Number of lines per cm = 4000 cm^-1; m = 2; θ = 30°; λ = ?
Number of lines per unit length
\(N=\cfrac { 4000 }{ 1\times { 10 }^{ -2 } } =4\times { 10 }^{ 5 }\) 
Equation for diffraction maximum in grating is, sinθ = Nmλ
After Rewriting, \(\lambda =\cfrac { sin\theta }{ Nm } \) 
Substituting,
\(\lambda =\cfrac { { \sin30 }^{ o } }{ 4\times { 10 }^{ 5 }\times 2 } =\cfrac { 0.5 }{ 4\times { 10 }^{ 5 }\times 2 } \) 
= \(\cfrac { 1 }{ 2\times 4\times { 10 }^{ 5 }\times 2 } =\cfrac { 1 }{ 16 \times 10^5} \)
λ = 6250 x 10^-10 m = 6205 Å

235 g of \(_{ 92 }^{ 235 }{ U }\) has 6.02 x 10²³ atoms. In one gram of \(_{ 92 }^{ 235 }{ U }\), the number of atoms is equal to \(\frac { 6.02\times { 10 }^{ 23 } }{ 235 } =2.56\times { 10 }^{ 21 }\)
So the number of atoms in 1 kg of \(_{ 92 }^{ 235 }{ U }\) = 2.56 x 10²¹ x 1000 = 2.56 x 10²⁴
Each \(_{ 92 }^{ 235 }{ U }\) nucleus releases 200 MeV of energy during the fission. The total energy released by 1kg of \(_{ 92 }^{ 235 }{ U }\) is
Q = 2.56 x 10²⁴ x 200MeV = 5.12 x 10²⁶ MeV
In terms of joules,
Q = 5.12 x  10²⁶ x 1.6 x 10^-13 J = 8.192 x 10¹³ J
In terms of Kilowatt hour,
Q = \(\frac { 8.192\times { 10 }^{ 13 } }{ 3.6\times { 10 }^{ 6 } } =2.27\times { 10 }^{ 7 }\) kWh

Area of the capacitor = A
Radius = R
Current in the wire connected to the plates I = 5 A
The electric field, between the plates of a parallel plate capacitor,
\(E=\frac{\sigma}{\varepsilon_0}  \)
\(E=\frac{Q}{A \varepsilon_0}\)
Q is the charge accumulated at the positive plate.
The flux of this field, \(\phi_E=\frac{Q}{A \varepsilon_0} \times A=\frac{Q}{\varepsilon_0}\)
Displacement current \(i_d=\varepsilon_0 \frac{d \phi_E}{d t}\)
\(=\varepsilon_0 \frac{d}{d t}\left(\frac{Q}{\varepsilon_0}\right)=i_c\)
\(\therefore \mathrm{i}_{\mathrm{d}}=5 \mathrm{~A} \quad\left(\because\right.\) The current through the capacitor i_c = 5 A)
Displacement current = 5 A

