The equations of two interferring waves are \(Y_1 = b cos \omega t\) and \(Y_2 = b cos (\omega t+\phi)\) respectively. Destructive interference will take place at the point of observation for the following value of \(\phi\) :–
1. 0°
2. 360°
3. 180°
4. 720°
View Answer
For destructive interference to occur, the phase difference between the two interfering waves must be an odd multiple of \(pi\) (i.e., \(180^0\), \(540^0 \), etc.). From the options, \(180^0 \) is correct.
The correct relation between frequencies of x-rays, \( \gamma\)-rays, heat rays and radio waves :
1. \(\gamma\)-rays < x-rays < heat rays = radio waves
2. \(\gamma\)-rays > x-rays > heat rays = radio waves
3. \(\gamma\)-rays > x-rays > heat rays > radio waves
4. \(\gamma\)-rays < x-rays < heat rays < radio waves
View Answer
The electromagnetic spectrum in order of decreasing wavelength (and thus increasing frequency) is: Radio waves < Heat rays (Infrared) < X-rays \nu_x > \nu_{\text{heat}} > \nu_{\text{radio}}\).
Two tuning forks A and B when sounded together produce \(4\text{ beats/s}\). When B is loaded with wax, the beat frequency remains same. If frequency of A is \(212\text{ Hz}\). The frequency of B before loading is:
1. 208 Hz
2. 212 Hz
3. 216 Hz
4. 220 Hz
View Answer
The frequency of B must be either \(212 + 4 = 216\text{ Hz}\) or \(212 - 4 = 208\text{ Hz}\). Loading B with wax decreases its frequency. For the beat frequency to remain \(4\text{ Hz}\), its frequency must drop from \(216\text{ Hz}\) to \(208\text{ Hz}\). Thus, the initial frequency of B is \(216\text{ Hz}\).
If a string fixed at both ends vibrates in three loops, the wavelength is \(30\text{ cm}\). The length of string is:
1. 30 cm
2. 45 cm
3. 60 cm
4. 90 cm
View Answer
For a string fixed at both ends vibrating in \(n\) loops, the length of the string is given by \(L = \frac{n\lambda}{2}\). Here, \(n = 3\) and \(\lambda = 30\text{ cm}\), so \(L = 3 \times \frac{30}{2} = 45\text{ cm}\).
Two waves represented by the following equations are travelling in the same medium: \(y_1 = 5 \sin 2\pi(75t – 0.25x)\) and \(y_2 = 10 \sin 2\pi(150t – 0.50x)\). The intensity ratio \(I_1/I_2\) of the two waves is:
1. 1 : 2
2. 1 : 4
3. 1 : 8
4. 1 : 16
View Answer
The intensity of a wave is proportional to the square of its amplitude and frequency, \(I \propto A^2 f^2\). Substituting \(A_1=5, f_1=75\) and \(A_2=10, f_2=150\) gives \(\frac{I_1}{I_2} = \left(\frac{5}{10}\right)^2 \left(\frac{75}{150}\right)^2 = \frac{1}{16}\).
The frequency of the first overtone of a closed pipe of length \(L_1\), is equal to that of the first overtone of an open pipe of length \(L_2\). The ratio of their lengths \((L_1 : L_2)\) is:
1. 2 : 3
2. 4 : 5
3. 3 : 5
4. 3 : 4
View Answer
The first overtone of a closed pipe of length \(L_1\) has frequency \(f_{c,1} = \frac{3v}{4L_1}\) and that of an open pipe of length \(L_2\) is \(f_{o,1} = \frac{v}{L_2}\). Equating the two frequencies gives \(\frac{3v}{4L_1} = \frac{v}{L_2}\), which simplifies to \(\frac{L_1}{L_2} = \frac{3}{4}\).
Given below are two statements:
Assertion (A): Sound would travel faster on a hot summer day than on a cold winter day.
Reason (R): Velocity of sound is directly proportional to the square root of its absolute temperature.
1. Both (A) and (R) are true and (R) is the correct explanation of (A).
2. Both (A) and (R) are true but (R) is not the correct explanation of (A).
3. (A) is true but (R) is false.
4. (A) is false but (R) is true.
View Answer
The speed of sound in a gas is \(v = \sqrt{\frac{\gamma RT}{M}}\), meaning \(v \propto \sqrt{T}\). Since temperature is higher on a hot summer day than a cold winter day, sound travels faster in summer. Both statements are true and Reason is the correct explanation.