JEE Physics Electricity Optics and Modern Physics MCQs Set 01

Answer: (b) \( \frac{8\pi r^2 E}{5\rho \ell} \)

Explanation:
1. \( R_1 = \frac{\rho (\ell/2)}{\pi r^2} = \frac{\rho \ell}{2\pi r^2} \).
2. \( R_2 = \frac{\rho (\ell/2)}{\pi (2r)^2} = \frac{\rho \ell}{8\pi r^2} \).
3. \( R = R_1 + R_2 = \frac{5\rho \ell}{8\pi r^2} \), so \( I = \frac{E}{R} = \frac{8\pi r^2 E}{5\rho \ell} \).

Teacher's Note:
1. Resistance varies inversely with the cross-sectional area, so doubling the radius cuts resistance to one fourth.
2. Conductors in one line carry the same current and their resistances add.

Answer: (c) \( q_1 \) negative, \( q_2 \) positive\, \( |q_1| > |q_2| \)

Explanation:
1. Just to the left of \( q_1 \) the field is positive (towards +x, i.e. towards \( q_1 \)), so \( q_1 \) is negative. Just to the right of \( q_2 \) the field is positive (away from \( q_2 \)), so \( q_2 \) is positive.
2. Between the charges the field is negative (towards \( q_1 \) and away from \( q_2 \)), which agrees.
3. Far to the right the field is negative: the net charge is negative and the field changes sign beyond \( q_2 \), so \( |q_1| > |q_2| \).

Teacher's Note:
1. A field pointing towards a charge shows it is negative\, pointing away shows it is positive.
2. A zero of the field outside the pair lies on the side of the smaller charge.

Answer: (b) i-q, ii-q, iii-p, iv-r

Explanation:
1. The capacitance becomes \( KC_0 \), so it increases: iii-p.
2. The battery is disconnected, so the charge stays constant: iv-r.
3. \( V = Q/C \) decreases: ii-q. Energy \( U = \frac{Q^2}{2C} \) decreases: i-q.
4. Matching: i-q, ii-q, iii-p, iv-r.

Teacher's Note:
1. When the battery stays connected the potential difference is fixed\, when it is removed the charge is fixed.
2. Use \( U = Q^2/2C \) when charge is constant.

Answer: (c) \( \frac{2v_0}{B_0\alpha} \)

Explanation:
1. \( \vec{v} \times \vec{B} = v_0 \hat{i} \times (-B_0 \hat{k}) = v_0B_0 \hat{j} \). For the electron the force \( = -e(\vec{v}\times\vec{B}) \) points along \( -\hat{j} \), so the circle's centre lies on the negative y-axis.
2. Radius \( r = \frac{mv_0}{eB_0} = \frac{v_0}{\alpha B_0} \), centre at \( (0, -r, 0) \).
3. The circle meets the y-axis again at \( (0, -2r, 0) \), so \( y = 2r = \frac{2v_0}{\alpha B_0} \).

Teacher's Note:
1. Use the right-hand rule and remember that the force on an electron is opposite to \( \vec{v} \times \vec{B} \).
2. The particle starts at the origin on the circle, so the diameter along the y-axis is 2r.

Answer: (d) zero

Explanation:
1. The x-z plane contains the axis (z-axis) of the loop.
2. By symmetry the magnetic field at every point of this plane lies in the plane itself (components along x and z only), so the component normal to the plane, \( B_y \), is zero.
3. Hence the flux through the x-z plane is zero.

Teacher's Note:
1. Flux depends on the component of B perpendicular to the surface.
2. For a loop, any plane that contains its axis carries field lines only along the plane.

Answer: (d) Both electric and magnetic field vectors are parallel to each other and perpendicular to the direction of propagation of the wave

Explanation:
1. Electromagnetic waves travel in vacuum, and E and B are in phase, with equal energy densities in the two fields.
2. The electric and magnetic fields are perpendicular to each other and both are perpendicular to the direction of propagation.
3. So statement (d), which says they are parallel to each other, is false.

