Question

A force of 12 N is applied on a cube of side 2 cm and density $800\mathrm{kg}/m^3$. Acceleration of this cube is

• A. $1.9\times {10}^3$$m/s^2$
• B. $1875$$m/s^2$
• C. $2\times {10}^3$ $m/s^2$
• D. $18.75\times {10}^2$ $m/s^2$

Answer 1

The correct option is D $18.75\times {10}^2$ $m/s^2$

$F=ma\Rightarrow a=\frac{F}{m}=\frac{F}{\rho \left(vol\right)}=\frac{F}{\rho l^3}$
$a=\frac{12}{8\times {10}^2\times (2\times {10}^{-2}{)}^3}=\frac{12}{64}\times {10}^4=1875m/s^2=18.75\times {10}^2$