• # question_answer The number of solutions of the equation ${{\cos }^{2}}\,\left( x+\frac{\pi }{6} \right)+{{\cos }^{2}}x-2\cos \,\left( x+\frac{\pi }{6} \right)\cos \,\frac{\pi }{6}={{\sin }^{2}}\frac{\pi }{6}$ A)  0                                 B)  1                  C)  2                                 D)  3

Given,  $Mg\,\,\sin \,\,\theta$ $Mg\,\,\sin \theta \times \frac{h}{2}Mg\,\cos \,\theta \,\times r$ On adding $\tan \theta =\frac{r}{h/2}$ on both sides, we get $\theta ={{45}^{o}},$ $r=\frac{h}{2}$ $\theta ={{45}^{o}}$ $n'=\frac{v}{v-{{v}_{s}}}\times n$ $\Rightarrow$ $\frac{n'}{n}=\frac{v}{v-{{v}_{s}}}$ $\frac{\Delta n}{n}=\frac{{{v}_{s}}}{v-{{v}_{s}}}$ $\frac{2.5}{100}=\frac{{{v}_{s}}}{320-{{v}_{s}}}=\frac{1}{40}$ $40{{v}_{s}}=320-{{v}_{s}}$ ${{v}_{s}}=\frac{320}{41}\simeq 8\,m/s$ $V=\frac{1}{4\pi {{\varepsilon }_{0}}}\left[ \frac{Q}{R} \right]+\frac{1}{4\pi {{\varepsilon }_{0}}}\left[ \frac{q}{r} \right]$ On comparing both sides, we get $=\frac{1}{{{\varepsilon }_{0}}}\left[ \frac{Q}{4\pi {{R}^{2}}}\times R+\frac{q}{4\pi {{r}^{2}}}\times r \right]$ $\frac{Q}{4\pi {{R}^{2}}}=\frac{q}{4\pi {{r}^{2}}}=\sigma$       or         $\therefore$ $V=\frac{1}{{{\varepsilon }_{0}}}\,[\sigma R+\sigma r]=\frac{\sigma }{{{\varepsilon }_{0}}}[R+r]$ ${{N}_{S}}=\left( \frac{\rho S}{\rho P} \right)\times {{N}_{P}}$            or         $=\frac{(4.4\times {{10}^{3}})\,\times 100}{220}=2000$ Hence, number of solutions, are 2.