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When capacitors are connected in series across a potential difference, there is a loss in their stored energy primarily due to the decrease in overall capacitance, which limits the amount of energy they can store, rather than due to charge flowing between the capacitors.

A Wheatstone bridge is specifically designed for the accurate measurement of small and unknown resistances. It operates on the principle of balancing two legs of a bridge circuit, enabling precise determination of resistance values. Potentiometer: Used to measure potential difference or electromotive force (emf), not to measure resistance directly. Rheostat: A variable resistor utilized to control current, rather than for measuring resistance. Wheatstone Bridge: The most effective method for measuring very small resistances.

C. Ammonium chloride. It is the electrolyte in the wet Leclanché cell (aqueous solution).Others: Nickel hydroxide: In Ni-Cd/Ni-MH batteries. Lead dioxide: Cathode in lead-acid batteries. Carbon: Conductivity enhancer, not electrolyte.

Parallel connection does not add the voltages; it only increases the current capacity. Thus, total e.m.f. E = E\(_1\) = E\(_2\) = E\(_3\).

Answer: B. 2.27A (I = \(\frac{\text{P}}{\text{V}}\) = \(\frac{500}{220}\) ≈ 2.27 A.)

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 \( Z \): \(Z = \sqrt{R^2 + X_L^2} = \sqrt{(45)^2 + (157.14)^2} = \sqrt{2025 + 24642.82} \approx \sqrt{26667.82} \approx 163.5 \, \Omega\).

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.

D. 1.25 m In a pinhole camera, the relationship between image size and object size is given by: \(\frac{\text{Image Size}}{\text{Object Size}} = \frac{v}{u}\) where \( v \) is the camera length. Given \( v = 25 \, \text{cm} = 0.25 \, \text{m} \). The image is one-fifth the object size, so the magnification is: \(\text{Magnification} = \frac{1}{5} = \frac{v}{u}\) Thus, we can rearrange this to find \( u \): \(u = 5v = 5 \times 0.25 = 1.25 \, \text{m}\).

In a kaleidoscope (two plane mirrors inclined at an angle ( \theta )): \(\text{Number of images of one object} = \frac{360^\circ}{\theta} - 1\) Given: \(\theta = 30^\circ\) \(\frac{360}{30}\) = 12 So, images formed by one object: 12 - 1 = 11 For 10 objects (tinsels): 10 x 11 = 110 images formed.