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No expanded working has been attached for this question yet. The verified answer is Option C.

No expanded working has been attached for this question yet. The verified answer is Option B.

Ions present in dilute sodium chloride solution are -H\(^-\), OH\(^-\) and Cl\(^-\) + Na\(^+\). When D.C is passed through it positive ions H+ and Na+ are attracted to the cathode while negative ions OH\(^-\) and CI\(^-\) are attracted to the anode.

When a D.C is a applied, the positive ions of the electrolyte are attracted to the cathode and the appropriate ions are discharged and deposited. But if an A.C. is used, the mass deposited during one half of the cycle will be completely removed during the next half cycle, hence there will be no electroplating.

When you insert a dielectric material (like glass) between the plates, the permittivity ε increases because the dielectric constant k of glass is greater than that of air or vacuum. The new capacitance when a dielectric is inserted is given C′ = \(\frac{\text{kεoA}}{\text{d}}\) where k is the dielectric constant of the glass, and ε\(_0\) is the permittivity of free space. Since k >1 for glass, inserting it increases the capacitance: C ′ = kC Thus, with the glass block inserted, the capacitance C′ will be greater than the original capacitance C.

let the internal resistance of the cell = r, therefore E = I (R+r) (i) E = 0.4(4+r), (ii) E = 0.2 (10+r) therefore 0.4 (4+r)= 0.2 (10+r) 1.6 + 0.4r = \(2.0 \times 0.2\text{r}\) 0.4r - 0.2r = 2.0 - 1.6 therefore 0.2r = 0.4 => r = \(\frac{0.4}{0.2}\) = 2Ω therefore r = 2Ω

I = \(\frac{V}{R}\) = \(\frac{E}{R+r}\) , if r = 2Ω V = \(\frac{3}{5}\)E => \(\frac{3E}{5R} = \frac{E}{R+2}\) i.e 3E(R+2) = 5RE 3RE + 6E = 5RE ( the Es will cancel out) i.e 3R + 6 = 5R 6 = 5R - 3R => 2R = 6 i.e R = \(\frac{6}{2}\) = 3Ω

E = \(\frac{V}{d}\) E is the electric field (in volts per meter, V/m), V is the potential difference (in volts, V), d is the distance between the two points (in meters, m) Given: Potential difference V = 10\(^7\) Height d = 4m That is, E = \(\frac{10^7}{4} = 2.5\times 10^6 NC^{-1}\)

If I = 2 Sin wt; but in a.c. circuit current I = I\(_0\) Sin wt; thus comparing the two equations, it implies that I\(_0\) = 2A; Again, the root means square (I(rms)) of an a.c. current is the value of the d.c. current that will dissipate the same amount of heat in a given resistance as the a.c. and I(rms) = \(\frac{I_0}{\sqrt{2}}\) = \(\frac{2.0A}{\sqrt{2}}\) = \(\sqrt{2}\)A (by rationalizing the denominator \(\frac{2.0A}{\sqrt{2}}\))

Transistors are versatile components used in electronic circuits mainly for two purposes: they can act as electronic switches, controlling the flow of current in a circuit, and they can amplify electrical signals, making them essential in audio devices, radios, and various electronic applications. Thus, option A correctly describes the main functions of a transistor.

I = \(\frac{E}{R + r}\) For parallel of 5\Omega and 10\Omega \(\frac{1}{R}\) = \(\frac{1}{10}\) + \(\frac{1}{5}\) = \(\frac{1+2}{10}\) = \(\frac{3}{10}\) ∴R = \(\frac{10}{3}\) Total R = 3\(\frac{1}{3}\) + 3 x 1 = 7\(\frac{1}{3}\) ∴I = 6\(\div\) 7\(\frac{1}{3}\) = \(\frac{6}{1}\) x \(\frac{3}{22}\) = \(\frac{18}{22}\) = \(\frac{9}{11}\) ∴I = \(\frac{9}{11}\)

0.0 E since one of the electric fields points in the opposite direction (left)

for a parallel R and X, when R = 5\Omega, Effective resistance R = \(\frac{5X}{5 + X}\) and when R = 2\Omega Effective resistance R = \(\frac{2X}{2 + X}\) Thus since E = I(R+r), and if r = 0 then E = IR ∴(i) E = 3(\(\frac{5X}{(5+x)}\)); (ii) E = 6(\(\frac{2X}{(2+x)}\)) ∴3(\(\frac{5X}{(5+X)}\)) = 6(\(\frac{2X}{(2+X)}\)) => (\(\frac{5X}{(5+X)}\)) = 2(\(\frac{2x}{(2+X)}\)) ∴ (\(\frac{5X}{(5+X)}\)) = (\(\frac{4X}{(2+X)}\)) 5x(2+x) = 4x(5+x) 10x + 5x\(^2\) = 20x + 4x\(^2\) 5x\(^2\) - 4x\(^2\) = 20x - 10x => x\(^2\) = 10x ∴x\(^2\) - 10x = 0 x(x - 10) = 0 ∴ x = 0 or 10 ∴ x = 10\Omega

Ir.m.s = Vr.m.s/Z But Z = \sqrt R2 + XL2 = \sqrt 42 + 32 = \sqrt 16 + 9 = \sqrt 25 ∴Ir.m.s = 240/5 = 48.0A

R = \(\frac{v^2}{P} = \frac{220^2}{100} = 806.7\Omega\)