(a) Focal length of the magnifying glass, f = 5 cm
Least distance of distance vision, d = 25 cm
Closest object distance = u
Image distance, v = -d = -25 cm
According to the lens formula, we have:
\( \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \)
\(\frac{1}{u}=\frac{1}{v}-\frac{1}{f} \)
\(=\frac{1}{-25}-\frac{1}{5}=\frac{-5-1}{25}=-\frac{6}{25} \)
\(\therefore u=-\frac{25}{6}=-4.167 \mathrm{~cm} \)
Hence, the closest distance at which the person can read the book is 4.167 cm.
For the object at the farthest distant (u’), the image distance (v′) = ∞
According to the lens formula, we have:
\( \frac{1}{f}=\frac{1}{\mathrm{v}^{\prime}}-\frac{1}{\mathrm{u}^{\prime}} \)
\(\frac{1}{\mathrm{u}}=\frac{1}{\infty}-\frac{1}{5}=-\frac{1}{5} \)
\(\therefore \mathrm{u}^{\prime}=-5 \mathrm{~cm}\)
Hence, the farthest distance at which the person can read the book is 5 cm.
(b) Maximum angular magnification is given by the relation:
\( \alpha_{\max }=\frac{d}{|u|} \)
\(=\frac{25}{\frac{25}{6}}=6 \)
Minimum angular magnification is given by the relation:
\( \alpha_{\min }=\frac{d}{|u \prime|} \)
\(=\frac{25}{5}=5\)

(i) Honesty, helpfulness, responsible behaviour towards students and concern for the students to create interest in the subject.
(ii) Magnetic declination, magnetic inclination and horizontal component of earth's magnetic field.

Charge located at the origin, q = 8 mC = 8 x 10^-3 C
The magnitude of the charge taken from the point P to R and then to Q, q₁ = 2 x 10^-9 C
Here OP= d₁= 3 cm = 3 x 10^-2 m
OQ = d₂= 4 cm = 4 x 10^-2 m
Potential at the point P, V₁ \(=\frac{q}{4 \pi \epsilon_{0} d_{1}}\) 

Potential at the point Q, V₁ \(=\frac{q}{4 \pi \epsilon_{0} d_{1}}\)
The work done (W) is independent of the path
Therefore, W = q₁[V₁ – V₂]
\(=V_{2}=q_{1}\left[\frac{q}{4 \pi \epsilon_{0} d_{2}}-\frac{q}{4 \pi \epsilon_{0} d_{1}}\right]\)
\(=V_{2}=\frac{q q_{1}}{4 \pi \epsilon_{0}}\left[\frac{1}{d_{2}}-\frac{1}{d_{1}}\right]\)
Where, \(\frac{1}{4 \pi \epsilon_{0}}=9 \times 10^{9} \mathrm{Nm}^{2} \mathrm{C}^{-2}\)
Therefore,
\(W=9 \times 10^{9} \times 8 \times 10^{-3} \times\left(-2 \times 10^{-9}\right)\left[\frac{1}{4 \times 10^{-2}}-\frac{1}{3 \times 10^{-2}}\right]\)
= -144 x 10^-3 x (-100/12)
= 1.2  Joule
Therefore, the work done during the process is 1.2 J

(i) (a) Declination (D), (b) Angle of dip or angle of inclination (I), (c) Horizontal component of earth's field (H_E)
(ii) At the equator

(i) i = 52\(\unicode{xb0} \), when prism is adjusted at an angle of minimum deviation, then angle of incidence is equal to the angle of emergence.
Hence, r = 0
This ray pass unrefracted at AC interface and reaches AB interface. Here, we can see angle of incidence becomes 30\(\unicode{xb0} \).
Thus, applying snell's law, \(\frac{\sin 30^{\circ}}{\sin e}=\frac{\mu_{a}}{\mu_{g}}=\frac{1}{\sqrt{3}}\)
\(\sin e=\sqrt{3} \times \sin 30^{\circ}=\frac{\sqrt{3}}{2}\)
Thus, e = 60\(\unicode{xb0} \)

Yes, but motion is simple harmonic only when charge - q is not very far from the centre of ring on its axis. Otherwise motion is periodic, but not simple harmonic in nature.

