JEE Physics Electricity Optics and Modern Physics MCQs Set 02

Answer: (a) 1.24 nm

Explanation:
1. The electron reaches the target with kinetic energy \(K = eV = 20\times10^3\) eV.
2. The photon energy is 5% of this, \(E = 0.05 \times 20\times10^3 = 10^3\) eV.
3. \(\lambda = \dfrac{hc}{E} = \dfrac{1240\ \text{eV nm}}{10^3\ \text{eV}} = 1.24\) nm.

Teacher's Note:
1. Use \(hc = 1240\) eV nm to move quickly between photon energy and wavelength.
2. The photon takes only 5% of the electron's energy, not all of it.

Answer: (b) 56

Explanation:
1. The curve of binding energy per nucleon against mass number rises steeply for light nuclei, reaches a broad maximum of about 8.8 MeV near \(A = 56\) (iron region), and then falls slowly for heavy nuclei.
2. So the most tightly bound nuclei have mass numbers near 56.

Teacher's Note:
1. This is why fusion of light nuclei and fission of heavy nuclei both release energy: they move towards the peak.
2. Fe-56 is among the most stable nuclei.

Answer: (a) 2 : 1

Explanation:
1. \(\dfrac1{16} = \left(\dfrac12\right)^4\), so 8 h is 4 half-lives of X and \(T_X = 2\) h.
2. \(\dfrac1{256} = \left(\dfrac12\right)^8\), so 8 h is 8 half-lives of Y and \(T_Y = 1\) h.
3. \(T_X : T_Y = 2 : 1\).

Teacher's Note:
1. Write the remaining fraction as a power of \(\tfrac12\) to read off the number of half-lives.
2. A larger fraction left means a longer half-life.

Answer: (b) 1 \(\Omega\)

Explanation:
1. Use \(r = R\left(\dfrac{l_1 - l_2}{l_2}\right)\), where \(l_1\) is the balancing length for the open cell and \(l_2\) the length with the shunt \(R\).
2. \(r = 2\left(\dfrac{150 - 100}{100}\right) = 1\ \Omega\).

Teacher's Note:
1. The balance length is proportional to the potential difference across the cell.
2. With the shunt connected, the terminal voltage drops by the voltage lost across \(r\).

Answer: (b) 484 \(\Omega\)

Explanation:
1. The resistance is fixed by the rating: \(R = \dfrac{V^2}{P} = \dfrac{220^2}{100} = 484\ \Omega\).
2. Operating at a different voltage does not change this resistance (the power would become 25 W).

Teacher's Note:
1. Resistance is a property of the filament, so use the rated voltage and power to find it.
2. The operating voltage only changes the current and the power.

Answer: (b) 36 mA

Explanation:
1. Branch 3 is reverse biased, so it carries no current.
2. Branch 1: \(50 + 150 = 200\ \Omega\). Branch 2: \(50 + 50 = 100\ \Omega\). Their parallel resistance is \(\dfrac{200 \times 100}{300} = \dfrac{200}{3}\ \Omega\).
3. Total resistance \(= 100 + \dfrac{200}{3} = \dfrac{500}{3}\ \Omega\), so \(I = \dfrac{6}{500/3} = 0.036\) A = 36 mA.

Teacher's Note:
1. A reverse-biased diode with infinite backward resistance behaves as an open circuit.
2. Combine the parallel branches first and then add the series resistor.

Answer: (a) collision of minority charge carriers with atoms of the crystal

Explanation:
1. Under a large reverse bias, minority carriers gain enough kinetic energy to knock out electrons from covalent bonds when they collide with atoms.
2. The new carriers are accelerated in turn and produce more carriers in a chain, so the reverse current rises sharply.

Teacher's Note:
1. Avalanche breakdown happens in lightly doped junctions with a wide depletion layer.
2. Zener breakdown, in contrast, comes from a very thin depletion layer and a strong field.

Answer: (d) 1 for every combination of A and B

Explanation:
1. The first NAND output is \(\overline{\bar A\cdot B} = A + \bar B\). The second is \(\overline{A\cdot\bar B} = \bar A + B\).
2. \(Y = (A + \bar B) + (\bar A + B) = (A + \bar A) + (B + \bar B) = 1 + 1 = 1\).
3. So Y is 1 for all inputs.

Teacher's Note:
1. Apply De Morgan's laws to each NAND gate before combining the outputs.
2. A term and its complement in an OR always give 1.

Answer: (d) \(-z\) axis

Explanation:
1. For the wire at \(x=-d\), the origin lies to its right. By the right-hand rule, a current along \(+y\) gives a field into the plane there, that is along \(-z\).
2. For the wire at \(x=+d\), the current is along \(-y\) and the origin lies to its left. The field there is also into the plane, along \(-z\).
3. The two fields add, so B at O is along \(-z\).

Teacher's Note:
1. Opposite currents produce fields that add in the region between the wires.
2. Use the right-hand grip rule on each wire separately and then add the vectors.

Answer: (c) the reflected ray rotates through an angle \(2\theta\)

Explanation:
1. If the mirror turns through \(\theta\), the normal turns through \(\theta\), so the angle of incidence changes by \(\theta\).
2. The reflected ray makes the same angle with the normal on the other side, so it turns through \(\theta + \theta = 2\theta\).

Teacher's Note:
1. The incident ray is fixed, so only the reflected ray moves.
2. This is the principle behind the mirror galvanometer's optical lever.

Answer: (c) \(c/b\)

Explanation:
1. For a real image, \(|m| = \dfrac{f}{u - f}\), so \(\left|\dfrac1m\right| = \dfrac{u}{f} - 1\).
2. This is a straight line of slope \(\dfrac1f\). Its intercept on the \(u\) axis is \(f\) (where \(1/|m| = 0\)).
3. The slope from the graph is \(b/c\), so \(\dfrac1f = \dfrac bc\) and \(f = \dfrac cb\).

