Section A
[1 Marks]
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(A) Spherical surface
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(B) Paraboloidal surface
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(C) Plane
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(D) Ellipsoidal surface
Explanation: For a group of charges with a non-zero net charge (Σq ≠ 0), the behavior of the electric field at large distances can be approximated as that of a point charge. Thus, the equipotential surfaces tend to be spherical, similar to those of a single charge, as they are concentric spherical surfaces centered around the net charge of the group.
[1 Marks]
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(A) remain stationary
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(B) move along z-axis
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(C) move along y-axis
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(D) move along x-axis
Explanation: The loop will remain stationary because the magnetic field is uniform and perpendicular to the plane of the loop. According to the Lorentz force law, the force on each segment of the loop due to the magnetic field is balanced by the opposite forces on the other segments, resulting in a net force of zero.
[1 Marks]
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(A) 15.0 V
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(B) 12.5 V
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(C) 17.5 V
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(D) 22.5 V
Explanation: To find the induced emf (ε), we can use the formula ε = -L(dI/dt). Here, L = 15 mH = 0.015 H, and dI/dt = (final current - initial current) / time = (4 A - 0 A) / 0.004 s = 1000 A/s. Thus, ε = -0.015 H * 1000 A/s = -15 V. The negative sign indicates the direction of the induced emf, but the magnitude is 15 V, so the correct answer is 15.0 V.
[1 Marks]
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(A) both l and A are decreased
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(B) l is decreased and A is increased
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(C) both l and A are increased
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(D) l is increased and A is decreased
Explanation: The self-inductance L of a solenoid is given by the formula L = µr µ0 n² A l. In this formula, increasing the area A (cross-section) will lead to an increase in L, while decreasing the length l also leads to an increase in L as inversely related to l. Thus, option 'l is decreased and A is increased' will result in an increase in self-inductance.
[1 Marks]
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(A) Infrared rays
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(B) Gamma rays
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(C) Radio waves
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(D) Microwaves
Explanation: Gamma rays have the highest frequency among the options provided. According to the context, gamma rays lie in the upper frequency range of the electromagnetic spectrum, while infrared rays, microwaves, and radio waves have lower frequencies. This categorization is based on the increasing wavelength from gamma rays to radio waves.
[1 Marks]
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(A) 4
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(B) 1/2
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(C) 1/4
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(D) 2
Explanation: The distance of closest approach for charged particles is inversely proportional to the charge of the particle. A proton has a charge of +1e while an alpha particle has a charge of +2e. Hence, the distance of closest approach of the proton (d_p) relative to the alpha particle (d_a) can be given by d_p/d_a = (Z1*Z2)/ (Z'1*Z'2) = 1/(2) = 1/2. Therefore, the correct ratio is 1/2.
[1 Marks]
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(A) increases by 87.5%
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(B) decreases by 87.5%
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(C) increases by 43.75%
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(D) decreases by 43.75%
Explanation: In Bohr's model, the period of revolution T is proportional to n3. When the electron moves from n = 2 to n = 1, the period changes from T₂ to T₁ such that T₂ / T₁ = (2 / 1)3 = 8. Thus, T₁ = T₂ / 8, which means the period decreases by (1 - 1/8) * 100% = 87.5%. Therefore, the period of revolution decreases by 87.5% during this transition.
[1 Marks]
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(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
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(B) Assertion (A) is false and Reason (R) is also false.
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(C) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
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(D) Assertion (A) is true, but Reason (R) is false.
Explanation: Assertion (A) is false because electrons in the conduction band actually have higher energy than those in the valence band. Reason (R) is true; however, it does not explain Assertion (A) since the position of the donor energy level being above the valence band does not make the electrons in the conduction band have lesser energy.
[1 Marks]
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(A) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
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(B) Assertion (A) is false and Reason (R) is also false.
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(C) Assertion (A) is true, but Reason (R) is false.
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(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). The photoelectric effect indeed supports the particle nature of light, as photons are quantized packets of energy that can eject electrons from materials. However, the photoelectric current is proportional to the intensity of light, and while frequency does play a role in determining if photoemission occurs (i.e., it must exceed a threshold frequency), it does not directly describe the relationship of current to frequency in terms of particle nature.
[1 Marks]
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(A) Assertion (A) is false and Reason (R) is also false.
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(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
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(C) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
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(D) Assertion (A) is true, but Reason (R) is false.
Explanation: Assertion (A) is true, but Reason (R) is false. While it is true that a convex lens can appear to disappear in a liquid if the refractive index of the lens matches that of the liquid, the reason provided is misleading. The lens does not literally disappear; it becomes less visible due to the matching refractive indices, causing light to refract similarly through both the lens and the liquid.
Section B
[2 Marks]
Answer: To find the focal length of the objective lens (f0) in an astronomical telescope, we use the magnification formula m = L/f0 + d/fe. Given that m = 24 and L = 150 cm. Assuming that fe is the focal length of the eyepiece. We rearrange to find f0. If we assume fe is small compared to L, we can simplify to f0 = L/m, resulting in 150 cm / 24 = 6.25 cm (approximately).
