3 Marks Question

Question 13 Marks

Explain use of juction diode as full wave rectifier.

Answer

→ The circuit diagram of the full-wave rectifier is shown in the figure. In full wave rectifier, two $p-n$ junction diodes are used.
→ In this type of rectifier, the rectified output voltage is obtained during both the positive as well as negative half of ac cycle. Hence, it is known as full-wave rectifier.
→ As shown in fig., the $p$-side of the two diodes are connected to the ends of the secondary of the transformer. The $n$-side of the diodes are connected together and the output is taken between this common point of diodes and the mid-point of the secondary of the transformer. So for a full wave rectifier the secondary of the transformer is provided with a centre tapping and so it is called centre-tap transformer.
→ As can be seen from fig. (c), the voltage rectified by each diode is only half the total secondary voltage. Each diode rectifies only for half the cycle, but the two do so for alternate cycles. Thus the output between their common terminals and the centre tap of the transformer becomes a full-wave rectifier output.
→ Suppose the input voltage to $A$ with respect to centre tap at any instant is positive. At that instant, voltage B being out of phase should be negative. In this case, diode $D_1$ gets forward biased and conducts, while $D_2$ gets reverse biased and does not conduct. Hence, as shown in fig. c, output current is obtained between two terminals of $R_L$ during this half-cycle.
→ During the other half-cycle, voltage at $A$ is negative and voltage at $B$ is positive. In this case diode $D_1$ is in reverse bias condition and $D_2$ is in forward bias. Hence, in this part of cycle, $D_2$ conducts and output voltage is obtained.
→ Thus, we get output voltage during both positive as well as negative half of the cycle.

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Question 23 Marks

A given coin has a mass of 3.0 g . Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of ${ }_{29}^{63} Cu$ atoms (of mass 62.92960 u )

Answer

→ Nucleus of copper contains 29 protons and number of neutrons $N = A - Z$
$
=34
$
→ Mass Defect $\Delta M =\left[ Z m_p+ N m_n\right]-m({ }_{29}^{63} Cu)$
$
\begin{array}{l}
\therefore \Delta M=[29 \cdot 1.007825+34 \cdot 1.008665]-62.92960 \\
\therefore \Delta M=[29.226925+34.29461]-62.92960 \\
\therefore \Delta M=0.591935 u
\end{array}
$
→ Binding Energy
$
\begin{array}{l}
E_b=\Delta M c^2 \\
\therefore E_b=0.591935 \times 931.5 \\
\therefore E_b=551.39 MeV
\end{array}
$
→ An energy of 551.39 MeV is required to separate protons and neutrons in a copper nucleus.
→ Number of atoms in 3 g copper coin,
Mass of $C u$ Number of atoms in $C u$
$
\begin{array}{l}
63 g 6.022 \cdot 10^{23} \\
\therefore 3 g(?)
\end{array}
$
→ Number of atoms in coin,
$
N=\frac{3 \times 6.022 \times 10^{23}}{63}=2.87 \cdot 10^{22}
$
→ The energy required to separate all the neutrons and protons from the copper of 3 g .
$
\begin{array}{l}
E=E_b \cdot N \\
\therefore E=551.39 \cdot 2.87 \cdot 10^{22} MeV \\
\therefore E=1582.4893 \cdot 10^{22} MeV \\
\therefore E=1582.4893 \cdot 10^{22} \cdot 10^6 \cdot 1.6 \cdot 10^{-19} J \\
\therefore E=2531.98 \cdot 10^9 J \\
\therefore E=2.53 \cdot 10^9 J
\end{array}
$

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Question 33 Marks

The work function of caesium is 2.14 eV . Find
(a) The threshold frequency for caesium.
(b) The wavelength of incident light if photocurrent is brought to zero by a stopping potential of $0.60 V .\left( h =6.625 \times 10^{-34} Js \right)$