C_p = C_b + C_c
\(C_{P}=6+2=8 \mu \mathrm{F} \)
\(\frac{1}{C_{s}}=\frac{1}{C_a}+\frac{1}{C_p}=\frac{1}{C_d} \)
\(C_{s}=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{3}{8} \)
\(\therefore C_{s}=\frac{8}{3} \mu \mathrm{F} \)
Total capacitance \(C_{s}=\frac{8}{3} \times 10^{-6} \mathrm{~F} \)
Total Charge, \(Q=C_{s} V=\frac{8}{3} \times 10^{-6} \times 9 \)
\(Q_{a}=24 \mu C\)
(a) Charge on capacity \(Q_{a}=24 \mu C\)         .....(1)
Charge on capacitor \(Q_{b}=24 \times \frac{6}{8}=18 \mu \mathrm{C} \)             ..............(2)
Charge on capacitor  \(Q_{c}=24 \times \frac{2}{8}=6 \mu \mathrm{C} \)              ..............(3)
Charge on capacitor \(Q_{d}=24 \times \frac{8}{8}=24 \mu \mathrm{C} \)              ..............(4)
(b) Potential difference across \(C_{a} \ is\ V_{a}=\frac{Q_{a}}{C_{a}} \)
\(=\frac{24}{8}=3 \mathrm{~V} \)      ...(5)
Potential difference across \(C_{b}\ is \ V_{b}=\frac{Q_{b}}{C_{b}} \)
\(=\frac{18}{6}=3 \mathbf{V}\)       ........(6)
Potential difference across \(C_{c}\ is \ V_{c}=\frac{Q_{c}}{C_{c}}=\frac{6}{2}=3 \mathrm{~V} \)       ......(7)
Potential difference across \(C_{d}\ is \ V_{d}=\frac{Q_{d}}{C_{d}} \)
\(=\frac{24}{8}=3 \mathbf{V} \)          ....(8)
(c) Energy stored in each capacitor \(U=\frac{1}{2} C V^{2}\)
Energy stored in \(\mathrm{C}_{\mathrm{a}} \text { is } U_{a}=\frac{1}{2} C_{a} V_{a}^{2}\)
\(U_{\mathrm{a}}=\frac{1}{2} \times 8 \times 10^{-6} \times 3 \times 3=36 \mu \mathrm{J}\)   ....(9)
Energy stored in \(C_{b}\ is \ U_{b}=\frac{1}{2} C_{b} V_{b}^{2} \)
\(U=\frac{1}{2} \times 6 \times 10^{-6} \times 3 \times 3 \)
\(=27 \mu \mathrm{J} \)         ........(10)
Energy stored in C_e is \(U_{c} =\frac{1}{2} C_{c} V_{c}^{2} \)
\(=\frac{1}{2} \times 2 \times 10^{-6} \times 3 \times 3=9 \mu \mathrm{J} \)          ......(11)
Energy stored in C_d is \(U_{d} =\frac{1}{2} C_{d} V_{d}^{2} \)
\(U_d=\frac{1}{2} \times 8 \times 10^{-6} \times 3 \times 3 \)
\(=36 \times 10^{-6} \mathrm{~J}=36 \mu \mathrm{J} \)      .........(12)

\(W =\left(V_{b}-V_{a}\right) q\left[where\ V=\frac{K Q}{r}\right] \)
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.

R = 10 Ω, I = 5 A, t = 5 minutes = 5 x 60 s
H = I² R t
= 5² x 10 x 5 x 60
= 25 x 10 x 300
= 25 x 3000
= 75000 J (or) 75 kJ

When the water molecules are aligned in the direction of the electric field, it has minimum potential energy. The work done to rotate the dipole from θ = 0° to 90° is equal to the potential energy difference between these two configurations.
W = ΔU = U(90°) - U(0°)
From the equation U =−pE cosθ = −\(\hat p.\hat E\) ,
we write U = − pE cosθ, Next, we calculate the work done to rotate one water molecule from θ = 0° to 90°.
For one water molecule
W = - pE cos90^o + pE cos0^o  = pE
W=  6.3 x 10^-30 x 3 x 10⁵ = 18.9 x 10^-25J
For 10²² water molecules, the total work done is
W_tot = 18.9 x 10^-25 x 10²² = 18.9 x 10^-3J

(a) To find N₀, we have to find the number of \(_{ 7 }^{ 13 }{ N }\) atoms in 2.6μg. The atomic mass of nitrogen is 13. Therefore, 13 g of  \(_{ 7 }^{ 13 }{ N }\) contains Avogadro number (6.02 x 10²³) of atoms.
In 1 g, the number of \(_{ 7 }^{ 13 }{ N }\) atoms present is equal to \(\frac { 6.02\times { 10 }^{ 23 } }{ 13 } \). So the number of \(_{ 7 }^{ 13 }{ N }\) atoms present in 2.6μg is
\({ N }_{ 0 }=\frac { 6.02\times { 10 }^{ 23 } }{ 13 } \times 2.6\times { 10 }^{ -6 }=12.04\times { 10 }^{ 16 }\) atoms
(b) To find the initial activity R₀, we have to evaluate decay constant λ
\(\lambda =\frac { 0.6931 }{ { T }_{ 1/2 } } =\frac { 0.6931 }{ 10\times 60 } =1.155\times { 10 }^{ -3 }{ s }^{ -1 }\)
Therefore
R₀ = λN₀ = 1.155 x 10^-3 x 12.04 x 10¹⁶
= 13.90 x 10 ¹³ decays/s
= 13.90 x 10 ¹³ Bq
In terms of a curie,
\({ R }_{ 0 }=\frac { 13.90\times { 10 }^{ 13 } }{ 3.7\times { 10 }^{ 10 } } =3.75\times { 10 }^{ 3 }Ci\)
since 1Ci = 3.7 x 10¹⁰Bq
(c) Activity after 2 hours can be calculated in two different ways:
Method 1: R = R₀ e^–λt
At t = 2 hr = 7200 s
R = 3.75 x 10³ x e^{-7200 x 1.155 x 10–3}
R = 3.75 x 10³ x 2.4 x 10^–4 = 0.9 Ci
Method 2: \(R={ \left( \frac { 1 }{ 2 } \right) }^{ n }{ R }_{ 0 }\)
Here \(n=\frac { 120min }{ 10min } =12\)
\(R={ \left( \frac { 1 }{ 2 } \right) }^{ 12 }\times 3.75\times { 10 }^{ 3 }\) ≈ 0.9 Ci
(d) mean life ፒ = \(\frac{T_{1/2}}{0.6931}=\frac{10\times60}{0.6931}\)
= 865.67 s

