172155
The velocity of a travelling wave of frequency $500 \mathrm{~Hz}$ is $360 \mathrm{~m} / \mathrm{s}$. The minimum distance between two points having a phase difference of $60^{\circ}$ is
1 $10 \mathrm{~cm}$
2 $12 \mathrm{~cm}$
3 $36 \mathrm{~cm}$
4 $18 \mathrm{~cm}$
Explanation:
B Given that, Frequency $(\mathrm{f})=500 \mathrm{~Hz}$ Velocity $(\mathrm{v})=360 \mathrm{~m} / \mathrm{sec}$. Phase difference $\Delta \phi=\pi / 3$ Distance between two particles is given by Path difference, $\Delta \mathrm{x}=\frac{\lambda}{2 \pi} \times \Delta \phi$ $\Delta \mathrm{x}=\frac{(\mathrm{v} / \mathrm{f})}{2 \pi} \times \Delta \phi$ $\Delta \mathrm{x}=\frac{360 / 500}{2 \pi} \times \frac{\pi}{3}$ $\Delta \mathrm{x}=0.12 \mathrm{~m}=12 \mathrm{~cm}$ $\therefore \lambda=\frac{\mathrm{v}}{\mathrm{f}}$
Tripura-27.04.2022
WAVES
172156
The distance between two successive minima of a transverse wave is $2.7 \mathrm{~m}$. Five crests of the wave pass a given point along the direction of travel $15.0 \mathrm{~s}$. The speed of the wave is
1 $0.9 \mathrm{~m} / \mathrm{s}$
2 $1.2 \mathrm{~m} / \mathrm{s}$
3 $0.5 \mathrm{~m} / \mathrm{s}$
4 $2.4 \mathrm{~m} / \mathrm{s}$
Explanation:
A Given, Distance between two successive minima of a transverse wave $(\lambda)=2.7 \mathrm{~m}$ Number of passing waves crests $=5$ Time period $=15 \mathrm{sec}$. Frequency $(f)=\frac{\text { Number of waves'crests }}{\text { Time period }}$ $\mathrm{f}=\frac{5}{15}=\frac{1}{3} \sec ^{-1}$ We know that, speed $\mathrm{v}=\lambda \mathrm{f}$ $\mathrm{v}=2.7 \mathrm{~m} \times \frac{1}{3} \mathrm{sec}^{-1}$ $\mathrm{v}=0.9 \mathrm{~m} \mathrm{sec}^{-1}$
TS EAMCET 18.07.2022
WAVES
172288
The speed of wave in a medium is $1500 \mathrm{~m} / \mathrm{s}$. If 3600 waves pass through a point in $1 \mathrm{~min}$ in this medium, then the wavelength of wave is
1 $25 \mathrm{~m}$
2 $50 \mathrm{~m}$
3 $0.41 \mathrm{~m}$
4 $90000 \mathrm{~m}$
Explanation:
A Given that, $\mathrm{v}=1500 \mathrm{~m} / \mathrm{sec}$ $\mathrm{f}=3600 / 60=60 \mathrm{~Hz}$ $\mathrm{v}=\mathrm{f} \lambda$ $\lambda=\frac{\mathrm{v}}{\mathrm{f}}$ $\lambda=\frac{1500}{60}=25 \mathrm{~m}$
172155
The velocity of a travelling wave of frequency $500 \mathrm{~Hz}$ is $360 \mathrm{~m} / \mathrm{s}$. The minimum distance between two points having a phase difference of $60^{\circ}$ is
1 $10 \mathrm{~cm}$
2 $12 \mathrm{~cm}$
3 $36 \mathrm{~cm}$
4 $18 \mathrm{~cm}$
Explanation:
B Given that, Frequency $(\mathrm{f})=500 \mathrm{~Hz}$ Velocity $(\mathrm{v})=360 \mathrm{~m} / \mathrm{sec}$. Phase difference $\Delta \phi=\pi / 3$ Distance between two particles is given by Path difference, $\Delta \mathrm{x}=\frac{\lambda}{2 \pi} \times \Delta \phi$ $\Delta \mathrm{x}=\frac{(\mathrm{v} / \mathrm{f})}{2 \pi} \times \Delta \phi$ $\Delta \mathrm{x}=\frac{360 / 500}{2 \pi} \times \frac{\pi}{3}$ $\Delta \mathrm{x}=0.12 \mathrm{~m}=12 \mathrm{~cm}$ $\therefore \lambda=\frac{\mathrm{v}}{\mathrm{f}}$
