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25 Must-Do Numericals

CBSE Physics XII — Chapter-wise, with step-by-step solutions
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Ch 1-2 Electrostatics

Q1
Three point charges $q_1 = +2\,\mu\text{C}$, $q_2 = -3\,\mu\text{C}$, and $q_3 = +4\,\mu\text{C}$ are placed along a straight line. $q_2$ is at the origin, $q_1$ is at $x = -10\,\text{cm}$, and $q_3$ is at $x = +20\,\text{cm}$. Find the net force on $q_2$.
3 Marks
Given
  • $q_1 = +2 \times 10^{-6}\,\text{C}$, at $x = -0.10\,\text{m}$
  • $q_2 = -3 \times 10^{-6}\,\text{C}$, at origin
  • $q_3 = +4 \times 10^{-6}\,\text{C}$, at $x = +0.20\,\text{m}$
Formula
$$F = \frac{1}{4\pi\varepsilon_0}\,\frac{|q_a\,q_b|}{r^2}, \quad k = 9 \times 10^9\,\text{N m}^2\text{C}^{-2}$$
Solution
Step 1: Force on $q_2$ due to $q_1$. Since $q_1 > 0$ and $q_2 < 0$, the force is attractive, directed toward $q_1$ (i.e., in the $-x$ direction).
$$F_{12} = k\,\frac{|q_1||q_2|}{r_{12}^2} = 9\times10^9 \times \frac{2\times10^{-6}\times3\times10^{-6}}{(0.10)^2}$$ $$= 9\times10^9 \times \frac{6\times10^{-12}}{0.01} = 5.4\,\text{N} \quad (\text{in } -x \text{ direction})$$
Step 2: Force on $q_2$ due to $q_3$. Since $q_3 > 0$ and $q_2 < 0$, the force is attractive, directed toward $q_3$ (i.e., in the $+x$ direction).
$$F_{32} = k\,\frac{|q_3||q_2|}{r_{32}^2} = 9\times10^9 \times \frac{4\times10^{-6}\times3\times10^{-6}}{(0.20)^2}$$ $$= 9\times10^9 \times \frac{12\times10^{-12}}{0.04} = 2.7\,\text{N} \quad (\text{in } +x \text{ direction})$$
Step 3: Net force on $q_2$:
$$F_{\text{net}} = F_{32} - F_{12} = 2.7 - 5.4 = -2.7\,\text{N}$$
$\vec{F}_{\text{net}} = 2.7\,\text{N}$ in the $-x$ direction (toward $q_1$)
Tip: Always draw a diagram showing force directions on the target charge. Be careful with signs — attractive forces pull toward the source charge, repulsive forces push away. Convert cm to m before substituting.
Q2
Two charges $q_1 = +5\,\mu\text{C}$ and $q_2 = -3\,\mu\text{C}$ are placed 20 cm apart. Find the electric potential at a point P that is 10 cm from $q_1$ and 15 cm from $q_2$, along the line joining them (between the charges). Also find the work done in bringing a charge $q_3 = +1\,\mu\text{C}$ from infinity to point P.
3 Marks
Given
  • $q_1 = +5 \times 10^{-6}\,\text{C}$, $q_2 = -3 \times 10^{-6}\,\text{C}$
  • $r_1 = 0.10\,\text{m}$ (distance from $q_1$ to P)
  • $r_2 = 0.15\,\text{m}$ (distance from $q_2$ to P)
  • $q_3 = +1 \times 10^{-6}\,\text{C}$
Formula
$$V = \frac{1}{4\pi\varepsilon_0}\sum\frac{q_i}{r_i}, \qquad W = q_3 \cdot V_P$$
Solution
Step 1: Potential at P due to $q_1$:
$$V_1 = k\,\frac{q_1}{r_1} = 9\times10^9 \times \frac{5\times10^{-6}}{0.10} = 4.5 \times 10^5\,\text{V}$$
Step 2: Potential at P due to $q_2$:
$$V_2 = k\,\frac{q_2}{r_2} = 9\times10^9 \times \frac{-3\times10^{-6}}{0.15} = -1.8 \times 10^5\,\text{V}$$
Step 3: Total potential at P:
$$V_P = V_1 + V_2 = 4.5\times10^5 - 1.8\times10^5 = 2.7 \times 10^5\,\text{V}$$
Step 4: Work done in bringing $q_3$ from infinity to P:
$$W = q_3 \cdot V_P = 1\times10^{-6} \times 2.7\times10^5 = 0.27\,\text{J}$$
$V_P = 2.7 \times 10^5\,\text{V}$,   $W = 0.27\,\text{J}$
Tip: Potential is a scalar — add algebraically with sign. Unlike force, you do NOT need to resolve directions. The sign of $q$ matters in the formula for V.
Q3
A parallel plate capacitor of capacitance $20\,\mu\text{F}$ is charged to a potential difference of $100\,\text{V}$. A dielectric slab of dielectric constant $K = 5$ is then inserted between the plates. Find the energy stored in the capacitor (a) if the battery remains connected, and (b) if the battery is disconnected before inserting the dielectric.
5 Marks
Given
  • $C_0 = 20\,\mu\text{F} = 20 \times 10^{-6}\,\text{F}$
  • $V_0 = 100\,\text{V}$
  • $K = 5$
Formula
$$U = \frac{1}{2}CV^2 = \frac{Q^2}{2C}, \quad C' = KC_0$$
