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Electric current is a fundamental quantity in physics while the rest are not.

At constant pressure, volume ∝ absolute temperature (Charles’ law). \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\) \(\frac{200}{273} = \frac{V_2}{546}\) v\(_2\) = \(\frac{200 \times 546}{273}\) = 400cm\(^3\)

The pressure exerted by the water at the bottom of the reservoir is given by: \(P = \frac{F}{A}\) Given: \( F = 5000 \, \text{N} \) (weight of water), \( L = 10 \, \text{m} \) (length), \( B = 5 \, \text{m} \) (breadth) Calculate the area \( A \): \(A = L \times B = 10 \, \text{m} \times 5 \, \text{m} = 50 \, \text{m}^2\) Calculate the pressure \( P \): \(P = \frac{F}{A} = \frac{5000 \, \text{N}}{50 \, \text{m}^2} = 100 \, \text{Pa}\)

Resistance thermometers respond faster because they have small sensor mass and use direct electrical detection. Liquid-in-glass and gas thermometers are slower due to thermal expansion and larger thermal inertia, often taking minutes to equilibrate.

Using Charles' law: \(\frac{V_1}{T_1} = \frac{V_2}{T_2} \quad \text{(T in Kelvin, constant P)}\) Calculate \( T_1 \): \(T_1 = 57 + 273 = 330 \, \text{K}\) Since \( V_2 = \frac{1}{5} V_1 \) (volume shrinks to one-fifth), we find \( T_2 \): \(T_2 = T_1 \times \frac{V_2}{V_1} = 330 \times \frac{1}{5} = 66 \, \text{K}\) In ^\circC, this is: \(T_2 = 66 - 273 = -207 \, ^\circC\)

100º → 100 70º → x -20º → 0 Using the interpolation method \(\frac{70 -(-20)}{100 -(-20)}\) = \(\frac{x - 0}{100 - 0}\) \(\frac{90}{120} = \frac{x}{100}\) 120x = 90 x 100 = 9000 x = \(\frac{9000}{120}\) = 75.0ºC

L = 2 m; \alpha = 1.8 \times 10\(^{-5}\) K\(^{-1}\)); \Delta\theta = 100 K \alpha = \(\frac{\DeltaL }{ (L\Delta\theta)}\) \DeltaL = \alpha L \Delta \theta \DeltaL = 1.8 \times 10\(^{-5}\) x 2 x 100 = 3.6 x 10\(^{-3}\) \(\therefore\) \(\Delta\) L = 0.0036m.

P\(_1\) = 20 m + 10 m = 30 m = 3000 cm V\(_1\) = 2 cm\(^3\) P\(^2\) = 10 m = 1000 cm V\(_2\) = ? P\(_1\)V\(_1\) = P\(_2\)V\(_2\) 3000 \times 2 = 1000 \times V\(_2\) V\(_2\) = \(\frac{3000 \times 2}{1000}\) ∴ V\(_2\) = 6 cm\(^3\)

P\(_1\) = 200 Nm\(^{-2}\); P₂ = 100 Nm\(^{-2}\); T\(_1\) = 127 + 273 = 400 K; T\(_2\) = -73 + 273 = 200 K; \(\frac{V_2}{V_1}\) = ? \(\frac{P_1V_1 }{ T_1} = \frac{P_2V_2 }{T_2}\) \(\frac{V_2}{V_1}\) = \(\frac{P_1 T_2}{P_2 T_1}\) \(\frac{V_2}{V_1}\) = \(\frac{200 \times 200}{100 \times 400}\) = \(\frac{1}{1}\) = 1: 1

