Question

30-VIN_2 PRESSURE
A cube of side 0.2 m rests on the floor, as shown. Given that the cube has a mass of 50 kg.
the pressure exerted by the cube on the floor is (Take g = 10 N kg) (NS0-2014)
50 kg
02 m
(A) 25 Nm?
(B) 250 Nm
(C) 1250 N m2
(D) 12500 Nm
The​

Answer 1

Appropriate Question:

A cube of side 0.2 m rests on the floor, as shown. Given that the cube has a mass of 50 kg. The pressure exerted by the cube on the floor is (Take g = 10 N/kg).

Answer:

Option D (12500 N/m²)

Explanation:

As per the provided information in the given question, we have :

• Side of the cube = 0.2 m
• Mass of cube (m) = 50 kg
• Acceleration due to gravity (g) = 10 N/kg

We are asked to calculate the pressure exerted by the cube on the floor.

As we know that,

$\longmapsto \bf { P = \dfrac{F}{A} }$

• P denotes pressure
• F denotes force
• A denotes area in contact

Calculating force :

⇒ F = ma

• F denotes force
• m denotes mass
• a denotes acceleration

⇒ F = 50 kg × 10 N/kg

⇒ F = 500 N

∴ Force applied is 500 N.

Calculating area in contact :

The cube's base surface area is in the contact. So, we need to find the area of its base which is square in shape.

⇒ Area of square = Side × Side

⇒ Area in contact = 0.2 m × 0.2 m

⇒ Area in contact = 0.04 m²

∴ Area in contact is 0.04 m².

Calculating the pressure :

$\longmapsto \rm { P = \dfrac{F}{A} }$

$\longmapsto \rm { P = \dfrac{500 \; N}{0.04 \; m^2} }$

$\longmapsto \rm { P = 12500 \; N.m^{-2} }$

$\longmapsto \bf { P = 12500 \; Pa }$

∴ Pressure exerted is 12500 Pa or 12500 N/m².

Option D is correct.