**Sound from a small loudspeaker L reaches a point X by two paths which differ in length by 1.8m. When the frequency of the sound is gradually increased, the resultant intensity at X reaches a maxima when the frequency is 1000 Hz.**

**Given the velocity of sound is 360 m s ^{-1}, at what next higher frequency will a maximum be detected?**

**1200 Hz****1400 Hz****1600 Hz****1800 Hz**

**Show/Hide Answer**

Path difference = 1.8m

Given f = 1000 Hz, v = 360 m s

^{-1}

And since a maximum is detected at X,

Path difference = nλ

1.8 = n x (360/1000)

n = 5

Thus, for the next maximum to be produced, n’ = n +1 = 6

n’λ’ = 1.8

λ’ = 0.30m

So, the frequency f’ that produce λ’ = (360/0.30) = 1200 Hz

Answer: 1

**A parallel beam of white light (range of wavelengths from 4.5 x 10 ^{-7} m to 7.5 x 10^{-7} m) is incident normally on a diffraction grating. Red light in the second order spectrum is observed to be diffracted through an angle of 60° from the direction of the incident beam.**

**How many lines per metre are there on the grating?**

**5.8 x 10**^{5}**9.6 x 10**^{5}**11.6 x 10**^{5}**19.2 x 10**^{5}

**Show/Hide Answer**

d sin θ = mλ

Most deviated wavelength is 7.5 x 10

^{-7}m, m = 2 and θ = 60

Therefore, d = 1.7 x 10^{-6} m

In one metre, number of lines is 1/1.7x 10^{-6} = 5.8 x 10^{5}

Answer: 1

**A taut wire is set into resonance with a node at either end and another single node at the centre of the wire. Which one of the following statements is not correct?**

**The wavelength of the wave on the wire is equal to the length of the wire.****All points to one side of the centre vibrate in phase with one another.****Any two points on either side of the centre have a phase difference of 90°.****Two points equidistant from the centre on either side of the centre have the same amplitude of vibration.**

**Show/Hide Answer**

Any two points either side of the centre have a phase difference of 180°.

Answer: 3

**A small pipe opened at both ends is partially submerged in water as shown below.**

**A tuning fork vibrating at frequency 850 Hz is placed over the top of the pipe. The pipe is slowly raised until the first loud sound is heard when the length of pipe above water is L.**

**The experiment is repeated with another tuning fork of unknown frequency f and first loud sound is heard when the length of pipe above the water is 2L.**

**What is the value of f? You may assume the speed of sound is 340 m s ^{-1}.**

**2550 Hz****1700 Hz****425 Hz****213 Hz**

**Show/Hide Answer**

¼λ = L for 1

^{st}resonance for both frequency

v = fλ

since v is constant, therefore f inversely proportional to λ and λ proportional to L. Hence, f inversely proportional to L.

f

_{2}/ f

_{1}= L

_{1}/ L

_{2}

f

_{2}/ f

_{1}= 1/2

f

_{2}= 425 Hz

Answer: 3

Hello, sorry to be a bother, but there seems to be a formatting error. Answers are in plain view at all times.

Thanks for pointing it out! I’ve fixed the problem. 🙂