Answer

$
\begin{array}{c}
\varphi_0=2.14 eV \\
V_0=0.60 V \\
v_0=? \\
\lambda=?
\end{array}
$
→ (a)The cut off or threshold frequency
$
\begin{array}{l}
\varphi_0=h v_0 \\
\therefore v_0=\frac{\phi_0}{h} \\
=\frac{2.14 \times 1.6 \times 10^{-19}}{6.625 \times 10^{-34}} \\
v_0=0.5163 \times 10^{15} Hz \\
v_0=5.16 \times 10^{14} Hz
\end{array}
$
→ (b) Einstein's photoelectric equation is
$\begin{array}{l} K _{\max }=h v-\varphi_0 \\ \text { but } K _{\max }=e V_0 \\ \therefore e V_0=\frac{h c}{\lambda}-\varphi_0( C =v \lambda) \\ \therefore \frac{h c}{\lambda}=e V_0+\varphi_0 \\ \therefore \frac{h c}{\lambda}=\left(1.6 \times 10^{-19} \times 0.60\right)+\left(2.14 \times 1.6 \times 10^{-19}\right) \\ \therefore \frac{h c}{\lambda}=1.6 \times 10^{-19}(0.60+2.14)\end{array}$
$\therefore \frac{h c}{\lambda}=4.384 \times 10^{-19}$
$\therefore \lambda=\frac{6.625 \times 10^{-34} \times 3 \times 10^8}{4.384 \times 10^{-19}}$
$\begin{array}{l}\therefore \lambda=4.54 \times 10^{-7} m \\ =454 \times 10^{-9} m \\ \lambda=454 nm\end{array}$

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Question 43 Marks

Derive $\frac{n_1}{-u}+\frac{n_2}{v}=\frac{n_2-n_1}{ R }$ for refraction at a spherical surface.

Answer

→ As shown in figure, a point like object $O$ is placed on the principal axis of the spherical surface. A spherical surface has centre of curvature ' $C$ ' and radius of curvature $R$.
→ Rays emerge from a medium having refractive index $n_1$. Here, OM and ON are the incident rays.
→ They refract in a medium having refractive index $n_2$. Here NI and MI are the refractive rays. As a result, image I of the point object $O$ is obtained.
→ Assume that the aperture of the spherical surface is small compared to the object distance, image distance and radius of curvature, so that the angles can be taken small.
→ Since the aperture of the surface is assumed to be small here, NM will be taken to be nearly equal to the length of the perpendicular from the point N on the principal axis.
→ From figure,
$
\begin{array}{l}
\tan \angle NOM \approx \angle NOM=\frac{MN}{OM}...(1) \\
\tan \angle NCM \approx \angle NCM=\frac{MN}{MC} \ldots(2) \\
\tan \angle NIM \approx \angle NIM=\frac{MN}{MI} \ldots(3)
\end{array}
$
→ For $\triangle N O C, i$ is the exterior angle.
Therefore,
$
i=\angle NOM+\angle NCM
$
Substituting values from equation (1) and equation (2),
$
\therefore i=\frac{MN}{OM}+\frac{MN}{MC} \ldots(4)
$
→ From figure for $\triangle NIC , \angle NCM$ is the exterior angle.
$
\begin{array}{l}
\therefore \angle NCM=r+\angle NIM \\
r=\angle NCM-\angle NIM
\end{array}
$
$
\therefore r=\frac{MN}{MC}-\frac{MN}{MI} \ldots(5)
$
→ By applying Snell's law at point N ,
$
\begin{array}{l}
n_1 \sin i=n_2 \sin r \\
\text { But, } \sin i \approx i \\
\sin r \approx r \\
\therefore n_1 i=n_2 r
\end{array}
$
→ Substituting $i$ and $r$ from equation (4) and equation (5),
$
\begin{array}{l}
\therefore n_1\left(\frac{MN}{OM}+\frac{MN}{MC}\right)=n_2\left(\frac{MN}{MC}-\frac{MN}{MI}\right) \\
\therefore \frac{n_1}{OM}+\frac{n_1}{MC}=\frac{n_2}{MC}-\frac{n_2}{MI} \\
\therefore \frac{n_1}{OM}+\frac{n_2}{MI}=\frac{n_2}{MC}-\frac{n_1}{MC} \\
\therefore \frac{n_1}{OM}+\frac{n_2}{MI}=\frac{n_2-n_1}{MC}
\end{array}
$
→ But from figure, applying Cartesian sign convention,
$
\begin{array}{l}
OM=-u, MI=v \text { and } MC=R \\
\therefore-\frac{n_1}{u}+\frac{n_2}{v}=\frac{n_2-n_1}{R}
\end{array}
$
→ Above equation gives us a relation between object and image distance in terms of refractive index of the medium and the radius of curvature of the curved spherical surface.

