Physics MCQs for NEET — Practice Questions with Answers

3. Flux can’t change in a superconducting loop.

$∆\mathrm{\varphi }=2{\mathrm{\pi R}}^2.\mathrm{B}$

    Initially current was zero, so self flux was zero.

$\therefore$ Finally $\mathrm{Li}=2{\mathrm{\pi R}}^2\times \mathrm{B}$

$\mathrm{i}=\frac{2{\mathrm{\pi R}}^2\times \mathrm{B}}{\mathrm{L}}$

$$\begin{gathered} 4. \mathrm{Current} \mathrm{developed} \mathrm{with} \mathrm{time} \mathrm{in} \mathrm{a} \mathrm{coil} \mathrm{of} \mathrm{inductance} \\ \mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}\left(1-{\mathrm{e}}^{-\mathrm{t}/\mathrm{\tau }}\right) \mathrm{where} \mathrm{\tau }=\mathrm{L}/\mathrm{R} \\ \mathrm{we} \mathrm{have} \mathrm{\tau }=\frac{8.4 \mathrm{mH}}{6\mathrm{\Omega }}=1.4 \mathrm{ms} \\ \mathrm{Hence}, 1 \mathrm{A}=\left(\frac{12 \mathrm{V}}{6\mathrm{\Omega }}\right) \left(1-{\mathrm{e}}^{-\mathrm{t}/1.4 \mathrm{ms}}\right) \\ \mathrm{or} {\mathrm{e}}^{-\mathrm{t}/1.4 \mathrm{ms}} =1-\frac{1}{2}=\frac{1}{2} \\ \mathrm{or} -\mathrm{t}/1.4 \mathrm{ms} =1\mathrm{n}\left(\frac{1}{2}\right)=-0.693 \\ \mathrm{or} \mathrm{t}=\left(1.4\times 0.693\right)\mathrm{ms} =0.97 \mathrm{ms}\approx 1 \mathrm{ms}. \end{gathered}$$

$$\begin{gathered} 2. \mathrm{\varphi }=\mathrm{NBA} \\ {\mathrm{\varphi }}_1=50\times \mathrm{\pi }\times {\left(3\times {10}^{-2}\right)}^2\times 0.1 =141.3\times {10}^{-4} \mathrm{Wb} \\ {\mathrm{\varphi }}_2=50\times \mathrm{\pi }\times {\left(3\times {10}^{-2}\right)}^2\times 0.35 =494.5\times {10}^{-4} \mathrm{Wb} \\ \therefore \mathrm{e}=\frac{\mathrm{d\varphi }}{\mathrm{dt}}=17.7 \mathrm{volt} \end{gathered}$$

$$\begin{gathered} 1. \mathrm{L}=\frac{\mathrm{N\varphi }}{\mathrm{i}} \\ 8\times {10}^{-3}=\frac{400\times \mathrm{\varphi }}{5\times {10}^{-3}} \\ \Rightarrow \mathrm{\varphi }=\frac{40\times {10}^{-6}}{400} \mathrm{Wb} = {10}^{-7} \mathrm{Wb} \Rightarrow \mathrm{\varphi } = \frac{4\mathrm{\pi }\times {10}^{-7}}{4\mathrm{\pi }} \mathrm{Wb} \\ \Rightarrow \mathrm{\varphi }=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }} \mathrm{Wb} \Rightarrow \mathrm{\varphi }\approx 0.1 {\mathrm{\mu }}_0 \mathrm{Wb} \end{gathered}$$

$$\begin{gathered} 2. \mathrm{I}={\mathrm{I}}_0 \left(\mathrm{l}-{\mathrm{e}}^{-\mathrm{t}/\mathrm{\tau }}\right) \mathrm{where} \mathrm{\tau } \to \mathrm{time} \mathrm{constant} \\ \therefore \frac{3}{4}{\mathrm{I}}_0={\mathrm{I}}_0\left(\mathrm{l}-{\mathrm{e}}^{-\mathrm{t}/\mathrm{\tau }}\right) \\ \Rightarrow \frac{3}{4}=\mathrm{l}-{\mathrm{e}}^{-\mathrm{t}/\mathrm{\tau }} \Rightarrow {\mathrm{e}}^{-\mathrm{t}/\mathrm{\tau }}=\frac{1}{4} \\ \Rightarrow \frac{-\mathrm{t}}{\mathrm{\tau }}\mathrm{In} \mathrm{e} =\mathrm{In}\frac{1}{4} \Rightarrow \frac{-4}{\mathrm{\tau }} =-2 \mathrm{In} 2 \Rightarrow \mathrm{\tau }=\frac{2}{\mathrm{In} 2} \end{gathered}$$

$$\begin{gathered} 2. \mathrm{\varphi }=\left({\mathrm{t}}^3-2\mathrm{t}\right)\times {10}^{-3} \\ \frac{\mathrm{d\varphi }}{\mathrm{dt}}=\left(3{\mathrm{t}}^2-2\right)\times {10}^{-3} \\ {\overline{)\frac{\mathrm{d\varphi }}{\mathrm{dt}}}}_{\mathrm{t}=2}=\left(3\times 4-2\right)\times {10}^{-3} \mathrm{Wb}/\mathrm{s} ={10}^{-2} \mathrm{Wb}/\mathrm{s} \\ \mathrm{e}=-\mathrm{N}\frac{\mathrm{d\varphi }}{\mathrm{dt}}=-100\times {10}^{-2} \mathrm{V} =-1\mathrm{V} \end{gathered}$$

