium  is a type of condition in which all the forces are balanced in a given system or surrounding.

In this question we have been given a mass where four forces which are balanced.We will be using the he condition of equilibrium.

Weight=m=3 kg

Density=ρ=0.8

Relative density= σ =10

Tension of the string= T

Taking gravitational force=g= 10

For equilibrium :

T+Fₐ =Weight

Here,

Fₐ= ρ* V * g

Weight=σ * V * g

now,

T=V*g*[ 1- (σ / ρ) ]

T=m*g*[ 1- (σ / ρ) ]

T=3*10*[ 1- (0.8 / 10) ]

T=30*0.92

T=27.6 Newtons

Answer 2

given,

mass of Metallic sphere in air , m = 3kg

relative density of liquid = 0.8

so, density of liquid , $\sigma$ = 0.8 g/cm³

relative density of metallic sphere = 10

so, density of metallic sphere, $\rho$ = 10 g/cm³

now from Archimedes principle,

tension in string = apparent weight = $V(\rho-\sigma)g$

where V is volume of metallic sphere,
so, V = Mass of metallic sphere /density

e.g,. Tension in the string = $\frac{M}{\rho}(\rho-\sigma)g$

= $M\left(1-\frac{\sigma}{\rho}\right)g$

= 3 × (1 - 0.8/10) × 10

= 30 × (1 - 0.08)

= 30 × 0.92

= 27.6 N