Solution
Initial energy:
$$U_0 = \tfrac{1}{2}C_0 V_0^2 = \tfrac{1}{2}\times 20\times10^{-6}\times(100)^2 = 0.1\,\text{J}$$
New capacitance: $C' = KC_0 = 5 \times 20 = 100\,\mu\text{F}$
(a) Battery connected — voltage stays at $V_0 = 100\,\text{V}$:
$$U_a = \tfrac{1}{2}C'V_0^2 = \tfrac{1}{2}\times100\times10^{-6}\times(100)^2 = 0.5\,\text{J}$$
Energy increases by factor $K$ (battery supplies extra charge).
(b) Battery disconnected — charge stays at $Q_0 = C_0 V_0 = 2 \times 10^{-3}\,\text{C}$:
$$U_b = \frac{Q_0^2}{2C'} = \frac{(2\times10^{-3})^2}{2\times100\times10^{-6}} = \frac{4\times10^{-6}}{2\times10^{-4}} = 0.02\,\text{J}$$
Energy decreases by factor $K$ (energy used to pull dielectric in).
(a) $U = 0.5\,\text{J}$   (b) $U = 0.02\,\text{J}$
Tip: The key distinction: battery connected keeps V constant (C changes, Q changes); battery disconnected keeps Q constant (C changes, V changes). Energy formula choice depends on which quantity is constant.
Q4
Three capacitors of capacitances $C_1 = 2\,\mu\text{F}$, $C_2 = 3\,\mu\text{F}$, and $C_3 = 6\,\mu\text{F}$ are connected such that $C_1$ and $C_2$ are in series, and their combination is in parallel with $C_3$. This arrangement is connected across a $12\,\text{V}$ battery. Find the equivalent capacitance, total charge, and charge on each capacitor.
3 Marks
Given
  • $C_1 = 2\,\mu\text{F}$, $C_2 = 3\,\mu\text{F}$, $C_3 = 6\,\mu\text{F}$
  • $C_1$ and $C_2$ in series; combination in parallel with $C_3$
  • $V = 12\,\text{V}$
Formula
$$\text{Series: } \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2}, \qquad \text{Parallel: } C_{eq} = C_s + C_3$$
Solution
Step 1: Series combination of $C_1$ and $C_2$:
$$\frac{1}{C_s} = \frac{1}{2} + \frac{1}{3} = \frac{5}{6} \implies C_s = \frac{6}{5} = 1.2\,\mu\text{F}$$
Step 2: Parallel combination with $C_3$:
$$C_{eq} = C_s + C_3 = 1.2 + 6 = 7.2\,\mu\text{F}$$
Step 3: Total charge from battery:
$$Q_{\text{total}} = C_{eq} \times V = 7.2\times10^{-6}\times12 = 86.4\,\mu\text{C}$$
Step 4: Charge on $C_3$ (parallel, so full voltage):
$$Q_3 = C_3 \times V = 6 \times 12 = 72\,\mu\text{C}$$
Step 5: Charge on $C_1$ and $C_2$ (series, same charge):
$$Q_1 = Q_2 = C_s \times V = 1.2 \times 12 = 14.4\,\mu\text{C}$$
$C_{eq} = 7.2\,\mu\text{F}$,   $Q_{\text{total}} = 86.4\,\mu\text{C}$,   $Q_1 = Q_2 = 14.4\,\mu\text{C}$,   $Q_3 = 72\,\mu\text{C}$
Tip: In series, charge is the same on each capacitor. In parallel, voltage is the same. Always identify which capacitors share charge vs. which share voltage before calculating.
Q5
An electric dipole with dipole moment $p = 4 \times 10^{-9}\,\text{C m}$ is placed along the $x$-axis. Find the electric field and potential at a point on the axis at a distance $r = 20\,\text{cm}$ from the centre of the dipole (far-field approximation, $r \gg 2a$).
3 Marks
Given
  • $p = 4 \times 10^{-9}\,\text{C m}$
  • $r = 0.20\,\text{m}$
  • Point is on the axial line
Formula
$$E_{\text{axial}} = \frac{1}{4\pi\varepsilon_0}\,\frac{2p}{r^3}, \qquad V_{\text{axial}} = \frac{1}{4\pi\varepsilon_0}\,\frac{p}{r^2}$$
Solution
Step 1: Electric field on the axis:
$$E = \frac{1}{4\pi\varepsilon_0}\,\frac{2p}{r^3} = 9\times10^9 \times \frac{2\times4\times10^{-9}}{(0.20)^3}$$ $$= 9\times10^9 \times \frac{8\times10^{-9}}{8\times10^{-3}} = 9\times10^9 \times 10^{-6} = 9000\,\text{V/m}$$
Step 2: Potential on the axis:
$$V = \frac{1}{4\pi\varepsilon_0}\,\frac{p}{r^2} = 9\times10^9 \times \frac{4\times10^{-9}}{(0.20)^2}$$ $$= 9\times10^9 \times \frac{4\times10^{-9}}{0.04} = 9\times10^9 \times 10^{-7} = 900\,\text{V}$$
$E_{\text{axial}} = 9000\,\text{V/m}$,   $V_{\text{axial}} = 900\,\text{V}$
Tip: For the equatorial point, $E_{\text{eq}} = \frac{kp}{r^3}$ (half of axial) and $V_{\text{eq}} = 0$. Remember: axial E is twice equatorial E; equatorial V is always zero. These formulas are valid only when $r \gg 2a$.

