⋋₂ = 600 x 10^-9 m, ⇒ d = 1 x 10^-3 m, D =2m
n⋋ = dsinθ
For first dark fringe
n = 1 For minimum
d sinθ = ⋋, here
CO = Nc (approximately)

From Fig, sinθ \(=\frac{x/2}{D} =x/2D\)
\(sin \theta =\frac{ \lambda}{d} \)
\(\lambda/d=x/2D\)
\(\therefore x =\frac{2D\lambda}{d}=\frac{2 \times 2 \times 600\times 10^{-9}}{ 10^{-3}} \)
\(x=2.4 \times10^{-3}m =2.4\mathrm{~mm} \)

Magnetic moment of a bar magnet = \({ \vec { { p } } }_{ m }\)

First, When the bar magnet is cut into two pieces along the axis of the magnet. New magnetic pole strength is \(\frac{q_{m}}{2}\), but magnetic length is 2l
Second, Each piece is further cut into two pieces along the axis of the magnet, we get 
New magnetic pole strength is, 
\(q_{m}^{{\prime}}=\frac{q_{m}}{4}\)
But, magnetic length = 2l.

So, new magnetic moment of each piece is
\(\vec{p_m}_{\text {new }} =q'_{m} \times2 l \)
\(=\frac{q_{m}}{4} \times 2 l \)
\(=\frac{1}{4}\left(q_{m} \times 2 l\right) \)
\(\because \overrightarrow{\mathbf{p}}_{\mathrm{m}_{\text {new }}} =\frac{1}{4} (\vec{p_{m})} \)    \((\because{p}_{m} =(q_{\mathrm{m}} \times 2 l) )\)

Power factor will be maximum, when Φ = 0 i.e., \(\tan ^{-1}\left(\frac{X_L-X_C}{R}\right)=0\)
\(\therefore X_L=X_C \Rightarrow L \omega=\frac{1}{C \omega}  \)
\(\therefore \omega=\frac{1}{2 \pi \sqrt{L C}}\)
∴ Current I be max \(I_m=\frac{V_m}{R}\)
Hence power factor of a RLC series circuit becomes maximum, when
(i) \(X_L=X_C\)
(ii) Current \(I_m=\frac{V_m}{R}\) will be maximum
(iii) Frequency \(\omega_r=\frac{1}{2 \pi \sqrt{L C}}\)

Magnetic field B = 1 T
Mass of the proton, m_p = 1.67 x 10⁻²⁷kg
Charge of the proton, q = 1.60 x 10⁻¹⁹C
\(f=\frac { qB }{ { { 2\pi m }_{ p } } } =\frac { \left( 1.60\times { 10 }^{ -19 } \right) \left( 1 \right) }{ 2\left( 3.14 \right) \left( 1.67\times { 10 }^{ -27 } \right) } \)
= 15.3 x 10⁶ Hz = 15.3 MHz

The magnetic field due to current in the upper semicircle and lower semicircle of the circular coil are equal in magnitude but opposite in direction. Hence, the net magnetic field at the center of the loop (at point O) is zero \(\overset { \rightarrow }{ B } =\overset { \rightarrow }{ 0 } \).

\(W =\left(V_{b}-V_{a}\right) q\left[where\ V=\frac{K Q}{r}\right] \)
To find V_b :
\(V_b=\frac{kQ}{r_3}+\frac{kQ}{r_4}\)
\(V_{b} =\frac{K \times 10 \times 10^{-6}}{\sqrt{50 \times 10^{-4}}}+\frac{K \times 10 \times 10^{-6}}{\sqrt{250 \times 10^{-4}}} \)
\(V_{b} =\frac{K \times 10^{-5}}{\sqrt{50} \times 10^{-2}}+\frac{K \times 10^{-5}}{\sqrt{250} \times 10^{-2}} \)
\(=K \times 10^{-3}\left[\frac{1}{\sqrt{50}}+\frac{1}{\sqrt{250}}\right] \)
\(=9 \times 10^{9} \times 10^{-3}\left[\frac{1}{\sqrt{50}}+\frac{1}{\sqrt{250}}\right] \)
V_b = 1842002 V
To find V_a :
\(V_a=\frac{kQ}{r_1}+\frac{kQ}{r_2}\)
\(V_{a} =\frac{K \times 10 \times 10^{-6}}{\sqrt{5 \times 10^{-2}}}+\frac{K \times 10 \times 10^{-6}}{\sqrt{15 \times 10^{-2}}} \)
V_a = 2400000 V
\(\therefore W_{D} =\left(V_{b}-V_{a}\right) q=(1842002-2400000)( -2 \times 10^{-6} )\)
W = +1.12J
Positive sign implies that to move the charge - 2 μC external work is required.