Micro waves

The resistivity of a metallic conductor is given by
ρ = ρ₀[1 + ∝(T-T₀)]
Where, ρ₀ = Resistivity at reference temperature
T₀ = Reference temperature
∝ = Coefficient of resistivity
From the above relation, we can say that the graph between resistivity of a conductor with temperature is straight line. But, at temperatures much lower than 273 K ( i.e. 0°C), the graph deviates considerably from a straight line as shown in the figure.

When a charged particle moving on a circular path of radius R in a uniform magnetic field, the Lorentz magnetic force F (= qvB) acting on the particle, provides the required centripetal force for its circular motion. It means the Lerentz force acts along the radius towards the centre of circular path. While moving on a circular path, the small displacement \(\overset { \rightarrow }{ dr } \) of the charged particle is always perpendicular to Lerentz force, i.e., \(\theta ={ 90 }^{ o }\) , therefore work done
\(dW=\overset { \rightarrow }{ F } .\overset { \rightarrow }{ dr } =F\quad dr\quad cos{ 90 }^{ o }=0.\) 
Since the velocity of the charged particle, moving on a circular path is acting tangentially to the path whose direction is changing continuously in circular motion of the particle, therefore the momentum of the particle is changing.

To have a uniform, strong and radial magnetic field.

Conductivity \(\sigma =\frac{1}{\rho}=\frac{ne^2 \tau}{m}\)
Relaxation time,  \(\tau = \frac{m\sigma}{ne^2}, i.e., \tau \propto \frac{\sigma}{n}\)
\(\therefore \tau_A:\tau_B:\tau_C = \frac{\sigma}{2n}:\frac{2\sigma}{n}:\frac{2\sigma}{2n}= \frac{\sigma}{2n}:\frac{2\sigma}{n}:\frac{\sigma}{n}\)
Thus, \(\tau_B>\tau_C>\tau_A.\) So average collision time for material B is maximum.

Given: u = -8 cm,f = +12 cm
\(\because \frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{12}-\frac{1}{8} \Rightarrow v=-24 \mathrm{~cm} \)
\((i) m=\frac{v}{u}=\frac{-24}{-8}=3 \)
\((ii) A_{o}=1 \mathrm{~mm}^{2}\)
∵ Areal magnification \(=\frac{A_{I}}{A_{o}}=m^{2}\)
\(\Rightarrow A_{I}=A_{o} \times m^{2}=1 \mathrm{~mm}^{2} \times(3)^{2}=9 \mathrm{~mm}^{2}\)
∴ The area of each square in the virtual image is equal to 9 mm².

Given: \( M=60 \mathrm{Am}^{2}, B_{H}=40 \times 10^{-6} \mathrm{Wbm}^{-2}, \tau=1.2 \times 10^{-3} \mathrm{Nm} \)
\(\Rightarrow \sin \theta=\frac{\tau}{M B_{H}}=\frac{1.2 \times 10^{-3}}{60 \times 40 \times 10^{-6}}=0.5 \Rightarrow \theta=30^{\circ}\)

Here, length, 2a = 2 cm = 2 x 10^-2 m,
9 = 60°,\(\tau\) = 8.\(\sqrt{3}\) Nm
Charge, Q = 4nC = 4 x 10^-9 C, U = ?
As we know that, = \(\tau\) Q(2a) E sin \(\theta\)
 \(\Rightarrow\)Electric field,
\(E=\frac{\tau}{Q(2 a) \sin \theta}=\frac{8 \sqrt{3}}{4 \times 10^{-9} \times 2 \times 10^{-2} \times \sin 60^{\circ}} \mathrm{N} / \mathrm{C}\)
\(\therefore\) Potential energy,U = - pE cos\(\theta\)  = - Q(2a) E cos\(\theta\) 
\(=-4 \times 10^{-9} \times 2 \times 10^{-2} \times \frac{8 \sqrt{3} \times \cos 60^{\circ}}{4 \times 10^{-9} \times 2 \times 10^{-2} \times \sin 60^{\circ}}\)
\(=-\frac{8 \sqrt{3}}{\sqrt{3}}=-8 \mathrm{~J}\)

Current sensitivity of a galvanometer is defined as the deflection in galvanometer per unit current. Its SI unit is radian/ampere.