Ch 3 Current Electricity

Q6
A copper wire of cross-sectional area $A = 1.0 \times 10^{-6}\,\text{m}^2$ carries a current of $1.5\,\text{A}$. If the number density of free electrons is $n = 8.5 \times 10^{28}\,\text{m}^{-3}$, calculate the drift velocity of electrons. Also find the current density.
2 Marks
Given
  • $A = 1.0 \times 10^{-6}\,\text{m}^2$
  • $I = 1.5\,\text{A}$
  • $n = 8.5 \times 10^{28}\,\text{m}^{-3}$
  • $e = 1.6 \times 10^{-19}\,\text{C}$
Formula
$$I = neAv_d \implies v_d = \frac{I}{neA}, \qquad J = \frac{I}{A}$$
Solution
Step 1: Drift velocity:
$$v_d = \frac{I}{neA} = \frac{1.5}{8.5\times10^{28}\times1.6\times10^{-19}\times1.0\times10^{-6}}$$ $$= \frac{1.5}{8.5\times1.6\times10^{3}} = \frac{1.5}{1.36\times10^{4}} = 1.1 \times 10^{-4}\,\text{m/s}$$
Step 2: Current density:
$$J = \frac{I}{A} = \frac{1.5}{1.0\times10^{-6}} = 1.5 \times 10^{6}\,\text{A/m}^2$$
$v_d = 1.1 \times 10^{-4}\,\text{m/s} \approx 0.11\,\text{mm/s}$,   $J = 1.5 \times 10^6\,\text{A/m}^2$
Tip: Drift velocity is extremely small (~0.1 mm/s). Don't confuse it with the speed of the electric signal (which is near the speed of light). Be careful with powers of 10 in the denominator.
Q7
In a meter bridge experiment, the balance point is found at $39.5\,\text{cm}$ from end A when a known resistance of $12.5\,\Omega$ is in the right gap. Find the unknown resistance. If the known resistance is known to $\pm 0.1\,\Omega$ and the balance point to $\pm 0.5\,\text{cm}$, estimate the percentage error in the unknown resistance.
3 Marks
Given
  • $l = 39.5\,\text{cm}$, so $100 - l = 60.5\,\text{cm}$
  • $S = 12.5\,\Omega$
  • $\Delta S = 0.1\,\Omega$, $\Delta l = 0.5\,\text{cm}$
Formula
$$\frac{R}{S} = \frac{l}{100-l} \implies R = S\,\frac{l}{100-l}$$
Solution
Step 1: Unknown resistance:
$$R = 12.5 \times \frac{39.5}{60.5} = 12.5 \times 0.6529 = 8.16\,\Omega$$
Step 2: Percentage error:
$$\frac{\Delta R}{R} = \frac{\Delta S}{S} + \frac{\Delta l}{l} + \frac{\Delta l}{100-l}$$ $$= \frac{0.1}{12.5} + \frac{0.5}{39.5} + \frac{0.5}{60.5}$$ $$= 0.008 + 0.01266 + 0.00826 = 0.02892$$
Percentage error $\approx 2.9\%$
$R = 8.16\,\Omega$,   Percentage error $\approx 2.9\%$
Tip: For error analysis, use the rule: if $R = S \cdot \frac{l}{100-l}$, take log and differentiate. The balance point should ideally be near the middle (50 cm) to minimise error. This is a frequently asked CBSE question.
Q8
In a potentiometer experiment, the balance point for a cell of EMF $1.25\,\text{V}$ is at $35.0\,\text{cm}$. When the cell is shunted with an external resistance of $10\,\Omega$, the balance point shifts to $28.0\,\text{cm}$. Find the internal resistance of the cell.
3 Marks
Given
  • $l_1 = 35.0\,\text{cm}$ (open circuit)
  • $l_2 = 28.0\,\text{cm}$ (with shunt $R = 10\,\Omega$)
  • $\varepsilon = 1.25\,\text{V}$
Formula
$$r = R\left(\frac{l_1 - l_2}{l_2}\right)$$
Solution
$$r = 10 \times \frac{35.0 - 28.0}{28.0} = 10 \times \frac{7.0}{28.0} = 10 \times 0.25 = 2.5\,\Omega$$
$r = 2.5\,\Omega$
Tip: This formula works because $\frac{\varepsilon}{V} = \frac{l_1}{l_2}$ and $r = R\left(\frac{\varepsilon}{V} - 1\right)$. The EMF value ($1.25\,\text{V}$) is not needed in the final formula — it cancels out. This is a classic 3-mark CBSE numerical.
Q9
A battery of EMF $12\,\text{V}$ and internal resistance $2\,\Omega$ is connected to an external resistance $R$. Find the value of $R$ for which the power dissipated in $R$ is maximum. What is this maximum power?
3 Marks
Given
  • $\varepsilon = 12\,\text{V}$
  • $r = 2\,\Omega$ (internal resistance)
Formula
$$P = I^2 R = \frac{\varepsilon^2 R}{(R+r)^2}, \qquad P_{\text{max}} \text{ when } R = r$$
Solution
Step 1: For maximum power transfer, $R = r = 2\,\Omega$.
Step 2: Current at maximum power:
$$I = \frac{\varepsilon}{R + r} = \frac{12}{2 + 2} = 3\,\text{A}$$
Step 3: Maximum power:
$$P_{\text{max}} = I^2 R = (3)^2 \times 2 = 18\,\text{W}$$
Alternatively: $P_{\text{max}} = \frac{\varepsilon^2}{4r} = \frac{144}{8} = 18\,\text{W}$
$R = 2\,\Omega$,   $P_{\text{max}} = 18\,\text{W}$
Tip: Maximum power transfer theorem: $R = r$. At this point, only 50% of total power is useful (rest is lost in internal resistance). The shortcut formula $P_{\text{max}} = \frac{\varepsilon^2}{4r}$ saves time.