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Question 53 Marks

A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which $R =3 \Omega, L=25.48 mH$ and $C =796 \mu F$. Find
(a) The impedance of the circuit.
(b) The phase difference between the voltage across the source and the current.
(c) The power dissipated in the circuit.

Answer

$
\begin{array}{l}
V_m=283 V \\
R=3 \Omega \\
C=796 \mu F \\
\nu=50 Hz \\
L=25.48 mH
\end{array}
$
→ (a) Impedence of the circuit ( Z ),
→ Inductive reactance $\left( X _{ L }\right)$
$
\begin{array}{l}
X_{L}=\omega L=2 \pi v L \\
\therefore X_{L}=2 \times 3.14 \times 50 \times 25.48 \times 10^{-3} \\
\therefore X_{L}=8000.72 \times 10^{-3} \\
\therefore X_{L}=8 \Omega
\end{array}
$
→ Capacitive reactance $\left( X _{ C }\right)$
$
\begin{array}{l}
X_{C}=\frac{1}{\omega C}=\frac{1}{2 \pi v C} \\
\therefore X_{C}=\frac{1}{2 \times 3.14 \times 50 \times 796 \times 10^{-6}} \\
\therefore X_{C}=\frac{1000000}{249944} \\
\therefore X_{C}=4 \Omega \\
→ Z=\sqrt{R^2+\left(X_{C}-X_{L}\right)^2} \\
\therefore Z=\sqrt{3^2+(4-8)^2} \\
\therefore Z=5 \Omega
\end{array}
$
(b) Phase difference ( $\varphi$ )

$
\begin{array}{l}
\tan \varphi=\frac{X_{C}-X_{L}}{R} \\
\tan \varphi=\frac{4-8}{3} \\
\tan \varphi=-\frac{4}{3} \\
\tan \varphi=-1.3333 \\
\varphi=-53.1^{\circ} \\
(\because \tan (-\theta)=-\tan \theta)
\end{array}
$
Note : Here $\varphi$ is negative. So the current in the circuit is lagging behind the voltage between two terminals of the source.
(c) Power dissipated in the circuit :
$
\begin{array}{l}
P=I^2 R \\
\text { But } I=\frac{I_m}{\sqrt{2}}
\end{array}
$
$
\begin{array}{l}
\therefore I=\frac{V_{m}}{Z \sqrt{2}} \\
\therefore P=\frac{V_m^2}{Z^2(2)} \cdot R \\
\therefore P=\frac{(283)^2 \times 3}{25 \times 2} \\
\therefore P=4800 W
\end{array}
$
(d) Power factor,
$
\begin{array}{l}
\cos \varphi=\cos \left(-53.1^{\circ}\right)(\because \cos (-\theta)=\cos \theta) \\
=\cos 53.1^{\circ} \\
=0.6
\end{array}
$

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Question 63 Marks

Find magnetic field on the axis of a circular current carrying loop.

Answer

→ As shown in the figure, a steady current I is flowing through a conducting loop of radius R .

