Section A
[1 Marks]
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(A) 2R
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(B) R/8
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(C) R/2
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(D) R
Explanation: The resistance R of a wire is given by R = ρ * (length) / (cross-sectional area). Since Q has half the length of P, length_Q = L/2. The diameter of Q is twice that of P, so radius_Q = 2 * radius_P. The cross-sectional area A is proportional to the square of the diameter, so A_Q = π * (2d)^2 / 4 = 4 * A_P. Therefore, resistance of Q is R_Q = ρ * (L/2) / (4 * A_P) = (1/8) * (ρ * L / A_P) = R / 8.
[1 Marks]
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(A) 0.45 C
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(B) 0.30 C
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(C) 3.0 mC
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(D) 1.5 C
Explanation: When the coil is inserted into the magnetic field, the change in magnetic flux induces an emf, which causes a current and thus a charge to flow. The total change in magnetic flux linkage is N × B × A since the coil is initially out of field (zero flux) and is instantaneously inserted (flux changes to N × B × A). Here, N = 100 turns, B = 90 mT = 0.09 T, A = 0.05 m².
Change in flux linkage = 100 × 0.09 × 0.05 = 0.45 Wb.
The induced emf E = -d(φ)/dt, but since insertion is instant, the charge Q induced can be found by Q = Change in flux linkage / Resistance = 0.45 / 1.5 = 0.3 C.
However, since the insertion is considered instantaneous, and the calculation needs the total charge induced, correctly, the charge induced Q = emf × time / resistance, but with instantaneous insertion Q = ΔΦ / R = 0.45 / 1.5 = 0.3 C. But this corresponds to 0.3 C which is 300 mC.
Considering the problem context and options, it appears there may be a miscalculation if 3.0 mC is correct. To find the precise value for the charge induced (Q), the formula is Q = N × B × A / R
= 100 × 0.09 × 0.05 / 1.5 = 0.3 C.
Since 0.3 C is an option, but options are 1.5 C, 3.0 mC, 0.30 C, 0.45 C, the correct answer is 0.30 C.
Therefore, the correct option is 0.30 C.
[1 Marks]
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(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
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(B) Assertion (A) is true, but Reason (R) is false.
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(C) Both Assertion (A) and Reason (R) are false.
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(D) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
Explanation: Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). When a magnet is moved into a coil with a closed circuit, an induced current is generated that creates a magnetic field opposing the motion of the magnet, following Lenz's law. This opposing force makes it difficult to move the magnet into the coil, confirming both the assertion and the reason.
[1 Marks]
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(A) X-rays, microwaves, UV radiation
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(B) X-rays, UV radiation, microwaves
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(C) UV radiation, microwaves, X-rays
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(D) Microwaves, UV radiation, X-rays
Explanation: The correct option is 'X-rays, UV radiation, microwaves'. This is because, according to the electromagnetic spectrum, X-rays have a shorter wavelength and higher energy compared to UV radiation, which also has higher energy than microwaves, which possess the longest wavelength of the three. The relationship between wavelength and photon energy indicates that as wavelength decreases, energy increases.
[1 Marks]
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(A) H/4
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(B) 4H/3
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(C) H
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(D) 3H/4
Explanation: The apparent depth (h1) when viewed from above is calculated as h2 (real depth) divided by the refractive index (n) of the medium. Here, h2 is H and n is 4/3. Therefore, h1 = H / (4/3) = 3H/4. Hence, the correct answer is 3H/4.
[1 Marks]
Explanation: The correct figure will show that the binding energy per nucleon is nearly constant (around 8.0 MeV) for nuclei with mass numbers between 30 and 170, as mentioned in the context. It should also illustrate that the binding energy has a maximum around A = 56. This reflects the stability of medium-sized nuclei.
[1 Marks]
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(A) the barrier height and the depletion layer width both increase.
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(B) the barrier height increases and the depletion layer width decreases.
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(C) the barrier height and the depletion layer width both decrease.
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(D) the barrier height decreases and the depletion layer width increases.
Explanation: When a p-n junction diode is forward biased (p-side connected to positive terminal and n-side connected to negative), the external voltage reduces the potential barrier at the junction. This decrease in barrier height allows charge carriers to cross the junction more easily, reducing the width of the depletion region. Hence, both the barrier height and the depletion layer width decrease under forward bias, resulting in increased current flow.
[1 Marks]
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(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
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(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
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(C) Assertion (A) is false and Reason (R) is also false.
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(D) Assertion (A) is true, but Reason (R) is false.
Explanation: Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A). The deflection of the galvanometer is indeed directly proportional to the current, but this relationship arises from the principles of electromagnetism governing the galvanometer and not solely due to the coil being suspended in a magnetic field.
[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 and Reason (R) is the correct explanation of Assertion (A).
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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, but Reason (R) is not the correct explanation of Assertion (A).
Explanation: Assertion (A) is false because the potential energy (U) of an electron in a hydrogen atom is negative, as shown in the context where U = -e^2 / (4πε0 r). Reason (R) is also false since the total energy of the electron in a hydrogen atom is also negative (E = -e^2 / (8πε0 r)). Therefore, both the assertion and reason are incorrect.
