172728
The frequency of tuning fork is $500 \mathrm{~Hz}$ and the velocity of sound in air is $300 \mathrm{~m}_{\mathrm{s}}^{-1}$. The distance travelled by sound while the fork executes 100 oscillations per second is
1 $45 \mathrm{~m}$
2 $60 \mathrm{~m}$
3 $30 \mathrm{~m}$
4 $50 \mathrm{~m}$
Explanation:
B Given, (f) $=500 \mathrm{~Hz}$ $\mathrm{T}=\frac{1}{\mathrm{f}}=\frac{1}{500}$ $\mathrm{T}$ is for 1 vibration $\therefore \mathrm{T}$ for 100 vibrations in $\mathrm{T} \times 100$ $\mathrm{T}=\frac{100}{500}$ Distance travelled $\mathrm{d}=\mathrm{vt}$ $=300 \times \frac{100}{500}=60 \mathrm{~m}$
AP EAMCET-23.09.2020
WAVES
172824
How many times more intense is a $60 \mathrm{~dB}$ sound than a $30 \mathrm{~dB}$ sound ?
1 1000
2 2
3 100
4 4
Explanation:
A $\mathrm{L}=10 \log _{10} \frac{\mathrm{I}}{\mathrm{I}_{0}}$ $60=10 \log _{10} \frac{\mathrm{I}_{1}}{\mathrm{I}_{0}}$ $\Rightarrow \quad \frac{\mathrm{I}_{1}}{\mathrm{I}_{0}}=10^{6}$ Similarly, $\quad 30=10 \log _{10} \frac{\mathrm{I}_{2}}{\mathrm{I}_{0}}$ $\frac{\mathrm{I}_{2}}{\mathrm{I}_{0}}=10^{3}$ Dividing Eq. (i) by Eq. (ii), we get $\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=1000$
Karnataka CET-2008
WAVES
172713
$Y_{1}=0.25 \sin (k x-316 t)$ and $y_{2}=0.25 \sin (k x-$ $310 t)$ are two waves travelling in the same direction when they superpose the number of beats produced
1 $\frac{3}{\pi}$
2 3
3 $3 \pi$
4 6
Explanation:
A $\mathrm{Y}_{1}=0.25 \sin (\mathrm{kx}-316 \mathrm{t})$ $\mathrm{Y}_{2}=0.25 \sin (\mathrm{kx}-310 \mathrm{t})$ $\text { Comparing } \mathrm{Y}=\mathrm{A} \sin (\omega \mathrm{t}-\mathrm{kx})$ $\omega_{1}=316 \mathrm{rad} / \mathrm{s}, 2 \pi \mathrm{f}_{1}=316$ $\mathrm{f}_{1}=\frac{316}{2 \pi}$ $\omega_{2}=310 \mathrm{rad} / \mathrm{s}, 2 \pi \mathrm{f}_{2}=310$ $\mathrm{f}_{2}=\frac{310}{2 \pi}$ $\therefore$ Beat frequency or number of beats produced per second $f_{1}-f_{2}=\frac{316}{2 \pi}-\frac{310}{2 \pi}=\frac{1}{2 \pi}[316-310]$ $=\frac{6}{2 \pi}=\frac{3}{\pi}$
Tripura-27.04.2022
WAVES
172715
The velocity of sound in a gas, in which two wavelengths $4.08 \mathrm{~m}$ and $4.16 \mathrm{~m}$ produces 40 beats in $12 \mathrm{~s}$, will be:
1 $282.8 \mathrm{~ms}^{-1}$
2 $175.5 \mathrm{~ms}^{-1}$
3 $353.6 \mathrm{~ms}^{-1}$
4 $707.2 \mathrm{~ms}^{-1}$
Explanation:
D Wavelength $\lambda_{1} \& \lambda_{2}$ are $4.08 \mathrm{~m}$ and $416 \mathrm{~m}$ We know, Number of beats produced per second $\mathrm{b}=\mathrm{n}_{1}-\mathrm{n}_{2}$ $\mathrm{b}=\frac{\mathrm{v}}{\lambda_{1}}-\frac{\mathrm{v}}{\lambda_{2}}$ $\frac{40}{12}=\frac{\mathrm{v}}{4.08}-\frac{\mathrm{v}}{4.16}$ $\mathrm{v}=\frac{40}{12} \times \frac{4.08 \times 4.16}{0.08}$ $\mathrm{v}=707.2 \mathrm{~m} / \mathrm{s}$
JEE Main-28.06.2022
WAVES
172716
A tuning fork of frequency $200 \mathrm{~Hz}$ is in unison with a sonometer wire. How many beats per second will be heard if the tension of the wire is increased by $2 \%$ ?