Tripura-27.04.2022
WAVES
172156
The distance between two successive minima of a transverse wave is $2.7 \mathrm{~m}$. Five crests of the wave pass a given point along the direction of travel $15.0 \mathrm{~s}$. The speed of the wave is
1 $0.9 \mathrm{~m} / \mathrm{s}$
2 $1.2 \mathrm{~m} / \mathrm{s}$
3 $0.5 \mathrm{~m} / \mathrm{s}$
4 $2.4 \mathrm{~m} / \mathrm{s}$
Explanation:
A Given, Distance between two successive minima of a transverse wave $(\lambda)=2.7 \mathrm{~m}$ Number of passing waves crests $=5$ Time period $=15 \mathrm{sec}$. Frequency $(f)=\frac{\text { Number of waves'crests }}{\text { Time period }}$ $\mathrm{f}=\frac{5}{15}=\frac{1}{3} \sec ^{-1}$ We know that, speed $\mathrm{v}=\lambda \mathrm{f}$ $\mathrm{v}=2.7 \mathrm{~m} \times \frac{1}{3} \mathrm{sec}^{-1}$ $\mathrm{v}=0.9 \mathrm{~m} \mathrm{sec}^{-1}$
TS EAMCET 18.07.2022
WAVES
172288
The speed of wave in a medium is $1500 \mathrm{~m} / \mathrm{s}$. If 3600 waves pass through a point in $1 \mathrm{~min}$ in this medium, then the wavelength of wave is
1 $25 \mathrm{~m}$
2 $50 \mathrm{~m}$
3 $0.41 \mathrm{~m}$
4 $90000 \mathrm{~m}$
Explanation:
A Given that, $\mathrm{v}=1500 \mathrm{~m} / \mathrm{sec}$ $\mathrm{f}=3600 / 60=60 \mathrm{~Hz}$ $\mathrm{v}=\mathrm{f} \lambda$ $\lambda=\frac{\mathrm{v}}{\mathrm{f}}$ $\lambda=\frac{1500}{60}=25 \mathrm{~m}$
172155
The velocity of a travelling wave of frequency $500 \mathrm{~Hz}$ is $360 \mathrm{~m} / \mathrm{s}$. The minimum distance between two points having a phase difference of $60^{\circ}$ is
1 $10 \mathrm{~cm}$
2 $12 \mathrm{~cm}$
3 $36 \mathrm{~cm}$
4 $18 \mathrm{~cm}$
Explanation:
B Given that, Frequency $(\mathrm{f})=500 \mathrm{~Hz}$ Velocity $(\mathrm{v})=360 \mathrm{~m} / \mathrm{sec}$. Phase difference $\Delta \phi=\pi / 3$ Distance between two particles is given by Path difference, $\Delta \mathrm{x}=\frac{\lambda}{2 \pi} \times \Delta \phi$ $\Delta \mathrm{x}=\frac{(\mathrm{v} / \mathrm{f})}{2 \pi} \times \Delta \phi$ $\Delta \mathrm{x}=\frac{360 / 500}{2 \pi} \times \frac{\pi}{3}$ $\Delta \mathrm{x}=0.12 \mathrm{~m}=12 \mathrm{~cm}$ $\therefore \lambda=\frac{\mathrm{v}}{\mathrm{f}}$
Tripura-27.04.2022
WAVES
172156
The distance between two successive minima of a transverse wave is $2.7 \mathrm{~m}$. Five crests of the wave pass a given point along the direction of travel $15.0 \mathrm{~s}$. The speed of the wave is
1 $0.9 \mathrm{~m} / \mathrm{s}$
2 $1.2 \mathrm{~m} / \mathrm{s}$
3 $0.5 \mathrm{~m} / \mathrm{s}$
4 $2.4 \mathrm{~m} / \mathrm{s}$
Explanation:
A Given, Distance between two successive minima of a transverse wave $(\lambda)=2.7 \mathrm{~m}$ Number of passing waves crests $=5$ Time period $=15 \mathrm{sec}$. Frequency $(f)=\frac{\text { Number of waves'crests }}{\text { Time period }}$ $\mathrm{f}=\frac{5}{15}=\frac{1}{3} \sec ^{-1}$ We know that, speed $\mathrm{v}=\lambda \mathrm{f}$ $\mathrm{v}=2.7 \mathrm{~m} \times \frac{1}{3} \mathrm{sec}^{-1}$ $\mathrm{v}=0.9 \mathrm{~m} \mathrm{sec}^{-1}$
