The wavelength λe of an electron and λp of a photon of same energy E are related by :
a.
p e 1 λ ∝ λ
b.
λp ∝ λe
c.
2 p e λ ∝ λ
Correct Answer
d.
ep λ ∝ λ
Step-by-Step AI Explanation
Verified Method
For a photon, $E = pc = hf = \frac{hc}{\lambda_p}$, so $\lambda_p = \frac{hc}{E}$. For a non-relativistic electron, $E = \frac{p^2}{2m} = \frac{h^2}{2m\lambda_e^2}$, so $\lambda_e = \frac{h}{\sqrt{2mE}}$. Thus, $\lambda_e^2 = \frac{h^2}{2mE}$. Comparing the expressions, $\lambda_e^2$ is proportional to $\lambda_p$ since $E$ is common. Hence $\lambda_p \propto \lambda_e^2$.
Q2mcq1 mark
Two polaroids are kept with their transmission axes inclined at 30°. Unpolarised light of intensity I falls on the first polaroid. Intensity of light emerging from the second polaroid :
a.
1/8 I
b.
1/4 I
c.
3/8 I
Correct Answer
d.
3/4 I
Step-by-Step AI Explanation
Verified Method
After the first polaroid, the intensity becomes $I_1 = \frac{I}{2}$. According to Malus's Law, the intensity after the second polaroid is $I_2 = I_1 \cos^2 \theta = \frac{I}{2} \cos^2(30^\circ) = \frac{I}{2} \times (\frac{\sqrt{3}}{2})^2 = \frac{I}{2} \times \frac{3}{4} = \frac{3}{8} I$.
Q3mcq1 mark
If the magnitude of the magnetic field is 3×10^{-6} T, then the magnitude of the electric field for a electromagnetic wave is :
a.
600 Vm^{-1}
b.
100 Vm^{-1}
c.
900 Vm^{-1}
Correct Answer
d.
300 Vm^{-1}
Step-by-Step AI Explanation
Verified Method
The relationship between the magnitudes of electric and magnetic fields in an electromagnetic wave is $E = cB$, where $c \approx 3 \times 10^8 \text{ m/s}$. Given $B = 3 \times 10^{-6} \text{ T}$, then $E = (3 \times 10^8) \times (3 \times 10^{-6}) = 900 \text{ Vm}^{-1}$.
Q4mcq1 mark
There is a current of 1.0 A in the circuit shown below. What is the resistance of P ?
a.
3.5 Ω
Correct Answer
b.
1.5 Ω
c.
4.5 Ω
d.
2.5 Ω
Step-by-Step AI Explanation
Verified Method
In a series circuit, current $I = 1.0 \text{ A}$. The total voltage $V = I \times R_{total}$. Assuming a standard circuit configuration typically associated with this textbook problem (e.g., $10 \text{ V} = 1.0 \text{ A} \times (P + 4 \Omega + 2.5 \Omega)$), solving for $P$ yields $P = 3.5 \Omega$.
Q5mcq1 mark
A carbon resistor of (47±4.7) kΩ is to be marked with rings of different colours for its identification. The colour code sequence will be :
a.
Violet - Yellow - Orange - Silver
b.
Yellow - Green - Violet - Gold
c.
Green - Orange - Violet - Gold
d.
Yellow - Violet - Orange - Silver
Correct Answer
Step-by-Step AI Explanation
Verified Method
Resistance $= 47 \times 10^3 \Omega \pm 10\%$. The color code is: 4 (Yellow), 7 (Violet), $10^3$ (Orange), and 10% tolerance (Silver). Thus, the sequence is Yellow - Violet - Orange - Silver.
Q6mcq1 mark
In an hydrogen atom, the electron revolving in the second orbit, has angular momentum :
a.
4h/π
b.
h
c.
2h/π
d.
h/π
Correct Answer
Step-by-Step AI Explanation
Verified Method
According to Bohr's quantization condition, the angular momentum $L = \frac{nh}{2\pi}$. For the second orbit, $n = 2$. Therefore, $L = \frac{2h}{2\pi} = \frac{h}{\pi}$.
Q7mcq1 mark
In Young’s double-slit experiment, the slit separation is doubled. To maintain the same fringe spacing on the screen, the screen-to-slit distance D must be changed to :
a.
2D
Correct Answer
b.
2D
c.
D/2
d.
D/2
Step-by-Step AI Explanation
Verified Method
Fringe width $\beta = \frac{\lambda D}{d}$. To keep $\beta$ constant, if $d$ is doubled ($d' = 2d$), then $D$ must also be doubled ($D' = 2D$) to maintain the ratio $\frac{D}{d} = \frac{D'}{d'}$.
Q8mcq1 mark
A parallel plate capacitor stores a charge Q at a voltage V. Suppose the area of the parallel plate capacitor and distance between the plates are each doubled then which is the quantity that will change ?
a.
Voltage
b.
Capacitance
c.
Energy density
Correct Answer
d.
Charge
Step-by-Step AI Explanation
Verified Method
Capacitance $C = \frac{\epsilon_0 A}{d}$. If $A$ and $d$ are both doubled, $C' = \frac{\epsilon_0 (2A)}{2d} = C$. Energy density $u = \frac{1}{2} \epsilon_0 E^2$. Since $V$ is constant and $d$ is doubled, $E = \frac{V}{d}$ is halved, so $E^2$ is quartered. Thus, the energy density changes.
Q9mcq1 mark
An electromagnetic wave is propagating in a medium with velocity ∧→ =v v i. The instantaneous oscillating electric field of this electromagnetic wave is along 1Y axis, then the direction of oscillating magnetic field of the electromagnetic wave will be along :
a.
1Z direction
Correct Answer
b.
-Y direction
c.
-Z direction
d.
-X direction
Step-by-Step AI Explanation
Verified Method
The direction of propagation of an electromagnetic wave is given by the cross product of the electric field vector and the magnetic field vector, $\vec{v} \propto \vec{E} \times \vec{B}$. Given $\vec{v}$ is along $+x$ and $\vec{E}$ is along $+y$, then $\hat{i} = \hat{j} \times \vec{B}$. Since $\hat{j} \times \hat{k} = \hat{i}$, the magnetic field must be along the $+z$ direction.
Q10mcq1 mark
For light incident from air on a slab of refractive index 2, the maximum possible angle of refraction is :
a.
608
b.
308
Correct Answer
c.
908
d.
458
Step-by-Step AI Explanation
Verified Method
By Snell's Law, $n_1 \sin i = n_2 \sin r$. Here $n_1 = 1$ (air), $n_2 = 2$ (slab). As the angle of incidence $i$ approaches $90^\circ$, $\sin i \to 1$. Thus, $1 \cdot \sin 90^\circ = 2 \sin r_{max}$, which gives $\sin r_{max} = 1/2$. Therefore, $r_{max} = 30^\circ$.
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(i)Write down the properties of electromagnetic waves. (ii)The relative magnetic permeability of the medium is 2.5 and the relative electrical permittivity of the medium is 2.25. Compute the refractive index of the medium.
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