→ The loop is placed in such a way that it lies in the $y-z$ plane and the $X$-axis passing through its axis.
→ A point P lies at a distance $x$ on the X -axis from its origin. We want to calculate the magnetic field at the point P .
→ Consider a current element $I d \vec{l}$ from the loop shown in figure. The magnitude of the magnetic field due to this element is,
$
d B=\frac{\mu_0}{4 \pi} \cdot \frac{|I d \vec{l} \times \vec{r}|}{r^3} \ldots(1)
$
→ But $I d \vec{l} \perp \vec{r}$ because $I d \vec{l}$ is in the $y z$ plane and the position vector $(\vec{r})$ is in $x y$ plane.
$
\begin{array}{l}
\therefore d B=\frac{\mu_0}{4 \pi} \cdot \frac{I d l r \sin 90}{r^3} \\
\therefore d B=\frac{\mu_0}{4 \pi} \cdot \frac{I d l}{r^2} \ldots(2)
\end{array}
$
→ From the figure, $r^2= R ^2+x^2$. Hence,
$
d B=\frac{\mu_0}{4 \pi} \cdot \frac{I d l}{\left(R^2+x^2\right)}
...(3)$
→ The magnetic field has two components at point P
(i) Perpendicular component $\left(d B_{\perp}=d B \sin \theta\right)$
→ When the perpendicular components are summed to get the net magnetic field, they cancel each other and the result is zero(ii) Parallel component $\left(d B_{\|}=d B \cos \theta\right)$
→ The parallel components are summed up to get the net magnetic field, so it can be obtained by integrating $d B_x=$ $d B \cos \theta$ over the loop.
→ $
d B(x)=d B \cos \theta
$
$\therefore d B(x)=\frac{\mu_0}{4 \pi} \cdot \frac{ l d l}{ R ^2+x^2} \cdot \cos \theta \ldots$ (4) $(\because$ from equation (3))
→ From the figure,
$
\begin{array}{l}
\cos \theta=\frac{R}{\left(x^2+R^2\right)^{\frac{1}{2}}} \\
\therefore d B(x)=\frac{\mu_0}{4 \pi} \cdot \frac{I d l}{R^2+x^2} \cdot \frac{R}{\left(R^2+x^2\right)^{\frac{1}{2}}} \\
\therefore d B(x)=\frac{\mu_0}{4 \pi} \cdot \frac{I d l \cdot R}{\left(R^2+x^2\right)^{\frac{3}{2}}}
\end{array}
$
→ The resultant magnetic field.
$
\begin{array}{l}
B=\oint_{d B(x)=} \frac{\mu_0 IR}{4 \pi\left(R^2+x^2\right)^{\frac{3}{2}}} \oint_{d l} \\
\therefore B=\frac{\mu_0 IR}{4 \pi\left(R^2+x^2\right)^{\frac{3}{2}}}(2 \pi R)\left(\because \oint_{d l=2 \pi R)}\right. \\
\therefore B=\frac{\mu_0 IR^2}{2\left(R^2+x^2\right)^{\frac{3}{2}}}
\end{array}
$
→ In vector form,
$
\overrightarrow{B}=\frac{\mu_0 IR^2}{2\left(R^2+x^2\right)^{\frac{3}{2}}} \cdot \hat{i}
$
→ To obtain the magnetic field at the centre of the loop $x=0$
$
\therefore B=\frac{\frac{\mu_0 IR^2}{2 R^3}=\frac{\mu_0 I}{2 R}, ~}{2 R}
$
→ If there are N turns, then
$
\overrightarrow{B}=\frac{\mu_0 NIR^2}{2\left(R^2+x^2\right)^{\frac{3}{2}}}
$

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Question 73 Marks

Considering drift velocity of electron, prove that $\sigma=\frac{n e^2}{m} r$.

Answer

→ A conductor of cross- sectional area $\vec{A}$ is shown in the figure. The electric field inside the conductor is $\vec{E}$.
→ Due to this electric field, there will be net flow of charges across any area of the conductor.
→ Because of the drift, distance travelled by electron in time $\Delta t$ is $\left|\overrightarrow{ v _d}\right| \Delta t$.
→ Suppose the number of free electrons per unit volume in metal is $n$, then the number of electrons passing through the area $A$ is $N$
$
=n A\left|\overrightarrow{v_d}\right|_{\Delta t.}
$
→ The total charge flowing through the cross - sectional area in time $\Delta t$ is -neA $\left|\overrightarrow{ v }_d\right|_{\Delta t} \ldots$ (1)
→ Here, electric field $\vec{E}$ is directed towards the left as a result the total electric charge passing through the surface in the direction of $\vec{E}$, will be equal to the negative value of above equation (1).
$
\begin{array}{l}
\therefore q=-\left(-n e A\left|\vec{v}_d\right| \Delta t\right)...(2)\\ \therefore q=n e A^{\left|\overrightarrow{v_d}\right|} \Delta t
\end{array}
$
→ The amount of charge crossing the area $\overrightarrow{ A }$ in time $\Delta t$ is by definition I $\Delta t$ (where I is the magnitude of the current).
→ Hence,
$
\begin{array}{l}
\therefore I \Delta t=n e A^{\left|\overrightarrow{v_d}\right|} \Delta t \\
\therefore I=n e A\left|\overrightarrow{v}_d\right|
\ldots(3)\end{array}
$
→ but current density $j=\frac{ I }{ A }$
$
\begin{array}{l}
I=j A \\
\therefore j A=n e A\left|\overrightarrow{v_d}\right|_{\left(\because \text { from } e q^{n}(3)\right)}
\end{array}
$
$
\begin{array}{l}
\therefore j=n e\left|\overrightarrow{v_d}\right| \ldots \text { (4) } \\
\therefore j=n e\left(\frac{e E}{m}\right) \cdot \tau\left(\because\left|\overrightarrow{v_d}\right|=\frac{e E}{m} \tau\right) \\
\therefore j=\frac{n e^2 E}{m} \tau \ldots \text { (5) }
\end{array}
$
→ Writing above equation (5) in vector form
$
\vec{j}=\frac{n e^2 \tau}{m} \cdot \overrightarrow{E}
$
→ Now comparing above equation with $\vec{j}=\sigma \overrightarrow{ E }$ we get
$
\begin{array}{l}
\therefore \sigma \overrightarrow{E}=\frac{n e^2 \tau}{m} \cdot \overrightarrow{E} \\
\therefore \sigma=\frac{n e^2 \tau}{m} \ldots(6)
\end{array}
$
→ Resistivity of conductor is reciprocal of conductivity
$
\begin{array}{l}
Q=\frac{1}{\sigma} \\
\therefore Q=\frac{m}{n e^2 \tau}
...(7)\end{array}
$