\(\mathrm{T}_{\frac{1}{2}}=5.01 \text { days} \), \(\mathrm{R}_{o}=1 μ \mathrm{Ci}=3.7 \times 10^{10} \times 10^{-6} \)  decays per second
\(if\ \mathrm{t}=1\ year\ \mathrm{R}= ? \), \(\tau=? \), \(\lambda=? \), \(\mathrm{~N}_{o}=? \)
a) \(\mathrm{R}=\mathrm{R}_{o} \mathrm{e}^{-\lambda t} \)
\(\mathrm{R}=\mathrm{R}_{o} \mathrm{e}^{-\frac{0.6931}{\mathrm{~T}_{1 / 2} }t} \)
\(\mathrm{R}=\mathrm{R}_{o} \mathrm{e}^{-\frac{0.6931}{5.01} \times 365} \)
\(\mathrm{R}=\mathrm{R}_{o} \mathrm{e}^{-50.5}=\mathrm{R}_{o} \times 1.17 \times 10^{-22} \)
\(\mathrm{R}=1 \mu \mathrm{Ci} \times 1.17 \times 10^{-22}=1.17 \times 10^{-22} \mu \mathrm{Ci} \)
\(\mathrm{R}=1.17 \times 10^{-22} \mu \mathrm{Ci} \)
b) \(\lambda=\frac{0.6931}{T_{\frac{1}{2}}}=\frac{0.6931}{5.01}=0.1383 \text { day }^{-1} \)
\(\lambda=\frac{0.6931}{T_{\frac{1}{2}}}=\frac{0.6931}{5.01 \times 24 \times 60 \times 60}=1.6 \times 10^{-6} \mathrm{~s}^{-1} \)
\(\lambda=1.6 \times 10^{-6} \mathrm{~s}^{-1} \)
c) \(\tau=\frac{1}{\lambda}=\frac{1}{0.1383}=7.23 \text { days } \)
\(\tau=7.23 \text { days } \)
\(R_{o}=\lambda N_{o} \)
\(N_{o}=\frac{R_{o}}{\lambda}=\frac{3.7 \times 10^{10} \times 10^{-6}}{1.6 \times 10^{-6}}=2.3 \times 10^{10} \)
\(N_{o}=2.3 \times 10^{10} \)

For A, Half life of  \(, \mathrm{T}_{\mathrm{A}} \) = 20 minutes, n = 4
For B, Half life of  \(, T_{B}\) = 40 minutes, n = 2
\(\mathrm{N}_{01}=\mathrm{N}_{02}=\mathrm{N}_{0}\)
sample left A, \(\frac{N_{1}}{N_{0}}=\left(\frac{1}{2}\right)^{n}=\left(\frac{1}{2}\right)^{4}=\frac{1}{16}\)
A-sample decayed \(=1-\frac{N_{1}}{N_{0}}=1-\frac{1}{16}=\frac{15}{16}\)
sample Ieft B, \(\frac{N_{1}}{N_{0}}=\left(\frac{1}{2}\right)^{n}=\left(\frac{1}{2}\right)^{2}=\frac{1}{4}\)
B-sample decayed \(=1-\frac{N_{2}}{N_{0}}=1-\frac{1}{4}=\frac{3}{4}\)
Ratio of decayed number of A and B \(=\frac{\frac{15}{16}}{\frac{1}{4}}=\frac{15}{16} \times \frac{4}{3}=\frac{5}{4}\)
Ratio of decayed number of A and |B = 5:4