$$\begin{gathered} 2. \mathrm{I}={\mathrm{I}}_0\left(1-{\mathrm{e}}^{-\mathrm{Rt}/\mathrm{L}}\right) \\ {\mathrm{I}}_0=\frac{\mathrm{E}}{\mathrm{R}}\left(\mathrm{Steady} \mathrm{current}\right) \\ \mathrm{When} \mathrm{t}=\infty \\ {\mathrm{I}}_{\infty }=\frac{\mathrm{E}}{\mathrm{R}}\left(1-{\mathrm{e}}^{-\infty }\right) =\frac{15}{10} =1.5 \\ {\mathrm{I}}_1=1.5\left(1-{\mathrm{e}}^{-\mathrm{R}/\mathrm{L}}\right) =1.5\left(1-{\mathrm{e}}^{-2}\right) \\ \therefore \frac{{\mathrm{I}}_{\infty }}{{\mathrm{I}}_1}=\frac{1}{1-{\mathrm{e}}^{-2}}=\frac{{\mathrm{e}}^2}{{\mathrm{e}}^2-1} \end{gathered}$$

$$\begin{gathered} 1. \mathrm{\varphi }=\mathrm{Li} \Rightarrow \mathrm{NBA}=\mathrm{Li} \\ \mathrm{Since} \mathrm{magnetic} \mathrm{field} \mathrm{at} \mathrm{the} \mathrm{centre} \mathrm{of} \mathrm{circular} \mathrm{coil} \mathrm{carrying} \mathrm{current} \mathrm{is} \mathrm{given} \mathrm{by} \\ \mathrm{B}=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\times \frac{2\mathrm{\pi Ni}}{\mathrm{r}} \\ \therefore \mathrm{N}.\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}. \frac{2\mathrm{\pi Ni}}{\mathrm{r}}. {\mathrm{\pi r}}^2=\mathrm{Li} \Rightarrow \mathrm{L}=\frac{{\mathrm{\mu }}_0{\mathrm{N}}^2\mathrm{\pi r}}{2} \\ \mathrm{Hence} \mathrm{self} \mathrm{inductance} \mathrm{of} \mathrm{a} \mathrm{coil} \frac{4\mathrm{\pi }\times {10}^{-7}\times 500\times 500\times \mathrm{\pi }\times 0.05 }{2} =25 \mathrm{mH} \end{gathered}$$

1.  Magnetic induction in the solenoid

     $\mathrm{B} = {\mu }_0\mathrm{ni}$

     Magnetic flux linked with the coil

     $\mathrm{\varphi } = \mathrm{NAB} = \mathrm{NA}{\mu }_0\mathrm{ni}$

$$\begin{gathered} \mathrm{M}=\frac{\mathrm{\varphi }}{\mathrm{i}}=\frac{\mathrm{N} \mathrm{A} {\mathrm{\mu }}_0\mathrm{n} \mathrm{i}}{\mathrm{i}}={\mathrm{NA\mu }}_0\mathrm{n} \\ \mathrm{M}=100\times \mathrm{\pi }(1 \times {10}^{-2}{)}^2\times 4\mathrm{\pi }\times {10}^{-7}\times 800= 316\times {10}^{-7} \mathrm{H}=0.0316 \mathrm{mH}. \end{gathered}$$

$$\begin{gathered} 3. \mathrm{The} \mathrm{magnetic} \mathrm{field} \mathrm{at} \mathrm{a} \mathrm{distance} \mathrm{x} \mathrm{from} \mathrm{the} \mathrm{wire} \mathrm{B}\left(\mathrm{x}\right)=\frac{{\mathrm{\mu }}_0\mathrm{i}}{2\mathrm{\pi x}} \\ \mathrm{EMF} \mathrm{induced} \mathrm{in} \mathrm{an} \mathrm{element} \mathrm{of} \mathrm{length} \mathrm{dx} \mathrm{at} \mathrm{a} \mathrm{distance} \mathrm{x} \mathrm{from} \mathrm{wire}=\mathrm{Bvdx} \\ \therefore \mathrm{Total} \mathrm{EMF} \mathrm{induced} \mathrm{in} \mathrm{the} \mathrm{rod} \\ \mathrm{E}=\underset{0.01}{\overset{0.2}{\int }}\mathrm{Bv} \\ \mathrm{dx}=\underset{0.01}{\overset{0.2}{\int }}\frac{{\mathrm{\mu }}_0\mathrm{iv}}{2\mathrm{\pi x}} \\ \mathrm{dx}=\frac{{\mathrm{\mu }}_0\mathrm{iv}}{2\mathrm{\pi }} \underset{0.01}{\overset{0.2}{\int }}\frac{1}{\mathrm{x}} \mathrm{dx} \\ \mathrm{E}=\frac{{\mathrm{\mu }}_0\mathrm{iv}}{2\mathrm{\pi }}{\left[{\log }_{\mathrm{e}} \mathrm{x}\right]}_{0.01}^{0.2} \\ = \frac{{\mathrm{\mu }}_0\mathrm{iv}}{2\mathrm{\pi }}\left[{\log }_{10}\left(0.2\right)-{\log }_{10}\left(0.01\right)\right]\times 2.303 \\ \mathrm{E}=\frac{4\mathrm{\pi }\times {10}^{-7}\times 5\times 10}{2\mathrm{\pi }}\left[1.301\right]\times 2.303 \\ =2.99\times {10}^{-5} \mathrm{V}\approx 30\mathrm{\mu V} \end{gathered}$$