Question

Masses of $1g$, $2g$, $3g$.....$100g$ are suspended from the $1cm$, $2cm$, $3cm$.....$100cm$ marks of a light metre scale. The system will be supported in equilibrium at :

• A. $60cm$
• B. $67cm$
• C. $55cm$
• D. $72cm$

Answer 1

The correct option is B $67cm$

We are given that masses $1g,2g,3g.......100g$ are suspended from the $1cm,2cm,3cm.........100cm$ marks on a light metre scale.

But, we know that by law of moments, the equilibrium will be established at a distance $r$ from origin.

Considering the origin at $x=0$.

We have $\frac{\left(100\right)(100+1)(200+1)}{6}=\frac{\left(100\right)(100+1)}{2}r$.

$r=\frac{201}{3}=67cm$

Note:

We have applied formula for series the sum of square of natural numbers $=\frac{\left(n\right)(n+1)(2n+1)}{6}$

and formula for series of sum of natural numbers $=\frac{\left(n\right)(n+1)}{2}$.