N = 5 turns
A = 2 x 10^-2 m²
B = 4 x 10^-2 Wb m^-2
K = 4 x 10^-9 N m deg^-1
(a) Current sensitivity
\({ I }_{ s }=\frac { NAB }{ K } =\frac { 5\times 2\times { 10 }^{ -2 }\times 4\times { 10 }^{ -2 } }{ 4\times 10^{ -9 } } \)
= 10⁶ divisions per ampere
\(I\mu A=\) 1microambire =10^-6ampere
Therefore,
\({ I }_{ s }={ 10 }^{ 6 }\frac { div }{ A } =1\frac { div }{ { 10 }^{ -6 }A } =1\frac { div }{ \mu A } \)
\({ I }_{ s }=1div{ \left( \mu A \right) }^{ -1 }\)
(b) Voltage sensitivity
\({ V }_{ s }=\frac { \theta }{ V } =\frac { 50div }{ 25mv } =2\times { 10 }^{ 3 }{ div \ V }^{ -1 }\)
(c) The resistance of the galvanometer is
\({ R }_{ g }=\frac { { I }_{ s } }{ { v }_{ s } } =\frac { { 10 }^{ 6 }\frac { div }{ A } }{ { 2\times }10^{ 6 }\frac { div }{ V } } =0.5\times { 10 }^{ 3 }\frac { V }{ A } =0.5k\Omega \)

\(N_1  =4 \times 10^4, \mathrm{~N}_2=100, \mathrm{I}=2 \mathrm{~A}, \mathrm{~A}=4 \times 10^{-4} \mathrm{~m}^2  \)
\(\mathrm{t}  =0.04 \mathrm{~s}, \mathrm{e}=?  \)
\(\mathrm{~B}  =\mu_0 \mathrm{n}_1 \mathrm{I}=\frac{\mu_0 \mathrm{~N}_1 \mathrm{I}}{l}=\frac{4 \pi \times 10^{-7} \times 4 \times 10^4 \times 2}{1}  \)
\(\mathrm{~B}  =100.48 \times 10^{-3} \mathrm{~T}  \)
\(\phi  =\mathrm{BA}=100.48 \times 10^{-3} \times 4 \times 10^{-4}  \)
\(\therefore \phi_1  =\phi_2=\phi=401.92 \times 10^{-7} \mathrm{wb}  \)
\(\mathrm{d} \phi  =\phi_1-\left(-\phi_2\right)=2 \phi=2 \times 401.92 \times 10^{-7}  \)
\(\therefore \mathrm{d} \phi  =803.84 \times 10^{-7} \mathrm{wb}\)
EMF induced in the coil,
\(e=-N_2 \frac{d \phi}{d t}  \)
\(e=\frac{-100 \times 803.84 \times 10^{-7}}{0.04}=-200.96 \times 10^{-3} \mathrm{~V}\)
\(e=-0.2 \mathrm{~V}\)

The magnetic field of a solenoid (inside) is
\({ B }_{ L,N }=\mu _{ ° }\frac { NI }{ L } \)
(a) length of the solenoid becomes twice and fixed number of turns
L → 2L (length becomes twice)
N → N (number of turns remains constatnt)
The magnetic field is
\({ B }_{ L,N }=\mu _{ ° }\frac { NI }{ 2L } =\frac { 1 }{ 2 } { B }_{ L,N }\)
(b) both the length of the solenoid and number of turns are doubled
L → 2L (length becomes twice)
N → 2N (number of turns becomes twice)
The magnetic field is
\({ B }_{2 L,2N }=\mu _{ ° }\frac { 2NI }{ 2L } ={ B }_{ L,N }\)
(c) the number of turns becomes twice but the fixed length of the solenoid
L → L (length is fixed)
N → 2N (number of turns becomes twice)
The magnetic field is
\({ B }_{ L,2N }=\mu _{ ° }\frac { 2NI }{ L } ={ 2B }_{ L,N }\)
From the above results,
\({ B }_{ L,2N }>{ B }_{ 2L,2N }>{ B }_{ 2L,N }\)
Thus, strength of the magnetic field is increased when we pack more loops into the same length for a given current.

