172226
The angle between particle velocity and wave velocity in a transverse wave is
1 zero
2 $\pi / 4$
3 $\pi / 2$
4 $\pi$
Explanation:
C In a transverse wave the particles of the medium vibrate about their mean positions in a direction perpendicular to the direction of wave propagation. Here, the particle velocity is given by $\frac{d y}{d t}$ and wave velocity is given by $\frac{\mathrm{dx}}{\mathrm{dt}}$. Hence, the angle between particle velocity and wave velocity in a transverse wave is $\frac{\pi}{2}$.
UPSEE - 2008
WAVES
172228
The equation of a wave is represented by $y=10^{-4} \sin \left(100 t-\frac{x}{10}\right) m$, then the velocity of wave will be
1 $100 \mathrm{~m} / \mathrm{s}$
2 $4 \mathrm{~m} / \mathrm{s}$
3 $1000 \mathrm{~m} / \mathrm{s}$
4 $10 \mathrm{~m} / \mathrm{s}$
Explanation:
C Given equation. $y=10^{-4} \sin \left(100 t-\frac{x}{10}\right) m$ Comparing the given equation with general equation $\mathrm{y}=\mathrm{a} \sin 2 \pi\left(\frac{\mathrm{t}}{\mathrm{T}}-\frac{\mathrm{x}}{\lambda}\right)$ $\begin{aligned} \frac{2 \pi}{\mathrm{T}}=100, \frac{2 \pi}{\lambda} =\frac{1}{10} \\ \text { So, } \quad \mathrm{T}=\frac{2 \pi}{100}, \quad \lambda =20 \pi \\ \text { We know, } \quad \mathrm{v} =\frac{\lambda}{\mathrm{T}} \\ \mathrm{v} =\frac{20 \pi}{2 \pi / 100} \\ \mathrm{v} =1000 \mathrm{~m} / \mathrm{s}\end{aligned}$
BITSAT-2007
WAVES
172231
The phase velocity $\left(v_{P}\right)$ of travelling wave is
A The wave is, $\mathrm{y}=\mathrm{A} \sin (\omega \mathrm{t}-\mathrm{kx})$ Wave will appear to travel at the phase velocity $\left(v_{p}\right)$. It is given in terms of waves angular velocity $\omega$ and wave number is $\mathrm{k}$ $\mathrm{v}_{\mathrm{p}}=\frac{\omega}{\mathrm{k}}$
CG PET- 2009
WAVES
172248
A boat anchor is rocked by waves whose crests are $100 \mathrm{~m}$ apart and velocity is $25 \mathrm{~m} / \mathrm{sec}$. The boat bounces up once in every :
1 $2500 \mathrm{~s}$
2 $75 \mathrm{~s}$
3 $4 \mathrm{~s}$
4 $0.25 \mathrm{~s}$
Explanation:
C Given, $\lambda=100 \mathrm{~m}, \mathrm{v}=25 \mathrm{~m} / \mathrm{s}$ We know that, Time period $(\mathrm{T})=\frac{\lambda}{\mathrm{V}}$ $\mathrm{T}=\frac{100}{25}$ $\mathrm{~T}=4 \mathrm{sec}$
AIIMS-2006
WAVES
172249
A stone thrown into still water, creates a circular wave pattern moving radially outwards. If $r$ is the distance measured from the centre of the pattern. The amplitude of the wave aries as :
1 $\mathrm{r}^{-1 / 2}$
2 $\mathrm{r}^{-1}$
3 $\mathrm{r}^{-2}$
4 $\mathrm{r}^{-3 / 2}$
Explanation:
A For circular wave pattern Wave intensity, $\mathrm{I} \propto \frac{1}{\mathrm{r}}$ or $\mathrm{r}^{-1}$ Amplitude, $\mathrm{A} \propto \sqrt{\mathrm{I}}$ $\mathrm{A} \propto \sqrt{\frac{1}{\mathrm{r}}}$ $\mathrm{A} \propto \mathrm{r}^{-1 / 2}$
172226
The angle between particle velocity and wave velocity in a transverse wave is
1 zero
2 $\pi / 4$
3 $\pi / 2$
4 $\pi$
Explanation:
C In a transverse wave the particles of the medium vibrate about their mean positions in a direction perpendicular to the direction of wave propagation. Here, the particle velocity is given by $\frac{d y}{d t}$ and wave velocity is given by $\frac{\mathrm{dx}}{\mathrm{dt}}$. Hence, the angle between particle velocity and wave velocity in a transverse wave is $\frac{\pi}{2}$.
