Question

A block of iron of mass 15 kg and of dimensions
12cm x 8cm x 10 cm is kept on a surface on its base of side 12 cm x 10 cm. Calculate:
i) thrust and
ii) pressure exerted on the surface. ( 1 kgf = 10 N)

Answer 1

Answer:

Explanation:

Given,

Mass of iron block, m = 7.5 kg.

Dimensions = 12 cm × 8 cm × 10 cm

To Find,

(i) Thrust, and,

(ii) Pressure exerted on the surface.

Formula to be used,

We know that Thrust = Force so,

Thrust = Force = mg

Pressure = Thrust/Area

Solution,

Putting all the values, we get

Thrust = mg

⇒ Thrust = 7.5 × 10

⇒ Thrust = 75 N

Hence, the thrust is 75 N.

Now, the pressure exerted on the surface,

At 1st we have to find the area of the surface,

Area = 12 × 8 = 96 cm²

by converting it to m², we get

Area = 0.0096 m².

Now, putting pressure formula, we get

Pressure = Thrust/Area

⇒ Pressure = 75/0.0096

⇒ Pressure = 7812.5 Pa

Hence, the pressure exerted on the surface is 7812.5 Pa.

Answer 2

$\huge\bold{\underline{\underline{Answer!!!}}}$

Given,

• Mass of iron block = 15kg
• Dimensions of the iron block = 12cm × 8cm × 10cm
• Base of the iron block = 12 × 10 cm

To find,

• Thrust
• Pressure exerted on the surface.

Formulae used,

• Thrust = Force = m × g
• Pressure = Thrust/area

Solution,

According to the question, first we have to find out the thrust.

• Thrust = m × g

${\implies}$ = 15 × 9.8 (gravitaional acceleration)

${\implies}$ = 147 Nm²

Therefore, the thrust is 147 Nm².

__________________________________

Secondly, we have to find out the pressure exerted on the surface.

• Area = 12 × 8 = 96 cm

According to the unit of pressure, we have to convert 96 cm into m²

= 0.0096 m²

• Preasure = Thrust/Area

${\implies}$ = 147/0.0096

${\implies}$ = 15,312.5 Pa

Therefore, the pressure exercted on the surface is 15,312.5 Pa