A rectangular block measures \(40 cm \times 25 cm \times 5 cm\) and is made of a material of density \(7800 kg / m ^{3}\). Calculate the pressure the block exerts on the floor when it stands on the smallest of its surfaces.
A. \(3.12 \times 10^{3} N / m ^{2}\) B. \(3.00 \times 10^{3} N / m ^{2}\) C. \(1.95 \times 10^{4} N / m ^{2}\) D. \(3.12 \times 104 N / m ^{2}\)
Correct Answer: D
Explanation
Volume of the block It is stated in the question that it stands on the smallest of its surfaces. The smallest area is \((25 \times 5) cm ^{2}\) \(=125 cm ^{2}=\left[125 /(100)^{2}\right] m ^{2}\) \(=0.0125 m ^{2}\) Pressure \(=\frac{\text { force }}{\text { area }}\) $$ \begin{aligned} &=\frac{390 N }{0.0125 m ^{2}}=31200 Nm ^{-2} \\ &=3.12 \times 10^{4} Nm ^{-2} \end{aligned} $$
