Section A
[1 Marks]
-
(A) 250 V
-
(B) 200 V
-
(C) 100 V
-
(D) 50 V
Explanation: To keep the particle suspended between the plates, the gravitational force must be balanced by the electric force. The weight of the particle is given by W = mg = 5 mg * 9.81 m/s² = 49.05 * 10^-3 N. The electric force due to the electric field E is given by F = qE = q(V/d). Setting these equal allows us to solve for V, leading to the correct option of 200 V.
[1 Marks]
-
(A) 6 Ω
-
(B) 8/3 Ω
-
(C) 2 Ω
-
(D) 16/3 Ω
Explanation: The correct answer is 8/3 Ω. In the provided context, the circuit arrangement suggests that the internal resistances and the setup allow for a calculation of equivalent resistance using the formulas given in the mentioned equations. By combining resistances in parallel, the effective resistance between points A and B can be determined, resulting in 8/3 Ω.
[1 Marks]
-
(A) (1/150) s
-
(B) (1/100) s
-
(C) (1/50) s
-
(D) (1/200) s
Explanation: The current I = I₀ cos(100πt) reaches its maximum value when cos(100πt) = 1, and it becomes zero when cos(100πt) = 0. The first instance of this occurs at t = 1/(200) seconds, as the frequency component 100π in the cosine function indicates that the period of the function is T = 2π/(100π) = 0.02 seconds. Consequently, the current takes a quarter of the period (T/4 = 0.02/4 = 0.005 seconds) to go from maximum to zero. Hence, the least time the current takes to decrease from its maximum value to zero is (1/200) s.
[1 Marks]
-
(A) inductive only
-
(B) inductive and capacitive respectively
-
(C) capacitive only
-
(D) capacitive and inductive respectively
Explanation: The correct answer is 'inductive and capacitive respectively'. Before resonance occurs, at low frequencies, the circuit behaves predominantly capacitively (current leads voltage), while after resonance, at high frequencies, the circuit behaves predominantly inductively (voltage leads current). Resonance occurs when the inductive reactance (XL) equals the capacitive reactance (XC), resulting in the circuit being purely resistive at that point.
[1 Marks]
-
(A) 0.8 mV
-
(B) 0.2 mV
-
(C) 1.6 mV
-
(D) 0.1 mV
Explanation: The induced emf (ε) can be calculated using the formula ε = Blv. Here, B (the magnetic field) is 0.4 G, which is equivalent to 0.4 × 10^-4 T (since 1 G = 10^-4 T), l (the length of the rod) is 0.5 m (converted from 50 cm), and v (the velocity) is 10 m/s. Plugging in the values, we get ε = (0.4 × 10^-4 T) × (0.5 m) × (10 m/s) = 0.2 mV. Therefore, the correct answer is 0.2 mV.
[1 Marks]
-
(A) 3.20 × 10¹⁴ Hz
-
(B) 4.80 × 10¹⁵ Hz
-
(C) 4.20 × 10¹⁵ Hz
-
(D) 1.18 × 10¹⁴ Hz
Explanation: To find the frequency (ν) of a photon, we can use the formula E = hν, where E is the energy in joules and h is Planck's constant (6.626 x 10^-34 J·s). First, we convert the energy from electronvolts to joules: 1.326 eV = 1.326 × 1.6 × 10^-19 J = 2.1216 × 10^-19 J. Using the formula ν = E/h, we calculate ν = (2.1216 × 10^-19 J) / (6.626 x 10^-34 J·s) = 3.20 × 10¹⁴ Hz. Therefore, the correct option is 3.20 × 10¹⁴ Hz.
[1 Marks]
-
(A) electrons only
-
(B) holes and few electrons
-
(C) holes only
-
(D) electrons and few holes
Explanation: When germanium, a tetravalent semiconductor, is doped with boron, a trivalent element, it produces a p-type semiconductor. In this situation, boron creates holes, which become the majority charge carriers, while the few remaining electrons from the germanium structure serve as minority carriers. Therefore, the correct answer is 'holes and few electrons.'
[1 Marks]
-
(A) n = 4 to n = 3
-
(B) n = 3 to n = 2
-
(C) n = 2 to n = 1
-
(D) n = 3 to n = 1
Explanation: The emitted photon will have the maximum wavelength in the transition from n = 4 to n = 3. Wavelength is inversely proportional to frequency, which is determined by the energy difference between the two levels. The larger the difference in energy levels, the higher the frequency and the shorter the wavelength of the emitted photon. Therefore, the transition n = 4 to n = 3 has the smallest energy difference, resulting in the longest wavelength among the given options.
[1 Marks]
-
(A) both the potential barrier height and the width of depletion layer increase.
-
(B) both the potential barrier height and the width of depletion layer decrease.
-
(C) the potential barrier height decreases and the width of depletion layer increases.
-
(D) the potential barrier height increases and the width of depletion layer decreases.
Explanation: When a p-n junction diode is forward biased, the potential barrier height decreases and the width of the depletion layer decreases. This is due to the applied voltage that effectively reduces the barrier height, allowing for increased current flow, as stated in the provided context.
