This question was previously asked in

DSSSB TGT Natural Science Male Subject Concerned - 29 Sept 2018 Shift 2

Option 2 : 7.5 gm/cm^{3}

General Awareness Sectional Test 1

9741

40 Questions
40 Marks
15 Mins

__Concept:__

**Archimedes' principle**states that when a body immersed in a fluid, whether fully or partially submerged, the upward buoyant force that is exerted on it, is equal to the weight of the fluid that the body displaces.**Buoyant force:**When an object is submerged in fluid fully or partially, an upward force is exerted on the body. This force is known as the buoyant force.

**F**_{buoyant}** = ρ**_{f} **V**_{f }**g **

Where ρ_{f} is the **density of the fluid** in which the object is submerged, V_{f} is the volume of the displaced fluid (or volume of the object that is submerged inside the fluid) and g is the gravitational acceleration.

**Apparent loss of weight in an object, when submerged, is equal to the buoyant force.**

__Calculation:__

Let V be the volume of the solid.

Given relative density of the liquid = 0.9

Weight of the solid in air = 50 gf

Weigth of the solid in liquid = 44 gf

We know that,

The loss in weight of the solid when immersed in liquid = The weight of the liquid displaced. ----- (1)

Weight of the liquid displaced = Volume of the liquid displaced × Relative density of the liquid

= V × 0.9 ---- (2)

(2) in (1) gives:

Loss in weight of the solid when immersed in liquid = 0.9 V ---- (3)

Loss in weight of the solid when immersed in liquid = (50 - 44) gf = 6 gf

Substituting in (3) we get:

6 = 0.9 V

⇒ V = 6 /0.9 = 6.66 cm^{3}

**(1) Therefore the volume of the solid = 6.66 cm ^{3}**

Density of the solid = Mass / volume

= 50 gf / 6.66 cm^{3}

__⇒____ 7.5 g/cm ^{3}__