(a) Acceleration of the electron \(a=\frac{F}{m}\)
Electrostatic force F = - eE
\(a=\frac{F}{m}=\frac{-e E \hat{j}}{m}\)   ... ( 1)
(b) Let velocity of the electron be \(\vec{v}\)
v = u + at        ...(2)
Here, \( u=v_{0} \hat{i} \), \(a=\frac{-e \bar{E}}{m} \hat{j}\)
Substituting these values in the equation (2) we get
\(\therefore \vec{v}=\nu_{0} \hat{i}-\frac{e E}{m} t \hat{j}\)
(c) Displacement \(\vec{r}\) represents position,
\( s=u t+\frac{1}{2} a t^{2} \) ...(3)
Let, \(s=\hat{r} \) and Here, \(u=v_{0} \hat{i} \), \(a=\frac{-e E}{m} \hat{j}\)
Substituting these values in the equation (3), we get,
\(\hat{r}=v_{0} t \hat{i}-\frac{1}{2} \frac{e E}{m} t^{2} \hat{j}\)

The electric flux through the rectangular area
\({ \Phi }_{ E }=\vec { E } .\vec { A } \) = EA cosθ = 100 x 5 x 10 x 10^-4 x cos60°
⇒ \({ \Phi }_{ E }\) = 0.25 Nm²C^-1
For θ = 0°
\({ \Phi }_{ E }=\vec { E } .\vec { A } \)= EA
= 100 x 5 x 10 x 10^-4 = 0.5 Nm²C^-1

An alpha particle contains 2 protons and 2 neutrons. Therefore, the mass M of the alpha particle is 4 times that of a proton (m_p) (or a neutron) and its charge q is twice that of a proton (+e).
The de Broglie wavelength associated with it is
\(\lambda=\frac { h }{ \sqrt { 2MqV } } =\frac { h }{ \sqrt { 2\times (4m_{ p })\times (2e)\times V } } \)
\(=\frac { 6.626\times { 10 }^{ -34 } }{ \sqrt { 2\times 4\times 1.67\times 10^{ -27 }\times 2\times 1.6\times { 10 }^{ -19 }\times 400 } } \)
\(=\frac { 6.626\times 10^{ -34 } }{ 4\times 20\times { 10 }^{ -23 }\sqrt { 1.67\times 1.6 } } \) = 0.00507 \(\mathring { A }\)

\(\lambda=6000Å=6000 \times 10^{-10} \mathrm{~m} ; \mathrm{V}=0.8 \mathrm{v} \)
\(\mathrm{k} \cdot \mathrm{E}=\mathrm{hv}-\phi \)
\(\mathrm{eV}_o=\mathrm{hv}-\phi=\frac{\mathrm{hc}}{\lambda}-\phi \)
\((i) v=\frac{c}{\lambda}=\frac{3 \times 10^{8}}{6000 \times 10^{-10}}=5 \times 10^{14} \mathrm{~Hz} \)
\((ii)\ \mathrm{E}=\frac{\mathrm{hc}}{\lambda}=\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{6000 \times 10^{-10}}=3.313 \times 10^{-19} J\)
\(\mathrm{E}=\frac{3.313 \times 10^{-19}}{1.6 \times 10^{-19}}=2.07 \mathrm{eV} \)
\((iii) \ \mathrm{E}=\mathrm{hv}-\mathrm{W}\Rightarrow\mathrm{W}=\mathrm{h} v-\mathrm{E} \)
\(\mathrm{E}=\mathrm{eV_o}=1.6 \times 10^{-19} \times 0.8=1.2 8 \times10^{-19}J\)
\(\mathrm{hv}=6.626 \times 10^{-34} \times 5 \times 10^{14}=3.313 \times 10^{-19}J \)
\(\mathrm{~W}=\frac{(3.313-1.28) \times 10^{-19}}{1.6 \times 10^{-19}}=1.270 \mathrm{eV} \)
W = 1.27 eV
\((iv) \ \mathrm{W}=\mathrm{h} \mathrm{v}_{0} \)
\(v_{0}=\frac{W}{h}=\frac{2.033 \times 10^{-19}}{6.626 \times 10^{-34}}=3.07 \times 10^{14} \mathrm{~Hz} \)
\((v)\ \mathrm{E}=\mathrm{eV}_o=\frac{0.8 \times 1.6 \times 10^{-19}}{1.6 \times 10^{-19}}=\mathbf{0 . 8} \mathrm{eV}\)

