Question

The centroidal co - ordinate $\overline{x}$ for the shaded area will be m. The outer boundary of the shaded area is that of a circle having a radius of 1 m.

• A. 0.16928

Answer 1

The correct option is A 0.16928

Angle subtended by shaded area at centre $=({90}^{\circ }-{20}^{\circ })+{90}^{\circ }+{30}^{\circ }={190}^{\circ }$
$\therefore$ Total area of sector $=\left(\frac{\theta }{360}\right)\times \pi r^2$
$=\left(\frac{190}{360}\right)\times \pi \times 1^2=1.658m^2$


$\overline{x}=\frac{\int \overline{x}d.A}{\int dA}$
$=\frac{{\int }_{-{30}^{\circ }}^{{160}^{\circ }}\left(\frac{2}{3}R\cos \theta \right)\times \frac{1}{2}\times R\times \left(Rd\theta \right)}{1.658}=\frac{R^3{\int }_{-{30}^{\circ }}^{{160}^{\circ }\left(\cos \theta d\theta \right)}}{3\times 1.658}$
$=\frac{1^3[\sin \theta {]}_{-{30}^{\circ }}^{{160}^{\circ }}}{3\times 1.658}=\frac{\sin {160}^{\circ }-\sin (-{30}^{\circ })}{3\times 1.658}=0.16928m$