Practice

Q1. A circular coil of 100 turns, radius 10 cm, carries a current of 1 A. What is the magnetic field at the centre of the coil? ($\\mu_0 = 4\\pi \\times 10^{-7}$ T m A$^{-1}$)

A. $6.28 \times 10^-4

Answer: A

• $B = \\dfrac{\\mu_0 n I}{2R} = \\dfrac{4\\pi \\times 10^{-7} \\times 100 \\times 1}{2 \\times 0.10} = \\dfrac{4\\pi \\times 10^{-5}}{0.2} = 2\\pi \\times 10^{-4} \\approx 6.28 \\times 10^{-4}$ T. Difficulty: medium

Q2. In Young’s double-slit experiment using light of wavelength $\\lambda = 600$ nm, the fringe width is 2 mm. If the distance between the slits is doubled, the new fringe width will be:

A. 4 mm B. 2 mm C. 0.5 mm D. 1 mm

Answer: D

• Fringe width $\\beta = \\dfrac{\\lambda D}{d}$, where $D$ is the distance to the screen and $d$ is the slit separation. If $d$ is doubled, $\\beta$ is halved. New fringe width $= \\dfrac{2}{2} = 1$ mm. Difficulty: medium

Q3. A convex lens of focal length 20 cm is placed in contact with a concave lens of focal length 30 cm. What is the power of the combination?

A. $-1.67$ D B. $+5.00$ D C. $+0.60$ D D. $+1.67$ D

Answer: D

• $P_1 = \\dfrac{1}{f_1} = \\dfrac{1}{0.20} = +5$ D (convex). $P_2 = \\dfrac{1}{f_2} = \\dfrac{1}{-0.30} = -3.33$ D (concave). $P_{\\text{net}} = P_1 + P_2 = 5 - 3.33 = +1.67$ D. The equivalent focal length is $f = 1/P = 1/1.67 \\approx 60$ cm (convex). Difficulty: medium

Q4. In a single-slit diffraction experiment, the width of the central maximum is found to be 4 mm. If the slit width is halved, the width of the central maximum becomes:

A. 2 mm B. 16 mm C. 8 mm D. 4 mm

Answer: C

• The width of the central maximum in single-slit diffraction is $2y = \\dfrac{2\\lambda D}{a}$, where $a$ is the slit width. If $a$ is halved, $2y$ doubles. New width $= 2 \\times 4 = 8$ mm. Narrower slit produces broader diffraction pattern. Difficulty: medium

Q5. The work function of a metal surface is 2 eV. If light of wavelength 400 nm is incident on it, what is the maximum kinetic energy of the emitted photoelectrons? ($h = 6.63 \\times 10^{-34}$ J s, $c = 3 \\times 10^8$ m/s, $1$ eV $= 1.6 \\times 10^{-19}$ J)

A. $3.105$ eV B. $1.105$ eV C. $0.895$ eV D. $2.000$ eV

Answer: B

• Energy of photon: $E = \\dfrac{hc}{\\lambda} = \\dfrac{6.63 \\times 10^{-34} \\times 3 \\times 10^8}{400 \\times 10^{-9}} = 4.97 \\times 10^{-19}$ J $= \\dfrac{4.97 \\times 10^{-19}}{1.6 \\times 10^{-19}} = 3.105$ eV. By Einstein’s equation: $K_{\\max} = E - \\phi = 3.105 - 2.0 = 1.105$ eV. Difficulty: medium

Q6. The de Broglie wavelength associated with an electron accelerated through a potential difference of 100 V is approximately: ($m_e = 9.1 \\times 10^{-31}$ kg, $e = 1.6 \\times 10^{-19}$ C, $h = 6.63 \\times 10^{-34}$ J s)

A. 0.123 nm B. 1.23 nm C. 0.0123 nm D. 12.3 nm

Answer: A

• $\\lambda = \\dfrac{h}{\\sqrt{2m_e eV}} = \\dfrac{6.63 \\times 10^{-34}}{\\sqrt{2 \\times 9.1 \\times 10^{-31} \\times 1.6 \\times 10^{-19} \\times 100}} = \\dfrac{6.63 \\times 10^{-34}}{\\sqrt{2.912 \\times 10^{-47}}} = \\dfrac{6.63 \\times 10^{-34}}{5.396 \\times 10^{-24}} = 1.228 \\times 10^{-10}$ m $= 0.1228$ nm $\\approx 0.123$ nm. Difficulty: hard

Q7. In a p-n junction diode, the potential barrier is approximately 0.3 V for Ge and 0.7 V for Si. If a Si diode is forward biased with a voltage of 0.5 V applied, what is the current through the diode?

A. Moderately high B. Very high C. Cannot be determined D. Nearly zero

Answer: D

• For a forward-biased Si diode, the applied voltage must exceed the knee voltage (barrier potential) of approximately 0.7 V before significant current flows. Since 0.5 V < 0.7 V, the current is negligibly small (nearly zero). The diode is technically forward biased but below the threshold for conduction. Difficulty: hard

Q8. In a transistor, the current gain $\\beta = 99$. If the base current $I_B = 20$ $\\mu$ A, the collector current $I_C$ is:

A. 2.00 mA B. 0.20 mA C. 1.98 mA D. 0.99 mA

Answer: C

• $\\beta = \\dfrac{I_C}{I_B}$, so $I_C = \\beta \\cdot I_B = 99 \\times 20 \\times 10^{-6} = 1980 \\times 10^{-6}$ A $= 1.98$ mA. The emitter current $I_E = I_C + I_B = 1.98 + 0.02 = 2.00$ mA. Difficulty: medium

Q9. In an LED (Light Emitting Diode), the colour of emitted light depends on:

A. The band gap energy of the semiconductor material B. The intensity of forward current C. The applied reverse voltage D. The physical size of the diode’,

Answer: A

• The colour of light emitted by an LED is determined by the energy of photons released during electron-hole recombination across the band gap: $E = hc/\\lambda$. A wider band gap produces shorter wavelength (higher energy) light. For example, GaAs ($E_g \\approx 1.43$ eV) emits infrared; GaAsP ($\\approx 1.9$ eV) emits red; GaN ($\\approx 3.4$ eV) emits blue/violet. Difficulty: hard

Q3. A body of mass 2 kg experiences a net force of 10 N. What is its acceleration?

A. 5 m/s^2 B. 20 m/s^2 C. 0.2 m/s^2 D. 12 m/s^2

Answer: A

• Newton’s second law gives a = F/m = 10/2 = 5 m/s^2. Option B multiplies instead of divides; option D adds instead of dividing. Difficulty: easy

Q4. The SI unit of electric charge is the coulomb. One coulomb equals the charge transported by a current of one ampere in:

A. one second B. one hour C. one minute D. one millisecond

Answer: A

• By definition 1 C = 1 A s. The coulomb is the ampere-second; option B would give 3600 C. Difficulty: easy

Q5. Which phenomenon demonstrates the transverse nature of light?

A. Interference B. Diffraction C. Polarisation D. Refraction

Answer: C

• Only transverse waves can be polarised, since polarisation restricts oscillation to a plane perpendicular to propagation. Interference and diffraction occur for both longitudinal and transverse waves. Difficulty: medium

Q6. A wire of resistance 12 ohm is cut into three equal parts and the parts are connected in parallel. The equivalent resistance is:

A. 4 ohm B. 36 ohm C. 1.33 ohm D. 12 ohm

Answer: C

• Each part has resistance 12/3 = 4 ohm. Three equal resistors R in parallel give R/3, so 4/3 = 1.33 ohm. Difficulty: medium