Ch 4 Moving Charges & Magnetism

Q10
A proton ($m = 1.67 \times 10^{-27}\,\text{kg}$, $q = 1.6 \times 10^{-19}\,\text{C}$) enters a uniform magnetic field of $0.20\,\text{T}$ perpendicular to its velocity of $5 \times 10^{6}\,\text{m/s}$. Find the radius of the circular path and the time period of revolution.
3 Marks
Given
  • $m = 1.67 \times 10^{-27}\,\text{kg}$
  • $q = 1.6 \times 10^{-19}\,\text{C}$
  • $B = 0.20\,\text{T}$
  • $v = 5 \times 10^{6}\,\text{m/s}$
Formula
$$r = \frac{mv}{qB}, \qquad T = \frac{2\pi m}{qB}$$
Solution
Step 1: Radius of circular path:
$$r = \frac{mv}{qB} = \frac{1.67\times10^{-27}\times5\times10^{6}}{1.6\times10^{-19}\times0.20}$$ $$= \frac{8.35\times10^{-21}}{3.2\times10^{-20}} = 0.261\,\text{m} \approx 26.1\,\text{cm}$$
Step 2: Time period:
$$T = \frac{2\pi m}{qB} = \frac{2\pi\times1.67\times10^{-27}}{1.6\times10^{-19}\times0.20}$$ $$= \frac{1.049\times10^{-26}}{3.2\times10^{-20}} = 3.28 \times 10^{-7}\,\text{s}$$
$r \approx 26.1\,\text{cm}$,   $T \approx 3.28 \times 10^{-7}\,\text{s}$
Tip: The time period is independent of velocity (and radius). This is the principle behind the cyclotron. If the particle enters at an angle $\theta$ to $\vec{B}$, use $v\sin\theta$ as the perpendicular component and the path becomes helical.
Q11
Two long straight parallel wires, separated by $d = 10\,\text{cm}$, carry currents $I_1 = 5\,\text{A}$ and $I_2 = 8\,\text{A}$ in the same direction. Find the force per unit length between them and state whether it is attractive or repulsive.
2 Marks
Given
  • $I_1 = 5\,\text{A}$, $I_2 = 8\,\text{A}$
  • $d = 0.10\,\text{m}$
  • Currents in the same direction
Formula
$$\frac{F}{L} = \frac{\mu_0}{2\pi}\,\frac{I_1 I_2}{d}, \quad \mu_0 = 4\pi\times10^{-7}\,\text{T m/A}$$
Solution
$$\frac{F}{L} = \frac{4\pi\times10^{-7}}{2\pi}\times\frac{5\times8}{0.10} = 2\times10^{-7}\times\frac{40}{0.10}$$ $$= 2\times10^{-7}\times400 = 8 \times 10^{-5}\,\text{N/m}$$
Since currents are in the same direction, the force is attractive.
$\frac{F}{L} = 8 \times 10^{-5}\,\text{N/m}$ (attractive)
Tip: Same direction = attractive, opposite direction = repulsive. This is the basis of the definition of the ampere. Remember: $\frac{\mu_0}{2\pi} = 2 \times 10^{-7}$ is a useful shortcut.
Q12
A circular coil of 200 turns and radius $10\,\text{cm}$ carries a current of $5\,\text{A}$. Find the magnetic field at the centre of the coil.
2 Marks
Given
  • $N = 200$, $R = 0.10\,\text{m}$, $I = 5\,\text{A}$
Formula
$$B = \frac{\mu_0 N I}{2R}$$
Solution
$$B = \frac{4\pi\times10^{-7}\times200\times5}{2\times0.10}$$ $$= \frac{4\pi\times10^{-7}\times1000}{0.20} = \frac{4\pi\times10^{-4}}{0.20} = 2\pi\times10^{-3}$$ $$= 6.28 \times 10^{-3}\,\text{T} = 6.28\,\text{mT}$$
$B = 6.28 \times 10^{-3}\,\text{T} \approx 6.28\,\text{mT}$
Tip: Don't confuse the formula for a coil ($\frac{\mu_0 NI}{2R}$) with that for a solenoid ($\mu_0 nI$). For a point on the axis at distance $x$ from the centre, use $B = \frac{\mu_0 NIR^2}{2(R^2+x^2)^{3/2}}$.

