Question

The tip of a drawing pin has an area of 0.01 mm^2. Find the pressure exerted by it if a force of 20 N
is applied by it
of its tyres is 120 Pa. then find the​

Answer 1

Answer:

The tip of a drawing pin has an area of 0.01 mm2. Find the pressure exerted by it if a force of 20 N is applied on it. Ans: Given: F = 20 N, A = 0.01 mm2 = 0.01 × 10-3 m2 = 10-5 m2 P = F ÷ A = 20 ÷ 10-5 = 2×106 Pa.

Answer 2

Given,

The tip of a drawing pin has an area of 0.01 mm^2.

The force applied by it is 20 N.

To Find,

The pressure exerted by it =?

Solution,

We can find the pressure exerted by the drawing pin as follows:

It is given that a  force of 20 N is applied drawing pin with an area of 0.01 mm^2. We have to find the pressure exerted.

We know that the pressure exerted by an object is given as:

$Pressure, P = \frac{Force }{Area }$

In the given question,

$Force = 20 N\\Area = 0.01 mm^{2}$

Since the area of the drawing pin is not in S.I. units, unit conversion is required.

Using unit conversion,

$1 mm^{2} = 10^{-6} m^{2}$

Therefore,

$0.01 mm^{2} = 0.01*10^{-6} m^{2}$

Now,

$Force = 20 N, Area = 0.01*10^{-6}$ $m^{2}$

Substituting the given values in the formula for pressure,

$P = \frac{20}{0.01 * 10^{-6} }$

$P = 2000* 10^{6}$ $N/m^{2}$

$P = 2$ x $10^{9}$ $N/m^{2}$

Hence, the pressure exerted by it is 2 x 10^9 N/m^2.