Question

The pressure exerted by the weight of a cubical block of side 5 cm on the surface is 12 pascal calculate the weight that is force of the block.

Answer 1

Answer:-

Required weight, that is force on the block is 0.03 N .

Explanation:-

We have :-

→ Side of the cubical box = 5 cm

→ Pressure exerted = 12 Pa

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Firstly, let's convert the unit of side of the box from cm to m .

⇒ 1 cm = 0.01 m

⇒ 5 cm = 5(0.01)

⇒ 0.05 m

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Area of contant of the cubical block :-

= (side)²

= (0.05)²

= 0.0025 m²

Now we know that :-

Pressure = Force/Contact Area

⇒ 12 = Force/0.0025

⇒ Force = 12(0.0025)

⇒ Force = 0.03 N

Answer 2

Answer:

Given :-

• The pressure exerted by the weight of a cubical block of side 5 cm on the surface is 12 Pascal.

To Find :-

• What is the force of the block.

Formula Used :-

$\clubsuit$ Area of cubical block :

$\longmapsto \sf\boxed{\bold{\pink{Area =\: {(a)}^{2}}}}$

where,

• a = Side

$\clubsuit$ Pressure formula :

$\longmapsto \sf\boxed{\bold{\pink{Pressure =\: \dfrac{Force}{Area}}}}$

Solution :-

First, we have to convert cm to m :

Given :

• Side = 5 cm

So,

$\implies \sf Side =\: 5 cm$

$\implies \sf Side =\: \dfrac{5}{100} m$

$\implies \sf\bold{\purple{Side = 0.05\: m}}$

Now, we have to find the area of the cubical block :

Given :

• Side = 0.05 m

According to the question by using the formula we get,

$\implies \sf Area =\: {(0.05)}^{2}$

$\implies \sf Area =\: \dfrac{5}{100} \times \dfrac{5}{100}$

$\implies \sf Area =\: \dfrac{25}{10000}$

$\implies \sf\bold{\green{Area =\: 0.0025\: {m}^{2}}}$

Now, we have to find the pressure :

Given :

• Area = 0.0025 m²
• Force = 12 Pascal

According to the question by using the formula we get,

$\implies \sf 12 =\: \dfrac{Force}{0.0025}$

By doing cross multiplication we get,

$\implies \sf =\: Force =\: 0.0025 \times 12$

$\implies \sf Force =\: \dfrac{25}{10000} \times 12$

$\implies \sf Force =\: \dfrac{3\cancel{00}}{100\cancel{00}}$

$\implies \sf Force =\: \dfrac{3}{100}$

$\implies \sf\bold{\red{Force =\: 0.03\: N}}$

$\therefore$ The force of the block is 0.003 .