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Question 83 Marks

Two charge $5 \times 10^{-8} C$ and $-3 \times 10^{-8} C$ are located 20 cm apart. At what points(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

Answer

(a)
$\begin{aligned} q_1 & =5 \times 10^{-8} C \\ q_2 & =-3 \times 10^{-8} C \end{aligned}$

→ Suppose, here the positive charge ' $q_1$ ' is on the origin and negative charge ' $q_2$ ' is on the X -axis, towards the RHS of the origin.
→ Suppose, electric potential at point P is zero. Point P is $x cm$ away from charge $q_1$.
$
\begin{array}{l}
\therefore \frac{k q_1}{x \times 10^{-2}}+\frac{k q_2}{(20-x) \times 10^{-2}}=0 \\
\therefore \frac{k\left(5 \times 10^{-8}\right)}{x \times 10^{-2}}-\frac{k\left(3 \times 10^{-8}\right)}{(20-x) \times 10^{-2}}=0 \\
\therefore \quad \frac{k\left(5 \times 10^{-8}\right)}{x \times 10^{-2}}=\frac{k\left(3 \times 10^{-8}\right)}{(20-x) \times 10^{-2}} \\
\therefore \quad \frac{5}{x}=\frac{3}{20-x} \\
\therefore 100-5 x=3 x \\
\therefore 100=8 x \\
\therefore x=12.5 cm
\end{array}
$
(b)

→ As shown in fig, electric potential at point $P ^{\prime}$ is zero.
$
\begin{array}{l}
\therefore \quad \frac{k q_1}{x^{\prime} \times 10^{-2}}+\frac{k q_2}{\left(x^{\prime}-20\right) \times 10^{-2}}=0 \\
\therefore \quad \frac{k\left(5 \times 10^{-8}\right)}{x^{\prime} \times 10^{-2}}-\frac{k\left(3 \times 10^{-8}\right)}{\left(x^{\prime}-20\right) \times 10^{-2}}=0 \\
\therefore \quad \frac{k\left(5 \times 10^{-8}\right)}{x^{\prime} \times 10^{-2}}=\frac{k\left(3 \times 10^{-8}\right)}{\left(x^{\prime}-20\right) \times 10^{-2}} \\
\therefore \quad \frac{5}{x^{\prime}}=\frac{3}{x^{\prime}-20} \\
\therefore 5 x^{\prime}-100=3 x^{\prime} \\
\therefore 5 x^{\prime}-3 x^{\prime}=100 \\
\therefore 2 x^{\prime}=100 \\
\therefore x^{\prime}=50 cm
\end{array}
$
→ Hence, electric potential will be zero at distance 12.5 cm from $q_1$ (positive charge) and at 50 cm from $q_1$.

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Question 93 Marks

Find electric field of an electric dipole at any point on the axis of dipole. Also write one special case.

Answer

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Physics 3 Marks Question

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