JEE Main & Advanced Sample Paper JEE Main Sample Paper-43

  • 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

    Correct Answer: C

    Solution :

     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.

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