question_answer

The pressure exerted by the weight of a cubical block of a side 3 cm on the surfaces 5 Pa. What will be the weight (or force) of the block?

Answer:

\[P=5Pa=5N/{{m}^{2}}\] Side of cube \[(a)=3\text{ }cm=\frac{3}{100}m=0.03m\] Area of cube \[(A)={{a}^{2}}=0.03\text{ }m\times 0.03\text{ }m\] \[=9\times {{10}^{4}}{{m}^{2}}\] we know that, \[\therefore \] \[P=\frac{F}{A}\] \[\Rightarrow \] \[F=P\times A\] \[=5N/{{m}^{2}}\times 9\times {{10}^{4}}{{m}^{2}}=45\times {{10}^{4}}N\]