Physics MCQs for NEET — Practice Questions with Answers

4. The induced emf in the other coil (coil 2) is ${\mathrm{e}}_2=-\mathrm{M}\frac{{\mathrm{di}}_1}{\mathrm{dt}}=-\mathrm{M}\frac{∆{\mathrm{i}}_1}{∆\mathrm{t}} =\mathrm{M}\frac{\left({\mathrm{i}}_2-{\mathrm{i}}_1\right)}{∆\mathrm{t}}=\frac{2\left(0-10\right)}{0.1}=200 \mathrm{V}$

$$\begin{gathered} 4. \mathrm{e}=-\mathrm{L}\frac{\mathrm{di}}{\mathrm{dt}} \\ \mathrm{But} \mathrm{e}=4\mathrm{V} \mathrm{and} \frac{\mathrm{di}}{\mathrm{dt}}=\frac{0-1}{{10}^{-3}}=-1/{10}^{-3} \\ \therefore \frac{-1}{{10}^{-3}}\left(-\mathrm{L}\right)=4 \\ \Rightarrow \mathrm{L}=4\times {10}^{-3} \mathrm{henry} \end{gathered}$$

4. Let i be current flowing through the inductance, then flux linked with the circuit $\mathrm{\varphi }\propto \mathrm{i} \mathrm{or} \mathrm{\varphi }=\mathrm{Li}$

$\mathrm{e}=-\frac{\mathrm{d\varphi }}{\mathrm{dt}}=-\mathrm{L}\frac{\mathrm{di}}{\mathrm{dt}} \left(\mathrm{emf}\right)$

    Work done against back emf e in time dt and current i is $\mathrm{dW}=-\mathrm{eidt}=\mathrm{L}\frac{\mathrm{di}}{\mathrm{dt}} \mathrm{idt}=\mathrm{L} \mathrm{idi}$

$\therefore \mathrm{W}=\mathrm{L} {\int }_0^{\mathrm{i}}\mathrm{i} \mathrm{di}=\frac{1}{2} {\mathrm{Li}}^2$

1. Given : Voltage in primary ${\mathrm{V}}_{\mathrm{p}}$ = 200 volt

               Current in primary ${\mathrm{i}}_{\mathrm{p}}$ = 2 amp

               Voltage in secondary ${\mathrm{V}}_{\mathrm{s}}$ = 2000 volt

     The relation for the current in the secondary is

$$\begin{gathered} \frac{{\mathrm{V}}_{\mathrm{s}}}{{\mathrm{V}}_{\mathrm{p}}}=\frac{{\mathrm{i}}_{\mathrm{p}}}{{\mathrm{i}}_{\mathrm{s}}} \Rightarrow \frac{2000}{200}=\frac{2}{{\mathrm{i}}_{\mathrm{s}}} \\ \mathrm{or}, {\mathrm{i}}_{\mathrm{s}}=\frac{2\times 200}{2000}=0.2 \mathrm{amp}. \end{gathered}$$

$$\begin{gathered} 4. \mathrm{\varphi }\propto \frac{1}{{\mathrm{d}}^3} \Rightarrow \frac{{\mathrm{\varphi }}_2}{{\mathrm{\varphi }}_1} ={\left(\frac{{\mathrm{d}}_1}{{\mathrm{d}}_2}\right)}^3 \\ \Rightarrow {\mathrm{\varphi }}_2={\left(\frac{80}{40}\right)}^3\times 4\times {10}^{-5}=32\times {10}^{-5} \mathrm{weber} \\ \therefore ∆\mathrm{\varphi }=28\times {10}^{-5} \mathrm{weber}‐\mathrm{turns} \\ \therefore \mathrm{e}=\frac{∆\mathrm{\varphi }}{∆\mathrm{t}}=\frac{-28\times {10}^{-5}}{8\times {10}^{-2}}=-3.5\times {10}^{-3} \mathrm{V} \\ \mathrm{Thus} \mathrm{emf} \mathrm{produced} =3.5\times {10}^{-3} \mathrm{V} \end{gathered}$$

$$\begin{gathered} 4. \mathrm{Speed} \mathrm{of} \mathrm{train} = \frac{72\times 1000}{3600}=20 \mathrm{m}/\sec \\ \mathrm{Since} \mathrm{unit} \mathrm{of} \mathrm{Earth}\text{'}\mathrm{s} \mathrm{field} \mathrm{is} \mathrm{A}/\mathrm{m}, \mathrm{it} \mathrm{is} \mathrm{magnetic} \mathrm{intensity} \\ \mathrm{Vertical} \mathrm{component} \mathrm{of} \mathrm{earth}\text{'}\mathrm{s} \mathrm{field} \\ \mathrm{Magnetic} \mathrm{intensity} \mathrm{V}=\mathrm{H} \tan \mathrm{\theta }=0.345\times \tan 30° =0.199 \mathrm{A}/\mathrm{m} \\ \mathrm{Magnetic} \mathrm{field} \mathrm{B} = {\mathrm{\mu }}_0\times \mathrm{V} \\ \therefore \mathrm{B}={\mathrm{\mu }}_0\times 0.199 \\ \hspace{0.17em}\mathrm{Hence} \mathrm{induced} \mathrm{emf} \\ \mathrm{e}=4\mathrm{\pi }\times {10}^{-7}\times 0.199\times 1.7\times 20 \\ = 849.8\times {10}^{-6} \mathrm{V} \\ = 850 \mathrm{\mu } \mathrm{V} \end{gathered}$$

