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

Change of state (e.g., melting, boiling, freezing) occurs at constant temperature, but is always accompanied by: Change in volume (usually significant) and Change in heat content (latent heat absorbed or released)

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

Protons and neutrons have approximately the same mass, about 1.67 \times 10\(^{-24}\) grams

\(\frac{1}{v}\) + \(\frac{1}{u}\) = \(\frac{1}{f}\) \(\frac{1}{v}\) = +ve real image distance where; U = +ve real object distance f = real focus = 10 m = magnification M = 2 = \(\frac{v}{u}\) 2u = V \(\frac{1}{2u}\) + \(\frac{1}{u}\) = \(\frac{1}{10}\) 2u = 30 u = 15cm v = 2u = 2 x 15 = 30cm

The liquid having a higher thermal conductivity means that the liquid allows heat to pass through itself quickly hence a shorter time to respond.

Extension & Load Extension = K constant \(\frac{\text{load}}{0.56cm}\) = \(\frac{X}{20g}\) X = 2.24cm

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

for transformer, Es = Ns, Es = e.m.f of secondary Ep = Np, Ep = e.m.f of primary NS = number of turns in secondary Ns = number of turns in primary \(\frac{30}{300}\) = \(\frac{x}{100}\) 300X = 3000X = 10 volts

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

H \(\alpha\) I2RT

Relative density of solid = \(\frac{\text{weight of solid}}{\text{weight of an equal volume of water}}\) \(\frac{4.8}{4.8 -3.2}\) = \(\frac{4.8}{2}\) = 2.4 Ratio of density of liquid = \(\frac{\text{weight of liquid kerosine}}{\text{weight of an equal volume of water}}\) = \(\frac{4.8 - 3.2}{4.8 - 2.8}\) = \(\frac{1.6}{2}\) = 0.8 = \(\frac{\text{density of solid}}{\text{density of kerosine}}\) = \(\frac{2.4}{0.8}\) = 3

The intercept on either of the axes of the graph of \(\frac{1}{u}\) against \(\frac{1}{v}\) gives the value of \(\frac{1}{f}\). Hence, the reciprocal of this intercecpt gives the value of the focal length.

Calculate the weight of the body: \(W = mg = 6 \, \text{kg} \times 10 \, \text{m/s}^2 = 60 \, \text{N}\) The limiting frictional force balances the component of weight acting down the incline: \(F = W \sin \theta\) Substituting the known values: \(30 = 60 \sin \theta\) Solving for \( \sin \theta \): \(\sin \theta = \frac{30}{60} = \frac{1}{2}\) The angle \( \theta \) corresponding to \( \sin \theta = \frac{1}{2} \) is: \(\theta = 30^\circ\) The angle of inclination \( \theta \) is \(30^\circ\).

When K is open, the total resistance R = 8\(\Omega\) + 2\(\Omega\) = 10\(\Omega\) I = \(\frac{V}{R}\) = \(\frac{6}{10}\) = 0.6A When K is closed, we have \(\frac{1}{R}\) = \(\frac{1}{8}\) + \(\frac{1}{8}\) ⇒ \(\frac{1}{R}\) = \(\frac{1}{4}\) ⇒ R = 4\(\Omega\) The total resistance R = 4\(\Omega\) + 2\(\Omega\) = 6\(\Omega\) I = \(\frac{6}{6}\) = 1A Hence, I = 1A - 0.6A = 0.4A increase