D = 25 cm;
The magnifying lens must be a convex lens of positive focal length
f = 5 cm;
For closest object distance u', the image distance, v is, –25 cm. (near point, v = –D)
For farthest object distance u', the corresponding image distance, v' is infinity.
(a) To find closest distance between lens and book, we can use lens equation,\(\cfrac { 1 }{ v } -\cfrac { 1 }{ u } =\cfrac { 1 }{ f } \) 
Rewriting for closest object distance \(\cfrac { 1 }{ u } =\cfrac { 1 }{ v } -\cfrac { 1 }{ f } \) 
Substituting,
\(\cfrac { 1 }{ u } =\cfrac { 1 }{ -25 } -\cfrac { 1 }{ 5 } =\cfrac { 1 }{ 25 } =\cfrac { 1 }{ 5 } \left( \cfrac { -1-5 }{ 25 } \right) =-\cfrac { 6 }{ 25 } \)
\(u=-\cfrac { 25 }{ 6 } =4.167cm\)
The closest distance between the lens and the book is, u = –4.167 cm
To find farthest object distance, lens equation is, \(\cfrac { 1 }{ v' } -\cfrac { 1 }{ u' } =\cfrac { 1 }{ f' } \)
Rewriting for farthest object distance,\(\cfrac { 1 }{ u' } =\cfrac { 1 }{ v' } -\cfrac { 1 }{ f' } \)
Substituting,\(\cfrac { 1 }{ u' } =\cfrac { 1 }{ \infty } -\cfrac { 1 }{ 5 } ;u=-5cm\) 
The farthest distance at which the person can keep the book is, u' = -5 cm.
(b) To find magnification in near point focusing, \(m=1+\cfrac { D }{ f } =1+\cfrac { 25 }{ 5 } =6\)
To find magnification in normal focusing, \(m=\cfrac { D }{ f } =\cfrac { 25 }{ 5 } =5\)

λ₁ = 560 nm = 560 x 10^-9m;
λ₁ = 420 nm = 420 x 10^-9m;
D = 1 m; d = 3 mm = 3 x 10^-3m
Here, n and λ are inversely proportional for a given y.
Here, n^th order bright fringe of longer wavelength λ₁ coincides with (n+1)^th order bright fringe of shorter wavelength λ₂.
Equation for n^th bright fringe is, \({ y }_{ n }=n\cfrac { \lambda D }{ d } \) 
Here, \(n\cfrac { { \lambda }_{ 1 }D }{ d } =(n+1)\cfrac { { \lambda }_{ 2 }D }{ d } \) (as λ₁>λ₂)
\({ n\lambda }_{ 1 }=\left( n+1 \right) { \lambda }_{ 2 }\) (or) \(\cfrac { { \lambda }_{ 1 } }{ { \lambda }_{ 2 } } =\cfrac { (n+1) }{ n } ;1+\frac{1}{n}=\frac{\lambda_1}{\lambda_2}\) 
\(1+\cfrac { 1 }{ n } =\cfrac { 560\times { 10 }^{ -9 } }{ 420\times { 10 }^{ -9 } } \) (or) \(1+\cfrac { 1 }{ n } =\cfrac { 4 }{ 3 } \) 
\(\frac{1}{n}=\frac{1}{3}\) (or) n = 3
Thus, the 3^rd bright fringe of λ₁ and the 4^th bright fringe of λ₂ coincide at the least distance y.
The least distance from the central fringe where the bright fringes of the two wavelengths coincide is, \(y_{n}=n \frac{\lambda D}{d}\)  
\(y_{n}=3 \times \frac{560 \times 10^{-9} \times 1}{3 \times 10^{-3}}=560 \times 10^{-6} \mathrm{~m}\) 
\(y_{n}=0.560 \times 10^{-3} \mathrm{~m}=0.560 \mathrm{~mm}\) 

\(\beta =50 \)
\(V_{\beta E} =600 \mathrm{mV} \)
\(=0.6 \mathrm{~V} \)
\(\mathrm{V}_{\mathrm{B}} =\mathrm{V}_{\mathrm{E}}-\mathrm{V}_{\mathrm{EB}} \)
\(\mathrm{V}_{\mathrm{B}} =3-0.6 \)
\(=2.4 \mathrm{~V} \)
\(\mathrm{I}_{\mathrm{B}} =\frac{\mathrm{V}_{\mathrm{B}}}{R_B}=\frac{2.4}{60 \times 10^3}=40 \mu \mathrm{A} \)
\(\mathrm{I}_{\mathrm{C}} =\beta \mathrm{I}_{\mathrm{B}}=50 \times 40 \mu \mathrm{A} =2 \mathrm{~mA} \)
\(V_C=R_FI_C=500 \times 2 \times10^{-3}=1 V\)
\(V_{EC}=V_E-V_C\)
\(V_{EC}=V_E-V_C\)
\(V_{EC}=3-1=2V\)

