Explanation

(a) Explanation of wave-particle duality of light: Light can act as a particle (by way of Photon) and also simultaneously).

(b) Determination of Planck's Constant

h = π P = 4.2 x 10\(^{-11}\) x 1.6 x 10\(^{-23}\) as waves

= 6.72 x 10\(^{-34}\) JS

Explanation

(a) Lightening in fluorescent tubes operate on the principle of electric discharge through low pressure gas.

(b) Colour of light from fluorescent tube depend on (i) Nature of the coating inside the tube (ii) Nature of the gas in the tube.

Correct Answer: A. silvered surfaces

Correct Answer: A. 18.57NC−1

Correct Answer: D. 10.0 ms^-1

Correct Answer: B. step-down transformer

Explanation

In a step down transformer, there are more turns in the primary coil than in the secondary coil and more current flows in the secondary circuit than in the primary.

Correct Answer: A) 30 s

Correct Answer: C. 3s^-1

Explanation

F = \(\frac{v}{\lambda} = \frac{30}{10} = 3s^{-1}\)

Correct Answer: B. decrease

Explanation

(a) Capacitance of a capacitor is the ratio of the charge Q on either plate of the capacitor to the p.d in volts between the plates. Mathematically,

\(Capacitance = \frac{Q}{V}\)

Q - Charge ; V - voltage.

(b) Three factors on which the capacitance of a parallel - plate capacitor depends on are: (1) Common area of the plates (2) Dielectric between the plates (3) Separation between the plates.

(c) Work done W = QV

Average work done = \(\frac{1}{2} QV\)

But \(V = \frac{Q}{C}\)

\(\therefore W = \frac{Q}{2} \times \frac{Q}{C} = \frac{Q^{2}}{2C}\)

\(\therefore \text{Energy stored W = } \frac{Q^{2}}{2C}\)

(d)

\(\frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}}\)

= \(\frac{1}{6} + \frac{1}{4} = \frac{10}{24}\)

\(C = 2.4 \mu F\)

\(Q = \frac{24}{10} \times 100 \times 10^{-6} = 240 \mu C\)

= \(240 \times 10^{-6} C\)

(ii) p.d across 4\(\mu F, V_{4} = \frac{Q}{C}\)

\(V_{4} = \frac{240 \times 10^{-6}}{4 \times 10^{-6}} = 60V\)

p.d across 6\(\mu F, V_{6} = \frac{Q}{C}\)

= \(\frac{240 \times 10^{-6}}{6 \times 10^{-6}} = 40V\)

(iii) Energy stored = \(\frac{Q^{2}}{2C}\)

= \(\frac{(240 \times 10^{-6})^{2}}{2 \times 2.4 \times 10^{-6}}\)

= \(1.2 \times 10^{-2} J\)