2004 - JAMB Physics Past Questions and Answers - page 4

Pressure = Force / Area = Weight of the man / Area of his feet

=> 2.8 x 10³ = Weight of the man / 4 x 10^-2

∴ Weight of the man = 2.8 x 10³ x 4 x 10^-2

= 112N

From the relation refractive index = Sin ( \left(\frac{A+Dm}{2}\right) )

Where A = refractive angle = 60⁰, Sin( \left(\frac{A}{2}\right) )

( \implies 1.4 = \frac{\text{Sin}\left(\frac{60^0 + Dm}{2}\right)}{\text{Sin}\left(\frac{60}{2}\right)} \

= \frac{\text{Sin}\left(\frac{60 + Dm}{2}\right)}{\text{Sin}30^0} \

= \frac{\text{Sin}\left(\frac{60 + Dm}{2}\right)}{0.5}\

\text{Therefore} 1.4 \times 0.5 = \text{Sin}\left(\frac{60 + Dm}{2}\right) \

\text{Therefore} \left(\frac{60 + D_m}{2}\right)= \text{Sin}^{-1} 0.7000 \

= 44^0 12^1 \

60^0 + D_m = 88^0 24^1 \

D_m = 88^0 24^1 - 60^0 \

= 28^0 24 \

29^0 )

E = V + v

where E = emf of cell.

V = terminal p.d

v = loss voltage of the cell.

But from ohm's law V=IR ; v = Iv

( \implies ) lost voltage = V = Ir

= 4 x 0.5

= 2.0v

Original length = 20m

first new length = 20.01m

first increase in length = 20.01m - 20m = 0.01m

Second new length = 20.05m

second increase in length = 20.05 - 20m = 0.05m

From hooke's law = ( \frac{F_1}{e_1} = \frac{F_2}{e_2} \implies \frac{50}{0.01} = \frac{F_2}{0.05} \

\implies F_2 = \frac{50 \times 0.05}{0.01} = 250N )

Average life / Half life (^Tl/₂) = ^ln2/_λ = ^0.693/_10^-7

= 0.693 x 10⁷

= 6.93 x 10⁸

Current (I) in the circuit = ^V/_Z
But z =

√ R² + (X_L - X_C)² 

=

√ 8² + (16 – 10) ² 

=

√ 64 + 36 

=

√ 100  = 10Ω

∴ Current (I) = ^V/_Z = ¹²/₁₀ = 1.2 A

Temperature coefficient of resistance is given by

And for metals where resistance increase with temperature rise, temperature coefficient of resistance is positive; on the other hand in semi conductors, where resistance decrease with temperature rise, temperature coefficient of resistance is negative.