9702/May Jun/11/2015/Q26

A wave of amplitude a has an intensity of 3.0 Wm^–2.

What is the intensity of a wave of the same frequency that has an amplitude 2a?

A 4.2 Wm^-2

B 6.0 Wm^–2

C 9.0 Wm^–2

D 12 Wm^–2

Solution:

Answer: D.

Intensity is directly proportional to the square of the amplitude.

I = ka²

3 Wm^-2 = ka²

x = k(2a)²

x = 12ka² / ka²

   = 12Wm^-2

Reference: PYQ - May/Jun 2015 Paper 11 Q26

9702/May Jun/11/2015/Q3

When a constant braking force is applied to a vehicle moving at speed v, the distance d moved by the vehicle in coming to rest is given by the expression

d = kv²

where k is a constant.

When d is measured in metres and v is measured in metres per second, the constant has a value of k₁.

What is the value of the constant when the distance is measured in metres, and the speed is measured in kilometres per hour?

A 0.0772 k₁                       B 0.278 k₁                         C 3.60 k₁                            D 13.0 k₁

Solution:

Answer: A.

d = kv²

When speed v is in ‘m/s’ and distance d is in ‘m’, the value of k is ‘k₁’.

k₁ = d / v²

We want to know the value of k when v is in ‘km/h’ and d in ‘m’.

Converting m/s to km/h

1 m/s means that in 1s - - -> 1m

1 h = 3600 s

1s - - -> 1m

1h - - > 3600 m = 3.6 km

Thus, 1 m/.s is equal to 3.6 km/h. Let the new value of speed = v₂.

v₂ = 3.6 v

New value of k = d / v₂² = d / (3.6v)2 = (1 / 3.6²) (d / v²)

But d / v₂ = k₁  

New value of k = (1 / 3.6²) k₁ = 0.0772 k₁

Reference: PYQ - May/Jun 2015 Paper 11 Q3

9702/May Jun/11/2015/Q38

A wire RST is connected to another wire XY as shown.

Each wire is 100 cm long with a resistance per unit length of 10 Ω m^–1.
What is the total resistance between X and Y?
A 3.3 Ω
B 5.0 Ω
C 8.3 Ω
D 13.3 Ω

Solution:

Answer: C.

Resistance per unit length of wire = 10 Ω m^–1

Resistance of wire RST = 1m × 10 Ω m^–1 = 10 Ω

Since the wire XY is 100cm, the sum of XR + TY = 100 – 50 = 50cm

Sum of resistance of (XR+TY) = 0.5m × 10 Ω m^–1 = 5 Ω

Resistance of wire RT = 0.5m × 10 Ω m^–1 = 5 Ω

Hence the arrangement is effectively a 10 Ω resistor RST in parallel with a 5 Ω resistor RT, and in series with another 5 Ω resistor (XR+TY).

Effective resistance of parallel combination = [1/10 + 1/5]^-1 = 3.3 Ω

The total resistance is therefore 5 Ω + 3.3 Ω = 8.3 Ω.

Reference: PYQ - May/Jun 2015 Paper 11 Q38

9702/May Jun/11/2015/Q29

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?

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.

Reference: PYQ - May/Jun 2015 Paper 11 Q29