Question

A brick weighting 40N having dimensions of 25cm x 10cm x 5cm kept on a table with its three different sides one by one calculate the pressure exerted in each of the three cases

Answer 1

Given:

Weight of brick, F= 40 N

Dimensions of brick= 25cm×10cm×5cm

To Find:

Pressure in the all the cases in which brick is kept

Solution:

We know that,

• Pressure P exerting on certain surface is given by

$\pink{\boxed{\bf{P=\frac{Force\ or\ Weight}{Area}}}}$

$\rule{190}{1}$

Case-1: When dimension of surface in contact is 25cm×10cm

Let the area of the surface be A₁

So,

$\longrightarrow\rm{A_{1}=25\times10=250\ cm^{2}=0.025\ m^{2}}$

Let the pressure on the given surface be P₁

So,

$\longrightarrow\rm{P_{1}=\dfrac{F}{A_{1}}}$

$\longrightarrow\rm{P_{1}=\dfrac{40}{0.025}}$

$\longrightarrow\rm\green{P_{1}=1600\ Pa}$

$\rule{190}{1}$

Case-2: When dimension of surface in contact is 25cm×5cm

Let the area of the surface be A₂

So,

$\longrightarrow\rm{A_{2}=25\times5=125\ cm^{2}=0.0125\ m^{2}}$

Let the pressure on the given surface be P ₂

So,

$\longrightarrow\rm{P_{2}=\dfrac{F}{A_{2}}}$

$\longrightarrow\rm{P_{2}=\dfrac{40}{0.0125}}$

$\longrightarrow\rm\green{P_{2}=3200\ Pa}$

$\rule{190}{1}$

Case-3: When dimension of surface in contact is 10cm×5cm

Let the area of the surface be A₃

So,

$\longrightarrow\rm{A_{3}=10\times5=50\ cm^{2}=0.005\ m^{2}}$

Let the pressure on the given surface be P ₃

So,

$\longrightarrow\rm{P_{3}=\dfrac{F}{A_{3}}}$

$\longrightarrow\rm{P_{3}=\dfrac{40}{0.005}}$

$\longrightarrow\rm\green{P_{3}=8000\ Pa}$

Answer 2

Answer:

here the pressure is

F/A

40/25=1.6

40/10=4

40/5=8

Explanation: