Question

Calculate the mass of a ring with linear mass distribution $\lambda ={\lambda }_{0}si{n}^2\theta$ as shown in diagram. ${\lambda }_0$ = density at $\theta ={90}^{\circ }$.

• A. ${\lambda }_0\pi R$
• B. $\frac{1}{2}{\lambda }_0\pi R$
• C. $\frac{1}{3}{\lambda }_0\pi R$
• D. $\frac{2}{5}{\lambda }_0\pi R$

Answer 1

The correct option is A

${\lambda }_0\pi R$

$dm=\lambda Rd\theta ={\lambda }_{0}si{n}^2\theta Rd\theta m=\int dm={\int }_0^{2\pi }{\lambda }_{0}si{n}^2\theta Rd\theta =\frac{{\lambda }_{0}R}{2}{\int }_0^{2\pi }(1-cos2\theta )d\theta$
Integrating we get, $m={\lambda }_0\pi R$