1 1
2 2
3 3
4 4
Explanation:
B Given that, frequency (f) $=200 \mathrm{~Hz}$ We know that, $\therefore \quad \mathrm{f} \propto \sqrt{\mathrm{T}}$ $\frac{\Delta \mathrm{f}}{\mathrm{f}}=\frac{1}{2} \frac{\Delta \mathrm{T}}{\mathrm{T}}$ $\frac{\Delta \mathrm{f}}{\mathrm{f}}=\frac{1}{2} \times 0.02$ $\Delta \mathrm{f}=\mathrm{f} \times \frac{1}{2} \times 0.02$ $\Delta \mathrm{f}=200 \times \frac{1}{2} \times 0.02$ $\Delta \mathrm{f}=2 \mathrm{~Hz}$ Hence, No. of beats heard per second $=2 \mathrm{~Hz}$
172728
The frequency of tuning fork is $500 \mathrm{~Hz}$ and the velocity of sound in air is $300 \mathrm{~m}_{\mathrm{s}}^{-1}$. The distance travelled by sound while the fork executes 100 oscillations per second is
1 $45 \mathrm{~m}$
2 $60 \mathrm{~m}$
3 $30 \mathrm{~m}$
4 $50 \mathrm{~m}$
Explanation:
B Given, (f) $=500 \mathrm{~Hz}$ $\mathrm{T}=\frac{1}{\mathrm{f}}=\frac{1}{500}$ $\mathrm{T}$ is for 1 vibration $\therefore \mathrm{T}$ for 100 vibrations in $\mathrm{T} \times 100$ $\mathrm{T}=\frac{100}{500}$ Distance travelled $\mathrm{d}=\mathrm{vt}$ $=300 \times \frac{100}{500}=60 \mathrm{~m}$
AP EAMCET-23.09.2020
WAVES
172824
How many times more intense is a $60 \mathrm{~dB}$ sound than a $30 \mathrm{~dB}$ sound ?
1 1000
2 2
3 100
4 4
Explanation:
A $\mathrm{L}=10 \log _{10} \frac{\mathrm{I}}{\mathrm{I}_{0}}$ $60=10 \log _{10} \frac{\mathrm{I}_{1}}{\mathrm{I}_{0}}$ $\Rightarrow \quad \frac{\mathrm{I}_{1}}{\mathrm{I}_{0}}=10^{6}$ Similarly, $\quad 30=10 \log _{10} \frac{\mathrm{I}_{2}}{\mathrm{I}_{0}}$ $\frac{\mathrm{I}_{2}}{\mathrm{I}_{0}}=10^{3}$ Dividing Eq. (i) by Eq. (ii), we get $\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=1000$
Karnataka CET-2008
WAVES
172713
$Y_{1}=0.25 \sin (k x-316 t)$ and $y_{2}=0.25 \sin (k x-$ $310 t)$ are two waves travelling in the same direction when they superpose the number of beats produced
1 $\frac{3}{\pi}$
2 3
3 $3 \pi$
4 6
Explanation:
A $\mathrm{Y}_{1}=0.25 \sin (\mathrm{kx}-316 \mathrm{t})$ $\mathrm{Y}_{2}=0.25 \sin (\mathrm{kx}-310 \mathrm{t})$ $\text { Comparing } \mathrm{Y}=\mathrm{A} \sin (\omega \mathrm{t}-\mathrm{kx})$ $\omega_{1}=316 \mathrm{rad} / \mathrm{s}, 2 \pi \mathrm{f}_{1}=316$ $\mathrm{f}_{1}=\frac{316}{2 \pi}$ $\omega_{2}=310 \mathrm{rad} / \mathrm{s}, 2 \pi \mathrm{f}_{2}=310$ $\mathrm{f}_{2}=\frac{310}{2 \pi}$ $\therefore$ Beat frequency or number of beats produced per second $f_{1}-f_{2}=\frac{316}{2 \pi}-\frac{310}{2 \pi}=\frac{1}{2 \pi}[316-310]$ $=\frac{6}{2 \pi}=\frac{3}{\pi}$
