JAMB Physics Past Questions & Answers - Page 414

2,066.

Marching soldiers crossing a suspension bridge are usually advised to break their steps to avoid damaging the bridge owing to

A.

oscillation

B.

resonance

C.

swinging

D.

vibration

Correct answer is B

Marching soldiers crossing a suspension bridge are usually advised to break their steps to avoid damaging the bridge owing to resonance. The steps of the marching soldiers can set the bridge into vibration, and when the frequency of the bridge is equal to that of the steps of the soldiers, the resonance occurs,and at this resonance, the bridge vibrates violently with maximum amplitude, and can collapse.

2,067.

A man exerts a pressure of 2.8 x 103Nm-2 on the ground and has 4 x 10-2m2 of his feet in contact with the ground. The weight of the man is

A.

102N

B.

70N

C.

140N

D.

112N

Correct answer is D

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

2,068.

Transverse waves can be distinguished from longitudinal waves using the characteristic of

A.

diffraction

B.

refraction

C.

polarization

D.

reflection

Correct answer is C

No explanation has been provided for this answer.

2,069.

Metal rods of length 20m each are laid end to end to form a bridge at 25oC. What gap will be provided between consecutive rails for the bridge to withstand 75oC?
[Linear expansivity of the material = 2.0 x 10-5K-1

A.

0.20m

B.

0.22m

C.

0.25m

D.

0.02m

Correct answer is D

Safety gap = expansion of the rod

= \(\alpha \times L_0 \times \Delta t\)

where \(\alpha\) = linear expansivity of the material

\(L_0\) = Initial length of the material

\(\Delta t\) = Change in temperature

\(\therefore\) The safety gap = \(2 \times 10^{-5} \times 20 \times (75 - 25)\)

= \(2 \times 10^{-5} \times 10^3\)

= \(2 \times 10^{-2} m\)

= 0.02 m

2,070.

A 50W electric heater is used to heat a metal block of mass 5kg. If in 10 minutes a temperature rise of 12oC is achieved, the specific heat capacity of the metal is

A.

400 J kg-1K-1

B.

500 J kg-1K-1

C.

390 J kg-1K-1

D.

130 J kg-1K-1

Correct answer is B

Heat supplied by the heater in 10 mins
= power x time
= 50 x (10 x 60)
= 30000 J
This heat is equal = MCΔt
= 5 x c x 12
= 60cJ
= 60C = 30,000
C = 30,000/60
C = 500 J kg-1K-1