JEE Physics Electricity Optics and Modern Physics MCQs Set 03

Answer: (c) \(\dfrac{q_0}{\sqrt2\,\epsilon_0}\)

Explanation:
1. Each end disc subtends a half-angle \(\theta\) at the charge with \(\tan\theta = R/R = 1\), so \(\theta = 45^\circ\).
2. Flux through one end \(= \dfrac{q_0}{2\epsilon_0}(1 - \cos45^\circ)\), so both ends take \(\dfrac{q_0}{\epsilon_0}\left(1 - \dfrac1{\sqrt2}\right)\).
3. Curved surface: \(\dfrac{q_0}{\epsilon_0} - \dfrac{q_0}{\epsilon_0}\left(1 - \dfrac1{\sqrt2}\right) = \dfrac{q_0}{\sqrt2\,\epsilon_0}\).

Teacher's Note:
1. Total flux through the closed cylinder is \(q_0/\epsilon_0\).
2. Subtract the flux through the two flat ends to get the curved part.

Answer: (b) 90 m/s

Explanation:
1. Energy conservation: \(\tfrac12 mv^2 = kq_1q_2\left(\dfrac1{r_1} - \dfrac1{r_2}\right)\).
2. \(kq_1q_2 = 9\times10^9\times10^{-6}\times10^{-3} = 9\) J m, so the energy released is \(9\times(1 - 0.1) = 8.1\) J.
3. \(v = \sqrt{\dfrac{2\times8.1}{2\times10^{-3}}} = \sqrt{8100} = 90\) m/s.

Teacher's Note:
1. Convert grams to kilograms and micro/milli coulombs to coulombs before using the formula.
2. The loss in electrostatic potential energy appears as kinetic energy.

Answer: (d) zero

Explanation:
1. By symmetry the three fields at the centre have equal magnitude and are \(120^\circ\) apart.
2. Three equal vectors at \(120^\circ\) to each other add up to zero.

Teacher's Note:
1. Equal vectors separated by \(120^\circ\) always form a closed triangle.
2. The same holds for any regular polygon of equal charges at its centre.

Answer: (c) 7 V

Explanation:
1. Net emf \(= 12 - 3 = 9\) V and the total resistance \(= 2+1+2+3+1 = 9\ \Omega\), so \(I = 1\) A.
2. Going from the earthed point up to B: \(V_B = 0 + 12 - I(2 + 1 + 2) = 12 - 5 = 7\) V.

Teacher's Note:
1. With cells in opposition subtract the emfs.
2. Potential rises by the emf when crossing a cell from its negative to its positive terminal and falls by IR across each resistor along the current.

Answer: (b) \(\chi = \mu_r - 1\)

Explanation:
1. \(B = \mu_0(H + M) = \mu_0 H(1 + \chi)\).
2. Also \(B = \mu_0\mu_r H\), so \(\mu_r = 1 + \chi\), that is \(\chi = \mu_r - 1\).

Teacher's Note:
1. Susceptibility is dimensionless and is positive for para and ferromagnets, negative for diamagnets.
2. \(\mu_r\) is the permeability relative to free space.

Answer: (b) clockwise, anticlockwise

Explanation:
1. Approaching: the north pole faces the loop, so the flux towards the observer increases. By Lenz's law the induced field points away from the observer, which is a clockwise current as seen by the observer.
2. After crossing: the loop is beside the south (trailing) end, and the flux towards the observer decreases. The induced field points towards the observer, an anticlockwise current.

Teacher's Note:
1. Lenz's law: the induced current opposes the change in flux.
2. The current reverses once the magnet has passed through the loop.

Answer: (c) \(60^\circ\)

Explanation:
1. \(\cos\phi = \dfrac{R}{Z} = \dfrac{10}{20} = \dfrac12\).
2. So \(\phi = 60^\circ\).

Teacher's Note:
1. The power factor \(\cos\phi\) equals R/Z in a series circuit.
2. At resonance \(Z = R\) and the phase difference is zero.

