A rubber cord hangs from a rigid support. A weight attached to its lower end is gradually increased from zero, and then gradually reduced to zero.

Question 1

A rubber cord hangs from a rigid support. A weight attached to its lower end is gradually

increased from zero, and then gradually reduced to zero.

The force-extension curve for contraction is below the force-extension curve for stretching.

What does the shaded area between the curves represent?

A the amount of elastic energy stored in the rubber

B the amount of thermal energy dissipated in the rubber

C the work done on the rubber cord during stretching

D the work done by the rubber cord during contraction

Reference: Past Exam Paper – June 2013 Paper 12 Q22

Solution:

Answer: B. 

The area under a force-extension graph gives us the energy (work).

Work is done on the rubber cord to stretch it. This is given by the area under the ‘stretching’ curve. In other words, this area can be considered to be the total energy input.

However, it is observed that, for rubber, as the weight attached is reduced, the ‘contraction’ curve is a bit below the ‘stretching curve’. The area under this curve represents the energy stored in the rubber. That is, it is the useful energy recovered.

 

It can be seen that this area is lower than the area under the ‘stretching’ curve. This indicates that some energy is lost (it is the amount of thermal energy dissipated in the rubber). This is represented by the shaded area between the 2 curves.

What is a correct statement of Ohm’s law?

Question 1

What is a correct statement of Ohm’s law?

A The potential difference across a component equals the current providing the resistance and other physical conditions stay constant.

B The potential difference across a component equals the current multiplied by the resistance.

C The potential difference across a component is proportional to its resistance.

D The potential difference across a component is proportional to the current in it providing physical conditions stay constant.

Reference: Past Exam Paper – June 2007 Paper 1 Q31

Solution:

Answer: D. 

Ohm’s law: V ∝ I

The potential difference across a component is proportional to the current in it providing physical conditions (such as temperature, pressure …) stay constant.

From this proportionality, we have = V = kI  where k is a constant.

We know that the constant k is the resistance.

A loudspeaker emitting sound of frequency f is placed at the open end of a pipe of length l which is closed at the other end. A standing wave is set up in the pipe.

Question 4

A loudspeaker emitting sound of frequency f is placed at the open end of a pipe of length l which is closed at the other end. A standing wave is set up in the pipe.

A series of pipes are then set up with either one or two loudspeakers of frequency f. The pairs of loudspeakers vibrate in phase with each other.

Which pipe contains a standing wave?

Reference: Past Exam Paper – June 2015 Paper 11 Q29

Solution:

Answer: D.

For a stationary wave (resonance) to be formed in the tube, there should be a node (zero amplitude) at the closed end and an antinode (maximum amplitude) at the open end or at the loudspeaker (which is at an open end).

Let’s assume that the frequency f produces the fundamental mode of vibration. Since the same frequency is used in all cases, the wavelength will be the same.

We are told that when the frequency is f, a stationary wave is formed in the pipe of length l. For the fundamental mode, the wave formed is a quarter of a wavelength.

λ / 4 = L          giving wavelength λ = 4L

The wavelength will be the same in all of the cases.

Consider choice A:

For a stationary wave, there should be an antinode at the loudspeaker and an antinode at the open end of the pipe. This corresponds to half a wavelength.

For this case, λ / 2 = L             giving wavelength λ = 2L

BUT from above, we know that the wavelength = 4L while the length of the pipe is only l.

Thus, this is not possible.

Consider choice B:

This is similar to choice A as there should be an antinode at both ends. This is also not possible.

Consider choice C:

Here, a node is formed at the closed end at an antinode at the loudspeaker. This corresponds to a quarter of a wave.

For this case, λ / 4 = 2L                      giving wavelength λ = 8L

This does not correspond to the wavelength (= 4L) obtained initially. Hence, this is not correct.

Consider choice D:

Here, an antinode should be at both ends for a stationary wave to be formed. This corresponds to a half a wavelength.

For this case, λ / 2 = 2L                      giving wavelength λ = 4L

 
This is the only case where the wavelength corresponds to the original case.