Question

A block of dimensions $l\times t\times h$ and uniform density ${\rho }_w$ rests on a rough floor. Wind blowing with speed V and of density ${\rho }_a$ falls perpendicularly on one face of dimension $l\times h$ of the block as shown in figure. Assuming that air is stopped when it strikes the wall and there is sufficient friction on the ground so that the block does not slide, then the minimum speed V for which the block is just about to topple is

• A. ${\left(\frac{{\rho }_{w}g}{{\rho }_{a}h}\right)}^{\frac{1}{2}}.t$
• B. ${\left(\frac{{\rho }_{a}g}{{\rho }_{w}h}\right)}^{\frac{1}{2}}.t$
• C. ${\left(\frac{g}{h}\right)}^{\frac{1}{2}}.t$
• D. None of these

Answer 1

The correct option is A ${\left(\frac{{\rho }_{w}g}{{\rho }_{a}h}\right)}^{\frac{1}{2}}.t$


$F=V\frac{dm}{dt}={\rho }_{a}lhV.V={\rho }_{a}lh{V}^2\{\because \frac{dm}{dt}={\rho }_{a}lhV\}$
Total torque due to air about point P is ${\rho }_{a}lh{V}^2\frac{h}{2}$
${\tau }_a=\frac{{\rho }_{a}l{h}^2{V}^2}{2}$;
Torque due to weight of block about point P is
${\tau }_w=mg.\frac{t}{2}={\rho }_{\omega }.l.h.t.g.\frac{t}{2}$;
For toppling, ${\tau }_a>{\tau }_{\omega }\Rightarrow V>{\left(\frac{{\rho }_{\omega }g}{{\rho }_{a}h}\right)}^{\frac{1}{2}}.t$