Physics Questions & Answers - Free Online Practice

4,211.

The equation of a wave traveling along the positive x-direction is given by;
y = 0.25 x 10\(^{-3}\) sin (500t - 0.025x).
Determine the angular frequency of the wave motion.

0.25 x 10^-3rad s^-1

0.25 x 10^-1rad s^-1

5.00 x 10²rad s^-1

2.50 x 10³rad s^-1

View answer & explanation

Correct answer is C

The general wave equation is given by;
y = A Sin(wt - 2πx/Wavelength)
where w = angular velocity

Thus comparing this equation with the equation
y = 0.25 x 10-3 Sin (500t - 0.025x)
That Amplitude = 0.25 x 10-3m
the angular velocity(w) = 500 (the coefficient of t in both equations)
∴ w = 500 rad s-1
= 5.00 x 102rad s-1

4,212.

Calculate the mass of ice that would melt when 2kg of copper is quickly transferred from boiling water to a block of ice without heat loss;
Specific heat capacity of copper = 400JKg^-1K^-1
Latent heat of fusion of ice = 3.3 x 10⁵JKg^-1

8/33kg

33/80kg

80/33kg

33/8kg

View answer & explanation

Correct answer is A

Heat given off by the copper = MCt
= 2 x 400 x (100 - 0)
Heat absorbed by the ice = ML
= M x 3.3 x 10⁵
Heat lost = Heat gained
∴ M x 3.3 x 10⁵ = 2 x 400 x 100
∴ M = (2 x 400 x 100)/3.3 x 10⁵
= 8/33Kg

4,213.

The temperature gradient across a copper rod of thickness 0.02m, maintained at two temperature junctions of 20°C and 80°C respectively is

3.0 x 10²Km^-1

3.0 x 10³Km^-1

5.0 x 10³Km^-1

3.0 x 10⁴Km^-1

View answer & explanation

Correct answer is B

Temperature Gradient = \(\frac{\Delta {\text{temp.}}}{\text{thickness}}\)

= \(\frac{80 - 20}{0.02}\)

= \(\frac{60}{0.02}\)

= 3000Km-1

= 3.0 x 103Km-1

4,214.

I. A liquid boils when its saturated vapour pressure is equal to the external pressure
II. Dissolved substances in pure water lead to increase in the boiling point.
III. When the external pressure is increased, the boiling point increases.
IV. Dissolved substances in pure water decrease the boiling point.

Which combination of the above are peculiarities of the boiling point of a liquid?

I, II and III

I, II, III and IV

I, II and IV

II, III and IV

View answer & explanation

Correct answer is A

No explanation has been provided for this answer.

4,215.

A piece of substance of specific head capacity 450JKg^-1K^-1 falls through a vertical distance of 20m from rest. Calculate the rise in temperature of the substance on hitting the ground when all its energies are converted into heat. [g = 10ms^-2]

2/9°C

4/9°C

9/4°C

9/2°C

View answer & explanation

Correct answer is B

Energy at the height = mgh
This energy is converted to heat energy - mcΔt
∴ From the principle of energy conversion,
mcΔt = mgh
m x 450 x Δt = m x 10 x 20
∴ Δt = m x 10 x 20/(m x 450)
= 200/450
= 4/9°C