Question

A body of 100cm^3 is completely immersed in water. Find the buoyant force acting on the body(g=10m/s^2). Density of water=1000kg/m^3.
(Please don't give unnecessary answers) ​

Answer 1

Answer :-

Buoyant force acting on the body is 1 Newton .

Explanation :-

We have :-

→ Volume of the body = 100 cm³

→ Gravitational acceleration (g) = 10 m/s²

→ Density of water (ρ) = 1000 kg/m³

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We know that, volume of water displaced (V) will be equal to the volume of the body. Firstly, let's convert it's unit.

⇒ 1 cm³ = 10⁻⁶ m³

⇒ 100 cm³ = 100 × 10⁻⁶ m³

⇒ 10⁻⁴ m³

Now, putting values in the formula of buoyant force/upthrust, we get :-

F = Vρg

⇒ F = 10⁻⁴ × 1000 × 10

⇒ F = 10⁻⁴ × 10³ × 10

⇒ F = 10⁻¹ × 10

⇒ F = (1/10) × 10

⇒ F = 1 N

Answer 2

To Find :

The buoyant force acting on the body in water .

Given :

• mass of Body, m = 100cm³
• density of water , p= 1000 kg/m³
• gravity acceleration, a = 10m/s²

Formula to be applied:

• 1 m³ = 10⁶ cm³
• Buoyant force = m×p×g

Solution :

given, mass of Body = 100 cm³

1 m³ = 10⁶ cm³.

so, 100 cm³ = 100/10⁶ m

or, 100 cm³ = 10²/10⁶ m

or, 100 cm³ = 10^ -4 m³

Density of water = 1000kg/m³

Gravity = 10 m/s²

so, Buoyant F = 10^-4 × 1000 × 10

$\tt{ \implies \: f = {10}^{ - 4} \times {10}^{3} \times {10}^{1}}$

$\tt{ \implies \: f = {10}^{ - 4} \times {10}^{4} }$

$\red{ \underline{ \boxed{ \tt{f = 1 \: n}}}}$

hence, The buoyant force = 1 N