JEE Physics Electricity Optics and Modern Physics MCQs Set 04

Answer: (c) zero

Explanation:
1. By symmetry the contributions of diametrically opposite elements of the ring cancel at the centre.
2. The field at the centre is zero, so the force on q is zero.

Teacher's Note:
1. Symmetry arguments often give zero field at the centre of uniform rings.
2. The force is zero but the potential is not.

Answer: (b) 40 unit

Explanation:
1. \(\phi = \vec E\cdot\vec S = (2\hat i + 4\hat j + 3\hat k)\cdot10\hat j\).
2. \(\phi = 40\) units.

Teacher's Note:
1. Flux is the dot product of the field and the area vector.
2. Only the component of the field parallel to the area vector contributes.

Answer: (c) 25 V

Explanation:
1. \(\Delta V = Ed\cos\theta\), which is at most \(10\times3 = 30\) V.
2. Only 25 V is possible, when the line joining the points is inclined to the field.

Teacher's Note:
1. The potential difference between two points cannot exceed field times separation.
2. It equals this maximum only when the points lie along a field line.

Answer: (d) \(\dfrac r4\)

Explanation:
1. At the closest approach the kinetic energy turns into potential energy: \(\tfrac12mv^2 = \dfrac{kqQ}{r}\).
2. With 2v the kinetic energy is 4 times as large, so \(\dfrac{kqQ}{r'} = 4\dfrac{kqQ}{r}\), giving \(r' = \dfrac r4\).

Teacher's Note:
1. Distance of closest approach varies inversely with the kinetic energy.
2. The speed is squared in the energy.

Answer: (a) V

Explanation:
1. The field between the sphere and the shell depends only on the charge Q of the inner sphere, and the field inside the conducting shell is zero.
2. Charge given to the shell changes the potential of both surfaces equally, so the potential difference stays V.

Teacher's Note:
1. Adding charge to the outer shell shifts all potentials inside it by the same amount.
2. Potential difference depends only on the field between the surfaces.

Answer: (a) \(-8.85\times10^{-7}\) C

Explanation:
1. Net outward flux \(= 4\times10^5 - 5\times10^5 = -10^5\) units.
2. \(q = \epsilon_0\phi = 8.85\times10^{-12}\times(-10^5) = -8.85\times10^{-7}\) C.

Teacher's Note:
1. Inward flux counts as negative in Gauss's law.
2. A net inward flux means a negative enclosed charge.

Answer: (a) 2 J

Explanation:
1. Heat generated equals the stored energy \(U = \dfrac12 CV^2\).
2. \(U = \dfrac12\times100\times10^{-6}\times(200)^2 = 2\) J.

Teacher's Note:
1. The value of the resistance does not change the total heat, only the time taken.
2. All the stored energy is dissipated in the resistor.

Answer: (b) 50 k\(\Omega\)

Explanation:
1. \(V = V_0 e^{-t/RC}\), so \(100 = 200\,e^{-70\text{ ms}/RC}\) and \(\dfrac{70\text{ ms}}{RC} = \ln 2 = 0.7\).
2. \(RC = 100\) ms.
3. \(R = \dfrac{100\times10^{-3}}{2\times10^{-6}} = 5\times10^{4}\ \Omega = 50\) k\(\Omega\).

Teacher's Note:
1. The time for the voltage to halve is \(RC\ln2\).
2. Check that the time constant is in seconds before dividing by C in farads.

Answer: (c) The charge on the capacitor is not conserved.

Explanation:
1. The battery is disconnected, so the charge Q stays constant; statement (c) is therefore incorrect.
2. \(C' = KC\), so \(V' = V/K\) and \(U' = \dfrac{Q^2}{2KC} = U/K\), which makes (a), (b) and (d) correct.

Teacher's Note:
1. With the battery disconnected, charge is fixed; with the battery connected, potential is fixed.
2. Decide which quantity stays constant before comparing energies.

Answer: (d) 80%

Explanation:
1. Charge is shared: common potential \(V' = \dfrac{2V}{2+8} = \dfrac V5\).
2. Final energy \(= \dfrac12(10)\left(\dfrac V5\right)^2 = \dfrac{V^2}{5}\) (in \(\mu\)J units) against initial \(\dfrac12(2)V^2 = V^2\).
3. Fraction lost \(= 1 - \dfrac15 = 80\%\).

Teacher's Note:
1. The loss fraction is \(\dfrac{C_2}{C_1+C_2}\) when one capacitor charged and one uncharged are joined.
2. The lost energy appears as heat and radiation.

Answer: (b) \(4\times10^{-2}\) J

Explanation:
1. Heat \(= \dfrac12 CV^2\).
2. \(= \dfrac12\times2\times10^{-6}\times(200)^2 = 4\times10^{-2}\) J.

Teacher's Note:
1. Always convert microfarad to farad.
2. Heat equals the electrostatic energy lost.

Answer: (b) increase by 0.2%

Explanation:
1. Volume is constant, so \(R \propto l^2\).
2. \(\dfrac{\Delta R}{R} = 2\dfrac{\Delta l}{l} = 2\times0.1\% = 0.2\%\) increase.

Teacher's Note:
1. For small stretches the percentage change in R is twice that in length.
2. The wire also gets thinner, which adds to the increase.

Answer: (b) \(\dfrac{2\sigma_1\sigma_2}{\sigma_1+\sigma_2}\)

Explanation:
1. Series resistances add: \(\dfrac{2l}{\sigma A} = \dfrac{l}{\sigma_1 A}+\dfrac{l}{\sigma_2 A}\).
2. So \(\dfrac2\sigma = \dfrac1{\sigma_1}+\dfrac1{\sigma_2}\) and \(\sigma = \dfrac{2\sigma_1\sigma_2}{\sigma_1+\sigma_2}\).

Teacher's Note:
1. Resistivities add in series for equal dimensions.
2. Conductivity is the reciprocal of resistivity.

Answer: (c) \(\dfrac{E_1r_2+E_2r_1}{r_1+r_2}\) and \(\dfrac{r_1r_2}{r_1+r_2}\)

Explanation:
1. For two cells in parallel, \(\dfrac1{r} = \dfrac1{r_1}+\dfrac1{r_2}\), so \(r = \dfrac{r_1r_2}{r_1+r_2}\).
2. \(\dfrac{E}{r} = \dfrac{E_1}{r_1}+\dfrac{E_2}{r_2}\), which gives \(E = \dfrac{E_1r_2+E_2r_1}{r_1+r_2}\).

Teacher's Note:
1. Parallel cells combine like parallel conductances weighted by emf.
2. The equivalent internal resistance is always smaller than either \(r_1\) or \(r_2\).

Answer: (b) 20

Explanation:
1. Shunt \(S = \dfrac{I_gG}{I - I_g}\).
2. \(S = \dfrac{2\times180}{20-2} = 20\ \Omega\).

Teacher's Note:
1. The shunt carries the extra current \(I - I_g\).
2. A shunt is always connected in parallel with the galvanometer.

Answer: (d) 4v

Explanation:
1. \(I = nAev_d\) with n and e unchanged, so \(v_d \propto \dfrac1A\).
2. Diameter halves, so area falls to one quarter and \(v_d' = 4v\).

Teacher's Note:
1. Area depends on the square of the diameter.
2. Thinner wires need faster drift for the same current.

Answer: (b) 1.6 A to the right

Explanation:
1. Positive ions moving right: \(3.2\times10^{18}\times2\times1.6\times10^{-19} = 1.024\) A.
2. Negative ions moving left also give current to the right: \(3.6\times10^{18}\times1.6\times10^{-19} = 0.576\) A.
3. Total \(= 1.6\) A to the right.

Teacher's Note:
1. Both kinds of carrier add to the current when they move in opposite directions.
2. Conventional current points along the motion of positive charge.

Answer: (b) inversely proportional to t

Explanation:
1. Length of the current path \(= L\) and cross-section \(= L t\).
2. \(R = \dfrac{\rho L}{Lt} = \dfrac{\rho}{t}\), which is independent of L and inversely proportional to t.

Teacher's Note:
1. Square sheets have a constant sheet resistance \(\rho/t\).
2. This is why thin films are specified in ohms per square.

Answer: (b) 1.03 \(\Omega\)

Explanation:
1. Open circuit voltmeter reading gives the emf \(E = 1.53\) V.
2. With the key closed, the terminal voltage \(V = 1.03\) V is the potential drop across R, so \(R = \dfrac{1.03}{1.0} = 1.03\ \Omega\).

Teacher's Note:
1. Terminal voltage equals the voltage across the external resistor.
2. The internal resistance here is \(0.5\ \Omega\).

Answer: (b) 1.5 V

Explanation:
1. \(E = 0.5(2+r)\) and \(E = 0.25(5+r)\).
2. Equating: \(1+0.5r = 1.25+0.25r\), so \(r = 1\ \Omega\).
3. \(E = 0.5\times3 = 1.5\) V.

Teacher's Note:
1. Write the emf equation once for each resistance and solve the pair.
2. The internal resistance appears in both equations.

Answer: (a) 2 m

Explanation:
1. Current in the potentiometer wire \(= \dfrac{5}{4.5+0.5} = 1\) A, so the potential drop along AB is \(4.5\) V.
2. Potential gradient \(= \dfrac{4.5}{3} = 1.5\) V/m.
3. At null point \(l = \dfrac{3}{1.5} = 2\) m, independent of \(r_1\) because no current flows through that cell.

Teacher's Note:
1. At the balance point no current flows through the test cell branch, so its resistance does not matter.
2. The null length depends only on the gradient and the emf being measured.

Answer: (b) 4450 \(\Omega\)

Explanation:
1. Current for 30 divisions \(= \dfrac{3}{50+2950} = 1\) mA.
2. Current for 20 divisions \(= \dfrac23\) mA, so total resistance \(= \dfrac{3}{(2/3)\times10^{-3}} = 4500\ \Omega\).
3. Series resistance \(= 4500-50 = 4450\ \Omega\).

Teacher's Note:
1. Deflection is proportional to current.
2. Subtract the galvanometer resistance from the total to get the series resistance.

Answer: (b) b and c

Explanation:
1. Maximum power in R occurs when \(R = r = 1\ \Omega\), and \(P_{max} = \dfrac{E^2}{4r} = \dfrac{16}{4} = 4\) W.
2. So 5 W and 8 W exceed the maximum and are not possible, while 2 W and 4 W are possible.

Teacher's Note:
1. Power delivered to the load can never exceed \(E^2/4r\).
2. Each power below the maximum is obtained for two different values of R.