question_answer

A nail has \[2\text{ }c{{m}^{2}}\] at one end \[\frac{1}{100}c{{m}^{2}}\] and at the other end. A force of 1000 g. wt is applied on the first end. Calculate the pressure acting on the wall?

Answer:

Thrust acting = Force applied \[F=1000\text{ }g.\text{ }wt.\] The area of contact of the nail with the wall \[(A)=1/100\text{ }c{{m}^{2}}\] Pressure (P) = ? We know, \[Pressure=\frac{Thrust\text{ }acting}{Area\text{ }of\text{ }contact}\] \[\therefore \]\[\frac{1,000}{1/100}=(1000\times 100)\] \[={{10}^{5}}g.wt\text{ }c{{m}^{2}}\]       \[(\because 1g.wt=980\text{ }dyne)\] \[=980\times {{10}^{5}}dyne\text{ }c{{m}^{2}}\] Note: Dont get confused as the area of both the ends of the nail are given, We need to take only the area of contact with the wall.