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T = 27 + 273 = 300 K, volume, V = 10-2 m3 molar gass constant, R = 8.3 Jmol- 1k - 1 number of mole, n = 1 PV = NRT i.e P = (NRT)/V = (1 x 8.3 x 300)/10-2 = 2.49 x 10\(^5\) Nm\(^-2\)

Metals P, Q original length lP, lQ Increase in length \DeltalP, \DeltalQ Linear expansivities \alphaP, \alphaQ ∴ \alphaP/\alphaQ = 2/3 ; and lP/lQ = 3/4 But increase in length, \Deltal = \alphal\Delta\theta ∴\DeltalP = \alphaPlP\Delta\theta \DeltalQ = \alphaQlQ\Delta\theta ∴ \DeltalP = \alphaPlP\Delta\theta = 2 x 3 = 1 \DeltalQ \alphaQlQ\Delta\theta 3 4 2 ∴ Ratio of increase in length of P to Q = 1:2

A saturated vapour is one which is in dynamic equilibrium with its own liquid; that is as many go from the liquid to the vapour state every second return back to liquid.

The acceleration of free fall due to gravity is independent of mass of the body. Therefore, ignoring air resistance, which affects feather more than the stone, the stone and the feather will both reach the ground at the same time

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

Since the charged rod is not in contact with the electroscope, the leaves divergence will diverge.

IN magnetisation by stroking, the last touched has a pole opposite to that of the magnet used. Therefore, the poles T and V are N and N respectively

H eat generated in a wire carrying current, H = 12 Rt. so, when voltage is high and current is low heat generated in the wire will be minimal. Therefore, high voltage and low current prevent overheating of the coil.

C = C1 + C2 + C3 = 2 + 4 + 8 = 14\muF V = 6V Energy = 1/2CV2 1/2 x 14 x 10-6 x 62 = 2.52 x 10-4 J

pumping out of the gas in the discharge tube reduces the pressure inside the tube, therefore, electricity is conducted at low pressure

I = E/(R+r) = 12/6 = 2A

In Faradays law of electrolysis M = It = Q Where M is the mass of substance deposit . Q is the quantity of charge. A graph of m against Q is a straight line passing through the origin with slope equal to the electrochemical equivalent.Z

r = resistance of the galvanometer l = current through galvanometer = 1mA or 0.01A V1 = p.d across the galvanometer = I X r = 0.01 X r : V1 = 0.01r V2 = p.d across the multiplier = 5 - 0.01r R = resistance of the multiplier = 498\Omega where R = \(\frac{V}{I}\) --> 498 = \(\frac{5 - 0.01r}{0.01}\) CROSS MULTIPLY --> 498 X 0.01 = 5 - 0.01r 0.01r = 5 - 4.98 ∴ r = \(\frac{0.02}{0.01}\) = 2\Omega

Energy consumed = power x time = 200W x 24hr = 4800wh = 4.8 kwh But 1 kwh cost N40 ∴ 4.8kwh costs 4.8 x 40 = N192

Total inductance,L = L1 + L2 = m 15 + 5 = 20mH = (20)H/1000 1 = 5A Energy of inductor, (1)/2 L I2 = (1)/2 X (20)/1000 X 5-2 = 0.25 J