Question 11:
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 12:
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 13:
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 14:
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.
Question 15:
easyThe fundamental frequency of a sonometer wire increases by 6 Hz. If its tension is increased by 44%, keeping the length constant. Then find this fundamental frequency:
Since frequency \(f \propto \sqrt{T}\), increasing tension by 44% makes \(f' = f \sqrt{1.44} = 1.2f\). Thus, the change in frequency is \(0.2f = 6 \text{ Hz}\), which gives \(f = 30 \text{ Hz}\).
Question 16:
easyGiven below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): The presence of moisture increases the velocity of sound in air.
Reason (R): Density of moist air is more than the density of dry air.
In the light of the above statements, the correct option is
Velocity of sound \(v = \sqrt{\frac{\gamma P}{\rho}}\). Since water vapor has a lower density than dry air, moist air has a lower density, raising the sound velocity. Thus, (A) is true but (R) is false.
Question 17:
easyA string is fixed at both ends and the vibrations of string is given by the equation \(y = 10sin(2x)cos(2t)\) where \(x, y\) are in cm and \(t\) is in second. Nearby node from left end at \(x = 0\), is at a distance
Nodes occur where the spatial amplitude term \(sin(2x) = 0\), which implies \(2x = n\pi\) or \(x = \frac{n\pi}{2}\). The closest node to the left end \(x=0\) (where \(n=1\)) is at \(x = \frac{\pi}{2}\text{ cm}\).
Question 18:
easyA steel wire \(0.50\text{ m}\) long has a mass of \(4.0 \times 10^{-3}\text{ kg}\). If the wire is under a tension of \(80\text{ N}\), the speed of transverse waves on the wire is
Linear mass density \(\mu = \frac{m}{L} = \frac{4.0 \times 10^{-3}}{0.50} = 8.0 \times 10^{-3}\text{ kg/m}\). Wave speed is \(v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{80}{8.0 times 10^{-3}}} = 100\text{ m/s}\).
Question 19:
easyA wave travelling in the positive x-direction having displacement amplitude along y-direction as 1 m, wavelength \(2\pi\text{ m}\) and frequency of \(\frac{1}{\pi}\text{ Hz}\) is represented by
Wave equation is \(y = A\sin(kx - \omega t)\). Here, \(A = 1\text{ m}\), \(k = \frac{2\pi}{\lambda} = 1\text{ m}^{-1}\), and \(\omega = 2\pi f = 2\text{ rad/s}\). Thus, \(y = \sin(x - 2t)\).
Question 20:
easyA closed organ pipe of length \(l = 2 \text{ m}\) is vibrating in \(2^{\text{nd}}\) overtone. The frequency of vibration if speed of sound is 340 m/s is
For a closed organ pipe, the frequency of the \(n^{\text{th}}\) overtone is \(f = (2n+1)\frac{v}{4l}\). For \(2^{\text{nd}}\) overtone (\(n=2\)), \(f = 5\frac{v}{4l} = \frac{5 \times 340}{4 \times 2} = 212.5 \text{ Hz}\).