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R.H = \(\frac{Partial Pressure of Water Vapour Present}{S.V.P at temperature of atmosphere}\) x 100 = \(\frac{10}{17.50}\) x 100 = 57.14% ≈ 57%

In vibration in a closed pipe/tube, the first resonance occurs at a position L = V/4f, When reading from the top of the tube, L\(_1\) = \(\frac{340}{4 \times 340}\) = 0.25m = 0.25m L\(_2\) = \(\frac{3V}{4f}\) L\(_2\) = \(\frac{3 \times 340}{4 \times 340}\) = 0.75m L\(_3\) = \(\frac{5V}{4f}\) L\(_3\) = \(\frac{5\times 340}{4\times 340}\) = 1.25m However, the effective length of the tube is 1.2m. Thus reading from below, the first position of resonance will be 1.20 - 0.25m = 0.95m

Form the lens formula = \(\frac{1}{\text{U}}\) + \(\frac{1}{\text{V}}\) = \(\frac{1}{\text{f}}\) => \(\frac{1}{10} + \frac{1}{25}\) = \(\frac{1}{\text{f}}\) ∴ \(\frac{(5 + 2)}{50}\) = \(\frac{1}{\text{f}}\) ∴ f = \(\frac{50}{7}\)cm Since the image is observed on a screen, it is a real image. Thus magnification M =\(\frac{\text{f}}{U - f}\) = \(\frac{\frac{50}{7}}{10 - \frac{50}{7}}\) ∴ M = \(\frac{\frac{50}{7}}{\frac{20}{7}}\) = \(\frac{50}{7}\) x \(\frac{7}{20}\) = 2.5 Or m = \(\frac{\text{v}}{\text{u}}\) = \(\frac{25}{10}\) = 2.5

Two thermal objects are said to be in a thermal equilibrium when the temperature of both objects are equal. Generally, heat flows from an object at a higher temperature to the one at a lower temperature until both temperature are the same, during which the two bodies are said to be in thermal equilibrium

In vibration in a closed pipe/tube as shown in the figure, fundamental note is produced at the first position of resonance. When the length of the air column L = λ/4; => λ = 4L. Since the open end of the tube is taken as an antinode; and closed end as the node. Thus the associated frequency is given by f = V/λ = V/4L ∴ f = V/4L

The increase in length of the steel rail is given by = α 10 = 11 x 10-6 x 20 x (40 -20) = 11 x 10-6 x 20 x 20 = 4.4 x 10-3 ∴ Final length will be original length plus increase in length = 20 + 4.4 x 10-3 = 20 + 0.0044 = 20.0044m

The number of images to be observed will be infinite

Fog is a weather phenomenon when water droplets condense in the air, often around dust or other particulate matter. This condensation significantly reduces visibility. Mist: Similar to fog, but generally has higher visibility.

In general intensity of a wave is the rate of flow of energy across unit area normal to the direction of propagation of the wave intensity α1/r2 where r is the distance away from the point source. Thus if the distance from the point source of the sound wave is doubled, then the intensity will be decreased by a factor of 4

mc\DeltaT = Pt → 5 x c x 30 = 2000 x 10 x 60 c = \(\frac{1200000}{150}\) c = 8000 = 8.0 x \(10^3J/kgºC^{-1}\) But, Heat capacity C = mass x specific heat capacity C = 5 x 8000 = 40000 = 4 x (10^4JºC^{-1}\) (OR) C = \(\frac{2000* 10* 60*}{30}\) = 40,000JºC\(^{-1}\)

Generally in thermal/heat conduction, if 1.2 x 106 J is the heat flow per seconds; 30KM-1 is the temperature gradient of the materials; and 400Wm-1K-1 the thermal conductivity of the material, then we have that: heat energy = thermal conductivity x temp. gradient x time x area i.e = 1.2 x 10^6 = 400 x 30 x1x Area => 400 x 30 x Area = 1.2 x 10^6 ∴ Area = (1.2 x 10^6)/(400 x 30) Area = 100m^2 or 1.0 x 10^2m2

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

From the above diagram, if each of the resistors can dissipate a maximum of 18W, the from the relation power, P = I2R, the current in the 2Ω series resistor is given by: I2 = P/R = 18/2 = 9 => I = √9 = 3A The effective parallel arrangement of the two 2Ω resistors: 1/R = 1/2 + 1/2 = 1Ω ∴ Total resistance in the circuit = 1 + 2 = 3Ω current flowing the circuit = 3A ∴ Maxi. power = P = I2R = 32 x 3 = 27W

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 C.