Teacher's Note:
1. The three vectors \( \vec{E}, \vec{B}, \vec{k} \) are mutually perpendicular.
2. Check each option against this basic geometry of EM waves.

Answer: (b) decreases with mass number at high mass numbers

Explanation:
1. Binding energy per nucleon peaks near mass number 56 and then falls slowly for heavier nuclei.
2. Splitting a heavy nucleus gives fragments of intermediate mass with higher binding energy per nucleon.
3. Energy is released because the binding energy per nucleon decreases with mass number at high mass numbers.

Teacher's Note:
1. Fission releases energy for heavy nuclei and fusion for light nuclei.
2. Study the shape of the binding-energy curve.

Answer: (a) \( \frac{1}{7\lambda} \)

Explanation:
1. \( N_A = N_0 e^{-8\lambda t},\ N_B = N_0 e^{-\lambda t} \).
2. \( \frac{N_A}{N_B} = e^{-7\lambda t} = e^{-1} \).
3. So \( 7\lambda t = 1 \) and \( t = \frac{1}{7\lambda} \).

Teacher's Note:
1. Take the ratio first\, the common \( N_0 \) cancels.
2. The exponents subtract, giving \( 8\lambda - \lambda = 7\lambda \).

Answer: (d) zero

Explanation:
1. Doubled wavelength \( = 6 \times 10^{-7} \) m.
2. This is greater than the threshold wavelength \( 5 \times 10^{-7} \) m, so the photon energy is less than the work function.
3. No photoelectrons are emitted, so the stopping potential is zero.

Teacher's Note:
1. Photoemission needs \( \lambda \le \lambda_0 \).
2. Compare the wavelength with the threshold before doing any calculation.

Answer: (c) 0.04 mA

Explanation:
1. Voltage across \( R_C \) \( = 15 - 7 = 8 \) V, so \( I_C = \frac{8}{2 \times 10^3} = 4 \) mA.
2. \( I_B = \frac{I_C}{\beta} = \frac{4\ \text{mA}}{100} = 0.04 \) mA.

Teacher's Note:
1. In the common-emitter circuit, current gain \( \beta = I_C / I_B \).
2. First find \( I_C \) from the voltage drop across the collector resistor.

Answer: (a) \( \frac{1}{5} \) m/s

Explanation:
1. Initially \( u = -9 \) m, \( f = -1 \) m: \( \frac{1}{v} = \frac{1}{f} - \frac{1}{u} = -1 + \frac19 \), so \( v = -1.125 \) m.
2. After 1 s the object is at \( u = -4 \) m: \( \frac1v = -1 + \frac14 \), so \( v = -1.333 \) m.
3. Shift of the image \( = 1.333 - 1.125 \approx 0.21 \) m in 1 s, so the average speed \( \approx 0.2 = \frac15 \) m/s.

Teacher's Note:
1. Apply the mirror formula at the start and at the end of the interval and divide the shift by the time.
2. The image moves much more slowly than the object when the object is far from the mirror.

Answer: (a) 40

Explanation:
1. Fringe width \( \beta = \frac{\lambda D}{d} \propto \lambda \); the width of the screen region is fixed, so the number of fringes \( N \propto \frac{1}{\beta} \propto \frac{1}{\lambda} \).
2. \( N_2 = 60 \times \frac{4000}{6000} = 40 \).

Teacher's Note:
1. A larger wavelength gives wider fringes, so fewer of them fit on the screen.
2. N times \( \lambda \) is constant.

Answer: (d) the intensity of light is twice maximum and twice zero

Explanation:
1. By Malus' law \( I = I_0 \cos^2\theta \).
2. In one full rotation of \( 360^\circ \), \( \cos^2\theta \) reaches its maximum value at \( 0^\circ \) and \( 180^\circ \), and zero at \( 90^\circ \) and \( 270^\circ \).
3. So the intensity is maximum twice and zero twice.

Teacher's Note:
1. Malus' law repeats every \( 180^\circ \).
2. For unpolarised light there would be no change at all.