2005 - JAMB Physics Past Questions and Answers - page 5

Generally, from Hares Experiment, density is inversely proportional to height

i.e h ∝1/d; → h₁d₁ = h₂d₂

∴ 8 x 10³ = h₂ x 800

= 10cm

From the heat exchange involved, the latent heat of vaporization, L can be deduced from the relation:

ML = 10⁶, where M = mass of the liquid.

∴ 2 x L = 10⁶

=> L = ¹⁰⁶/₂ = 5.0 x 10⁵J kg^-1

The effective capacitance in the circuit is given by:

1/C = ¹/₂ + ¹/₃ + ¹/₆ = ⁶/₆

therefore C = 1μF or

1.0 x 10^-6F.

Since C = Q/V, its implies that the charges Q, flowing in the circuit is given by Q = CV

= 1.0 x 10^-6 x 12

= 12.0 x 10^-6 Coulomb.

∴ P.d across the 6μF,

= 2.0 volts

For the two point charges, the magnitude of the electrostatic force between them is given by the equation:

F ∝ ¹/_r², and if the distance is increased by a factor of 4, that is new distance = 4r,

then, new force F ∝ ¹/_(4r)²

= ∝ ¹/_16r²

Thus it will be ¹/₁₆ of the former value

In general, the velocity of sound in air varies directly as the square root of temperature measured in kelvin.

That V ( \propto \sqrt{T} \implies V^2 \propto T. \

\text{Therefore} \frac{V^2_1}{T_1} = \frac{V^2_2}{T_2} \

\text{Thus Let } V_1 = 4m/s \

  T_1 = 10K \\

\text{Therefore } V_2 = 2V_1 = 8m/s \

\implies \frac{4^2}{10} = \frac{8^2}{T_2} \

T_2 = \frac{64 \times 10}{16} = 40K\

T_2 = 4T_1 )

Thus when the velocity of sound in air is doubled, it's absolute temperature will be quadrupled.

let the masses of the two objects be M1 and M2

and their distance apart = r

Therefore the force acting F1 = ( \frac{GM1M2}{r^2} )

for a new distance of r/2, we have a force of

F2 = ( \frac{GM2M2}{r/2} \

\text{Thus } F1 \times r^2 = GM1M1 \text{and} \

      F2 \times (r/2)^2 = GM1M1 \\

\text{Since } GM1M1 = GM1M1 \

\implies F1 \times r^2 = F2 \times (r/2)^2 \

\text{Therefore } 200 \times r^2 = F2 \times r^2/4 \

F2 = \frac{4 \times 200 \times r^2}{r^2} = 800N )

l =l1 + l2 = 3 + 6 = 9H

For Hooke's law F = Ke

( \implies F/e = K \

f1/e1=f2/e2 \

\text{Let the original length} = t_0 \

\text{therefore} e1 = (36 - t_0)cm ; e2 = (46 - t_0)cm \

\text{if f1} = 40N \text{f2} = 60N \

\text{Then } \frac{40}{36 - t_0} = \frac{60}{45 - t_0} \

\implies 40(45 - t_0) = 60(36 - t_0) \

\text{therefore } 1800 - 40t_0 \

 2160 - 60t_0 \\

60t_0 - 40t_0 = 2160 - 1800 \

20t_0 = 360  \  <br />

t_0 = 18cm \)