i = 77 sin 314t ; t = 2 ms = 2 x 10^-3 s
The general equation of an alternating current is i = I_m sinωt. On comparison,
(i) Peak current, I_m = 77A
(ii) Frequency, \(f=\frac { \omega }{ 2\pi } =\frac { 314 }{ 2\times 3.14 } =50Hz\)
(iii) Time period, \(T-\frac { 1 }{ f } =\frac { 1 }{ 50 } =0.02s\)
(iv) At t = 2 m s, Instantaneous current, i = 77sin(314 x 2 x 10⁻³)
\( =77 \sin \left(314 \times 2 \times 10^{-3} \times \frac{180^{\circ}}{3.14}\right) \)
\(=77 \sin 36^{\circ}=77 \times 0.5878 \)
= 45.26 A

f = 50Hz ; V_m = 20V
Instantaneous voltage, υ = V_m sinωt
= V_m sin2πvt
= 20sin(2π x 50)t = 20sin(100 x 3.14)t
υ = 20sin 314t
Time for one cycle, \(T=\frac { 1 }{ f } =\frac { 1 }{ 50 } =0.02s\)
= 20 x 10⁻³ s = 20ms
The wave form is given below

V_p = 12 V; V_s = 240 V
I_s = 50 mA; N_p = 100 turns
\(\frac { { V }_{ s } }{ { V }_{ P } } =\frac { { N }_{ s } }{ { N }_{ p } } =\frac { { I }_{ P } }{ { I }_{ S } } =K\)
Transformation ratio, K = \(\frac{240}{12}=20\)
The number of turns in the secondary
N_S = N_P x K = 100 x 20 = 2000
Primary current,
I_P = K x I_s = 20 x 50 mA = 1 A

G = 6.67 x 10^-11 Nm²kg^-2
Mass of Earth m_E = 5.9 x 10²⁴ kg
Mass of Moon m_m = 7.9 x 10²² kg
Gravitational force \(F_{g}=\frac{G m_{E} \times m_{M}}{r^{2}}\); Electro static force \(F_e=k\frac{q \times q}{r^2}\)
By equating the forces, \( k\frac{q \times q}{r^{2}}=G \cdot \frac{m_{E} \times m_{M}}{r^{2}} \)
\(\because k=\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \)
\( 4 \pi \varepsilon_{0} =0.11 \times 10^{-9} \)
\(q =\sqrt{4 \pi \varepsilon_{0} G m_{E} \cdot m_{M}} \)   ....(1)
\(=\sqrt{0.11 \times 10^{-9} \times 6.67 \times 10^{-11} \times 5.9 \times 10^{24} \times 7.9 \times 10^{22}}\)
\( q =\sqrt{34.2 \times 10^{26}} \)
q ≈ 5.85 x 10¹³ C
b) Suppose the distance (r) between Moon and Earth is halved, there is no change in the value of charge (q). Because from equation (1), q is independent of distance (r).

If the conductor has no net charge, then its motion is the same as usual projectile motion of a mass m which we studied in Kinematics (unit 2, vol-1 XI physics). Here, in this problem, in addition to downward gravitational force, the charge also will experience a downward uniform electrostatic force.
The acceleration of the charged ball due to gravity = -g\(\hat { j } \)
The acceleration of the charged ball due to uniform electric field =\(-\frac { qE }{ m } \hat { j } \)
The total acceleration of charged ball in downward direction \(\vec { a } =-\left( g+\frac { qE }{ m } \right) \vec { j } \)
It is important here to note that the acceleration depends on the mass of the object. Galileo’s conclusion that all objects fall at the same rate towards the Earth is true only in a uniform gravitational field. When a uniform electric field is included, the acceleration of a charged object depends on both mass and charge.
But still the acceleration a = \(\left( g+\frac { qE }{ m } \right) \) is constant throughout the motion. Hence we use kinematic equations to calculate the range, maximum height and time of flight. In fact we can simply replace g by \(g+\frac { qE }{ m } \) in the usual expressions of range, maximum height and time of flight of a projectile.

Note that the time of flight, maximum height, range are all inversely proportional to the acceleration of the object. Since \(\left( g+\frac { qE }{ m } \right) \) >g for charge +q, the quantities T, h_max, and R will decrease when compared to the motion of an object of mass m and zero net charge. Suppose the charge is –q, then \(\left( g-\frac { qE }{ m } \right) \)< g, and the quantities T, h_max and  R will increase. Interestingly the trajectory is still parabolic as shown in the figure.

i) Equivalent emf ξ_eq = 5 V
ii) Equivalent internal resistance,
\({ R }_{ eq }=\frac { r }{ n } =\frac { 0.5 }{ 4 } =0.125\Omega \)
iii) total current, \(I=\frac { \xi }{ { R }_{ 5 }+\frac { r }{ n } } \)
\(I=\frac { 5 }{ 10+0.125 } =\frac { 5 }{ 10.125 } \)
I ≈ 0.5 A 
iv) Potential difference across each cell
V = IR = 0.5 x 10 = 5 V
v) Current from each cell, \(I'=\frac { I }{ n } \)
\(I'=\frac { 0.5 }{ 4 } =0.125A\)