Tripura-27.04.2022
WAVES
172715
The velocity of sound in a gas, in which two wavelengths $4.08 \mathrm{~m}$ and $4.16 \mathrm{~m}$ produces 40 beats in $12 \mathrm{~s}$, will be:
1 $282.8 \mathrm{~ms}^{-1}$
2 $175.5 \mathrm{~ms}^{-1}$
3 $353.6 \mathrm{~ms}^{-1}$
4 $707.2 \mathrm{~ms}^{-1}$
Explanation:
D Wavelength $\lambda_{1} \& \lambda_{2}$ are $4.08 \mathrm{~m}$ and $416 \mathrm{~m}$ We know, Number of beats produced per second $\mathrm{b}=\mathrm{n}_{1}-\mathrm{n}_{2}$ $\mathrm{b}=\frac{\mathrm{v}}{\lambda_{1}}-\frac{\mathrm{v}}{\lambda_{2}}$ $\frac{40}{12}=\frac{\mathrm{v}}{4.08}-\frac{\mathrm{v}}{4.16}$ $\mathrm{v}=\frac{40}{12} \times \frac{4.08 \times 4.16}{0.08}$ $\mathrm{v}=707.2 \mathrm{~m} / \mathrm{s}$
JEE Main-28.06.2022
WAVES
172716
A tuning fork of frequency $200 \mathrm{~Hz}$ is in unison with a sonometer wire. How many beats per second will be heard if the tension of the wire is increased by $2 \%$ ?
1 1
2 2
3 3
4 4
Explanation:
B Given that, frequency (f) $=200 \mathrm{~Hz}$ We know that, $\therefore \quad \mathrm{f} \propto \sqrt{\mathrm{T}}$ $\frac{\Delta \mathrm{f}}{\mathrm{f}}=\frac{1}{2} \frac{\Delta \mathrm{T}}{\mathrm{T}}$ $\frac{\Delta \mathrm{f}}{\mathrm{f}}=\frac{1}{2} \times 0.02$ $\Delta \mathrm{f}=\mathrm{f} \times \frac{1}{2} \times 0.02$ $\Delta \mathrm{f}=200 \times \frac{1}{2} \times 0.02$ $\Delta \mathrm{f}=2 \mathrm{~Hz}$ Hence, No. of beats heard per second $=2 \mathrm{~Hz}$
172728
The frequency of tuning fork is $500 \mathrm{~Hz}$ and the velocity of sound in air is $300 \mathrm{~m}_{\mathrm{s}}^{-1}$. The distance travelled by sound while the fork executes 100 oscillations per second is
1 $45 \mathrm{~m}$
2 $60 \mathrm{~m}$
3 $30 \mathrm{~m}$
4 $50 \mathrm{~m}$
Explanation:
B Given, (f) $=500 \mathrm{~Hz}$ $\mathrm{T}=\frac{1}{\mathrm{f}}=\frac{1}{500}$ $\mathrm{T}$ is for 1 vibration $\therefore \mathrm{T}$ for 100 vibrations in $\mathrm{T} \times 100$ $\mathrm{T}=\frac{100}{500}$ Distance travelled $\mathrm{d}=\mathrm{vt}$ $=300 \times \frac{100}{500}=60 \mathrm{~m}$
AP EAMCET-23.09.2020
WAVES
172824
How many times more intense is a $60 \mathrm{~dB}$ sound than a $30 \mathrm{~dB}$ sound ?
