3 Marks Question

Question 13 Marks

A small flat search coil of area $5 cm^2$ with $140$ closely wound turns is placed between the poles of a powerful magnet producing magnetic field $0.09 \ T$ and then quickly removed out of the field region. Calculate
$a.$ change of magnetic flux through the coil, and
$b.$ emf induced in the coil.

Answer

$\text { i. Flux, } \phi= NBA \cos \theta$
$\theta=0^{\circ},$
Thus $\phi_1=$ NBA $=140 \times 0.09 \ \times 5 \times 10^{-4}=6.3 \times 10^{-3} Wb$
When coil is quickly removed, flux becomes zero i.e. $\phi_2=0$
Thus,
$\Delta \phi=\phi_2-\phi_1=0-6.3 \times 10^{-3}=-6.3 \times 10^{-3} Wb$
$ii.$ Let in time dt = 1s, the coil is quickly removed. So, induced emf,
$e =\frac{-d \phi}{d t}=-\frac{\left(-6.3 \times 10^{-3}\right)}{1}=6.3 \times 10^{-3} V$

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Question 23 Marks

i. Define mutual inductance.
ii. A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil?

Answer

ii. Mutual inductance of a pair of coils, $\mu=1.5 H$
Initial current, $I _1=0 A$
Final current $I _2=20 A$
Change in current will be: $\Delta I= 1 _2- I _1$
20 - 0 = 20 A
and we know,
$\Delta \phi= M \Delta I$
where, $\Delta \phi$ is change in magnetic flux
$\Delta \phi=1.5 \times 20=30 Wb$
Hence, change in the flux linkage will be 30 Wb.

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Question 33 Marks

In single slit diffraction, explain why the maxima at $\theta=\left(n+\frac{1}{2}\right)\left(\frac{\lambda}{a}\right)$ becomes weaker and weaker as an increases. State two important differences between interference and diffraction pattern.

Answer

As n increases. deviation of light from straight direction increases thus the light spread out more and which result in decreasing intensity of bright fringes.
In interference pattern the fringes are of equal width (dark and bright) and are of equal intensity and in diffraction the central maxima is brightest and its width is wider twice as compare to the other maxima and intensity of bands goes on decreasing as we move away from the Centre.

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Question 43 Marks

Using Bohr's total postulates, derive the expression for the total energy of the electron in the stationary states of hydrogen atom.

Answer

According to Bohr's postulates, in a hydrogen atom, a single electron revolves around a nucleus of charge $+e$. For an electron moving with a uniform speed in a circular orbit of a given radius, the centripetal force is provided by Coulomb force of attraction between the electron and the nucleus. The gravitational attraction may be neglected as the mass of electron and proton is very small. So,
$m v^2 / r = ke ^2 / r ^2\left(\right.$ where, $\left.k =1 / 4 \pi \varepsilon_0\right)$
or $m v^2= ke ^2 / r$.
where, $m =$ mass of electron, $r =$ radius of electronic orbit, $v =$ velocity of electron
Again, by Bohr's second postulates
$mvr = nh / 2 \pi$
where, $n=1,2,3, \ldots$ or $v=n h / 2 \pi m r$
Putting the value of $v$ in Eq. $(i)$
$m\left(\frac{n h}{2 \pi m r}\right)^2=\frac{k r^2}{r} \Rightarrow r=\frac{n^2 h^2}{4 \pi^2 k m e^2}$
Kinetic energy of electron ,
$E_K=\frac{1}{2} m v^2=\frac{k e^2}{2 r}\left(\because \frac{m v^2}{r}=\frac{k e^2}{r^2}\right)$
Using Eq$(ii)$, we get
$E_K=\frac{k \kappa^2}{2} \frac{4 \pi^2 k m c^2}{n^2 h^2}=\frac{2 \pi^2 k^2 m c^4}{n^2 h^2}$
Potential energy of electron,
$E_P=-\frac{k(e) \times(e)}{r}=-\frac{k x^2}{r}$
Using Eq$(ii)$, we get
$E_P=-k e^2 \times \frac{4 \pi^2 k m c^2}{n^2 h^2}=-\frac{4 \pi^2 k^2 m e^4}{n^2 h^2}$
Hence, total energy of the electron in the $n ^{ th }$ orbit
$E=E_P+E_K=-\frac{4 \pi^2 k^2 m e^4}{n^2 h^2}+\frac{2 \pi^2 k^2 m e^4}{n^2 h^2}$
$=-\frac{2 \pi^2 k^2 m e^4}{n^2 h^2}$

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Question 53 Marks

Deuteron is a bound state of a neutron and a proton with a binding energy $B =2.2\ MeV$. A $ \gamma-$ray of energy $E$ is aimed at a deuteron nucleus to try to break it into a $($neutron $+$ proton$)$ such that the $n$ and $p$ move in the direction of the incident $\gamma-$ray. If $E = B$, show that this cannot happen. Hence calculate how much bigger than $B$ must $E$ be for such a process to happen.

