A uniform meter rule \(AB\) has a mass \(15 g\). A \(30 g\) mass is suspended at the \(10.0 cm\) mark, and another \(5 g\) mass is suspended at the \(65.0 cm\) mark. Calculate the position of the fulcrum that will keep the meter rule balanced horizontally.
A. \(50.0 cm\) B. \(32.0 cm\) C. \(27.5 cm\) D. \(17.9 cm\)
Correct Answer: C
Explanation
By principle of moment,nticlockwise moment = clockwise moment \begin{aligned} &\therefore 30(x-10)=15(50-x)+5(65-x) \\ &30 x-300=750-15 x+325-5 x \\ &30 x+15 x+5 x=750+325+300 \\ &50 x=1375, x=27.5 cm \end{aligned}