Section B
[2 Marks]
Answer: The emf (E) of the battery can be calculated using the formulas derived from Ohm's law and circuit principles. For 2A current, V = E - Ir gives E = 5V + (2A)(r). For 4A current, V = E - Ir gives E = 4V + (4A)(r). From these, we can find two equations. Solving these equations yields E = 10V and r = 2.5Ω.
[2 Marks]
Answer: To find the width of the slit (a), we use the formula for the angle of the first minimum in single slit diffraction: a sin(θ) = nλ, where n=1 for the first minimum. Here, λ = 600 nm = 600 x 10^-9 m and θ = 30°. Thus, a sin(30°) = 1(600 x 10^-9). Since sin(30°) = 0.5, we have a x 0.5 = 600 x 10^-9. Therefore, the width of the slit a = (600 x 10^-9) / 0.5 = 1.2 x 10^-6 m or 1.2 μm.
Section C
[3 Marks]
Answer: To find the equivalent emf (E_eq) and internal resistance (r_eq) of two batteries in parallel, we can use the formulas:\n\n1. **Equations for equivalent emf and internal resistance**: \n 1/E_eq = 1/E_1 + 1/E_2 \n E_eq = (E_1*r_2 + E_2*r_1) / (r_1 + r_2)\n\n2. For the given values, \n E_1 = 3V, r_1 = 0.2Ω; E_2 = 6V, r_2 = 0.4Ω.\n \nFinding E_eq: \nE_eq = (3*0.4 + 6*0.2) / (0.2 + 0.4) = (1.2 + 1.2) / 0.6 = 4V.\n\nFinding r_eq: \n1/r_eq = 1/r_1 + 1/r_2 = 1/0.2 + 1/0.4 = 5 + 2.5 = 7.5 \n=> r_eq = 1/7.5 = 0.133Ω\n\nTo find current (I) through the 4Ω resistor, we use Ohm's Law: \nI = E_eq / (R + r_eq) = 4V / (4Ω + 0.133Ω) = 4 / 4.133 ≈ 0.967A.
[3 Marks]
Answer: When monochromatic light strikes a metal surface, photoelectrons are emitted with varying kinetic energy because they absorb photons of energy that can slightly differ due to their interactions with the metal lattice. The saturation current varies with intensity because a higher intensity means more photons are available, leading to more emitted photoelectrons. Moreover, as the wavelength increases (and frequency decreases), fewer electrons can absorb sufficient energy to escape. Beyond a certain wavelength, no photoemission occurs as the energy of incoming photons falls below the threshold energy required to free electrons.
[3 Marks]
Answer: To study the V-I characteristics of a p-n junction diode, a circuit arrangement is made which includes a power source, a diode, a voltmeter, and an ammeter connected in series. In forward bias, when the voltage is applied in the direction of conventional current, the current increases exponentially after reaching the threshold (cut-in) voltage. In reverse bias, the current is almost negligible until breakdown occurs, at a much higher voltage. Typical V-I characteristics exhibit a steep rise in forward bias while showing very little current in reverse bias. From these characteristics, we can glean information about the forward and reverse breakdown voltages, as well as the ideality factor of the diode, which indicates its efficiency.
Section D
[5 Marks]
Answer: To determine the electric fields at a distance of 3R from the centers of the shells, we can use the principle that the electric field produced by a charged spherical shell outside its surface acts as if all the charge were concentrated at its center. For shell A, the net charge is +6q; thus, the electric field E_A at 3R is k(6q)/(3R)². For shell B, the net charge is -4q, giving E_B = k(-4q)/(3R)². Shell C has a net charge of 14q, resulting in E_C = k(14q)/(3R)². The ordering of the magnitudes of E will be E_C > E_A > E_B due to their respective charges. For the semicircle, the work done can be derived from the concept of electric potential energy. Given the symmetrical nature and the charges involved, the work done is calculated as the potential difference multiplied by the charge moved, implying that the calculation must take into account the original configuration of the system with respect to potential.
[5 Marks]
Answer: Coherent sources are light sources that emit waves with a constant phase difference and the same frequency. They are critical for observing interference patterns because they maintain a stable phase relationship, allowing for consistent constructive and destructive interference. In contrast, lights from independent sources are incoherent as they might have varying phase relationships and different frequencies, leading to unpredictable interference patterns. For the two-slit arrangement, using the formula for fringe separation, the distance between adjacent bright fringes can be calculated as 1.5 mm, while the angular width of the first bright fringe is approximately 0.043 degrees.
[5 Marks]
Answer: A wavefront is defined as the surface formed by connecting all points that are in the same phase of a wave, such that all points on the wavefront vibrate together. When an incident plane wave strikes a convex lens, the central part of the wavefront is delayed more than its edges due to varying thickness of the lens. As the light passes through, the wavefront transforms from a planar shape to a spherical shape, converging at the focus point F. The diagram shows the incident planar wavefront approaching the lens, and the refraction creates a new wavefront that curves towards the focal point.\n\nFor the spherical glass ball, considering a distant light source, the incoming rays can be approximated as parallel. The glass ball refracts these rays according to the lens-maker's formula: 1/f = (n-1)(1/R1 - 1/R2). For a radius of 15 cm and a refractive index of 1.5, the effective focal length f can be calculated. A ray diagram will show these incoming parallel rays converging, forming a real image on the opposite side of the ball. The exact image position can be determined from the geometry of the setup. Overall, the incident rays converge at a point determined by the properties of the lens.