(i) The energy E remains constant for varying values of x and v. Differentiating E with respect to time, we get \(\frac { dE }{ dt } =\frac { 1 }{ 2 } \left( 2v\frac { dv }{ dt } \right) +\frac { 1 }{ 2 } k\left( \frac { dx }{ dt } \right) =0\)
or m \(\frac { { d }^{ 2 }x }{ { dt }^{ 2 } } +kx=0\)
since \(\frac { dx }{ dt } =\nu \) and \(\frac { dv }{ dt } =\frac { { d }^{ 2 }x }{ d{ t }^{ 2 } } \)
(ii) This is the differential equation of the oscillations of the spring-mass system. The general solution of an equation is of the form
x(t) = X_m cos (ωt + Φ)
where X is the maximum value of x(t), ω the angular frequency, and Φ the phase constant.
(iii) Similarly, the electromagnetic energy of the LC system is given by
\(U=\frac { 1 }{ 2 } { Li }^{ 2 }+\frac { 1 }{ 2 } \left( \frac { 1 }{ C } \right) { q }^{ 2 }\) = constant
Differentiating U with respect to time, we get
\(\frac { dU }{ dt } =\frac { 1 }{ 2 } L\left( 2i\frac { di }{ dt } \right) +\frac { 1 }{ 2C } \left( 2q\frac { dq }{ dt } \right) =0\)
(or) \(\frac { { d }^{ 2 }q }{ { dt }^{ 2 } } +\frac { 1 }{ C } q=0\) ....(1)
since \(i=\frac { dq }{ dt } \frac { di }{ dt } =\frac { { d }^{ 2 }q }{ d{ t }^{ 2 } } \)
(iv) The general solution of equation (1) is of the form
q(t) = Q_m cos (ωt + Φ)
(v) where Q_m is the maximum value of q(t), ω the angular frequency, and Φ the phase constant.

Diameter d 0.24 cm = 0.24 x 10^-2 m
radius \(r=\cfrac { d }{ 2 } =0.12\times { 1 }^{ -2 }m\)
Current, I = 10A
Current density \(J=\cfrac { 1 }{ A } =\cfrac { 1 }{ { \pi r }^{ 2 } } \)
= \(\cfrac { 10 }{ 3.14\times \left( 0.12\times { 10 }^{ -2 } \right) ^{ 2 } } \)
= 2.2 x 106 Am^-2

\(E=-\left[ \frac { \partial V }{ \partial x } \hat { i } +\frac { \partial V }{ \partial y } \hat { j } +\frac { \partial V }{ \partial z } \hat { k } \right] \)
\(\frac { \partial V }{ \partial x } =\frac { \partial }{ \partial x } (2x+3y-z)=2\)
\(\frac { \partial V }{ \partial y } =3;\frac { \partial V }{ \partial z } =-1\)
∴ Electric field, E \(\\ =-2\hat { i } -3\hat { j } +1\hat { k } \)

For the same potential V₀
The resistor at T_1,\({ R }_{ 1 }=\cfrac { { V }_{ 0 } }{ { I }_{ 1 } } \)
The resistor at T_{2, \({ R }_{ 2 }=\cfrac { { V }_{ 0 } }{ { I }_{ 2 } } \)}
I₁>I₂ (from the graph) \(\therefore { R }_{ 1 }<{ R }_{ 2 }\) Since the resistor of a metal increases with temperature \(\therefore { T }_{ 1 }<{ T }_{ 2 }\)

Given:
Total number of electrons in the slab,
N = 29 x e = 29 x 2 x 10²²
Number g electrons remvoed, n \(=\frac{q}{e}\)
\(n=\frac { 2\times { 10 }^{ -6 } }{ 1.6\times { 10 }^{ -19 } } \)
n 1.25 x 10¹³
∴ fraction of electrons removed
\(=\frac{No.of\ electrons\ removed\ (n)}{Total\ No.of \ electrons(N)}\)
\(\\ =\frac { 29\times 2\times { 10 }^{ 22 } }{ 1.25\times { 10 }^{ 13 } } =2.16\times { 10 }^{ -11 }\)

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.