2010 - JAMB Physics Past Questions and Answers - page 5

I = E/(R+r)

= 12/6 = 2A

¹/_R = ¹/₈ + ¹/₈ = ²/₈

∴ R = 4Ω
^l/_l1 = ^R/_R1

(x)/(100-x) = 4/8

4(100-x) = 8x

400-4x = 8x

400 = 8x + 4x = 12x

x = 400/12 = 33.3cm

¹/_R = ¹/₄ + ¹/₄ + ¹/₄ = ³/₄

R = ⁴/₃

I = ^V/_R

= 12 A

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

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

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

(\frac{2\sqrt{3}} + \sqrt{5}}{\sqrt{5} - \sqrt{3}) = (\frac{\sqrt{5}} + 2\sqrt{3}}{\sqrt{5} - \sqrt{3}) x (\frac{\sqrt{5}} + \sqrt{3}}{\sqrt{5} + \sqrt{3})

(\frac{5 + \sqrt{15} + 2\sqrt{15} + 2 \times 3}{(\sqrt{5} - \sqrt{3})^2)

= (\frac{5 + 3\sqrt{15} +6} {5 - 3))

= (\frac{11 + 3\sqrt{15}}{2})

The weight of a body is the resultant of earth's gravitational pull on it. The earth is not a perfect sphere, it is more flattened at the poles and bulges at the equator. Hence an object at the pole is closer to the earth's center than the one at the equator. The weight of a body at the poles should be more than that of when weighed at the equator.

Thus weight varies from place to place while the mass of a body being the quantity of matter it contains remains the same at different points on the earth.