1 1000
2 2
3 100
4 4
Explanation:
A $\mathrm{L}=10 \log _{10} \frac{\mathrm{I}}{\mathrm{I}_{0}}$ $60=10 \log _{10} \frac{\mathrm{I}_{1}}{\mathrm{I}_{0}}$ $\Rightarrow \quad \frac{\mathrm{I}_{1}}{\mathrm{I}_{0}}=10^{6}$ Similarly, $\quad 30=10 \log _{10} \frac{\mathrm{I}_{2}}{\mathrm{I}_{0}}$ $\frac{\mathrm{I}_{2}}{\mathrm{I}_{0}}=10^{3}$ Dividing Eq. (i) by Eq. (ii), we get $\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=1000$
Karnataka CET-2008
WAVES
172713
$Y_{1}=0.25 \sin (k x-316 t)$ and $y_{2}=0.25 \sin (k x-$ $310 t)$ are two waves travelling in the same direction when they superpose the number of beats produced
1 $\frac{3}{\pi}$
2 3
3 $3 \pi$
4 6
Explanation:
A $\mathrm{Y}_{1}=0.25 \sin (\mathrm{kx}-316 \mathrm{t})$ $\mathrm{Y}_{2}=0.25 \sin (\mathrm{kx}-310 \mathrm{t})$ $\text { Comparing } \mathrm{Y}=\mathrm{A} \sin (\omega \mathrm{t}-\mathrm{kx})$ $\omega_{1}=316 \mathrm{rad} / \mathrm{s}, 2 \pi \mathrm{f}_{1}=316$ $\mathrm{f}_{1}=\frac{316}{2 \pi}$ $\omega_{2}=310 \mathrm{rad} / \mathrm{s}, 2 \pi \mathrm{f}_{2}=310$ $\mathrm{f}_{2}=\frac{310}{2 \pi}$ $\therefore$ Beat frequency or number of beats produced per second $f_{1}-f_{2}=\frac{316}{2 \pi}-\frac{310}{2 \pi}=\frac{1}{2 \pi}[316-310]$ $=\frac{6}{2 \pi}=\frac{3}{\pi}$
Tripura-27.04.2022
WAVES
172715
The velocity of sound in a gas, in which two wavelengths $4.08 \mathrm{~m}$ and $4.16 \mathrm{~m}$ produces 40 beats in $12 \mathrm{~s}$, will be:
1 $282.8 \mathrm{~ms}^{-1}$
2 $175.5 \mathrm{~ms}^{-1}$
3 $353.6 \mathrm{~ms}^{-1}$
4 $707.2 \mathrm{~ms}^{-1}$
Explanation:
D Wavelength $\lambda_{1} \& \lambda_{2}$ are $4.08 \mathrm{~m}$ and $416 \mathrm{~m}$ We know, Number of beats produced per second $\mathrm{b}=\mathrm{n}_{1}-\mathrm{n}_{2}$ $\mathrm{b}=\frac{\mathrm{v}}{\lambda_{1}}-\frac{\mathrm{v}}{\lambda_{2}}$ $\frac{40}{12}=\frac{\mathrm{v}}{4.08}-\frac{\mathrm{v}}{4.16}$ $\mathrm{v}=\frac{40}{12} \times \frac{4.08 \times 4.16}{0.08}$ $\mathrm{v}=707.2 \mathrm{~m} / \mathrm{s}$
JEE Main-28.06.2022
WAVES
172716
A tuning fork of frequency $200 \mathrm{~Hz}$ is in unison with a sonometer wire. How many beats per second will be heard if the tension of the wire is increased by $2 \%$ ?
