(a) Moment of a force is the product of force and perpendicular distance from the turning point.

Moment = F x D
The S.I unit is Nm.~(b) Conditions: (i) The sum of forces in one direction must be equal to sum of forces in the opposite direction.
            (ii) Sum of clockwise forces must be equal to sum of anti-clockwise forces about the same point.

Sum of clockwise moment = Sum of anti- clockwise moment.
\(M \times 18 = 60 \times 24\)
Mass of metre rule \(M = \frac{60 \times 24}{18} = 80g\)

Clockwise moment = anti- clockwise moment
\(80 \times 48 - (13 + x) = 60 \times x\)
\(80 \times 35 - x = 60x\)
\(2800 - 80x = 60x \implies 2800 = 60x + 80x\)
\(2800 = 140x \implies x = 20cm\)
Distance of balance point from the zero end = 20cm + 13cm = 33cm.
(c) \(Efficiency = \frac{\text{Work output}}{\text{Work input}} \times 100%\)
= \(\frac{\text{Load x Distance moved by load}}{\text{Effort x Distance moved by effort}}\)
= \(M.A \times \frac{1}{V.R} \times 100%\)
(ii) \(80% = \frac{200 \times 3.0}{\text{Work done by effort}} \times100% \)
\(0.8 = \frac{600}{\text{Work done by effort}} \implies \text{Work done by effort} = \frac{600}{0.8} = 750J\)
~(c) \(Efficiency = \frac{\text{Work output}}{\text{Work input}} \times 100%\)
= \(\frac{\text{Load x Distance moved by load}}{\text{Effort x Distance moved by effort}}\)
= \(M.A \times \frac{1}{V.R} \times 100%\)
(ii) \(80% = \frac{200 \times 3.0}{\text{Work done by effort}} \times100% \)
\(0.8 = \frac{600}{\text{Work done by effort}} \implies \text{Work done by effort} = \frac{600}{0.8} = 750J\)
