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since currency and finger contain unlike molecules, the force involved is adhesion

Energy stored in a spring = (1)/2 K e2 = (1)/2 x 500(0.05)2 = 0.625J

R\(^2\) = P\(^2\) + Q\(^2\) = 12\(^2\) + 5\(^2\) = 144 + 25 = 169 ∴R = \(\sqrt{169}\) = 13N ∴ the angle between the forces is 90º

Cold water reduces the temperature of the hot water inside the can causing the contraction of the hot water. This reduces the pressure inside the can and the can collapses due to the external greater expression.

Temperature difference between hot and cold junctions of thermocouple is proportional to the emf. generated

The elastic potential energy of the balance wheel reduces as the wheel expands, therefore the clock will loose time.

V1 = 600cm3, T1 = 27 = 273 = 300k T2 = 35 + 273 = 308K, V2 = ? (V)1/T1 = (V)2/T2 = V2 = (V)1/T1 x X T2 = (600 X 308)/300 = 616 cm3

Relative Humidity = (S.V.P. of water at dew point)/S.V.P of water at air temperature =(18)/24 x 100% = 75%

At melting point there is no change temperature

Heat supplied = Heat gained P x t = C ∆ Ø (25 x 5 x 60)/10 = 750 JK-1

To find the primary current of the step-down transformer, we follow these steps: 1. Calculate the Power Input: The efficiency (\( \eta \)) of the transformer is given by: \(\eta = \frac{\text{Power Output}}{\text{Power Input}} \times 100\) Rearranging gives: \(\text{Power Input} = \frac{\text{Power Output}}{\eta}\) Given: - Power Output = 50 W - Efficiency = 80\% = 0.80 Thus, \(\text{Power Input} = \frac{50 \, \text{W}}{0.80} = 62.5 \, \text{W}\) 2. Calculate the Primary Current: The power input can also be expressed in terms of voltage and current: \(\text{Power Input} = V_p \times I_p\) Where: - \( V_p \) = primary voltage (given as 200 V) - \( I_p \) = primary current (what we want to find) Rearranging gives: \( I_p = \frac{\text{Power Input}}{V_p}\) Now substituting the values: \(I_p = \frac{62.5 \, \text{W}}{200 \, \text{V}} = 0.3125 \, \text{A}\) Thus, the primary current of the transformer is: 0.3125A

Left = L1 + L2 + L3 = 5mH + 10mH + 20mH = 35mH

Power supplied = I\(^2\) R = 20\(^2\) X 1 = 400 W Power dissipated = IV = 20 X 220 = 4400 W (Power supplied)/Power dissipated = 4400 W / 400 W = 11 / 1 or 11 : 1

The resistors are in series ∴ Reff = R1 + R2 + ..... = 4 + 4 + 4 + 4 + 4 + 4 = 24 \Omega

Energy consumed = power x time = (60 W x 10 + 40 W x 5) x 24 h = (600 W + 200 W ) x 24 h = 800 W x 24h = 19200 w h = 19.2 kWh