A = 108, Z  =47, N = 108 - 47 = 61
m_p = 1.007825 u, m_n = 1.008665 u, M = 107.905949 u
(a) \(\Delta \mathrm{m}=Z \mathrm{~m}_{\mathrm{P}}+\mathrm{Nm}_{\mathrm{n}}-\mathrm{M} \)
\(\Delta \mathrm{m}\) = (47 x 1 .007825 + 61 x 1 .008665 - 107 .905949)
\(\Delta \mathrm{m}\) = 47.367775 + 61.528565 -107.905949
\(\Delta \mathrm{m}\) = 108.89634 - 107.905949
\(\Delta \mathrm{m}\) = 0.990391 u
\(\mathrm{BE}=\Delta \mathrm{m} \times 931 \mathrm{MeV} \)
BE = 0.990391 x 931 MeV = 922.054 MeV
(c) \(\overline{\mathbf{B E}}=\frac{\mathbf{B E}}{\mathbf{A}} \)
\(\overline{\mathrm{BE}}=\frac{922.054}{108}=8.537 \mathrm{MeV}=8.5 MeV\)

Transformer ratio, \(\frac { { V }_{ s } }{ { V }_{ p } } =\frac { { N }_{ s } }{ N_p } \)
Here V_s = 11 V,  V_r = 220 V, P = 88 W
(i) Voltage Transformation ratio = \(\frac { { V }_{ s } }{ { V }_{ p } } =\frac { { 11 } }{ 220 } \)
\(K =\frac { { 1 } }{ 20 } \)
(ii) Current in the primary, \(I_p =\frac { { P } }{ V_p } \)
\(=\frac { { 88 } }{220 } =\frac{4}{10}=0.4 A\)
∴ I_p = 0.4 A

Amplitude of magnetic field B = 10^-5 T
Frequency, f = 10¹⁰ HZ
(i) Wavelength of the wave,
\({\lambda}=\frac{c}{f} =\frac{3 \times 10^8}{ 10^{10}} \)
= 3 x 10^{8 - 10}
Wavelength = 3 x 10^-2 m
(ii) Electric field E(x,t)\(\hat i\)
Angular frequency \(\omega =2 \pi f \)
\(\omega=2 \times 3.14 \times 10^{10}=6.28 \times 10^{10} rads^{-1}\)
\(k=\frac{2 \pi}{\lambda} =\frac{2 \times 3.14}{3 \times 10^{-2}} \)
\(=\frac{6.28}{3 \times 10^{-2}}=\frac{628}{3}=2.09 \times 10^{2} \)
k = 2.09 x 10²
(iii) E_o = B_oC
E_o = 10^-5 x 3 x 10⁸
E_o = 3 x 10³ V m^-1
The required expression for electric field is
\(\vec{E}(x, t)=E_o \sin \left(\frac{2 \pi}{\lambda} x-2 \pi f t\right) \hat{i} N C^{-1} \)
\(\vec{E}(x , t)=3 \times 10^{3} \sin \left(2.09 \times 10^{2} \mathrm{x}-6.28 \times 10^{10} \mathrm{t}\right) \hat(-{k}) N C^{-1} \)

The given equation is v = 10 sin ( 3π x 10⁴ t)
(i) At t = 0 s,
v = 10 sin0^o = 0
(ii) At t = 50 µs,
v = 10sin( 3π x 10⁴ x 50 x 10^-6)
= \(10 \sin \left(150 \pi \times 10^{-2} \times \frac{180^{\circ}}{\pi}\right)\) 
= 10sin (270^o) = 10 x -1
= -10V
(iii) At t = 75 µs,
\(v=10 \sin \left(3 \pi \times 10^{4} \times 75 \times 10^{-6}\right)\) 
= \(10 \sin \left(225 \pi \times 10^{-2} \times \frac{180^{\circ}}{\pi}\right)\) 
= 10sin (405^o) = 10sin45^o
= \(10 \times 1 / \sqrt{2}=7.07 \mathrm{~V}\) 

