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Using the equation, R = \(\frac{\rho l}{A}\) we conclude that the wire with the smallest diameter will have the highest resistance i.e. wire E

All the options are reasons why water is not a good thermometric substance, but the most critical one is its lack of uniform expansion. Water's non-linear expansion makes it difficult to correlate temperature changes with volume changes perfectly.

From Hooke's law, F = kx where x = extension, k = force constant K = 680Nm\(^{-1}\), F = 306N x = \(\frac{\text{F}}{\text{x}}\) = \(\frac{306}{680}\) ≈ 0.450m Fractional Change in Length: Given the fractional change is 1.5: 1.5 = \(\frac{\text{x}}{\text{L}}\) L = \(\frac{0.450}{1.5}\) = 0.30m.

k = \(\frac{Q.d}{A. ΔT .t}\) d = thickness = 3m A = 9m\(^2\) Q = 288kJ = 288000J, t = 7200s, ΔT = 90ºC/m → 90K/m x 3 = 270K k = \(\frac{288000 \times 3}{9 \times 270 \times 7200}\) k = 4.93 x 10\(^{-2}\)Wm\(^{-1}\)k\(^{-1}\)

3\(\alpha\) = \(\gamma\) \(\alpha\) = \(\frac{3.9\times 10^{-6}}{3}\) = 1.3 x 10\(^{-6}\) BUT \(\beta\) = 2\(\alpha\) \(\beta\) = 2 x 1.3 x 10\(^{-6}\) = 2.6 x 10\(^{-6}\)K\(^{-1}\)

Vacuum: It prevents conduction and convection because there are no air molecules to transfer heat. Silvered Walls: They minimize radiation by reflecting thermal radiation back.

Thermal expansion can affect accuracy and timekeeping in a watch's balance wheel, making it disadvantageous in this application.

The given graph illustrates the anomalous expansion of water, where water has its highest density and consequently its lowest volume at 4ºC

Electrical heat generated = quantity of heat required to change water from liquid to vapour I\(^2\) R t = mL I = current = 10A, R = resistance = 2\Omega, t = time = 20s, m = mass of water = 5 X 10\(^{-3}\)kg and L = specific latent heat of vaporization. 10\(^2\) x 2 x 20 = 5 X 10\(^{-3}\)L 4000 = 5 X 10\(^{-3}\)L L = \(\frac{4000}{5\times10^{-3}}\) = 800000 ⇔ 8.0 x 10\(^5\)Jkg\(^{-1}\)

Quantity of heat Q = mass of iron x specific heat capacity of iron x change in temperature. Q = mc\(\delta\)\(\theta\) m = 60g ⇒ 0.060, c = 460Jkg\(^{-1}\)k\(^{-1}\) and \(\delta\)\(\theta\) = 75 - 35 = 40ºC Q = 0.060 x 460 x 40 = 1104J.

Quantity of Heat = heat capacity x change in temperature Q = C\(_p\) x \(\Delta\)\(\theta\) Q = 400 x 20 = 8000J = 8kJ

The quantity of heat required to melt ice = mass of ice x specific latent heat Q = m L = 0.2 x 3.4 x 10\(^5\) = 6.8 x 10\(^4\)J

Heat gained by 25º C body = Heat lost by 35º C body M\(_c\) c \(\delta\)\(\theta\) = M\(_h\) c\(\delta\)\(\theta\) 70 x c x (f - 25) = 70 x c x (35 - f) where f = final temperature f - 25 = 35 - f (f and 70 cancel out) 2f = 35 + 25 = 60 f = \(\frac{60}{2}\) = 30ºC

specific heat capacity of water = 4200Jkg\(^{-1}\)k\(^{-1}\), Specific heat of vaporization of water = 2.26 x 10\(^6\)Jkg\(^{-1}\), mass of water = 0.02kg, water vapour temp. = 100ºC, water at 97ºC The amount of heat required = ML + Mc\Delta\(\theta\) = 0.02 x 2.26 x 10\(^6\) + 0.02 x 4200 x (100 - 97) = 45,200 + 252 = 45,452J.

Relative humidity = \(\frac{\text{S.V.P at dew point}}{\text{S.V.P at air temperature}}\) S.V.P at dew point = 7.3mmHg S.V.P at air temperature = 13.5mmHg R.H = \(\frac{7.3}{13.5}\) x 100 \(\approxeq\) 54%