[2 Marks]
Answer: A simple microscope creates a virtual image at the near point, allowing the angular size of the image and object to be the same while providing magnification. This occurs due to the lens's focal properties. Convex mirrors typically yield virtual images but can produce real images when their configuration and positioning allow, such as using multiple mirrors to focus light to a point, contrary to their usual behavior.
[2 Marks]
Answer: Circuit Diagram: A Wheatstone bridge is a diamond-shaped circuit with four resistors arranged on each side: R1, R2, R3, and R4. A galvanometer is connected between the two middle points B and D, and a voltage source is connected across points A and C.
Condition for No Current in Galvanometer: When the Wheatstone bridge is balanced, no current flows through the galvanometer. The condition is given by:
R1 / R2 = R3 / R4
This means the ratio of resistances in one pair of opposite arms is equal to the ratio in the other pair. Under this condition, the potential difference across the galvanometer is zero, so no current flows through it.
Section C
[3 Marks]
Answer: To determine the currents in branches AB, AC, and BC, we apply Kirchhoff's loop and junction rules. In loop ADBA, the equation -I1 R1 + I2 R2 = 0 indicates that the voltage drops are balanced by the voltage sources. Using another loop, CBDC, where we rearrange to meet I3 = I1 and I4 = I2, we find that I2 R4 - I1 R3 = 0. At junction a, the current entering (I3) equals the sum of currents leaving (I1 + I2). By solving these equations simultaneously, we can find the specific values for I1, I2, and I3, which provide the required currents in each branch of the circuit.
[3 Marks]
Answer: Two long straight parallel conductors carrying currents exert a magnetic force on each other due to the magnetic field generated by the currents. According to the Biot-Savart Law, the magnetic field (B) generated by a conductor at a distance (d) from it is proportional to the current (I) flowing through the conductor. If two conductors carry currents I1 and I2 in opposite directions, the force per unit length (f) between them can be derived using the formula: f = (μ₀/2π) * (I1 * I2/d). This shows that conductors with currents in opposite directions repel each other, unlike like currents which attract.
[3 Marks]
Answer: The circuit element 'X' in which the current leads the voltage by π/2 is a capacitor. The formula for its reactance is given by Xc = 1/(ωC), where ω is the angular frequency and C is the capacitance. Graphically, the capacitive reactance (Xc) decreases with an increase in frequency, appearing as a downward slope on a graph with frequency on the x-axis and reactance on the y-axis. In an AC circuit, a capacitor allows alternating current to pass while blocking direct current, meaning it can store and release energy cyclically. In a DC circuit, after charging, a capacitor eventually blocks current flow entirely.
[3 Marks]
Answer: To obtain the V-I characteristics of a p-n junction diode, two circuit arrangements are used: one for forward biasing and another for reverse biasing. In the forward bias setup, the positive terminal of the power supply connects to the p-side of the diode, allowing current to flow which increases rapidly after a threshold voltage (typically around 0.7V for silicon diodes). In reverse bias, the n-side is connected to the positive terminal, and the current remains very small (reverse saturation current) until breakdown occurs at high reverse voltages. The forward V-I characteristic shows exponential growth, while the reverse characteristic remains mostly constant until the breakdown region.
[3 Marks]
Answer: Energy band diagrams illustrate the differences between insulators, semiconductors, and conductors based on their electronic structures. Insulators have a large energy band gap (Eg > 3 eV) between the valence band and conduction band, meaning no electrons can be thermally excited to conduct electricity, resulting in no electrical conductivity. Semiconductors possess a smaller band gap, allowing for some electrons to jump to the conduction band with thermal energy, leading to limited conduction. Conductors have overlapping bands with no gap, allowing electrons to move freely, facilitating maximum conductivity.
Section D
[5 Marks]
Answer: (i) Consider a thin conducting spherical shell of radius R with a total charge +Q uniformly distributed on its surface. We need to find the electric field at a point at a distance r from the centre.
(a) Electric field inside the shell (r < R):
By Gauss's theorem, take a spherical Gaussian surface inside the shell with radius r. Since the shell is conducting and charges reside on surface, no charge is enclosed inside this Gaussian surface. Charge enclosed, q = 0.
According to Gauss's law, E * 4 π r2 = q/ε0 = 0 which implies E = 0
Therefore, the electric field inside the shell is zero.
(b) Electric field outside the shell (r > R):
Take a spherical Gaussian surface of radius r outside the shell. The enclosed charge q = +Q = σ * 4 π R2. Applying Gauss's law,
E * 4 π r2 = q/ε0
=> E = q/(4 π ε0 r2) = (σ * 4 π R2)/(4 π ε0 r2) = (σ R2)/(ε0 r2)
This shows the electric field outside the shell behaves as though all charge is concentrated at the centre.
(ii) For the same surface charge density σ:
Electric field due to a conducting plate (a charged infinite plane) is Econducting = σ/ε0
Electric field due to a nonconducting sheet (with uniform charge on both sides) is Enonconducting = σ/(2 ε0)
This happens because in a conducting plate, all charges reside only on one surface and the field lines emerge from one side, while in a nonconducting sheet, charges are spread on both sides and field lines emerge from both sides, effectively halving the field on each side.
Thus, electric field with conducting plate is twice that of a nonconducting sheet for the same σ.