Question

The tip of drawing pin has an area 1.0 x 10-8 sq. m. Find the pressure exerted if force applied is 10 N?

Answer 1

Given

• Area of drawing pin = 1.0×10^-8 m²
• Force applied = 10N

Find out

• Find the pressure exerted

Solution

$\implies\sf Pressure=\dfrac{Force}{Area} \\ \\ \\ \implies\sf P=\dfrac{10}{1.0\times{10^{-8}}} \\ \\ \\ \implies\sf P=\dfrac{10}{10^{-8}} \\ \\ \\ \implies\sf P=10^{1-(-8)} \\ \\ \\ \implies\sf P=10^{1+8} \\ \\ \\ \implies\sf P= 10^9N/m^2$

Hence, the pressure exerted on drawing pin is 10⁹ N/m²

Additional Information

• The force acting in a perpendicular direction to unit area of a surface is called pressure
• It's S.I unit is Pascal (N/m²)

Answer 2

GIVEN:

• Area of the tip of a drawing pin = $\sf{ 1.0 \times {10}^{-8} \: m^2}$
• Force applied = 10 N

TO FIND:

• What the pressure exerted by the pin ?

SOLUTION:

Let the Pressure exerted by the pin be 'P' N/m²

We know that the formula for finding the Pressure is:-

$\large\bf{\star \: Pressure = \dfrac{Force \: applied}{Area}\: \star}$

According to question:-

On putting the given values in the formula, we get

$\rm{\dashrightarrow P = \dfrac{10}{1.0 \times {10}^{-8} }}$

$\rm{\dashrightarrow P = \dfrac{10}{{10}^{-8}}}$

$\rm{\dashrightarrow P = 10 \times {10}^{-8}}$

$\rm{\dashrightarrow P = {10}^{1-(-8)}}$

$\rm{\dashrightarrow P = {10}^{1+8}}$

$\rm{\dashrightarrow P = 10^9 \: N/m^2 }$

❝ Hence, the pressure applied by the drawing pin is 10⁹ N/m² ❞

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✰ Extra Information ✰

➣ In honour of scientist Blaise Pascal , The S.I. unit of pressure is called ❛ Pascal ❜. It is denoted by ❛ Pa ❜.

➣ The pressure depends on two factors:

❶ Force applied

❷ Area of surface over which force acts.