1 1
2 2
3 3
4 4
Explanation:
B Given that, frequency (f) $=200 \mathrm{~Hz}$ We know that, $\therefore \quad \mathrm{f} \propto \sqrt{\mathrm{T}}$ $\frac{\Delta \mathrm{f}}{\mathrm{f}}=\frac{1}{2} \frac{\Delta \mathrm{T}}{\mathrm{T}}$ $\frac{\Delta \mathrm{f}}{\mathrm{f}}=\frac{1}{2} \times 0.02$ $\Delta \mathrm{f}=\mathrm{f} \times \frac{1}{2} \times 0.02$ $\Delta \mathrm{f}=200 \times \frac{1}{2} \times 0.02$ $\Delta \mathrm{f}=2 \mathrm{~Hz}$ Hence, No. of beats heard per second $=2 \mathrm{~Hz}$
172728
The frequency of tuning fork is $500 \mathrm{~Hz}$ and the velocity of sound in air is $300 \mathrm{~m}_{\mathrm{s}}^{-1}$. The distance travelled by sound while the fork executes 100 oscillations per second is
1 $45 \mathrm{~m}$
2 $60 \mathrm{~m}$
3 $30 \mathrm{~m}$
4 $50 \mathrm{~m}$
Explanation:
B Given, (f) $=500 \mathrm{~Hz}$ $\mathrm{T}=\frac{1}{\mathrm{f}}=\frac{1}{500}$ $\mathrm{T}$ is for 1 vibration $\therefore \mathrm{T}$ for 100 vibrations in $\mathrm{T} \times 100$ $\mathrm{T}=\frac{100}{500}$ Distance travelled $\mathrm{d}=\mathrm{vt}$ $=300 \times \frac{100}{500}=60 \mathrm{~m}$
AP EAMCET-23.09.2020
WAVES
172824
How many times more intense is a $60 \mathrm{~dB}$ sound than a $30 \mathrm{~dB}$ sound ?
1 1000
2 2
3 100
4 4
Explanation:
A $\mathrm{L}=10 \log _{10} \frac{\mathrm{I}}{\mathrm{I}_{0}}$ $60=10 \log _{10} \frac{\mathrm{I}_{1}}{\mathrm{I}_{0}}$ $\Rightarrow \quad \frac{\mathrm{I}_{1}}{\mathrm{I}_{0}}=10^{6}$ Similarly, $\quad 30=10 \log _{10} \frac{\mathrm{I}_{2}}{\mathrm{I}_{0}}$ $\frac{\mathrm{I}_{2}}{\mathrm{I}_{0}}=10^{3}$ Dividing Eq. (i) by Eq. (ii), we get $\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=1000$
Karnataka CET-2008
WAVES
172713
$Y_{1}=0.25 \sin (k x-316 t)$ and $y_{2}=0.25 \sin (k x-$ $310 t)$ are two waves travelling in the same direction when they superpose the number of beats produced
1 $\frac{3}{\pi}$
2 3
3 $3 \pi$
4 6
Explanation:
A $\mathrm{Y}_{1}=0.25 \sin (\mathrm{kx}-316 \mathrm{t})$ $\mathrm{Y}_{2}=0.25 \sin (\mathrm{kx}-310 \mathrm{t})$ $\text { Comparing } \mathrm{Y}=\mathrm{A} \sin (\omega \mathrm{t}-\mathrm{kx})$ $\omega_{1}=316 \mathrm{rad} / \mathrm{s}, 2 \pi \mathrm{f}_{1}=316$ $\mathrm{f}_{1}=\frac{316}{2 \pi}$ $\omega_{2}=310 \mathrm{rad} / \mathrm{s}, 2 \pi \mathrm{f}_{2}=310$ $\mathrm{f}_{2}=\frac{310}{2 \pi}$ $\therefore$ Beat frequency or number of beats produced per second $f_{1}-f_{2}=\frac{316}{2 \pi}-\frac{310}{2 \pi}=\frac{1}{2 \pi}[316-310]$ $=\frac{6}{2 \pi}=\frac{3}{\pi}$
