Question

A solid of length 12 cm is immersed 9 cm in water. The density of solid is A) 0.75 g cm–3
B) 0.57 g cm–3
C) 1.2 g cm–3
D) 0.9 g cm–3

Answer 1

Given:

A solid of length 12 cm is immersed 9 cm in water.

To find:

Density of substance ?

Calculation:

Let the cross-sectional area of the solid be A.

Now, according to the basic condition of floating, the weight of the body is equal and opposite to the buoyant force.

$\therefore \: mg = F_{b}$

$\implies\: (V_{total} \times \sigma)g = (V_{inside} \times \rho )\times g$

• $\sigma$ is density of solid and $\rho$ is density of liquid.

$\implies\: (A \times l_{total} \times \sigma)g = (A \times l_{inside} \times \rho )\times g$

$\implies\: l_{total} \times \sigma = l_{inside} \times \rho$

$\implies\: \sigma = \dfrac{ l_{inside} }{l_{total}}\times \rho$

$\implies\: \sigma = \dfrac{ 9 }{12}\times 1$

$\implies\: \sigma = \dfrac{ 3 }{4}$

$\implies\: \sigma = 0.75 \: g / cc$

So, density of solid is 0.75 g/cc.

Hope It Helps.

Answer 2

HELLO DEAR,

GIVEN:-A solid of length 12 cm is immersed 9 cm in water. The density of solid is

A) 0.75 g cm–3

B) 0.57 g cm–3

C) 1.2 g cm–3

D) 0.9 g cm–3

SOLUTION:-Length of solid = 12cm

Immersef length of solid = 9cm

Let the densityof solid be σ.

According to fluid mechanics, w w equal to the buoyant force flotation.

Therefore, weight of object in fluid = force of buoyancy.

mg. = Fb

⇒ mass of solid = volume of solid × density of solid

σ= m/ V

Therefore , mg = Fb

⇒( V total × σ ) g = ( V imerged × ρ ) g , where ρ is density of liquid

⇒( A × l total × σ ) = ( A × l imerged × ρ ) g

⇒l total × σ = l imerged× ρ

⇒σ = (l imerged / l total) ρ

⇒ σ =( 9/12) × 1

⇒ σ = 0.75 g/ cm³

Therefore , option ( A) 0.75 g / cm^3 is correct .

I HOPE IT'S HELP YOU DEAR,

THANKS.