[1 Marks]
-
(A) same atomic numbers
-
(B) same mass numbers
-
(C) same neutron number, but different atomic number
-
(D) different neutron number, and different mass number
Explanation: Isotones are defined as nuclides that have the same neutron number (N) but different atomic numbers (Z). This aligns with the context provided, which states that nuclides with the same N but different Z are called isotones.
[1 Marks]
-
(A) Assertion (A) is true, but Reason (R) is false.
-
(B) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
-
(C) Both Assertion (A) and Reason (R) are false.
-
(D) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
Explanation: Assertion (A) is true, but Reason (R) is false. The charged particle will indeed move in a helical path due to the magnetic force acting perpendicular to both the velocity and the magnetic field (which is incorrect in statement R). Thus while A accurately describes the motion, R incorrectly states that the magnetic force is along the velocity.
[1 Marks]
-
(A) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
-
(B) Assertion (A) is true, but Reason (R) is false.
-
(C) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
-
(D) Both Assertion (A) and Reason (R) are false.
Explanation: Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). Assertion (A) correctly describes the behavior of light in relation to the critical angle and the angle of the prism, while Reason (R) states that the refractive index depends on the angle of the prism, which is true. However, Reason (R) does not directly explain why the emergent ray grazes along the opposite face.
[1 Marks]
-
(A) Assertion (A) is true, but Reason (R) is false.
-
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
-
(C) Both Assertion (A) and Reason (R) are false.
-
(D) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
Explanation: Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). EM waves indeed do not require a medium for propagation, which is a defining characteristic of electromagnetic waves. While it is also true that electromagnetic waves are transverse waves, this property does not directly explain why they do not require a medium, as transverse waves can exist in certain media, unlike electromagnetic waves.
[1 Marks]
-
(A) Both Assertion (A) and Reason (R) are false.
-
(B) Assertion (A) is true, but Reason (R) is false.
-
(C) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
-
(D) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
Explanation: Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). The cut-off voltage is correctly defined in Assertion (A), and the threshold frequency defined in Reason (R) is also accurate; however, the threshold frequency does not directly explain what cut-off voltage is.
Section B
[2 Marks]
Answer: The terminal voltage (V) across the cell is given by the equation V = E - Ir, where I is the current. As R increases, the current I decreases, leading to an increase in terminal voltage. This creates a downward slope on the graph of V against R. The current I, given by I = E/(R + r), starts at a maximum when R is zero and decreases as R increases. This results in a downward trend in the graph of I against R.
[2 Marks]
Answer: The mirror equation is \(1/v + 1/u = 1/f\), where \(u\) is negative for convex mirrors. The magnification formula is \(m = -v/u\). For a convex mirror, since \(f\) is positive and \(u\) is negative, \(v\) is also positive but less than \(f\). Thus, the image distance \(v\) is between the pole and focus, and since \(|v| < |u|\), the image is diminished.
[2 Marks]
Answer: In an n-type semiconductor, the energy band diagram shows the conduction band (Ec) and valence band (Ev) with the donor energy level (Ed) just below Ec. At T > 0 K, electrons from Ed move to the conduction band, indicating increased conductivity. For a p-type semiconductor, Ev is the valence band, and the acceptor energy level (Ea) lies just above Ev. Electrons from the valence band can jump to Ea, creating holes that enhance conductivity.
[2 Marks]
Answer: Energy in stars is produced through nuclear fusion, where light nuclei combine to form a heavier nucleus, releasing energy. One example is the fusion of two hydrogen nuclei (1 1 H + 1 1 H → 2 1 H + e+ + ν), which generates energy of 0.42 MeV. Another example is the fusion of deuterium (2 1 H) nuclei to produce helium (2 1 H + 2 1 H → 3 2 H + n + 3. In stars, this process occurs at extremely high temperatures, driving stellar energy output.
[2 Marks]
Answer: To find the focal length of the third lens, we use the formula for the combination of thin lenses: 1/f = 1/f₁ + 1/f₂ + 1/f₃. Given f₁ = 10 cm (convex) and f₂ = -15 cm (concave) with the combination focal length f = 12 cm, we substitute: 1/12 = 1/10 - 1/15 + 1/f₃. Solving this gives us f₃ = 30 cm, which is a convex lens. If the lenses were thick, the effective focal length would require different calculations due to additional lens thickness considerations.
Section C
[3 Marks]
Answer: To find the currents flowing through each cell, we first determine the equivalent voltage (V_eq) across the cells and then apply Ohm's law. For cells in parallel, the voltage across each cell is the same. The total current is the sum of the individual currents through each cell, given by I = E/R. For cell A, I_A = (2 V) / (5 Ω) = 0.4 A; for cell B, I_B = (3 V) / (5 Ω) = 0.6 A; for cell C, I_C = (5 V) / (1 Ω) = 5 A. Therefore, the currents are I_A = 0.4 A, I_B = 0.6 A, and I_C = 5 A.
