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C. 50.40Kwh Convert power to kW (150 W = 0.15 kW), then multiply by time in hours: energy = 0.15 kW \times 336 h = 50.4 kWh.

The circuit is a simple series circuit with a 16 V battery (or voltage source) directly connected across a 2 \Omega resistor. The power (\(P\)) dissipated (developed) in the resistor is given by: \(P = \frac{V^2}{R}\) where \(V = 16 \, \text{V}\) and \(R = 2 \, \Omega\). Calculating the power: \(P = \frac{16^2}{2} = \frac{256}{2} = 128 \, \text{W}\)

For an A.C. current, \(I_{\text{r.m.s}} = \frac{I_0}{\sqrt{2}}\) where (I_0) is the maximum (peak) current. Given (I_0 = 7,\text{A}), \(I_{\text{r.m.s}} = \frac{7}{\sqrt{2}} \approx 4.95,\text{A}\) Answer: \(4.95 \text{A}\ \ (\approx 5\ \text{A})\)

Given: Inductance: \( L = 0.5 \, \text{H} \), Resistance: \( R = 45 \, \Omega \), Frequency: \( f = 50 \, \text{Hz} \) inductive reactance \( X_L \): \(X_L = 2 \pi f L = 2 \times \frac{22}{7} \times 50 \times 0.5 = \frac{1100}{7} \approx 157.14 \, \Omega\) The impedance \( Z \): \(Z = \sqrt{R^2 + X_L^2} = \sqrt{(45)^2 + (157.14)^2} = \sqrt{2025 + 24642.82} \approx \sqrt{26667.82} \approx 163.5 \, \Omega\).

The period of a wave is the reciprocal of its frequency. Therefore, period = \(\frac{1}{\text{frequency}}\) Substituting the given frequency, we get period = \(\frac{1}{ 50}\) Hz = 0.02 s.

V = I X\(_L\) X\(_L\) = \(\frac{\text{V }}{\text{I}}\) = \(\frac{240}{2}\) = 120 \Omega But X\(_L\) = 2\(\pi\)fL 120 = 2 x 2\(\pi\) x \(\frac{100}{\pi}\) x L 120 = 2 x 100 x L = 200L L = \(\frac{120}{200}\) = 0.6H

i.e V\(_L\) - V\(_C\) = -ve (V\(_C\) - V\(_L\) = +ve) From Tan \(\phi\) = \(\frac{V_L - V_C}{\text{R}}\) It shows that V\(_L\) - V\(_C\) will be inverted on the y-axis V\(_C\) > V\(_L\)

I = 0.05 A, p.d across R\(_L\) = V\(_L\) = I R\(_L\) = 0.05 \times 50 V\(_L\) = 2.5 V

X\(_C\) = \(\frac{1}{(2\pifC)}\) = \(\frac{1}{(2\pi \times (\frac{200}{\pi}) \times 25 \times 10^{-6})}\) = \(\frac{1}{(2 \times 200 \times 25 \times 10^{-6})}\) = \(\frac{1}{10^{-2}}\) = 10\(^2\) = 100 \Omega \(\therefore\) X\(_C\) = 100 \Omega

X\(_o\) equals PQ because it is the amplitude or maximum current.

R\(_L\) = 2\(\pi\)fL = 2 \(\pi\) x \(\frac{100}{\pi}\) x 0.25 = 50.0\(\Omega\)

Electrical conduction in semiconductors occurs via electrons and holes.

Answer: A \lambda = \(\frac{ln2}{T½ }\) ⇒ \(\frac{0.693}{138.5}\) ≈ 5.0 \times 10\(^{-3}\) s\(^{-1}\).

A. Electrophorus. It generates static electricity via induction and is not a radiation detector. The others are: Scintillation chamber: Detects radiation via light flashes. Geiger-Muller counter: A gas-filled detector for ionizing radiation. Film badge: Records cumulative radiation exposure on photographic film.

A. Less than 20 cm For a concave mirror, an erect image is virtual (behind the mirror) and forms when the object is placed between the pole and focus (\( u < f \)). Magnification: \(m = +4 \quad (\text{erect, so positive})\) Using the mirror formula: \(\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\) Given \( f = -20 \, \text{cm} \) (concave), \( u \) is negative. Since \( m = -\frac{v}{u} \): \(v = -m u = -(+4)u = -4u\) Substituting into the mirror formula: \(\frac{1}{-4u} + \frac{1}{u} = \frac{1}{-20}\) Combining the fractions: \(\frac{3}{4u} = \frac{1}{-20}\) Thus, solving for \( u \): \(3 \cdot -20 = 4u \quad \Rightarrow \quad u = -15 \, \text{cm}\) Object distance \( |u| = 15 \, \text{cm} < 20 \, \text{cm} \). Thus, it is less than 20 cm. OR A concave mirror only gives an erect, magnified image when the object is placed between the mirror and its focal point. Since the focal length is 20 cm, the object must be closer than 20 cm to the mirror.