Physics 2008 Jamb Examination

Heating a pure semiconductor generates more charge carriers (electrons and holes), thus increasing conductivity and reducing resistance.

The eyepiece in a compound microscope functions as a magnifier, enlarging the real image formed by the objective lens.

Angular magnification is given by $$M = 1 + \dfrac{D}{f} = 1 + \dfrac{25}{5} = 6.$$ Since the image is virtual, magnification is taken as negative, hence $$M = -6.$$

The chain conducts excess static charges safely to the ground, preventing sparks that could ignite fuel vapors.

Power supplied externally: $$P_{ext} = V_t I = 220 \times 20 = 4400\ \text{W}.$$ Power dissipated internally: $$P_{loss} = I^2 r = 20^2 \times 1 = 400\ \text{W}.$$ Ratio $$= \dfrac{4400}{400} = 11:1.$$ But since total emf power is 4800 W, external:supplied ratio is $$4400:400 = 11:1.$$ 

Total power
P = (10 × 60) + (5 × 40) = 600 + 200 = 800 W = 0.8 kW

For 24 h:
E = 0.8 × 24 = 19.2 kWh

Total solar eclipse depends on the relative motion of the earth, sun and moon (III) and the rectilinear propagation of light (IV).

Magnification: $$\dfrac{h_i}{h_o} = \dfrac{v}{u}.$$ With $$h_i=2, h_o=4, v=6,$$ we have $$\dfrac{2}{4} = \dfrac{6}{u} \Rightarrow u=12\ \text{cm}.$$

Mirror formula: $$\tfrac{1}{f} = \tfrac{1}{v} + \tfrac{1}{u},$$ with $$f=10, u=20.$$ So $$\tfrac{1}{v} = \tfrac{1}{10} - \tfrac{1}{20} = \tfrac{1}{20} \Rightarrow v=20\ \text{cm}.$$ Magnification $$m = v/u = 20/20 = 1.$$ Image height = $$5 \times 1 = 5\ \text{cm}.$$

Convex mirrors always form erect, virtual and diminished images, making them suitable for wide-field driving mirrors.