JAMB Physics Past Questions & Answers - Page 66

326.

The image of an object is located 6cm behind a convex mirror. If it's magnification is 0.6, calculate the focal length of the mirror

A.

3.75cm

B.

6.70cm

C.

10.00cm

D.

-15.00cm

Correct answer is D

V = -6cm

M = 0.6cm

Recall m = \(\frac{V}{U}\)

U = \(\frac{V}{M}\)

U = \(\frac{6}{0.6}\)

 U = 10

  \(\frac{1}{U}\) + \(\frac{1}{V}\) = \(\frac{1}{F}\)

 \(\frac{1}{10}\) + \(\frac{1}{-6}\) = \(\frac{1}{F}\)

\(\frac{-4}{60}\) = \(\frac{1}{F}\)

 F = \(\frac{60}{-4}\)

=  -15cm

327.
328.

Electrical power is transmitted at a high voltage rather than a low voltage because the amount of energy loss is due to

A.

Heat dissipation

B.

Production of eddy currents

C.

Excessive current discharge

D.

Excessive voltage discharge

Correct answer is A

The primary reason that power is transmitted at high voltages is to increase efficiency. As electricity is transmitted over long distances, there are inherent energy losses along the way. High voltage transmission minimizes the amount of power lost as electricity flows from one location to the next. How? The higher the voltage, the lower the current. The lower the current, the lower the resistance losses in the conductors. And when resistance losses are low, energy losses are low also. Electrical engineers consider factors such as the power being transmitted and the distance required for transmission when determining the optimal transmission voltage

329.

Which of the following instruments is most suitable for measuring the outside diameter of a narrow pipe in a few millimeters in diameter?

A.

Pair of calipers

B.

Meter rule

C.

Micrometer screw gauge

D.

Tape rule

Correct answer is A

No explanation has been provided for this answer.

330.

An object weighs 30N in air and 21N in water. The weight of the object when completely immersed in a liquid of relative density 1.4 is

A.

25.2N

B.

17.4N

C.

12.6N

D.

9.0N

Correct answer is B

Weight of water displaced = upthrust = 30 - 21 = 9N

  Mass of water displaced = \(\frac{9}{10}\) = 0.9kg

  Volume of object = 9 × 10\(^{-4}\)m\(^3\)

  = (9 × 10\(^{-4}\)) (1.4 ×103)

 = 1.26kg = 12N

  30 - 12.6 = 17.4N