WAEC Past Questions and Answers - Page 3229

16,141.

The energy required to separate the nucleus in an atom completely is known as

A.

mass defect of the atom

B.

electron binding energy

C.

excitation energy

D.

nuclear binding energy

Correct answer is D

No explanation has been provided for this answer.

16,142.

The circuit above illustrates a standard resistor R, voltage source E and a lamp L connected in series. If the temperature of R is increased, the brightness of the lamp will

A.

increase sharply

B.

decrease slightly

C.

remains the same

D.

increase gradually

Correct answer is B

No explanation has been provided for this answer.

16,143.

A transformer has 400 turns of wire in the primary coil and 40 turns in the secondary col. If the input voltage is 150 volts. Calculate the magnitude of the output voltage

A.

150V

B.

55V

C.

36V

D.

15V

Correct answer is D

\(\frac{ES}{EP} = \frac{Ns}{Np} = \frac{Es}{150} = \frac{40}{400}\)

Es = \(\frac{150 \times 40}{400}\)

= 15V

16,144.

A wire of length 100 cm has a resistance of 10\(\Omega\). If its cross-sectional area is 0.005cm2, determine its resistivity

A.

0.0005\(\Omega\)cm

B.

0.0015\(\Omega\)cm

C.

0.0016\(\Omega\)cm

D.

0.0700\(\Omega\)cm

Correct answer is A

e = \(\frac{RA}{L} = \frac{10 \times 0.005}{100} = 0.0005\)

16,145.

In the circuit diagram above, E is a battery of negligible internal resistance. If its emf is 9.0V. Calculate the current in the circuit 

A.

1.8A

B.

1.0A

C.

0.8A

D.

0.3A

Correct answer is B

\(\frac{1}{r} = \frac{1}{r_1} + \frac{1}{r_2}\)

= \(\frac{1}{5} + \frac{1}{20}\)

= \(\frac{4 + 1}{20} = \frac{5}{20}\)

r = \(\frac{20}{5}\)

= 4\(\Omega\)

R = R1 + R2

= 5 + 4

= 9\(\Omega\)

but V = IR

I = \(\frac{V}{R} = \frac{9}{9}\)

= 1A