Given that the Inductance of the inductor in the circuit is, L = 5.0 H
Given that the Capacitance of the capacitor in the circuit is , C = 80 μH = 80 x 10 ^- 6 F
Given that Resistance of the resistor in the circuit, R = 40 Ω
Value of Potential of the variable voltage supply, V = 230 V
(a) We know that the Resonance angular frequency can be obtained by the following relation :
\(\omega_{r}=\frac{1}{\sqrt{L C}} \omega_{r}=\frac{1}{\sqrt{5 x 80 x 10-6}} \omega_{r}=\frac{10^{3}}{20}=50 \mathrm{rad} / \mathrm{sec}\)
Thus, the circuit encounters resonance at a frequency of 50 rad/s.
(b) We know that the Impedance of the circuit can be calculated by the following relation :
\(Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}\)
At resonant condition,
X _L = X _C
Z = R = 40 Ω
At resonating frequency amplitude of the current can be given by the following relation :
\(I_{0}=\frac{V_{0}}{Z}\)
where,
V ₀ = peak voltage = \(\sqrt{2} V\)
Therefore,
\(I_{0}=\frac{\sqrt{2 V}}{Z}=\frac{\sqrt{2} \times 230}{40}=8.13 \mathrm{~A}\)
Thus, at resonant condition, the impedance of the circuit is calculated to be 40 Ω and the amplitude of the current is found to be  8.13 A
c) rms potential drop across the inductor in the circuit,
( V _L ) _rms = I x ω _r L
Where,
\(I_{r m s}=\frac{I_{0}}{\sqrt{2}}=\frac{\sqrt{2} V}{\sqrt{2} Z}=\frac{230}{40}=\frac{23}{4} A\)
Therefore, ( V _L ) _rms
\(\frac{23}{4} \times 50 \times 5=1437.5 \mathrm{~V}\)
We know that the Potential drop across the capacitor can be calculated with the following relation :
\(\left(V_{c}\right)_{r m s}=I \times \frac{1}{\omega_{r} C}=\frac{23}{4} \times \frac{1}{50 \times 80 \times 10^{-6}}=1437.5 V\)
We know that the Potential drop across the resistor can be calculated with the following relation :
\(\left(V_{R}\right)_{r m s}=I R=\frac{23}{4} \times 40=230 \mathrm{~V}\)
Now, Potential drop across the LC connection can be obtained by the following relation :
V _{L C}  = I ( X _L − X _C )
At resonant condition,
X _L = X _C
V _{L C} = 0
Therefore, it has been proved from the above equation that the potential drop across the LC connection is equal to zero at a frequency at which resonance occurs.

(a) Though the image size is bigger than the object, the angular size of the image is equal to the angular size of the object. A magnifying glass helps one see the objects placed closer than the least distance of distinct vision (i.e., 25 cm). A closer object causes a larger angular size. A magnifying glass provides angular magnification. Without magnification, the object cannot be placed closer to the eye. With magnification, the object can be placed much closer to the eye.
(b) Yes, the angular magnification changes. When the distance between the eye and a magnifying glass is increased, the angular magnification decreases a little. This is because the angle subtended at the eye is slightly less than the angle subtended at the lens. Image distance does not have any effect on angular magnification.
(c) The focal length of a convex lens cannot be decreased by a greater amount. This is because making lenses having very small focal lengths is not easy. Spherical and chromatic aberrations are produced by a convex lens having a very small focal length.
(d) The angular magnification produced by the eyepiece of a compound microscope is \(\left[ \left( \frac { 25 }{ { f }_{ e } } \right) +1 \right] \)
Where,
f_e = Focal length of the eyepiece
It can be inferred that if f_e is small, then angular magnification of the eyepiece will be large.
The angular magnification of the objective lens of a compound microscope is given as \(\frac { 1 }{ (|{ u }_{ o }|{ f }_{ o }) } \)
Where,
u_o = Object distance for the objective lens
f_o = Focal length of the objective
The magnification is large when u_o > f_o. In the case of a microscope, the object is kept close to the objective lens. Hence, the object distance is very little. Since u_o is small, f_o will be even smaller. Therefore, f_e and f_o are both small in the given condition.
(e) When we place our eyes too close to the eyepiece of a compound microscope, we are unable to collect much refracted light. As a result, the field of view decreases substantially. Hence, the clarity of the image gets blurred.
The best position of the eye for viewing through a compound microscope is at the eye-ring attached to the eyepiece. The precise location of the eye depends on the separation between the objective lens and the eyepiece.

Given, C₁ = C₂ = 600 pF
= 600 \(\times\)10^-12F
= 6 \(\times\)10^-10 F
V₁ = 200 V, V₂ = 0
\(\begin{aligned} \therefore \text { Energy lost } & =\frac{C_1 C_2\left(V_1-V_2\right)^2}{2\left(C_1+C_2\right)} \\ \end{aligned}\)
\(\begin{aligned} =\frac{\left(6 \times 10^{-10}\right)^2(200-0)^2}{2 \times 12 \times 10^{-10}} \end{aligned}\)
= 6 \(\times\)10^-6 J

Current in the wire, I = 50 A
A point is 2.5 m away from the East of the wire.
∴  Magnitude of the distance of the point from the wire, r = 2.5 m.
Magnitude of the magnetic field at that point is given by the relation, B \(=\frac{\mu_{0} 2 I}{4 \pi r}\)
Where,
μ₀ = Permeability of free space = 4π x 10^–7 T m A^–1
\(B=\frac{4 \pi \times 10^{-7} \times 2 \times 50}{4 \pi \times 2.5}\)
= 4 x 10 ^-6 T
The point is located normal to the wire length at a distance of 2.5 m. The direction of the current in the wire is vertically downward. Hence, according to the Maxwell’s right hand thumb rule, the direction of the magnetic field at the given point is vertically upward.