Answer: (b) lies between 2 and \(\sqrt2\)

Explanation:
1. At minimum deviation \(n = \dfrac{\sin\frac{A+\delta}{2}}{\sin\frac A2}\). With \(\delta = A\), \(n = \dfrac{\sin A}{\sin(A/2)} = 2\cos\dfrac A2\).
2. A prism can have \(0 < A < 90^\circ\) for the ray to refract through, so \(0 < A/2 < 45^\circ\) and \(2\cos(A/2)\) lies between \(\sqrt2\) and 2.

Teacher's Note:
1. The relation \(n = 2\cos(A/2)\) holds only when \(\delta_{min} = A\).
2. Cosine decreases with angle, so the larger A is, the smaller n is.

Answer: (b) infinity

Explanation:
1. Use \(\dfrac{n_2}{v} - \dfrac{n_1}{u} = \dfrac{n_2 - n_1}{R}\) with \(u = -2\) m, \(R = +2\) m.
2. For \(n_2 = 1.6\): \(\dfrac{1.6}{v} + 0.5 = 0.3\), so \(v = -8\) m (virtual image on the object side).
3. For \(n_2 = 2.0\): \(\dfrac{2}{v} + 0.5 = 0.5\), so \(\dfrac2v = 0\) and the image is at infinity.
4. One image is at 8 m and the other at infinity, so the separation is infinite.

Teacher's Note:
1. Apply the refraction formula separately for each medium, keeping the sign convention.
2. Here \(1/v = 0\) means the rays emerge parallel, so the image is at infinity.

Answer: (a) \(\sin^{-1}(3\times10^{-3})\)

Explanation:
1. Maxima occur at \(d\sin\theta = n\lambda\).
2. \(\sin\theta = \dfrac{3\times589\times10^{-9}}{0.589\times10^{-3}} = 3\times10^{-3}\).
3. So \(\theta = \sin^{-1}(3\times10^{-3})\).

Teacher's Note:
1. Convert nanometres and millimetres to metres consistently.
2. The nth maximum has path difference \(n\lambda\).

Answer: (c) 25 : 1

Explanation:
1. Intensity from a slit is proportional to its width, so \(I_1 : I_2 = 4 : 9\) and \(A_1 : A_2 = 2 : 3\).
2. \(\dfrac{I_{max}}{I_{min}} = \left(\dfrac{A_1 + A_2}{A_2 - A_1}\right)^2 = \left(\dfrac{5}{1}\right)^2 = 25\).

Teacher's Note:
1. Amplitude is the square root of intensity.
2. Take care not to use the width ratio directly as the amplitude ratio.

Answer: (b) 6.2 eV

Explanation:
1. Work function \(\phi = \dfrac{1240}{200} = 6.2\) eV.
2. Photon energy at 100 nm is \(\dfrac{1240}{100} = 12.4\) eV.
3. \(K_{max} = 12.4 - 6.2 = 6.2\) eV.

Teacher's Note:
1. Use \(E(\mathrm{eV}) = 1240/\lambda(\mathrm{nm})\).
2. The threshold wavelength gives the work function.

Answer: (c) 2 times

Explanation:
1. At closest approach the kinetic energy is converted into potential energy: \(K = \dfrac{k(2e)(Ze)}{r_0}\).
2. So \(r_0 \propto \dfrac1K\).
3. If K is halved, \(r_0\) doubles.

Teacher's Note:
1. Closest approach distance is inversely proportional to the initial kinetic energy.
2. It is for a head-on collision only.

Answer: (c) \(F_{pp} = F_{pn} = F_{nn}\)

Explanation:
1. The strong nuclear force is charge independent.
2. At the same separation it is equal for p-p, p-n and n-n pairs (the extra electrostatic repulsion between protons is not part of the nuclear force).

Teacher's Note:
1. Nuclear force is independent of charge, short ranged and strongly attractive at about 1 fm.
2. It is not the same as the Coulomb force.

Answer: (a) 5 mA

Explanation:
1. Voltage across the series resistor \(= 20 - 15 = 5\) V, so the total current \(= \dfrac{5}{250} = 20\) mA.
2. Load current \(= \dfrac{15}{1000} = 15\) mA.
3. Zener current \(= 20 - 15 = 5\) mA.

Teacher's Note:
1. The load voltage is held at the breakdown voltage of the zener.
2. The zener takes whatever current the load does not.