Ch 5-6 Magnetism & Electromagnetic Induction

Q13
A short bar magnet has a magnetic moment of $0.48\,\text{J T}^{-1}$. Find the magnetic field at a point (a) on the axial line at $20\,\text{cm}$ from the centre, and (b) on the equatorial line at $20\,\text{cm}$ from the centre.
3 Marks
Given
  • $M = 0.48\,\text{J T}^{-1}$
  • $r = 0.20\,\text{m}$
Formula
$$B_{\text{axial}} = \frac{\mu_0}{4\pi}\,\frac{2M}{r^3}, \qquad B_{\text{equat}} = \frac{\mu_0}{4\pi}\,\frac{M}{r^3}$$
Solution
(a) Axial:
$$B_a = 10^{-7}\times\frac{2\times0.48}{(0.20)^3} = 10^{-7}\times\frac{0.96}{0.008} = 10^{-7}\times120 = 1.2\times10^{-5}\,\text{T}$$
(b) Equatorial:
$$B_e = 10^{-7}\times\frac{0.48}{(0.20)^3} = 10^{-7}\times\frac{0.48}{0.008} = 10^{-7}\times60 = 6.0\times10^{-6}\,\text{T}$$
Note: $B_{\text{axial}} = 2 \times B_{\text{equat}}$, as expected.
$B_{\text{axial}} = 1.2 \times 10^{-5}\,\text{T}$,   $B_{\text{equat}} = 6.0 \times 10^{-6}\,\text{T}$
Tip: These formulas are identical to the electric dipole formulas (replace $\frac{1}{4\pi\varepsilon_0}$ with $\frac{\mu_0}{4\pi}$ and $p$ with $M$). The direction on the axial line is along $M$; on the equatorial line it is antiparallel to $M$.
Q14
A rectangular coil of 100 turns, area $0.04\,\text{m}^2$, rotates at $50\,\text{rev/s}$ in a uniform magnetic field of $0.10\,\text{T}$. Find the maximum EMF induced. Also write the expression for instantaneous EMF as a function of time.
3 Marks
Given
  • $N = 100$, $A = 0.04\,\text{m}^2$
  • $f = 50\,\text{rev/s}$, so $\omega = 2\pi f = 100\pi\,\text{rad/s}$
  • $B = 0.10\,\text{T}$
Formula
$$\varepsilon = NBA\omega\sin(\omega t), \qquad \varepsilon_0 = NBA\omega$$
Solution
Step 1: Maximum EMF:
$$\varepsilon_0 = NBA\omega = 100\times0.10\times0.04\times100\pi$$ $$= 100\times0.004\times100\pi = 0.4\times100\pi = 40\pi \approx 125.7\,\text{V}$$
Step 2: Instantaneous EMF:
$$\varepsilon(t) = 125.7\sin(100\pi t)\,\text{V}$$
$\varepsilon_0 = 40\pi \approx 125.7\,\text{V}$,   $\varepsilon(t) = 125.7\sin(100\pi\, t)$ V
Tip: The EMF is maximum when the plane of the coil is parallel to $\vec{B}$ (i.e., $\theta = 90^{\circ}$), not when it is perpendicular. Common error: confusing $f$ (rev/s) with $\omega$ (rad/s). Always convert: $\omega = 2\pi f$.
Q15
A solenoid of length $50\,\text{cm}$, radius $2\,\text{cm}$, and 500 turns carries a current of $3\,\text{A}$. Calculate the self-inductance of the solenoid and the energy stored in it.
3 Marks
Given
  • $l = 0.50\,\text{m}$, $r = 0.02\,\text{m}$
  • $N = 500$, $I = 3\,\text{A}$
  • $A = \pi r^2 = \pi\times(0.02)^2 = 1.2566\times10^{-3}\,\text{m}^2$
Formula
$$L = \mu_0 \frac{N^2 A}{l}, \qquad U = \frac{1}{2}LI^2$$
Solution
Step 1: Self-inductance:
$$L = 4\pi\times10^{-7}\times\frac{(500)^2\times1.2566\times10^{-3}}{0.50}$$ $$= 4\pi\times10^{-7}\times\frac{250000\times1.2566\times10^{-3}}{0.50}$$ $$= 4\pi\times10^{-7}\times\frac{314.15}{0.50} = 4\pi\times10^{-7}\times628.3$$ $$= 7.90\times10^{-4}\,\text{H} \approx 0.79\,\text{mH}$$
Step 2: Energy stored:
$$U = \frac{1}{2}\times7.90\times10^{-4}\times(3)^2 = \frac{1}{2}\times7.90\times10^{-4}\times9 = 3.56\times10^{-3}\,\text{J}$$
$L \approx 0.79\,\text{mH}$,   $U \approx 3.56 \times 10^{-3}\,\text{J}$
Tip: The formula uses $N^2$ (total turns squared), not $n^2$ (turns per unit length squared). If given $n$ instead of $N$, use $L = \mu_0 n^2 A l$ where $n = N/l$. Don't forget to compute the cross-sectional area $A = \pi r^2$.