$$\begin{gathered} 3. \frac{\mathrm{d\varphi }}{\mathrm{dt}}=10\mathrm{t}+6 \\ \mathrm{e}=-\frac{\mathrm{d\varphi }}{\mathrm{dt}}=-\left(10\mathrm{t}+6\right) \\ \mathrm{e}{|}_{\mathrm{t}=3} =-\left(10\times 3+6\right)=-36 \\ \mathrm{e}{|}_{\mathrm{t}=0} =-\left(10\times 0+6\right)=-6 \\ \frac{{\mathrm{e}}_{\mathrm{t}=3}}{{\mathrm{e}}_{\mathrm{t}=0}} =\frac{-36}{-6} =\frac{6}{1} \end{gathered}$$

2. As the plane is flying horizontally it will cut the vertical component of earth’s field ${\mathrm{B}}_{\mathrm{V}}$ . So the

    emf induced between its tips, $\mathrm{e} = {\mathrm{B}}_{\mathrm{V}}\mathrm{lv}$

    But as by definition of angle of dip,

$$\begin{gathered} \tan \mathrm{\varphi }=\frac{{\mathrm{B}}_{\mathrm{V}}}{{\mathrm{B}}_{\mathrm{H}}} \mathrm{i}.\mathrm{e}., {\mathrm{B}}_{\mathrm{V}}={\mathrm{B}}_{\mathrm{H}} \tan \mathrm{\varphi } \\ \mathrm{So} \mathrm{e}=\left({\mathrm{B}}_{\mathrm{H}} \tan \mathrm{\varphi }\right)\mathrm{lv}=2\times {10}^{-5}\times \sqrt{3}\times 250\times 20 \\ \mathrm{i}.\mathrm{e}., \mathrm{e}=\left(\sqrt{3}\right)\times {10}^{-1} \mathrm{V} =0.173 \mathrm{V} \end{gathered}$$

$$\begin{gathered} 1. \mathrm{L}=\frac{{\mathrm{\mu }}_0{\mathrm{N}}^2{\mathrm{\pi r}}^2}{\mathcal{l}} \\ \mathrm{Length} \mathrm{of} \mathrm{wire}=\mathrm{N} 2\mathrm{\pi r}=\mathrm{constant}\left(=\mathrm{C}, \mathrm{suppose}\right) \\ \therefore \mathrm{L}={\mathrm{\mu }}_0{\left(\frac{\mathrm{C}}{2\mathrm{\pi r}}\right)}^2\frac{{\mathrm{\pi r}}^2}{\mathcal{l}} \therefore \mathrm{L}\propto \frac{1}{\mathcal{l}} \\ \therefore \mathrm{Self} \mathrm{inductance} \mathrm{will} \mathrm{become} 2\mathrm{L}. \end{gathered}$$

$$\begin{gathered} 3. \mathrm{According} \mathrm{to} \mathrm{Faraday}’\mathrm{s} \mathrm{law} \mathrm{of} \mathrm{electromagnetic} \mathrm{induction}, {\mathrm{E}}_{\mathrm{induced}}=\frac{-∆\mathrm{\varphi }}{∆\mathrm{t}}=-\frac{\mathrm{B}\left({\mathrm{A}}_{\mathrm{f}}-{\mathrm{A}}_{\mathrm{i}}\right)}{∆\mathrm{t}} \\ \mathrm{Let} \mathrm{r} \mathrm{be} \mathrm{the} \mathrm{radius} \mathrm{of} \mathrm{circle} ; \mathrm{then} \mathrm{side} \mathrm{of} \mathrm{square} \mathrm{formed} =\frac{2\mathrm{\pi r}}{4}=\frac{\mathrm{\pi r}}{2} \\ \mathrm{Change} \mathrm{is} \mathrm{area} \mathrm{of} \mathrm{loop} = {\mathrm{A}}_{\mathrm{i}}-{\mathrm{A}}_{\mathrm{f}}={\mathrm{\pi r}}^2-{\left(\frac{\mathrm{\pi r}}{2}\right)}^2=\frac{\mathrm{\pi }\left(4-\mathrm{\pi }\right){\mathrm{r}}^2}{4} \\ \mathrm{Hence} \mathrm{average} \mathrm{emf} \mathrm{induced} = \frac{\mathrm{\pi }\left(4-\mathrm{\pi }\right){\mathrm{r}}^2}{4}.\frac{\mathrm{B}}{\mathrm{t}} \\ =\frac{\mathrm{\pi }\left(4-\mathrm{\pi }\right)\times {\left(0.1\right)}^2\times 100}{4\times 0.1} =6.75 \mathrm{volt}. \end{gathered}$$