(a) As \(V \propto \frac{1}{r}, V_{P}>V_{Q^{}}\) Thus, (V_P – V_Q) is positive. Also V_B is less negative than V_A . Thus, V_B > V_A or (V_B – V_A) is positive.
(b) A small negative charge will be attracted towards positive charge. The negative charge moves from higher potential energy to lower potential energy. Therefore the sign of potential energy difference of a small negative charge between Q and P is positive. Similarly, (P.E.)_A > (P.E.)_B and hence sign of potential energy differences is positive.
(c) In moving a small positive charge from Q to P, work has to be done by an external agency against the electric field. Therefore, work done by the field is negative.
(d) In moving a small negative charge from B to A work has to be done by the external agency. It is positive.
(e) Due to force of repulsion on the negative charge, velocity decreases and hence the kinetic energy decreases in going from B to A.

(a) The sizes of the atoms taken in Thomson’s model and Rutherford’s model have the same order of magnitude.
(b) In  the ground state of Thomson’s model, the electrons are in stable equilibrium. However, in Rutherford’s model, the electrons always experience a net force.
(c) A classical atom based on Rutherford’s model is doomed to collapse.
(d) An atom has a nearly continuous mass distribution in Thomson’s model, but has a highly non-uniform mass distribution in Rutherford’s model.
(e) The positively charged part of the atom possesses most of the mass in both the models.

Mutual inductance of a pair of coils, µ = 1.5 H
Initial current, I₁ = 0 A
Final current I₂ = 20 A
Change in current, 
Time taken for the change, t = 0.5 s
Induced emf, 
Where  is the change in the flux linkage with the coil.
Emf is related with mutual inductance as:

Equating equations (1) and (2), we get

Hence, the change in the flux linkage is 30 Wb.

In a nuclues of ₂₀Ca⁴⁰, 
Number of protons = 20
Number of neutrons = 40 - 20 = 20
Total mass of 20 protons and 20 neutrons
= 20m_p + 20 m_n = 20(m_p + m_n)
= 20(1.007825+1.008665) = 40.3298 u
Mass defect, \(\triangle m\) = 40.3298 - 39.962589 = 0.367211 u
Total binding energy = \(\Delta m \times c^2 \times \frac{931}{c^2} \mathrm{MeV}\)
= \(\Delta\)m \(\times\) 931 MeV
= 0.367211 \(\times\) 931
= 341.873441 MeV
E_b per nucleon, E_bn = \(\frac{341.873441}{40}\)
= 8.547 MeV/nucleon

It is given that a plane electromagnetic wave travels in vacuum along z-direction and the frequency of the electromagnetic wave is 30MHz.
We can say that electric field and magnetic field will be in x-plane because the electromagnetic wave travels along the z-direction and both fields are mutually perpendicular to each other.
The formula of the wavelength of a wave is,
λ=c/ν
Substitute the values in the above expression,
λ=(3×10⁸)/(30×10⁶)=10m.
Thus, the value of wavelength is 10 m and the direction of electric and magnetic fields will be in x-y plane.

Resistivity of a conductor \(\rho =\frac { m }{ { ne }^{ 2 }\tau } \) 
So as temperature changes n and \(\tau \) changes.

Metal with work function 5 eV has lower threshold wavelength, i.e. metal B.

The image of an object formed by a lens can be brought to a sharp focus on a unfixed screen by changing the focal length of the lens by any of the following methods:
(i) By placing another lens of suitable focal length in contact with the previous lens.
(ii) By immersing the given lens in a liquid of appropriate refractive index.

Distance between the image (screen) and the object, D = 90 cm
Distance between two locations of the convex lens, d = 20 cm
Focal length of the lens = f
Focal length is related to d and D as:
f = \(\frac { { D }^{ 2 }-{ d }^{ 2 } }{ 4D } \)
= \(\frac { { (90) }^{ 2 }-({ 20) }^{ 2 } }{ 4\times 90 } =\frac { 770 }{ 36 } =21.3\) cm
Therefore, the focal length of the convex lens is 21.39 cm.

Emf of the battery, E = 12 V
Internal resistance of the battery, r = 0.4 Ω
Maximum current drawn from the battery = I
According to Ohm’s law,
E = Ir
\(I=\frac{E}{r}\)
\(=\frac{12}{0.4}=30 A\)
The maximum current drawn from the given battery is 30 A.

Note that thermally generated electrons (ni ~10¹⁶m^–3) are negligibly small as compared to those produced by doping.
Therefore, ne \(\approx\) ND
Since n_e n_h = \(n_{i}^{2}\) , The number of holes
n_h = (2.25 x 10³² ) / (5 x10²²)
= ~ 4.5 x 10⁹ m^–3

(i) (b): Each half lens will form an image in the same plane. The optic axes of the lenses are displaced
\(\frac{1}{v}-\frac{1}{(-30)}=\frac{1}{20} ; v=60 \mathrm{~cm}\)
(ii) (a): Here \(f_{1}=20 \mathrm{~cm} ; f_{2}=?\)
F= 80 cm
\(\text { As } \frac{1}{f_{1}}+\frac{1}{f_{2}}=\frac{1}{F} \Rightarrow \frac{1}{f_{2}}=\frac{1}{F}-\frac{1}{f_{1}}\)
\(\frac{1}{f_{2}}=\frac{1}{80}-\frac{1}{20}=\frac{-3}{80}\)
\(f_{2}=\frac{-80}{3}=-26.7 \mathrm{~cm}\)
(iii) (b): The bubble behaves libe a diverging lens
(iv) (b): Convex lens is used in magnifying glass.
(v) (d)