For a charged particle moving on a magnetic field, \(\vec { F } \)= q(\(\vec { v } \)x\(\vec { B } \))
The work done by the magnetic field is
\(W=\int { \overset { \rightarrow }{ F } .{ d \vec r } =\int { \overset { \rightarrow }{ F } .\overset { \rightarrow }{ v } dt } } \)
\(W=q\int { \left( \overset { \rightarrow }{ v } \times \overset { \rightarrow }{ B } \right) } .\overset { \rightarrow }{ v } dt=0\)
Since \(\overset { \rightarrow }{ v } \times \overset { \rightarrow }{ B } \) is perpendicular to \(\vec { v } \) and hence \(\left( \overset { \rightarrow }{ v } \times \overset { \rightarrow }{ B } \right) .\overset { \rightarrow }{ v } =\overset { \rightarrow }{ 0 } \) 
This means that Lorentz force does no work on the particle. From work-kinetic energy theorem, (Refer section 4.2.6, XI th standard Volume I)
\(\frac { dw }{ dt } =p=0\)
Since \(\overset { \rightarrow }{ F } .\overset { \rightarrow }{ v } =0\Rightarrow \overset { \rightarrow }{ F } \ and \ \overset { \rightarrow }{ v } \) are perpendicular to each other. The angle between Lorentz force and velocity of the charged particle is 90^o. Thus Lorentz force changes the direction of the velocity but not the magnitude of the velocity. Hence Lorentz force does no work and also does not alter kinetic energy of the particle. 

The capacitors 1 μF and 3μF are connected in parallel and 6μF and 2 μF are also separately connected in parallel. So these parallel combinations reduced to equivalent single capacitances in their respective positions, as shown in the figure (b).
C_eq = 1 + 3 = 4μF
C_eq = 6 + 2 = 8μF
From the figure (b), we infer that the two 4 μF capacitors are connected in series and the two 8 μF capacitors are connected in series. By using formula for the series, we can reduce to their equivalent capacitances as shown in figure (c).
\(\frac { 1 }{ { C }_{ eq } } =\frac { 1 }{ 4 } +\frac { 1 }{ 4 } =\frac { 1 }{ 2 } \)  ⇒ C_eq = 2μF and
\(\frac { 1 }{ C_{ eq } } =\frac { 1 }{ 8 } +\frac { 1 }{ 8 } =\frac { 1 }{ 4 } \)  ⇒ C_eq = 4μF 
From the figure (c), we infer that 2μF and 4μF are connected in parallel. So the equivalent capacitance is given in the figure (d).
C_eq = 2 + 4 = 6μF
Thus the combination of capacitances in figure (a) can be replaced by a single capacitance 6 μF.

\(\frac { P }{ Q } =\frac { R }{ S } \), when the network is balanced
\(\frac { 500 }{ 800 } =\frac { x+400 }{ 1000 } \)
\(x+400=\frac { 5 }{ 8 } \times 1000\)
x + 400 = 625 
x = 625 – 400
x = 225 Ω

Equivalent emf of the combination
ξ_eq = nξ = 4 x 9 = 36 V
ii) Equivalent internal resistance r_eq = nr = 4 x 0.1 = 0.4 Ω
iii) Total current \(I=\frac { n\xi }{ R+nr } \)
\(=\frac { 4\times 9 }{ 10+(4\times 0.1) } \)
\(=\frac { 4\times 9 }{ 10+0.4 } =\frac { 36 }{ 10.4 } \)
I = 3.46 A
iv) Potential difference across external resistance V = IR = 3.46 x 10 = 34.6 V. The remaining 1.4 V is dropped across the internal resistance of cells.
v) Potential difference across each cell \(\frac { V }{ n } =\frac { 34.6 }{ 4 } =8.65V\)

Let the current flowing be I.
Let the number of turns per unit length of the solenoid be n
Magnetic field B \(\mathrm{B}=\frac{\mu_{o} N l I}{l}=0 \)
But \(\mathbf{n}=\frac{N}{l} \)
\(B=\mu_{o} n I \)
But the magpetic field at any point on the axis of the solenoid is
\(B=\frac{1}{2}\left(\mu_{o} n I\right)\)
On the top of the solenoid, at a point magnetic field is
\(\mathrm{B}_{1}=\frac{1}{2} \mu_{o} n I \cos \theta_{1}\)
On the bottom of the solenoid, the magnetic field at a point is
\(B_{2} =\frac{1}{2} \mu_{0} n I \cos \theta_{2} \)
Resultant magnetic field \(B =B_{1}-B_{2} \)
\(B =\frac{1}{2} \mu_{0} n I\left[\cos \theta_{1}-\cos \theta_{2}\right] \)

Charge Q = Current x Time interval
= I x t
At t = 0 s,  
dq = dI x t
= 5 x 0
dq = 0 C
At t = 2 s,    
dg = dI x t
=5 x 2
dq = 10 C
At t = 5 s,
dg = dl x t
=0 x 5
dq = 0 C
At t= 0 s, dg = 0 C; At t = 2 s, dg = 10 C; At t= 5 s, dg = 0 C.