Physics Questions & Answers - Free Online Practice

4,006.

When an alternating current given by I = 10sin (120π)t passes through a 12Ω resistor, the power dissipated in the resistor is

1200W

600W

120W

30W

View answer & explanation

Correct answer is A

An a.c current is usually expressed as I = Io sin wt; and comparing this with the given current I = 10sin (120π)t, => Io = 10.
and power in an a.c circuit is given as:
P = I²R
= (10² x 12)
= 1200W

4,007.

The maximum kinetic energy of the photoelectron emitted from a metal surface is 0.34eV. If the work function of the metal surface is 1.83eV,find the stopping potential.

2.17V

1.49V

1.09V

0.34V

View answer & explanation

Correct answer is D

The stopping potential is the potential required to stop the most energetic photo-emitted electron. From the relation e Vs = K.E,_(max) where e is the electron charge, and Vs is the stopping potential => e x Vs = 0.34eV
∴ Vs = (0.34eV) / e = 0.34V
∴ stopping potential = 0.34V

4,008.

The force on a charge moving with velocity v in a magnetic field B is half of the maximum force when the angle between v and B is

90^o

45^o

30^o

0^o

View answer & explanation

Correct answer is C

the force on a moving charge in a magnetic field is given by: F = q V β sin θ
and this force is maximum when θ = 90^o, during which sin 90^o = 1: and for sinθ = 0.5, θ = 30^o

4,009.

A circuit has an area of 0.4m² and consists of 50 loops of wire. If the loops are twisted and allowed to rotate at a constant angular velocity of 10 rad s^-1 in a uniform magnetic field of 0.4T, the amplitude of the induced voltage is

16V

20V

80V

View answer & explanation

Correct answer is D

The induced voltage = time rate of change of magnetic flux

i/e E = -	NΔ∅_B
	Δt

But ∅_B = Magnetic Flux
= Magnetic Induction (B) x (A)
= 0.4T x 0.4m²
= 0.16 volt second

i/e E = -	NΔ∅_B
	Δt

= 0.16 x 10 x 50
= 80 volts

4,010.

The count rate of a radioactive material is 800 count/min. If the half-life of the material is 4 days, what would the count rate be in 16 days later

200 count / min

100 count / min

50 count / min

25 count / min

View answer & explanation

Correct answer is C

116 days = 4 half - lives
Thus, after 4 half lives the fraction left of any radioactive materials is given by (¹/₂)⁴ = ¹/₁₆
∴ Count rate / amount left = ¹/₁₆ x ⁸⁰⁰/₁
= 50 count / min