For balanced Wheatstone bridge, there will be no deflection in the galvanometer
\(\frac { 4 }{ { R }_{ 1 } } =\frac { 6 }{ 9 } \Rightarrow { R }_{ 1 }=\frac { 4\times 9 }{ 6 } =6\Omega \)

For the equivalent circuit, when the Wheatstone bridge is balanced, there will be no deflection in the galvanometer
\(\therefore \frac { 12 }{ 8 } =\frac { 6 }{ { R }_{ 2 } } \Rightarrow { R }_{ 2 }=\frac { 6\times 8 }{ 12 } =4\Omega \\ \therefore \frac { { R }_{ 1 } }{ { R }_{ 2 } } =\frac { 6 }{ 4 } =\frac { 3 }{ 2 } \) 

Force on charge at x = 2 cm due to charge at x = 0 m and x = 4 cm are equal and opposite. They cancel.
Net force on charge at x = 2 cm is resultant of repulsive forces due to two charges at x = 8 cm and x = 16 cm.
\(F={q\times q\over 4\pi\epsilon_o}\times[{1\over(0.08-0.02)^2}+{1\over (0.16-0.02)^2}]\)
\(F={9\times10^9(20\times10^{-6})^2}{[{1\over (0.06)^2}+{1\over (0.14)^2}]}\)
\(F=1.2\times 10^3N\) 

(i) (a): \(m=\frac{f_{o}}{f_{e}}=7\)
\(f_{o}=7 f_{e}\)
In normal adjustment, distance between the lenses
\(f_{o}+f_{e}=40 \)
\(7 f_{0}+f_{e}=40 \Rightarrow f_{e}=\frac{40}{8}=5 \mathrm{~cm} \)
\(f_{o}=7 f_{e}=7 \times 5=35 \mathrm{~cm}\)
(ii) (d): \(m=-10 ; L=22 \mathrm{~cm}\)
\(\text { As } m=\frac{-f_{o}}{f_{e}} \Rightarrow-10=-\frac{f_{o}}{f_{e}}\)
\(f_{o}=10 f_{\mathrm{e}} \)
\(\text { As } L=f_{o}+f_{e} \)
\(22=10 f_{e}+f_{e}=11 f_{e} \)
\(\text { or } f_{e}=\frac{22}{11}=2 \mathrm{~cm}\)
\(f_{o}=10 f_{e}=20 \mathrm{~cm}\)
(iii) (d): Objective lens has larger focal length than eye-piece.
(iv) (d): Astronomial telescope is used to see stars, sun etc.
(v) (c) :f₀>>f_e

(i) (a): As, \(q E=m g \Rightarrow q=\frac{1.08 \times 10^{-14} \times 9.8}{1.68 \times 10^{5}}\)
\(=6.4 \times 10^{-19} \mathrm{C}\)
(ii) (a): \(q=n e \text { or } \Rightarrow n=\frac{6.4 \times 10^{-19}}{1.6 \times 10^{-19}}=4\)
(iii) (c) : For the drop to be stationary,
Force on the drop due to electric field = Weight of the drop
qE=mg
\(q=\frac{m g}{E}=\frac{1.6 \times 10^{-6} \times 10}{100}=1.6 \times 10^{-7} \mathrm{C}\)
Number of electrons carried by the drop is
\(n=\frac{q}{e}=\frac{1.6 \times 10^{-7} \mathrm{C}}{1.6 \times 10^{-19} \mathrm{C}}=10^{12}\)
(iv) (c)
(v) (d): Millikan's experiment confirmed that the charges are quantized, i.e., charges are small integer multiples of the base value which is charge on electron. The charges on the drops are found to be multiple of 4. Hence, the quanta of charge is 4 \(\mu \)C.

(b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

(c): Optical fiber is extremely thin (radius of few microns) and long strand of very fine quality glass or quartz. When light is incident at a small angle at one end, it gets refracted into the strands (or fibres) and incident on the interface of the fibres and the coating. The angle of incidence being greater than the critical angle, the ray of light undergoes total internal reflections. It suffers the internal reflection again and again, till the angle of incidence remains greater than the critical angle for fibre material with respect to coating. Due to successive total internal reflection there is no loss of intensity in optical fibres.

(b): The wires are parallel to each other but the direction of current in it is in same direction so they attract each other. If the current in the wires is in opposite direction then wires repel each other. When the currents are in opposite directions, the magnetic forces are reversed and the wires repels each other

(a): Inside the charged metallic sphere every point is at the same electric potential, hence \(W=q \Delta V=0\). But outside the sphere, there exists a potential gradient at every point, hence \(\dot{W} \neq 0\).