Physics NCERT Exemplar Solutions Class 11th Chapter Eight

$\mathrm{R}={\mathrm{R}}_0(1+\alpha \mathrm{\Delta }\mathrm{T})$

$\begin{array}{rr}& \Rightarrow 6.8=2[1+\alpha *(80-0\left)\right]\\ & \Rightarrow \alpha =\frac{3.4-1}{80}\\ & =0.03\\ & =3*{10}^{-2}{}^{?}{\mathrm{C}}^{-1}\end{array}$

${\stackrel{?}{\mathrm{B}}}_{\mathrm{P}}={\stackrel{?}{\mathrm{B}}}_{\text{upper wire}}\otimes +{\stackrel{?}{\mathrm{B}}}_{\text{semi-circle}}?+{\stackrel{?}{\mathrm{B}}}_{\text{lower-wire}}\otimes$

$B_P=-\frac{{\mu }_{0}i}{4\pi R}+\frac{{\mu }_{0}i}{4R}-\frac{{\mu }_{0}i}{4\pi R}$ $=\frac{{\mu }_0\mathrm{i}}{4\mathrm{R}}\left(1-\frac{2}{\pi }\right)$  pointing away from the page

Time period of satellite

$\mathrm{T}=2\pi \sqrt[]{\frac{{\mathrm{R}}^3}{\mathrm{G}\mathrm{M}}}$

$=2\pi \sqrt[]{\frac{{\mathrm{R}}^3}{\mathrm{G}\mathrm{d}\frac{4}{3}\pi {\mathrm{R}}^3}}$

$\Rightarrow \mathrm{T}=\sqrt[]{\frac{3\pi }{\mathrm{G}\mathrm{d}}}$ $\Rightarrow \frac{3\pi }{\mathrm{G}\mathrm{d}}={\mathrm{T}}^2$

$\left|\stackrel{?}{\mathrm{F}}\right|=\left|\mathrm{I}\right(\stackrel{?}{\mathrm{L}}*\stackrel{?}{\mathrm{B}}\left)\right|$

${\stackrel{?}{\mathrm{B}}}_{\mathrm{P}}={\stackrel{?}{\mathrm{B}}}_{\text{upper wire}}\otimes +{\stackrel{?}{\mathrm{B}}}_{\text{semi-circle}}?+{\stackrel{?}{\mathrm{B}}}_{\text{lower-wire}}\otimes$

$B_P=-\frac{{\mu }_{0}i}{4\pi R}+\frac{{\mu }_{0}i}{4R}-\frac{{\mu }_{0}i}{4\pi R}$ $=\frac{{\mu }_0\mathrm{i}}{4\mathrm{R}}\left(1-\frac{2}{\pi }\right)$ pointing away from the page

$i_{\text{series}}=\frac{E}{R_{\text{series}}}$

$\Rightarrow {\mathrm{i}}_{\text{series}}=\frac{\mathrm{E}}{10\mathrm{R}}$

${\mathrm{i}}_{\text{parallel}}=\frac{\mathrm{E}}{{\mathrm{R}}_{\text{Parallel}}}$

$=\frac{\mathrm{E}}{\mathrm{R}/10}$ ${\mathrm{i}}_{\text{parallel}}=\mathrm{n}*{\mathrm{i}}_{\text{series}}$

$\Rightarrow \frac{10\mathrm{E}}{\mathrm{R}}=\frac{\mathrm{n}\mathrm{E}}{10\mathrm{R}}$

$\Rightarrow \mathrm{n}=100$

$\mathrm{R}={\mathrm{R}}_0(1+\alpha \mathrm{\Delta }\mathrm{T})$

$\begin{array}{rr}& \Rightarrow 6.8=2[1+\alpha *(80-0\left)\right]\\ & \Rightarrow \alpha =\frac{3.4-1}{80}\\ & =0.03\\ & =3*{10}^{-2}{}^{?}{\mathrm{C}}^{-1}\end{array}$

${\mathrm{V}}_{\mathrm{P}}={\mathrm{V}}_{\mathrm{q}}+{\mathrm{V}}_{-\mathrm{q}}$

$\begin{array}{rr}& =\left(\frac{\mathrm{K}\mathrm{q}}{5-3}+\frac{\mathrm{K}(-\mathrm{q})}{5+3}\right)*{10}^2\\ & =\left(\frac{\mathrm{K}\mathrm{q}}{2}-\frac{\mathrm{K}\mathrm{q}}{8}\right)*{10}^2\\ & =\left(\frac{3\mathrm{K}\mathrm{q}}{8}\right)*{10}^2\end{array}$

From $x-t$ graph,

$\begin{array}{rr}& A=1,T=8\\ & \Rightarrow \omega =\frac{2\pi }{T}\\ & \Rightarrow \omega =\frac{\pi }{4}\\ & \text{at}t=2,x=1\\ & a=-{\omega }^{2}x\\ & \Rightarrow \mathrm{a}=\frac{-{\pi }^2}{16}*1\\ & \Rightarrow \mathrm{a}=\frac{-{\pi }^2}{16}\mathrm{m}/{\mathrm{s}}^2\end{array}$

Time period of satellite

$\mathrm{T}=2\pi \sqrt[]{\frac{{\mathrm{R}}^3}{\mathrm{G}\mathrm{M}}}$

$=2\pi \sqrt[]{\frac{{\mathrm{R}}^3}{\mathrm{G}\mathrm{d}\frac{4}{3}\pi {\mathrm{R}}^3}}$

$\Rightarrow \mathrm{T}=\sqrt[]{\frac{3\pi }{\mathrm{G}\mathrm{d}}}$ $\Rightarrow \frac{3\pi }{\mathrm{G}\mathrm{d}}={\mathrm{T}}^2$