TS EAMCET 18.07.2022
WAVES
172288
The speed of wave in a medium is $1500 \mathrm{~m} / \mathrm{s}$. If 3600 waves pass through a point in $1 \mathrm{~min}$ in this medium, then the wavelength of wave is
1 $25 \mathrm{~m}$
2 $50 \mathrm{~m}$
3 $0.41 \mathrm{~m}$
4 $90000 \mathrm{~m}$
Explanation:
A Given that, $\mathrm{v}=1500 \mathrm{~m} / \mathrm{sec}$ $\mathrm{f}=3600 / 60=60 \mathrm{~Hz}$ $\mathrm{v}=\mathrm{f} \lambda$ $\lambda=\frac{\mathrm{v}}{\mathrm{f}}$ $\lambda=\frac{1500}{60}=25 \mathrm{~m}$
172155
The velocity of a travelling wave of frequency $500 \mathrm{~Hz}$ is $360 \mathrm{~m} / \mathrm{s}$. The minimum distance between two points having a phase difference of $60^{\circ}$ is
1 $10 \mathrm{~cm}$
2 $12 \mathrm{~cm}$
3 $36 \mathrm{~cm}$
4 $18 \mathrm{~cm}$
Explanation:
B Given that, Frequency $(\mathrm{f})=500 \mathrm{~Hz}$ Velocity $(\mathrm{v})=360 \mathrm{~m} / \mathrm{sec}$. Phase difference $\Delta \phi=\pi / 3$ Distance between two particles is given by Path difference, $\Delta \mathrm{x}=\frac{\lambda}{2 \pi} \times \Delta \phi$ $\Delta \mathrm{x}=\frac{(\mathrm{v} / \mathrm{f})}{2 \pi} \times \Delta \phi$ $\Delta \mathrm{x}=\frac{360 / 500}{2 \pi} \times \frac{\pi}{3}$ $\Delta \mathrm{x}=0.12 \mathrm{~m}=12 \mathrm{~cm}$ $\therefore \lambda=\frac{\mathrm{v}}{\mathrm{f}}$
Tripura-27.04.2022
WAVES
172156
The distance between two successive minima of a transverse wave is $2.7 \mathrm{~m}$. Five crests of the wave pass a given point along the direction of travel $15.0 \mathrm{~s}$. The speed of the wave is
1 $0.9 \mathrm{~m} / \mathrm{s}$
2 $1.2 \mathrm{~m} / \mathrm{s}$
3 $0.5 \mathrm{~m} / \mathrm{s}$
4 $2.4 \mathrm{~m} / \mathrm{s}$
Explanation:
A Given, Distance between two successive minima of a transverse wave $(\lambda)=2.7 \mathrm{~m}$ Number of passing waves crests $=5$ Time period $=15 \mathrm{sec}$. Frequency $(f)=\frac{\text { Number of waves'crests }}{\text { Time period }}$ $\mathrm{f}=\frac{5}{15}=\frac{1}{3} \sec ^{-1}$ We know that, speed $\mathrm{v}=\lambda \mathrm{f}$ $\mathrm{v}=2.7 \mathrm{~m} \times \frac{1}{3} \mathrm{sec}^{-1}$ $\mathrm{v}=0.9 \mathrm{~m} \mathrm{sec}^{-1}$
TS EAMCET 18.07.2022
WAVES
172288
The speed of wave in a medium is $1500 \mathrm{~m} / \mathrm{s}$. If 3600 waves pass through a point in $1 \mathrm{~min}$ in this medium, then the wavelength of wave is
1 $25 \mathrm{~m}$
2 $50 \mathrm{~m}$
3 $0.41 \mathrm{~m}$
4 $90000 \mathrm{~m}$
Explanation:
A Given that, $\mathrm{v}=1500 \mathrm{~m} / \mathrm{sec}$ $\mathrm{f}=3600 / 60=60 \mathrm{~Hz}$ $\mathrm{v}=\mathrm{f} \lambda$ $\lambda=\frac{\mathrm{v}}{\mathrm{f}}$ $\lambda=\frac{1500}{60}=25 \mathrm{~m}$
