Advertisements
Advertisements
Question
The figure shows a brick of weight 2 kgf and dimensions 20 cm × 10 cm × 5 cm placed in three different positions on the ground. Find the pressure exerted by the brick in each case.

Advertisements
Solution
(a) Weight of bricks = Thrust = F = 2 kgf

Area of base = `20/100 xx 10/100 = 1/50 "m"^2`
Pressure exerted P = `"F"/"A" = "2 kgf"/(1/50 "m"^2) = 100 "kgf m"^-2`
Or
Area of base = area of top L × B
20 cm × 10 = 200 cm2
P = `"Thrust"/"A" = "2 kgf"/200 "cm"^2 = 1/100 = 0.01 "kgf cm"^-2`
(b)

area of base = area of top
= 5 cm × 10 cm = 50 cm2
Pressure exerted = P = `"F"/"A" = "2 kgf"/"50 cm"^2 = 0.04`
P = 0.04 kgf cm-2
(c)

Weight of brick
F = 2 kgf
Area of base = L × B
= 20 cm × 5 cm = 100 cm2
Pressure exerted = `"F"/"A" = 2/100 = 0.02 "kgf cm"^-2 `
APPEARS IN
RELATED QUESTIONS
A large He balloon of volume 1425 m3 is used to lift a payload of 400 kg. Assume that the balloon maintains constant radius as it rises. How high does it rise? [Take y0= 8000 m and `rho_"He"`= 0.18 kg m–3].
The unit of thrust is
Consider the situation of the previous problem. Let the water push the left wall by a force F1 and the right wall by a force F2.
An air bubble of radius 0.2 mm is situated just below the water surface. Calculate the gauge pressure. Surface tension of water = 7.2 × 10−2 N/m.
An empty plastic bottle closed with an airtight stopper is pushed down into a bucket filled with water. As the bottle is pushed down, there is an increasing force on the bottom. This is because
Match the following.
| 1. | Density | hρg |
| 2. | 1 gwt | Milk |
| 3. | Pascal’s law | `"Mass"/" Volume"` |
| 4. | Pressure exerted by a fluid | Pressure |
| 5. | Lactometer | 980 dyne |
The total force exerted by a body normal to the surface is called ______.
Pressure is directly proportional to the area of contact.
In petrol bunks, in what unit is tire pressure measured?
The shape and size of the solids do not easily change.
