JAMB Physics Past Questions & Answers - Page 453

2,261.

A stone of mass 1kg is dropped from a height of 10m above the ground and falls freely under gravity. Its kinetic energy 5m above the ground is then equal to

A.

its kinetic energy on the ground

B.

twice its initial potential energy

C.

its initial potential energy

D.

half its initial potential energy

Correct answer is D

At the height of 10m, its total energy is potential, given by P.E. = mgh = 1 x 10 x 10 = 100J.
But as the stone descends, its potential energy decreases, while its kinetic energy increases.
Half way during the fall, (h = 5m), its P.E. = 50J, and consequently, its K.E. should be 50J, such that from the principle of conservation of energy, the total energy of the stone at any instant should be conserved.

2,262.

The height at which the atmosphere ceases to exist is about 80km. If the atmospheric pressure on the ground level is 760mmHg, the pressure at a height of 20km above the ground level is

A.

380mmHg

B.

570mmHg

C.

190mmHg

D.

480mmHg

Correct answer is B

(p - 760)/(0 - 760) = (h - 0)/(80 - 0)
∴ (p - 760)/-760 = (20 - 0)/80
80(p-760) = -760(20-0)
∴80p - 60800 = -15200
80p = 60800 - 15200
80p = 45600
∴ p = 45600/80
p = 570mmHg

2,263.

Light of energy 5eV falls on a metal of work function 3eV and electrons are liberated. The stopping potential is

A.

1.7V

B.

2.0V

C.

8.0V

D.

15.0V

Correct answer is B

For a photoemitted electron,
eV = K.E.max, where V = stopping potential and K.E. = maximum kinetic energy of the photo emitted electron given by K.E. = Energy of incident rad - Work - function
= 5ev - 3ev = 2ev
∴ eV = 2eV
V = 2eV/e = 2V
∴ Stopping Potential = 2.0V

2,264.

A conductor of length 2m carries a current of 0.8A while kept in a magnetic field of magnetic flux density 0.5T. The maximum force acting on it is

A.

0.2N

B.

0.8N

C.

3.2N

D.

8.0N

Correct answer is B

The force on the conductor is maximum when the conductor lies perpendicular to the field during which the maximum force is given as;
F = ILB = 0.8 x 2 x 0.5 = 0.8N

2,265.

A cell of internal resistance 1Ω supplies current to an external resistor of 3Ω. The efficiency of the cell is

A.

25%

B.

33%

C.

50%

D.

75%

Correct answer is D

Efficiency = Output/input x 100/1
Efficiency = 3/(3+1) x 100/1
= 3/4 x 100/1 = 75%