Ch 7 Alternating Current

Q16
A series LCR circuit has $L = 2\,\text{H}$, $C = 32\,\mu\text{F}$, and $R = 10\,\Omega$. It is connected to a $230\,\text{V}$, $50\,\text{Hz}$ AC source. Find the impedance, current, and phase angle. Is the circuit capacitive or inductive?
5 Marks
Given
  • $L = 2\,\text{H}$, $C = 32\times10^{-6}\,\text{F}$, $R = 10\,\Omega$
  • $V_{\text{rms}} = 230\,\text{V}$, $f = 50\,\text{Hz}$
  • $\omega = 2\pi f = 100\pi\,\text{rad/s}$
Formula
$$Z = \sqrt{R^2 + (X_L - X_C)^2}, \quad I = \frac{V}{Z}, \quad \tan\phi = \frac{X_L - X_C}{R}$$
Solution
Step 1: Inductive reactance:
$$X_L = \omega L = 100\pi\times2 = 200\pi \approx 628.3\,\Omega$$
Step 2: Capacitive reactance:
$$X_C = \frac{1}{\omega C} = \frac{1}{100\pi\times32\times10^{-6}} = \frac{1}{3.2\pi\times10^{-3}} \approx 99.5\,\Omega$$
Step 3: Net reactance:
$$X_L - X_C = 628.3 - 99.5 = 528.8\,\Omega$$
Step 4: Impedance:
$$Z = \sqrt{(10)^2 + (528.8)^2} = \sqrt{100 + 279\,629} = \sqrt{279\,729} \approx 529.0\,\Omega$$
Step 5: Current:
$$I = \frac{V}{Z} = \frac{230}{529.0} \approx 0.435\,\text{A}$$
Step 6: Phase angle:
$$\tan\phi = \frac{528.8}{10} = 52.88 \implies \phi \approx 88.9^{\circ}$$
Since $X_L > X_C$, the circuit is inductive. Voltage leads current by $88.9^{\circ}$.
$Z \approx 529\,\Omega$,   $I \approx 0.435\,\text{A}$,   $\phi \approx 88.9^{\circ}$ (inductive)
Tip: If $X_L > X_C$: inductive (voltage leads). If $X_C > X_L$: capacitive (current leads). Remember the mnemonic "CIVIL" — in a Capacitor, I leads V; in an Inductor, V leads I.
Q17
An LCR circuit has $L = 5.0\,\text{H}$, $C = 80\,\mu\text{F}$, and $R = 40\,\Omega$ connected to a $230\,\text{V}$ variable frequency AC source. Find the resonant frequency and the current at resonance. Also find the quality factor (Q-factor) of the circuit.
3 Marks
Given
  • $L = 5.0\,\text{H}$, $C = 80\times10^{-6}\,\text{F}$, $R = 40\,\Omega$
  • $V_{\text{rms}} = 230\,\text{V}$
Formula
$$f_0 = \frac{1}{2\pi\sqrt{LC}}, \qquad Q = \frac{1}{R}\sqrt{\frac{L}{C}}$$
Solution
Step 1: Resonant frequency:
$$\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{5.0\times80\times10^{-6}}} = \frac{1}{\sqrt{4\times10^{-4}}} = \frac{1}{0.02} = 50\,\text{rad/s}$$ $$f_0 = \frac{\omega_0}{2\pi} = \frac{50}{2\pi} = 7.96\,\text{Hz}$$
Step 2: At resonance, $Z = R$:
$$I_0 = \frac{V}{R} = \frac{230}{40} = 5.75\,\text{A}$$
Step 3: Quality factor:
$$Q = \frac{1}{R}\sqrt{\frac{L}{C}} = \frac{1}{40}\sqrt{\frac{5.0}{80\times10^{-6}}} = \frac{1}{40}\sqrt{62500} = \frac{250}{40} = 6.25$$
$f_0 \approx 7.96\,\text{Hz}$,   $I_0 = 5.75\,\text{A}$,   $Q = 6.25$
Tip: At resonance: $X_L = X_C$, impedance $Z = R$ (minimum), current is maximum, and the phase angle is zero. A higher Q-factor means a sharper resonance peak and better selectivity.
Q18
A step-up transformer converts $230\,\text{V}$ to $2300\,\text{V}$. The primary coil has $100$ turns and the primary current is $2\,\text{A}$. Find the number of turns in the secondary, the secondary current (assuming ideal transformer), and the efficiency if the actual secondary current is $0.18\,\text{A}$.
3 Marks
Given
  • $V_p = 230\,\text{V}$, $V_s = 2300\,\text{V}$
  • $N_p = 100$, $I_p = 2\,\text{A}$
  • Actual $I_s = 0.18\,\text{A}$
Formula
$$\frac{V_s}{V_p} = \frac{N_s}{N_p} = \frac{I_p}{I_s}, \qquad \eta = \frac{V_s I_s}{V_p I_p}\times100\%$$
Solution
Step 1: Turns ratio and $N_s$:
$$\frac{N_s}{N_p} = \frac{V_s}{V_p} = \frac{2300}{230} = 10 \implies N_s = 10\times100 = 1000$$
Step 2: Ideal secondary current:
$$I_s(\text{ideal}) = \frac{I_p}{10} = \frac{2}{10} = 0.2\,\text{A}$$
Step 3: Efficiency:
$$\eta = \frac{V_s I_s(\text{actual})}{V_p I_p}\times100 = \frac{2300\times0.18}{230\times2}\times100 = \frac{414}{460}\times100 = 90\%$$
$N_s = 1000$,   $I_s(\text{ideal}) = 0.2\,\text{A}$,   $\eta = 90\%$
Tip: A transformer works only with AC (not DC). Efficiency is always $\leq 100\%$ due to copper losses ($I^2R$), eddy current losses, flux leakage, and hysteresis. For an ideal transformer, $\eta = 100\%$ and $P_p = P_s$.

