Question

A metallic rod of mass $20kg$ and of uniform thickness rests against a wall while the lower end of rod is in contact with rough floor. The rod makes an angle ${60}^0$ with floor. If the weight of rod produces a torque $150Nm$ about its lower end, the length of rod $(g=10m{s}^{-2})$ is:

• A. $1.5m$
• B. $2m$
• C. $3m$
• D. $4m$

Answer 1

Answer: (C)

$3m$

$Torque=mg\left(R\right)$ (as shown in figure about point of contact of rod and floor)
$=mg\left(\frac{l}{2}\cos {60}^{\circ }\right)$ (simply putting the given data)
$150=20\times 10\left(\frac{l}{4}\right)$
$l=3m$