Question

A drawing pin is pushed against a wooden table with a force 10 N. Calculate the pressure exerted by the pin at a point on the table, if area of the point is \[0.01\,{\text{c}}{{\text{m}}^2}\].
A. \[{10^5}\,{\text{Pa}}\]
B. \[{10^6}\,{\text{Pa}}\]
C. \[{10^7}\,{\text{Pa}}\]
D. \[{10^8}\,{\text{Pa}}\]

Answer 1

Hint:The pressure is the ratio of the applied force per unit cross-sectional area. Convert the given quantities in S.I units. Substitute the given quantities in the formula for pressure and calculate the pressure exerted by the pin.

Formula used:
Pressure, \[P = \dfrac{F}{A}\]
Here, F is the force and A is the cross-sectional area.

Complete step by step answer:
We have given that the pin is pushed against a table with a force of \[F = 10\,{\text{N}}\] and the area of cross-section of the point is \[A = 0.01\,{\text{c}}{{\text{m}}^2}\].Let’s convert the area of cross-section of the point from \[c{m^2}\] to \[{m^2}\] as follows.
\[A = \left( {0.01\,{\text{c}}{{\text{m}}^2}} \right){\left( {\dfrac{{{{10}^{ - 2}}\,{\text{m}}}}{{1\,{\text{cm}}}}} \right)^2}\]
\[ \Rightarrow A = 0.01 \times {10^{ - 4}}\,{{\text{m}}^2}\]
We know the definition of the pressure. The pressure is the ratio of the applied force per unit cross-sectional area. Therefore, we can express the pressure as,
\[P = \dfrac{F}{A}\]
Here, F is the force and A is the cross-sectional area.
Substituting \[F = 10\,{\text{N}}\] and \[A = 0.01\,{\text{c}}{{\text{m}}^2}\] in the above equation, we get,
\[P = \dfrac{{10}}{{0.01 \times {{10}^{ - 4}}}}\]
\[ \Rightarrow P = \dfrac{{10}}{{0.01 \times {{10}^{ - 4}}}}\]
\[ \therefore P = {10^7}\,{\text{Pa}}\]
Therefore, the pressure exerted by the pin on the point is \[{10^7}\,{\text{Pa}}\].

So, the correct answer is option C.

Note:To use the formula, \[P = \dfrac{F}{A}\], the force F should be in newton and the area should be in square meter in order to get the pressure in the pascal unit. The crucial step in the solution is to convert the area into square meters. To do so, multiply the area by \[{10^{ - 4}}\] to convert it into square meters. The derived unit of the pressure is \[{\text{N/}}{{\text{m}}^2}\] and therefore students can express the unit of pressure in \[{\text{N/}}{{\text{m}}^2}\] instead of pascal.