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JAMB Physics Past Questions & Answers - Page 87

431.

Under which of the following conditions is workdone

A.

A man supports a heavy food above his head with his hands

B.

A boy climbs onto a table

C.

A man pushes unto a table

D.

A woman holds a pot of water

Correct answer is B

 

For work to be done, distance must be involved because
workdone = force × distance

432.

Under what conditions are cathode rays produced in a discharge tube?

A.

High pressure and low voltage

B.

High pressure and high voltage

C.

Low pressure and low voltage

D.

Low pressure and high volatge

Correct answer is D

 

Cathode rays are produced in a discharge tube under Low pressure and high voltage

433.

Calculate the upthrust on an object of volume 50cm3 which is immersed in liquid of density 103 kgm-3 [g = 10ms-2]

A.

0.8N

B.

2.5N

C.

0.5N

D.

1.0N

Correct answer is C

Upthrust = change in weight

density = massvolume

Mass of liquid displaced = Density of Liquid × Volume;

= 103×50×106m3

= 0.05kg

(Note that the volume of 50cm3 was converted to m3 by multiplying by 106) i.e 50cm3=50×106m3

Since mass displaced = 0.05kg

Upthrust = mg = 0.05 × 10

= 0.5N

434.

Why do soldiers march disorderly while crossing a bridge?

A.

To prevent resonance on the bridge

B.

To set the bridge into resonance

C.

To make the b ridge collapse

D.

To spread their weight evenly on the bridge

Correct answer is A

Soldiers march disorderly while crossing a bridge to prevent resonance because when soldiers march in orderly manner across a bridge, they generate a rhythmic oscillation of wave on the bridge and at a certain point, would start oscillation to the same rhythm as that of the marching steps. This oscillation would reach a maximum peak when the bridge can no longer sustain its own strength and hence collapses. So for this reasons, soldier are ordered to march disorderly while crossing a bridge.

435.

Calculate the electric field intensity between two plates of potential difference 6.5V when separated by a distance of 35cm.

A.

18.57NC1

B.

53.06N C1

C.

2.28NC1

D.

0.80NC1

Correct answer is A

Electric Field Intensity (E) = vd

= potential differencedistance

= 6.5v35cm

= 6.535×102

= 18.57NC1