Teacher's Note:
1. Compare the equation with \(y = mx + c\) to read off the slope.
2. The \(u\)-intercept also equals \(f\), so for this graph \(a = c/b\) as well.

Answer: (b) \(\dfrac{q}{4\epsilon_0}\)

Explanation:
1. Charge \(q\) lies at the centre of a cube of side \(a\) that has the square as one face. The flux through that face is \(\dfrac16\dfrac{q}{\epsilon_0}\).
2. Place four such squares around vertex C to form a square of side \(2a\). The charge \(2q\) is then at the centre of a cube of side \(2a\), and the flux through this big face is \(\dfrac16\dfrac{2q}{\epsilon_0}\). By symmetry, the share of ABCD is one quarter: \(\dfrac{2q}{24\epsilon_0} = \dfrac{q}{12\epsilon_0}\).
3. Total flux \(= \dfrac{q}{6\epsilon_0} + \dfrac{q}{12\epsilon_0} = \dfrac{q}{4\epsilon_0}\).

Teacher's Note:
1. Use symmetry: a charge at the centre of a cube sends one sixth of its flux through each face.
2. Build a bigger cube around the second charge so that the square becomes a quarter of one face.

Answer: (a) \(4\times10^{-3}\) N/m\(^2\)

Explanation:
1. The electrostatic pressure on a charged conductor is \(\dfrac{\epsilon_0E^2}{2}\).
2. \(E = \dfrac VR = \dfrac{3000}{0.1} = 3\times10^4\) V/m.
3. Pressure \(= \dfrac12(8.85\times10^{-12})(3\times10^4)^2 \approx 4\times10^{-3}\) N/m\(^2\).

Teacher's Note:
1. The force on a surface element uses the average field \(E/2\), which gives the factor of one half.
2. The pull is always outward, whatever the sign of the charge.

Answer: (c) 1.0 mm

Explanation:
1. Fringe width \(\beta = \dfrac{\lambda D}{d} = \dfrac{5\times10^{-7}\times 2}{10^{-3}}\).
2. \(\beta = 10^{-3}\) m = 1.0 mm.

Teacher's Note:
1. Keep all lengths in metres before substituting.
2. The separation between adjacent bright lines is one fringe width.

Answer: (a) \(\dfrac{Dt}{3d}\)

Explanation:
1. The sheet introduces an extra optical path \((\mu - 1)t\). The fringe shift is \(\Delta x = \dfrac{(\mu - 1)\,t\,D}{\text{slit separation}}\).
2. The slit separation is \(2d\), so \(\Delta x = \dfrac{\tfrac23\,t\,D}{2d} = \dfrac{Dt}{3d}\).

Teacher's Note:
1. The slit separation in this problem is \(2d\), not \(d\).
2. The fringe system shifts towards the side where the sheet is placed.

Answer: (d) \(t = \dfrac1{600}\) s

Explanation:
1. The current is maximum when the phase equals \(\dfrac\pi2\).
2. \(100\pi t + \dfrac\pi3 = \dfrac\pi2\), so \(100\pi t = \dfrac\pi6\) and \(t = \dfrac1{600}\) s.

Teacher's Note:
1. Set the phase equal to \(\pi/2\) for the first maximum of a sine.
2. Use the smallest positive value of \(t\).

Answer: (d) 2

Explanation:
1. The charge that flows is the area under the \(i\)-\(t\) graph: \(q = \tfrac12 \times 4 \times 0.1 = 0.2\) C.
2. Induced charge \(q = \dfrac{\Delta\Phi}{R}\), so \(\Delta\Phi = qR = 0.2 \times 10 = 2\) Wb.

Teacher's Note:
1. The charge flowing through a coil depends only on the flux change and the resistance, not on the time taken.
2. Area under an \(i\)-\(t\) graph always gives charge.

Answer: (c) 0.6 W

Explanation:
1. The current through the series resistor is constant, equal to \(I_Z + I_L\).
2. The zener current is largest (50 mA) when the load current is zero, so the total current is 50 mA.
3. Maximum power \(= V_Z I_{Z(max)} = 12 \times 0.05 = 0.6\) W.

Teacher's Note:
1. Because the voltage across the zener is fixed, its power is simply \(V_Z\) times its current.
2. The zener current is greatest when the load draws the least current.

Answer: (b) they travel with the same speed in all media

Explanation:
1. Electromagnetic waves are transverse and are produced by accelerated charges.
2. In vacuum all of them travel at the speed of light, but in a medium the speed is \(c/n\) and depends on the medium (and on the frequency).
3. So the statement that the speed is the same in all media is wrong.

Teacher's Note:
1. The speed of light in vacuum is a universal constant, but in a material it is smaller.
2. Read 'travel with the speed of light' as referring to vacuum.

Answer: (a) \(\dfrac{CV}{4}\)

Explanation:
1. With the key open, the bank has capacitance \(3C\), in series with \(C\): \(C_{eq} = \dfrac{3C \cdot C}{4C} = \dfrac{3C}{4}\). Charge drawn \(q_1 = \dfrac{3CV}{4}\).
2. With the key closed, the bank is shorted, so only \(C\) remains and \(q_2 = CV\).
3. Charge flowing through the cell \(= q_2 - q_1 = CV - \dfrac{3CV}{4} = \dfrac{CV}{4}\).

Teacher's Note:
1. The charge through the cell is the change in the charge delivered to the series combination.
2. Parallel capacitors add; series combination uses the reciprocal rule.