Question

A cube of side $a$ rests on the floor as shown in the given figure. Given that the pressure exerted by this cube on the floor is $P$, what is the pressure exerted by another cube of the same material of side 4$a$?
(Take g = 10 $Nk{g}^{-1}$)

• A. 2$P$
• B. 16$P$
• C. 4$P$
• D. $P$

Answer 1

The correct option is C 4$P$

We know that pressure, $P$ = $\frac{Force\left(F\right)}{Area\left(A\right)}$

So, $\frac{P_2}{P_1}$ = $\frac{F_2}{A_2}$$\times$$\frac{A_1}{F_1}$ ------ (i)

Subscript 1 refers to the smaller cube and subscript 2 refers to the bigger cube.

Force exerted by the cube on the floor will be equal to the weight (mg) of the block.
Therefore, force exerted by smaller cube, $F_1$ = $m_{1}g$
Force exerted by bigger cube, $F_2$ = $m_{2}g$

We know that, Mass (m) = Density ($\rho$) $\times$ Volume (V)
Sice, both the cubes are made up of same materials, therefore density will be same.
Now, volume of the smaller cube, $V_1$ = $a^3$
Similarly, volume of the bigger cube, $V_2$ = ${\left(4a\right)}^3$
$V_2$ = $64a^3$
Therefore, $m_2$ = 64 $m_1$

So, force, $F_2$ = 64$m_{1}g$ = 64$F_1$
or, $\frac{F_2}{F_1}$ = 64 ------- (ii)

Now, $A_1$ = $a^2$ and $A_2$ = $(4a{)}^2$ = $16a^2$
So, $\frac{A_1}{A_2}$ = $\frac{1}{16}$ -----(iii)

On putting equation (ii) and (iii) in equation (i) we get,
$\frac{P_2}{P_1}$ = $64$ $\times$ $\frac{1}{16}$
$\frac{P_2}{P_1}$ = 4
$P_2$ = 4$P_1$
Since we have given, $P_1$ = $P$
Therefore, $P_2$ = 4$P$