Tripura-27.04.2022
WAVES
172715
The velocity of sound in a gas, in which two wavelengths $4.08 \mathrm{~m}$ and $4.16 \mathrm{~m}$ produces 40 beats in $12 \mathrm{~s}$, will be:
1 $282.8 \mathrm{~ms}^{-1}$
2 $175.5 \mathrm{~ms}^{-1}$
3 $353.6 \mathrm{~ms}^{-1}$
4 $707.2 \mathrm{~ms}^{-1}$
Explanation:
D Wavelength $\lambda_{1} \& \lambda_{2}$ are $4.08 \mathrm{~m}$ and $416 \mathrm{~m}$ We know, Number of beats produced per second $\mathrm{b}=\mathrm{n}_{1}-\mathrm{n}_{2}$ $\mathrm{b}=\frac{\mathrm{v}}{\lambda_{1}}-\frac{\mathrm{v}}{\lambda_{2}}$ $\frac{40}{12}=\frac{\mathrm{v}}{4.08}-\frac{\mathrm{v}}{4.16}$ $\mathrm{v}=\frac{40}{12} \times \frac{4.08 \times 4.16}{0.08}$ $\mathrm{v}=707.2 \mathrm{~m} / \mathrm{s}$
JEE Main-28.06.2022
WAVES
172716
A tuning fork of frequency $200 \mathrm{~Hz}$ is in unison with a sonometer wire. How many beats per second will be heard if the tension of the wire is increased by $2 \%$ ?
1 1
2 2
3 3
4 4
Explanation:
B Given that, frequency (f) $=200 \mathrm{~Hz}$ We know that, $\therefore \quad \mathrm{f} \propto \sqrt{\mathrm{T}}$ $\frac{\Delta \mathrm{f}}{\mathrm{f}}=\frac{1}{2} \frac{\Delta \mathrm{T}}{\mathrm{T}}$ $\frac{\Delta \mathrm{f}}{\mathrm{f}}=\frac{1}{2} \times 0.02$ $\Delta \mathrm{f}=\mathrm{f} \times \frac{1}{2} \times 0.02$ $\Delta \mathrm{f}=200 \times \frac{1}{2} \times 0.02$ $\Delta \mathrm{f}=2 \mathrm{~Hz}$ Hence, No. of beats heard per second $=2 \mathrm{~Hz}$
172728
The frequency of tuning fork is $500 \mathrm{~Hz}$ and the velocity of sound in air is $300 \mathrm{~m}_{\mathrm{s}}^{-1}$. The distance travelled by sound while the fork executes 100 oscillations per second is
1 $45 \mathrm{~m}$
2 $60 \mathrm{~m}$
3 $30 \mathrm{~m}$
4 $50 \mathrm{~m}$
Explanation:
B Given, (f) $=500 \mathrm{~Hz}$ $\mathrm{T}=\frac{1}{\mathrm{f}}=\frac{1}{500}$ $\mathrm{T}$ is for 1 vibration $\therefore \mathrm{T}$ for 100 vibrations in $\mathrm{T} \times 100$ $\mathrm{T}=\frac{100}{500}$ Distance travelled $\mathrm{d}=\mathrm{vt}$ $=300 \times \frac{100}{500}=60 \mathrm{~m}$
AP EAMCET-23.09.2020
WAVES
172824
How many times more intense is a $60 \mathrm{~dB}$ sound than a $30 \mathrm{~dB}$ sound ?