Answer: (c) (A, B, Y): (0,0,0), (0,1,0), (1,0,0), (1,1,0)

Explanation:
1. \(Y = (A\cdot B)\cdot(\bar A\cdot B)\).
2. \(A\cdot\bar A = 0\), so \(Y = 0\) for every combination of A and B.

Teacher's Note:
1. A variable and its complement can never both be 1.
2. Simplify with Boolean algebra before writing a truth table.

Answer: (b) \(-11\) C

Explanation:
1. Charge of the added electrons \(= 10^{20}\times1.6\times10^{-19} = 16\) C, and it is negative.
2. Net charge \(= 5 - 16 = -11\) C.

Teacher's Note:
1. Adding electrons adds negative charge.
2. Net charge is the algebraic sum of the charges.

Answer: (c) \(\dfrac{KQ}{r^3}\left(r^2 - R^2\right)^{1/2}\)

Explanation:
1. On the axis at a distance x from the centre, \(E = \dfrac{KQx}{(R^2 + x^2)^{3/2}}\).
2. The distance from any point of the ring is \(r = \sqrt{R^2 + x^2}\), so \(x = \sqrt{r^2 - R^2}\).
3. Hence \(E = \dfrac{KQ\sqrt{r^2 - R^2}}{r^3}\).

Teacher's Note:
1. Express the axial distance in terms of the given distance r.
2. The field at the centre is zero and it is maximum at \(x = R/\sqrt2\).

Answer: (d) \(-\dfrac{q}{2\pi^2\epsilon_0 r^2}\hat j\)

Explanation:
1. Take an element at angle \(\theta\) from the x-axis, with charge \(dq = \dfrac{q}{\pi r}\,r\,d\theta\). The x components cancel by symmetry.
2. \(E_y = -\int_0^{\pi}\dfrac{dq\sin\theta}{4\pi\epsilon_0 r^2} = -\dfrac{q}{4\pi^2\epsilon_0 r^2}\int_0^\pi\sin\theta\,d\theta = -\dfrac{q}{2\pi^2\epsilon_0 r^2}\).
3. The field points away from the positive charge, along \(-\hat j\).

Teacher's Note:
1. The field at the centre of a semicircular arc points away from the charge for a positive charge.
2. Use symmetry to cancel the components along the axis of symmetry's perpendicular.

Answer: (b) E on the LHS will have a contribution from all charges while q on the RHS will have a contribution from \(q_2\) and \(q_4\) only

Explanation:
1. The electric field at the surface is the total field of all charges, inside and outside.
2. The flux of the outside charges cancels over a closed surface, so only the enclosed charge \(q_2 + q_4\) appears in q on the RHS.

Teacher's Note:
1. Gauss's law uses the total field on the left but only the enclosed charge on the right.
2. Charges outside the surface produce zero net flux.

Answer: (b) \(\dfrac{E^2q^2t^2}{2m}\)

Explanation:
1. \(a = \dfrac{qE}{m}\) and \(v = at = \dfrac{qEt}{m}\).
2. \(K = \dfrac12mv^2 = \dfrac{q^2E^2t^2}{2m}\).

Teacher's Note:
1. Uniform acceleration from rest gives \(v = at\).
2. K also equals the work done, \(qE\cdot\tfrac12at^2\).

Answer: (b) The electrostatic field at the surface of a charged conductor must be tangential to the surface at any point

Explanation:
1. In electrostatic equilibrium the field inside a conductor is zero, there is no net charge in the bulk, and the potential is constant.
2. The field at the surface must be normal to the surface, since a tangential component would drive surface charges. So (b) is false.

Teacher's Note:
1. A conductor surface is an equipotential, so the field is perpendicular to it.
2. Charges reside only on the outer surface.

Answer: (b) the field is non-zero but the potential is zero

Explanation:
1. The potential at the centre is \(\dfrac{k(2q - q - q)}{r} = 0\).
2. The field of the \(-q\) charges adds to give a resultant along the direction of the \(2q\) charge, and the field from \(2q\) adds in the same direction, so the field does not vanish.

Teacher's Note:
1. Potential is a scalar and field is a vector.
2. Charges adding up to zero do not necessarily give zero field.