The apparent cubic expansion of mercury is the increase in its volume relative to the glass vessel (i.e., the volume of mercury that would overflow). Let: \( V_0 = 100 \, \mathrm{cm}^3 \) (initial volume of mercury), \( \gamma_m = 1.8 \times 10^{-4} \, \mathrm{K}^{-1} \) (real cubic expansivity of mercury), \( \gamma_g = 2.4 \times 10^{-5} \, \mathrm{K}^{-1} \) (cubic expansivity of glass), \( \Delta \theta = 100^\circ\mathrm{C} - 10^\circ\mathrm{C} = 90 \, \mathrm{K} \) (temperature rise). The coefficient of apparent expansion is \(\gamma_a = \gamma_m - \gamma_g = 1.8 \times 10^{-4} - 2.4 \times 10^{-5} = 1.56 \times 10^{-4} \, \mathrm{K}^{-1}. \) The apparent cubic expansion (change in volume) is \(\Delta V_a = V_0 \cdot \gamma_a \cdot \Delta \theta = 100 \times (1.56 \times 10^{-4}) \times 90.\) First compute \( 1.56 \times 10^{-4} \times 90 = 0.01404 \), then \(\Delta V_a = 100 \times 0.01404 = 1.404 \, \mathrm{cm}^3.\) (Alternatively: Real expansion of mercury = \( 100 \times 1.8 \times 10^{-4} \times 90 = 1.62 \, \mathrm{cm}^3 \). Expansion of glass vessel = \( 100 \times 2.4 \times 10^{-5} \times 90 = 0.216 \, \mathrm{cm}^3 \). Apparent expansion = \( 1.62 - 0.216 = 1.404 \, \mathrm{cm}^3 \).) Thus, the apparent cubic expansion of the mercury is \( 1.404 \, \mathrm{cm}^3 \).

Using Charles' law; \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\) \(\frac{546}{(0 + 273)} = \frac{V_2}{(10 + 273)}\) V\(_2\) = \(\frac{546 \times 283}{273}\) = 566 cm\(^3\)

Answer: D Given a rod of length \( L = 0.2 \, \text{m} \) with a temperature gradient of \( 416 \, \text{K/m} \) and one end at \( T_1 = 233^\circ \text{C} \): \(\Delta T = \text{gradient} \times L = 416 \, \text{K/m} \times 0.2 \, \text{m} = 83.2 \, \text{K}\) The temperature at the other end \( T_2 \) is: \(T_2 = T_1 + \Delta T = 233^\circ \text{C} + 83.2 \, \text{C} = 316.2^\circ \text{C}\) Thus, the temperature at the other end is \( 316.2^\circ \text{C}\).

Clouds are collections of condensed water droplets or ice crystals suspended in the air, but high above the ground (typically hundreds of meters to kilometers up). Fog is the same phenomenon (condensed suspended droplets, often with dust), but at ground level, reducing visibility. The question specifies "in the air" near the surface (implied by dust particles and suspension context), so fog is correct, not cloud.

a. The boiling point of a liquid is the temperature at which a (pure) liquid changes from the liquid to the gaseous state without a change in temperature when the atmospheric pressure is normal (760 mmHg) b. Aim: To determine the boiling point of a small quantity of a liquid Apparatus: J-tube, thermometer, water, beaker & source of heat. Procedure: The J-tube, as shown in the diagram, is heated in a beaker of water, and the water in the J-tube is monitored. It remains trapped until the water in the beaker is boiling. Then the water in the tube comes to the same level in each limb, showing that the pressure of the vapour in the closed limb is equal to the pressure of air outside. Conclusion: Equal levels in the two limb shows that the vapour pressure is equal to the atmospheric pressure, which is the boiling point of the liquid. The experiment is repeated, and the mean is taken. c. (i). Presence of impurities (ii). Pressure on the liquid (iii) Nature of liquid. d. According to the kinetic theory of matter, temperature depends on the average kinetic energy of the molecules. When pure water boils, the heat supplied is not used to increase the kinetic energy of the molecules. Instead, it is used to overcome the intermolecular forces holding the water molecules together in the liquid state. This energy, called latent heat of vaporization, separates the molecules to form steam. Since the kinetic energy of the molecules does not increase, the temperature remains constant until all the water has changed to steam.

Weight on Earth: \(W_e = 2.5 \times 10^3 \, \text{N}, \quad g_e = 10 \, \text{m/s}^2\) Mass: \(m = \frac{W_e}{g_e} = \frac{2500}{10} = 250 \, \text{kg}\) Moon's gravity is approximately \( \frac{1}{6} \) of Earth's: \(g_m = \frac{10}{6} \, \text{m/s}^2\) Weight on the Moon: \(W_m = m \times g_m = 250 \times \frac{10}{6} = \frac{2500}{6} \approx 416.67 \, \text{N}\) Thus, the weight on the Moon is: 416.67N