Question 21:
easyThe 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 \(\phi\) :–
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.
Question 22:
The correct relation between frequencies of x-rays, \( \gamma\)-rays, heat rays and radio waves :
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}}\).
Question 23:
moderateThe \(4^{\text{th}}\) overtone of a closed organ pipe is same as that of \(3^{\text{th}}\) overtone of an open pipe. The ratio of the length of the closed pipe to the length of the open pipe is:
The frequency of the \(4^{\text{th}}\) overtone (9th harmonic) of a closed pipe is \(f_c = \frac{9v}{4L_c}\). The frequency of the \(3^{\text{rd}}\) overtone (4th harmonic) of an open pipe is \(f_o = \frac{4v}{2L_o} = \frac{2v}{L_o}\). Equating the two, \(\frac{9v}{4L_c} = \frac{2v}{L_o} ⇒ \frac{L_c}{L_o} = \frac{9}{8}\).
Question 24:
easyTwo 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:
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 \(216\text{ Hz}\).
Question 25:
easyIf a string fixed at both ends vibrates in three loops, the wavelength is \(30\text{ cm}\). The length of string is:
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 \(L = 3 \times \frac{30}{2} = 45\text{ cm}\).
Question 26:
easyTwo 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:
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 \(\frac{I_1}{I_2} = \left(\frac{5}{10}\right)^2 \left(\frac{75}{150}\right)^2 = \frac{1}{16}\).
Question 27:
easyTwo waves having the intensities in the ratio of 9 : 1 produce interference. The ratio of maximum to minimum intensity is equal to:
The ratio of maximum to minimum intensity is given by \(frac{I_{\text{max}}}{I_{\text{min}}} = \left(\frac{\sqrt{I_1/I_2} + 1}{\sqrt{I_1/I_2} - 1}\right)^2\). Substituting \(\frac{I_1}{I_2} = 9\) yields \(\left(\frac{3 + 1}{3 - 1}\right\)^2 = 4\), which is \(4:1\).
Question 28:
easyThe 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:
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 \(\frac{L_1}{L_2} = \frac{3}{4}\).
Question 29:
easyEquation of a progressive wave is given by \( y = 0.2 \cos \pi(0.04t + 0.02x – \pi/6) \). The distance is expressed in cm and time in second. What will be the minimum distance between two particles having the phase difference of \( \pi/2 \)?
The wave number is \(k = 0.02\pi\text{ cm}^{-1}\). Since phase difference \(\Delta \phi = k \Delta x\), we have \(\Delta x = \frac{\Delta \phi}{k} = \frac{\pi/2}{0.02\pi} = 25\text{ cm}\).
Question 30:
easyGiven 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.
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.