1 1000
2 2
3 100
4 4
Explanation:
A $\mathrm{L}=10 \log _{10} \frac{\mathrm{I}}{\mathrm{I}_{0}}$ $60=10 \log _{10} \frac{\mathrm{I}_{1}}{\mathrm{I}_{0}}$ $\Rightarrow \quad \frac{\mathrm{I}_{1}}{\mathrm{I}_{0}}=10^{6}$ Similarly, $\quad 30=10 \log _{10} \frac{\mathrm{I}_{2}}{\mathrm{I}_{0}}$ $\frac{\mathrm{I}_{2}}{\mathrm{I}_{0}}=10^{3}$ Dividing Eq. (i) by Eq. (ii), we get $\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=1000$
Karnataka CET-2008
WAVES
172713
$Y_{1}=0.25 \sin (k x-316 t)$ and $y_{2}=0.25 \sin (k x-$ $310 t)$ are two waves travelling in the same direction when they superpose the number of beats produced
1 $\frac{3}{\pi}$
2 3
3 $3 \pi$
4 6
Explanation:
A $\mathrm{Y}_{1}=0.25 \sin (\mathrm{kx}-316 \mathrm{t})$ $\mathrm{Y}_{2}=0.25 \sin (\mathrm{kx}-310 \mathrm{t})$ $\text { Comparing } \mathrm{Y}=\mathrm{A} \sin (\omega \mathrm{t}-\mathrm{kx})$ $\omega_{1}=316 \mathrm{rad} / \mathrm{s}, 2 \pi \mathrm{f}_{1}=316$ $\mathrm{f}_{1}=\frac{316}{2 \pi}$ $\omega_{2}=310 \mathrm{rad} / \mathrm{s}, 2 \pi \mathrm{f}_{2}=310$ $\mathrm{f}_{2}=\frac{310}{2 \pi}$ $\therefore$ Beat frequency or number of beats produced per second $f_{1}-f_{2}=\frac{316}{2 \pi}-\frac{310}{2 \pi}=\frac{1}{2 \pi}[316-310]$ $=\frac{6}{2 \pi}=\frac{3}{\pi}$
Tripura-27.04.2022
WAVES
172715
The velocity of sound in a gas, in which two wavelengths $4.08 \mathrm{~m}$ and $4.16 \mathrm{~m}$ produces 40 beats in $12 \mathrm{~s}$, will be:
1 $282.8 \mathrm{~ms}^{-1}$
2 $175.5 \mathrm{~ms}^{-1}$
3 $353.6 \mathrm{~ms}^{-1}$
4 $707.2 \mathrm{~ms}^{-1}$
Explanation:
D Wavelength $\lambda_{1} \& \lambda_{2}$ are $4.08 \mathrm{~m}$ and $416 \mathrm{~m}$ We know, Number of beats produced per second $\mathrm{b}=\mathrm{n}_{1}-\mathrm{n}_{2}$ $\mathrm{b}=\frac{\mathrm{v}}{\lambda_{1}}-\frac{\mathrm{v}}{\lambda_{2}}$ $\frac{40}{12}=\frac{\mathrm{v}}{4.08}-\frac{\mathrm{v}}{4.16}$ $\mathrm{v}=\frac{40}{12} \times \frac{4.08 \times 4.16}{0.08}$ $\mathrm{v}=707.2 \mathrm{~m} / \mathrm{s}$
JEE Main-28.06.2022
WAVES
172716
A tuning fork of frequency $200 \mathrm{~Hz}$ is in unison with a sonometer wire. How many beats per second will be heard if the tension of the wire is increased by $2 \%$ ?
1 1
2 2
3 3
4 4
Explanation:
B Given that, frequency (f) $=200 \mathrm{~Hz}$ We know that, $\therefore \quad \mathrm{f} \propto \sqrt{\mathrm{T}}$ $\frac{\Delta \mathrm{f}}{\mathrm{f}}=\frac{1}{2} \frac{\Delta \mathrm{T}}{\mathrm{T}}$ $\frac{\Delta \mathrm{f}}{\mathrm{f}}=\frac{1}{2} \times 0.02$ $\Delta \mathrm{f}=\mathrm{f} \times \frac{1}{2} \times 0.02$ $\Delta \mathrm{f}=200 \times \frac{1}{2} \times 0.02$ $\Delta \mathrm{f}=2 \mathrm{~Hz}$ Hence, No. of beats heard per second $=2 \mathrm{~Hz}$
