Question

Find the ratio of the pressure exerted by a block of 200n when placed on a table top along its two different sides with dimensions 20 ×15 and 30× 15.​

Answer 1

${\textsf{\textbf{\underline{Given :-}}}}$

Force = 200 N

Area 1 = 20 × 15 m

Area 2 = 30 × 15

${\textsf{\textbf{\underline{To Find :-}}}}$

Ratio of pressure

${\textsf{\textbf{\underline{Solution :-}}}}$

We know that

P = F/A

In Case 1

Area = 20 × 15 = 300 m²

Force = 200 N

Pressure = 200/300

Pressure = 2/3 Pa

In Case 2

Area = 30 × 15 = 450 m²

Force = 200 N

Pressure = 200/450

Pressure = 20/45

Pressure = 4/9 Pa

Ratio = 2/3 : 4/9

2 × 3/3 × 3 : 4/9

6/9 : 4/9

6 : 4

3 : 2

Thus,

Ratio is 3 : 2

Answer 2

To Find :

The pressure exerted by the block on the table from different dimensions

__________________________

Given :

• Weight = 200N
• Dimension in first case = 20×15
• Dimensions in second case = 30×15

__________________________

We Know that:

• Pressure = Force/Area

__________________________

Solution :

Area in first case = 20×15

or, area = 300

Area in Second case = 30×15

or, area = 450

So, pressure in first case = 200/300

or, Pressure = 2/3 Pa

so, pressure in second case = 200/450

or, Pressure = 4/9

So, ratio of pressure = 2/3 : 4/9

or, Ratio = 2/3 × 9 : 4/9 × 9

or, Ratio = 2×3 : 4

or, Ratio = 6:4

so, Ratio of pressures = 3:2