[3 Marks]
Answer: The Biot-Savart Law in vector form states that the magnetic field dB at a point in space due to an infinitesimal current element Idl is given by dB = (μ₀/4π) * (I dl × r̂) / r², where r̂ is the unit vector pointing from the current element to the field point and r is the distance from the current element to the point. For the two coils, Coil A, with current I, produces a magnetic field at the center B₁ = (μ₀I)/(2R). Coil B, with current √3 I, produces B₂ = (μ₀√3I)/(2R). Since they are oriented perpendicular, the net magnetic field will be B_net = B₁ + B₂, directed along the axis defined by their orientation.
[3 Marks]
Answer: Faraday's law of electromagnetic induction states that the induced electromotive force (emf) in a circuit is directly proportional to the rate of change of magnetic flux through that circuit. Mathematically, this is expressed as ε = -dΦB/dt. The negative sign indicates the direction of induced emf, which opposes the change in magnetic flux, in accordance with Lenz's law. For example, when a magnet approaches a conducting loop, the induced current flows in a direction that opposes the magnet's motion, demonstrating conservation of energy, as it prevents energy depletion by opposing external changes.
[3 Marks]
Answer: Conduction current refers to the flow of charge carriers (like electrons) in a conducting medium, resulting in the creation of a magnetic field. Displacement current, on the other hand, arises from time-varying electric fields in regions where there are no physical charge carriers but still generates a magnetic field. A similarity between the two is that both contribute to the continuity equation of electric fields and magnetic fields. However, a key dissimilarity is that conduction current occurs in conducting materials, while displacement current occurs in non-conducting spaces, such as the gap between capacitor plates during charging. The concept of displacement current allows us to demonstrate electromagnetic waves; as an accelerating charge produces a varying electric field, it results in a changing magnetic field, propagating energy as electromagnetic waves through free space.
[3 Marks]
Answer: The work function (Φ) of a metal is the minimum energy required for an electron to escape from its surface, typically measured in electronvolts (eV). This energy directly correlates with the threshold frequency needed for photoelectric emission. By plotting a graph of stopping potential versus the frequency of incident radiation, we can determine the value of the work function—the y-intercept of the linear graph will give the work function in eV. In part (b), using the formula, E = hf, where h is Planck's constant (4.14 × 10^-15 eV. s) and f is the frequency derived from the stopping potential, we calculate the wavelength (λ) of light using the relation λ = c/f, giving us approximately 520 nm.
[3 Marks]
Answer: In a p-n junction, when p-type and n-type semiconductors are joined, electrons from the n-side diffuse into the p-side and recombine with holes, leading to the formation of a depletion layer. This layer consists of immobile ion-cores, creating a localized region devoid of charge carriers. The potential barrier is established due to this depletion layer, which prevents further diffusion of charge carriers until sufficient voltage is applied. Junction diodes permit current flow primarily during forward bias, making them ideal rectifiers since they allow current to flow efficiently in one direction.
Section D
[5 Marks]
Answer: (i) A moving coil galvanometer works on the principle that a current-carrying coil placed in a magnetic field experiences a torque. The coil, suspended between the poles of a permanent magnet, carries current which produces a magnetic field that interacts with the magnet's field causing the coil to rotate. This rotation is opposed by a spring that provides restoring torque. The angle of deflection is proportional to the current flowing through the coil.
Purpose of radial magnetic field: The radial magnetic field ensures that the coil experiences a constant torque for every position of the coil, making the deflection directly proportional to the current.
Purpose of soft iron core: The soft iron core increases the magnetic field strength, making the galvanometer more sensitive.
(ii) Current sensitivity of a galvanometer is defined as the deflection produced in the galvanometer needle per unit current passing through the coil. Mathematically, current sensitivity = deflection / current.
Increasing the current sensitivity may not necessarily increase the voltage sensitivity because voltage sensitivity depends on both current sensitivity and the resistance of the galvanometer. Even if current sensitivity increases, if the resistance remains high, the voltage sensitivity may not improve accordingly. Voltage sensitivity = current sensitivity * resistance of the galvanometer.
[5 Marks]
Answer: (i) An astronomical telescope is designed to observe distant objects. When the final image is formed at infinity, the rays from a distant object come in parallel and enter the objective lens. The objective lens converges these rays to form a real and inverted image at its focal point, which acts as the object for the eyepiece. The eyepiece then magnifies this image to form a virtual image at infinity, allowing comfortable viewing. The magnification power (m) is given by the formula m = (L/f0) * (L/fe), where L is the distance between the focal points of the lenses, f0 is the focal length of the objective, and fe is the focal length of the eyepiece. (ii) For the compound microscope, the total magnification (M) is 20 and the eyepiece magnification (me) is 5. The formula for total magnification can be expressed as M = mo * me, where mo is the magnification of the objective. Thus, mo = M/me = 20/5 = 4. Given the distance (L) between the objective and eyepiece is 14 cm and the least distance of distinct vision (d) is 25 cm, we can relate this to the focal lengths: L = f0 + fe, giving us f0 + fe = 14 cm. The equation for total magnification can also be written as M = (L/f0) + (d/fe). Substituting known values, we get 20 = (14/f0) + (25/fe). By solving these two equations simultaneously, we can find the values of f0 and fe.