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The bursting of water pipes in very cold weather occurs because water expands as it freezes. This expansion increases pressure inside the pipes, which can lead to bursting.

ºF = (ºC x 1.8) + 32 But ºC = 60º º F = (60 x 1.8) + 32 = 108 + 32 = 140ºF

As altitude rises, there is less air above a particular point, leading to a decrease in atmospheric pressure. Furthermore, the density of air diminishes with increased altitude, which also contributes to this phenomenon.

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