Answer

Binding energy of a deuteron is, $B=2.2 MeV$
Let kinetic energy and the momentum of neutron and proton be $K _n, K_{ p }$, and $P _n, P _{ p }$ respectively. From the conservation of energy,
$E - B = K _{ n }+ K _{ p }=\frac{p_m^2}{2 m}+\frac{p_p^2}{2 m}$
Now applying conservation of momentum is given by,
$ Pn _{ n }+ p _{ p }=\frac{E}{C}$
$As E = B , Eq . \text { (i) } p_n^2+p_p^2=0$
It only happens if $p _{ n }= p _{ p }$
So, the Eq. $(ii)$ cannot be satisfied and the process cannot take place.
Let us take $E = B + x$, where $x << B$ for the process to take place.
Putting the value of $p _{ n }$ from Eq. $(ii)$ in Eq. $(i),$ we get
or $2 p_p^2-\left(\frac{2 E}{c}\right) p_p+\left(\frac{E^2}{c^2}-2 m x\right)=0$
Solving the quadratic equation, we get
$P _{ p }=\frac{\frac{2 E}{c}+\sqrt{\frac{4 E^2}{c^2}-8\left(\frac{E^2}{c^2}-2 m x\right)}}{4}$
For the real value $p _{ p }$, the discriminant is positive, Hence
$\frac{4 E^2}{c^2}=8\left(\frac{E^2}{c^2}-2 m x\right)$
$16 mx =\frac{4 E^2}{c^2}$
$\Rightarrow x=\frac{E^2}{4 m c^2}$
But $X \ll B$ hence $E \ll B$
$\Rightarrow x=\frac{B^2}{4 m c^2}$

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Question 63 Marks

A beam of monochromatic radiation is incident on a photosensitive surface. Answer the following questions giving reasons.
i. Do the emitted photoelectrons have the same kinetic energy?
ii. Does the kinetic energy of the emitted electrons depend on the intensity of incident radiation?
iii. On what factors does the number of emitted photoelectrons depend?

Answer

i. Yes, all emitted photoelectrons have same kinetic energy as the kinetic energy of emitted photoelectrons depends upon frequency of the incident radiation for a given photosensitive surface.
ii. No, the kinetic energy of the emitted electrons does not depend on the intensity of incident radiation. If the intensity of the incident light radiation is increased, then the number of incident photons falling per second on the metal surface also increases but the energy of each photon remains the same as the frequency of the radiation is not changing. The maximum kinetic energy of a photoelectron depends only on the frequency of incident radiation and not on the intensity of incident radiation.
iii. The number of emitted photoelectrons depends only on intensity of incident light.

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Question 73 Marks

Draw the circuit diagram of a full wave rectifier. Explain its working principle. Show the input waveforms given to the diodes $D _1$ and $D _2$ and the corresponding output waveforms obtained at the load connected to the circuit.

Answer

Circuit:-

Principle:- Diode conducts only in forward biasing.
Input $D _1$
$D _2$

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Question 83 Marks

Two cells of emf $2E$ and $E$ and internal resistances $2r$ and $r$ respectively, are connected in parallel. Obtain the expressions for the equivalent emf and the internal resistance of the combination.

Answer

Given, emf of first cell $=2 E$
emf of second cell $= E$
Internal resistance of first cell $=2 r$
Intemal resistance of second cell $= r$
Net current, $I = I _1+ I _2 \ldots (i)$
$E_1=2 E$

For cell$-I \ V = V _{ A }- V _{ B }=2 E - I _1(2 r )$
$\Rightarrow I_1=\frac{2 E-V}{2 r}$
For cell$-II, \ V = V _{ A }- V _{ B }= E - I _2 r$
$\Rightarrow I_2=\frac{E-V}{r} \ldots (iii)$
$\because$ From Eqs. $(ii)$ and $(iii)$, substituting in Eq. $(i)$, we get
$I=\frac{2 E-V}{2 r}+\frac{E-V}{r}$
On rearranging the term, we get
$V=\frac{4 E}{3}-I\left(\frac{2 r}{3}\right)$
But for equivalent of combination,
$V = E _{ eq }- I \left( r _{ eq }\right)$
On comparing,
$E_{\text {eq }}=\frac{4 E}{3}, r_{\text {eq }}=\frac{2 r}{3}$

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