Ch 9 Ray Optics

Q19
An object is placed $30\,\text{cm}$ from a convex lens of focal length $20\,\text{cm}$. Find the position and nature of the image, and the magnification.
2 Marks
Given
  • $u = -30\,\text{cm}$ (object on left, sign convention)
  • $f = +20\,\text{cm}$ (convex lens)
Formula
$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}, \qquad m = \frac{v}{u}$$
Solution
Step 1: Apply lens formula:
$$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{20} + \frac{1}{(-30)} = \frac{1}{20} - \frac{1}{30}$$ $$= \frac{3-2}{60} = \frac{1}{60} \implies v = +60\,\text{cm}$$
Step 2: Magnification:
$$m = \frac{v}{u} = \frac{60}{-30} = -2$$
$v > 0$: image is on the other side (real). $m = -2$: image is inverted and twice the size of the object.
$v = +60\,\text{cm}$ (real, inverted),   $m = -2$ (magnified)
Tip: Always use the New Cartesian Sign Convention consistently. For a convex lens: $f > 0$. Object distance $u$ is always negative. If you get $v > 0$, image is real; $v < 0$ means virtual.
Q20
Two thin convex lenses of focal lengths $20\,\text{cm}$ and $-40\,\text{cm}$ are placed in contact. Find the equivalent focal length and the power of the combination. An object is placed $30\,\text{cm}$ from this combination. Find the image position.
3 Marks
Given
  • $f_1 = +20\,\text{cm}$, $f_2 = -40\,\text{cm}$
  • $u = -30\,\text{cm}$
Formula
$$\frac{1}{f_{eq}} = \frac{1}{f_1} + \frac{1}{f_2}, \qquad P = \frac{1}{f_{eq}(\text{in m})}$$
Solution
Step 1: Equivalent focal length:
$$\frac{1}{f_{eq}} = \frac{1}{20} + \frac{1}{-40} = \frac{1}{20}-\frac{1}{40} = \frac{2-1}{40} = \frac{1}{40}$$ $$f_{eq} = +40\,\text{cm} = 0.40\,\text{m}$$
Step 2: Power:
$$P = \frac{1}{0.40} = +2.5\,\text{D}$$
Step 3: Image position using lens formula with $f_{eq}$:
$$\frac{1}{v} = \frac{1}{f_{eq}} + \frac{1}{u} = \frac{1}{40} + \frac{1}{-30} = \frac{3-4}{120} = \frac{-1}{120}$$ $$v = -120\,\text{cm}$$
$v < 0$: image is virtual, erect, on the same side as the object.
$f_{eq} = 40\,\text{cm}$,   $P = +2.5\,\text{D}$,   $v = -120\,\text{cm}$ (virtual)
Tip: When lenses are in contact, powers add directly: $P = P_1 + P_2$. A positive equivalent power means the combination is converging. Always convert $f$ to metres when computing power in dioptres.
Q21
A glass prism has a refractive index of $\mu = 1.5$ and the angle of the prism is $A = 60^{\circ}$. Find the angle of minimum deviation. Also find the angle of incidence at minimum deviation.
3 Marks
Given
  • $\mu = 1.5$, $A = 60^{\circ}$
Formula
$$\mu = \frac{\sin\left(\frac{A + D_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$$
Solution
Step 1: Apply the prism formula:
$$1.5 = \frac{\sin\left(\frac{60+D_m}{2}\right)}{\sin 30^{\circ}} = \frac{\sin\left(\frac{60+D_m}{2}\right)}{0.5}$$
$$\sin\left(\frac{60+D_m}{2}\right) = 1.5\times0.5 = 0.75$$
$$\frac{60+D_m}{2} = \sin^{-1}(0.75) = 48.59^{\circ}$$
$$60 + D_m = 97.18^{\circ} \implies D_m = 37.18^{\circ} \approx 37.2^{\circ}$$
Step 2: At minimum deviation, $i = \frac{A + D_m}{2} = \frac{60 + 37.2}{2} = 48.6^{\circ}$
$D_m \approx 37.2^{\circ}$,   $i \approx 48.6^{\circ}$
Tip: At minimum deviation, the ray passes symmetrically through the prism: $i = e$ and $r_1 = r_2 = A/2$. This is the condition used to derive the formula. Make sure your calculator is in degree mode.

Ch 10 Wave Optics

Q22
In Young's double slit experiment, the slit separation is $d = 0.5\,\text{mm}$, the screen distance is $D = 1.2\,\text{m}$, and the wavelength of light used is $\lambda = 600\,\text{nm}$. Find the fringe width. If the entire setup is immersed in a liquid of refractive index $\mu = 1.5$, find the new fringe width.
3 Marks
Given
  • $d = 0.5\,\text{mm} = 5\times10^{-4}\,\text{m}$
  • $D = 1.2\,\text{m}$
  • $\lambda = 600\,\text{nm} = 6\times10^{-7}\,\text{m}$
  • $\mu = 1.5$ (refractive index of liquid)
Formula
$$\beta = \frac{\lambda D}{d}, \qquad \lambda' = \frac{\lambda}{\mu} \implies \beta' = \frac{\beta}{\mu}$$
Solution
Step 1: Fringe width in air:
$$\beta = \frac{\lambda D}{d} = \frac{6\times10^{-7}\times1.2}{5\times10^{-4}} = \frac{7.2\times10^{-7}}{5\times10^{-4}} = 1.44\times10^{-3}\,\text{m} = 1.44\,\text{mm}$$
Step 2: In the liquid, $\lambda' = \frac{\lambda}{\mu} = \frac{600}{1.5} = 400\,\text{nm}$:
$$\beta' = \frac{\lambda' D}{d} = \frac{\beta}{\mu} = \frac{1.44}{1.5} = 0.96\,\text{mm}$$
$\beta = 1.44\,\text{mm}$,   $\beta' = 0.96\,\text{mm}$
Tip: Fringe width decreases in a denser medium. If only one slit is covered by a thin film, the central fringe shifts but the fringe width does not change. For the position of $n^{\text{th}}$ bright fringe: $y_n = \frac{n\lambda D}{d}$.

