Question 31:
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 32:
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 33:
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 34:
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 35:
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 36:
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 (\(n=2\)), \(f = 5\frac{v}{4l} = \frac{5 \times 340}{4 \times 2} = 212.5 \text{ Hz}\).
Question 37:
easyA person hums in a well and finds strong resonance at frequencies \(180\text{ Hz}\), \(300\text{ Hz}\) and \(420\text{ Hz}\). The fundamental frequency of the well is (velocity of sound \(= 335\text{ m/s}\))
A well acts as a closed organ pipe. The resonant frequencies are odd harmonics: \(f_n = (2n-1)f_1\). The difference between consecutive resonant frequencies is \(300 - 180 = 120\text{ Hz}\), which corresponds to \(2f_1\). Thus, \(f_1 = 60\text{ Hz}\).
Question 38:
easyMatch List-I with List-II where list-I denotes nature and medium of wave and list-II denotes the expression for speed of wave. (All symbols have their usual meaning)
List-I
a. Transverse wave on a stretched string
b. Longitudinal wave in a metallic bar
c. Longitudinal wave in a fluid
List-II
(i) \( \sqrt{\frac{Y}{\rho}} \)
(ii) \( \sqrt{\frac{B}{\rho}} \)
(iii) \( \sqrt{\frac{T}{\mu}} \)
Choose the correct option:
Speed of a wave on a string is \( \sqrt{T/\mu} \) (a matches iii). Speed of a longitudinal wave in a metallic bar is \( \sqrt{Y/\rho} \) (b matches i). Speed of a longitudinal wave in a fluid is \( \sqrt{B/\rho} \) (c matches ii).
Question 39:
moderateA person hums in a well and finds strong resonance at frequencies \(180\text{ Hz}\), \(300\text{ Hz}\) and \(420\text{ Hz}\). The fundamental frequency of the well is (velocity of sound = \(335\text{ m/s}\))
The resonance frequencies form an odd-harmonic progression for a closed-end pipe: \((2n-1)f_0\). The difference between consecutive harmonics is \(2f_0 = 300 - 180 = 120\text{ Hz}\) which gives \(f_0 = 60\text{ Hz}\).
Question 40:
easyAssertion (A): When a pulse on string reflects from free end, the resultant pulse is formed in such a way that slope of string at free end is zero.
Reason (R): Zero resultant slope ensures that there is no force component perpendicular to string.
At a free end, there's no transverse force, so the slope \( \frac{\text{dy}}{\text{dx}} = 0 \). The resultant pulse is formed such that the free end is a displacement antinode. This implies zero slope, ensuring no transverse force. Both Assertion and Reason are true, and R correctly explains A.