JAMB Physics Past Questions & Answers - Page 371

1,851.

The ground stated energy for the hydrogen atom is 5.44 x 10-19J . If an electron drops from zero energy to ground state, calculate the frequency of the emitted radiation?

[h = 6.6 x 10-34 Js]

A.

5.0 x 1016 Hz

B.

2.0 x 1016 Hz

C.

5.0 x 1015 Hz

D.

2.0 x 1015 Hz

E.

No correct option

Correct answer is E

∆E = hf
f = (∆E)/h = (5.44 x 10-19)/6.6 x 10-34 = 8.24 x 1014

1,852.

In a discharge tube, most of the gas is pumped out so that electricity is conducted at?

A.

low voltage

B.

steady voltage

C.

high pressure

D.

low pressure

Correct answer is C

pumping out of the gas in the discharge tube reduces the pressure inside the tube, therefore, electricity is conducted at low pressure

1,853.

In alternating current circuit at resonance, the angle of lead or lag is?

A.

π°

B.

(π)/2°

C.

(π)/3°

D.

Correct answer is D

Resonance occurs whenever the phase angle of the circuit is zero, The only way that f = 0 is if XL = XC

1,854.

Two inductors of inductance 5 mll and 15 mll are connected in series and a current of 5A flows through them. the total energy stored in the inductor is?

A.

500.0J

B.

250.0J

C.

62.5J

D.

50.0J

Correct answer is E

Total inductance,L = L1 + L2 = m
15 + 5 = 20mH = (20)H/1000
1 = 5A
Energy of inductor, (1)/2 L I2
= (1)/2 X (20)/1000 X 5-2
= 0.25 J

1,855.

A galvanometer with full-scale deflection of 10 mA is to be converted to a voltmeter with full-scale deflection of 5v.  if a series resistance of 498Ω is used for the conversion, the resistance of the galvanometer is ?

A.

1Ω

B.

2Ω

C.

5Ω

D.

10Ω

Correct answer is B

r = resistance of the galvanometer

l = current through galvanometer = 1mA or 0.01A

V1 = p.d across the galvanometer = I X r = 0.01 X r

: V1 = 0.01r

V2 =  p.d across the  multiplier = 5 - 0.01r


R = resistance of the multiplier = 498Ω
 

where R = \(\frac{V}{I}\) --> 498 = \(\frac{5 - 0.01r}{0.01}\)

CROSS MULTIPLY --> 498 X 0.01 = 5 - 0.01r

0.01r = 5 - 4.98

∴ r = \(\frac{0.02}{0.01}\) = 2Ω