Ch 11 Dual Nature of Radiation & Matter

Q23
Light of wavelength $400\,\text{nm}$ is incident on a metal surface with work function $W_0 = 2.0\,\text{eV}$. Find: (a) the threshold frequency, (b) the maximum kinetic energy of emitted photoelectrons, and (c) the stopping potential.
3 Marks
Given
  • $\lambda = 400\,\text{nm} = 4\times10^{-7}\,\text{m}$
  • $W_0 = 2.0\,\text{eV} = 2.0\times1.6\times10^{-19} = 3.2\times10^{-19}\,\text{J}$
  • $h = 6.63\times10^{-34}\,\text{J s}$, $c = 3\times10^{8}\,\text{m/s}$
Formula
$$E = h\nu = \frac{hc}{\lambda}, \quad KE_{\max} = h\nu - W_0, \quad W_0 = h\nu_0, \quad eV_0 = KE_{\max}$$
Solution
(a) Threshold frequency:
$$\nu_0 = \frac{W_0}{h} = \frac{3.2\times10^{-19}}{6.63\times10^{-34}} = 4.83\times10^{14}\,\text{Hz}$$
(b) Energy of incident photon:
$$E = \frac{hc}{\lambda} = \frac{6.63\times10^{-34}\times3\times10^{8}}{4\times10^{-7}} = \frac{1.989\times10^{-25}}{4\times10^{-7}} = 4.97\times10^{-19}\,\text{J}$$ $$= \frac{4.97\times10^{-19}}{1.6\times10^{-19}} = 3.1\,\text{eV}$$
Maximum KE:
$$KE_{\max} = E - W_0 = 3.1 - 2.0 = 1.1\,\text{eV}$$
(c) Stopping potential:
$$eV_0 = KE_{\max} \implies V_0 = \frac{KE_{\max}}{e} = 1.1\,\text{V}$$
$\nu_0 = 4.83 \times 10^{14}\,\text{Hz}$,   $KE_{\max} = 1.1\,\text{eV}$,   $V_0 = 1.1\,\text{V}$
Tip: The shortcut $E(\text{eV}) = \frac{1240}{\lambda(\text{nm})}$ saves a lot of time. For $\lambda = 400\,\text{nm}$: $E = 1240/400 = 3.1\,\text{eV}$. The stopping potential in volts is numerically equal to $KE_{\max}$ in eV.

Ch 12-13 Atoms & Nuclei

Q24
A hydrogen atom makes a transition from $n = 4$ to $n = 2$. Find the wavelength of the emitted photon. Which series does this line belong to? Also find the maximum wavelength in this series.
3 Marks
Given
  • $n_i = 4$, $n_f = 2$
  • $R = 1.097 \times 10^7\,\text{m}^{-1}$ (Rydberg constant)
Formula
$$\frac{1}{\lambda} = R\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$$
Solution
Step 1: Wavelength for $n = 4 \to 2$:
$$\frac{1}{\lambda} = 1.097\times10^7\left(\frac{1}{4} - \frac{1}{16}\right) = 1.097\times10^7\times\frac{3}{16}$$ $$= 1.097\times10^7\times0.1875 = 2.057\times10^6\,\text{m}^{-1}$$
$$\lambda = \frac{1}{2.057\times10^6} = 4.86\times10^{-7}\,\text{m} = 486\,\text{nm}$$
This is the Balmer series ($n_f = 2$), visible region (blue-green line, $H_{\beta}$).
Step 2: Maximum wavelength in Balmer series ($n = 3 \to 2$):
$$\frac{1}{\lambda_{\max}} = R\left(\frac{1}{4}-\frac{1}{9}\right) = 1.097\times10^7\times\frac{5}{36} = 1.524\times10^6$$ $$\lambda_{\max} = 656\,\text{nm}$$
$\lambda = 486\,\text{nm}$ (Balmer series, $H_\beta$),   $\lambda_{\max} = 656\,\text{nm}$ ($H_\alpha$)
Tip: Maximum wavelength = minimum energy transition (adjacent levels). Minimum wavelength = $n = \infty$ to $n_f$ (series limit). Know the series: Lyman ($n_f = 1$, UV), Balmer ($n_f = 2$, visible), Paschen ($n_f = 3$, IR).
Q25
Calculate the mass defect and the binding energy per nucleon of $^{56}_{26}\text{Fe}$. Given: mass of $^{56}_{26}\text{Fe}$ atom $= 55.9349\,\text{u}$, mass of proton $= 1.00783\,\text{u}$, mass of neutron $= 1.00867\,\text{u}$, and $1\,\text{u} = 931.5\,\text{MeV/c}^2$.
5 Marks
Given
  • $Z = 26$ (protons), $N = 56 - 26 = 30$ (neutrons)
  • $M(^{56}\text{Fe}) = 55.9349\,\text{u}$
  • $m_p = 1.00783\,\text{u}$, $m_n = 1.00867\,\text{u}$
Formula
$$\Delta M = [Zm_p + Nm_n] - M_{\text{nucleus}}, \qquad BE = \Delta M \times 931.5\,\text{MeV}$$ $$\frac{BE}{A} = \frac{BE}{56}$$
Solution
Step 1: Expected mass of separated nucleons:
$$M_{\text{sep}} = 26\times1.00783 + 30\times1.00867$$ $$= 26.2036 + 30.2601 = 56.4637\,\text{u}$$
Step 2: Mass defect:
$$\Delta M = 56.4637 - 55.9349 = 0.5288\,\text{u}$$
Step 3: Total binding energy:
$$BE = 0.5288\times931.5 = 492.6\,\text{MeV}$$
Step 4: Binding energy per nucleon:
$$\frac{BE}{A} = \frac{492.6}{56} = 8.79\,\text{MeV/nucleon}$$
$\Delta M = 0.5288\,\text{u}$,   $BE = 492.6\,\text{MeV}$,   $\frac{BE}{A} = 8.79\,\text{MeV/nucleon}$
Tip: $^{56}\text{Fe}$ has the highest BE/nucleon (~8.79 MeV) among all elements, making it the most stable nucleus. Use atomic masses (which include electron masses) consistently — the electron masses cancel out since $m_p$ given here is actually the hydrogen atom mass.