UPSEE - 2008
WAVES
172228
The equation of a wave is represented by $y=10^{-4} \sin \left(100 t-\frac{x}{10}\right) m$, then the velocity of wave will be
1 $100 \mathrm{~m} / \mathrm{s}$
2 $4 \mathrm{~m} / \mathrm{s}$
3 $1000 \mathrm{~m} / \mathrm{s}$
4 $10 \mathrm{~m} / \mathrm{s}$
Explanation:
C Given equation. $y=10^{-4} \sin \left(100 t-\frac{x}{10}\right) m$ Comparing the given equation with general equation $\mathrm{y}=\mathrm{a} \sin 2 \pi\left(\frac{\mathrm{t}}{\mathrm{T}}-\frac{\mathrm{x}}{\lambda}\right)$ $\begin{aligned} \frac{2 \pi}{\mathrm{T}}=100, \frac{2 \pi}{\lambda} =\frac{1}{10} \\ \text { So, } \quad \mathrm{T}=\frac{2 \pi}{100}, \quad \lambda =20 \pi \\ \text { We know, } \quad \mathrm{v} =\frac{\lambda}{\mathrm{T}} \\ \mathrm{v} =\frac{20 \pi}{2 \pi / 100} \\ \mathrm{v} =1000 \mathrm{~m} / \mathrm{s}\end{aligned}$
BITSAT-2007
WAVES
172231
The phase velocity $\left(v_{P}\right)$ of travelling wave is
A The wave is, $\mathrm{y}=\mathrm{A} \sin (\omega \mathrm{t}-\mathrm{kx})$ Wave will appear to travel at the phase velocity $\left(v_{p}\right)$. It is given in terms of waves angular velocity $\omega$ and wave number is $\mathrm{k}$ $\mathrm{v}_{\mathrm{p}}=\frac{\omega}{\mathrm{k}}$
CG PET- 2009
WAVES
172248
A boat anchor is rocked by waves whose crests are $100 \mathrm{~m}$ apart and velocity is $25 \mathrm{~m} / \mathrm{sec}$. The boat bounces up once in every :
1 $2500 \mathrm{~s}$
2 $75 \mathrm{~s}$
3 $4 \mathrm{~s}$
4 $0.25 \mathrm{~s}$
Explanation:
C Given, $\lambda=100 \mathrm{~m}, \mathrm{v}=25 \mathrm{~m} / \mathrm{s}$ We know that, Time period $(\mathrm{T})=\frac{\lambda}{\mathrm{V}}$ $\mathrm{T}=\frac{100}{25}$ $\mathrm{~T}=4 \mathrm{sec}$
AIIMS-2006
WAVES
172249
A stone thrown into still water, creates a circular wave pattern moving radially outwards. If $r$ is the distance measured from the centre of the pattern. The amplitude of the wave aries as :
1 $\mathrm{r}^{-1 / 2}$
2 $\mathrm{r}^{-1}$
3 $\mathrm{r}^{-2}$
4 $\mathrm{r}^{-3 / 2}$
Explanation:
A For circular wave pattern Wave intensity, $\mathrm{I} \propto \frac{1}{\mathrm{r}}$ or $\mathrm{r}^{-1}$ Amplitude, $\mathrm{A} \propto \sqrt{\mathrm{I}}$ $\mathrm{A} \propto \sqrt{\frac{1}{\mathrm{r}}}$ $\mathrm{A} \propto \mathrm{r}^{-1 / 2}$
172226
The angle between particle velocity and wave velocity in a transverse wave is
1 zero
2 $\pi / 4$
3 $\pi / 2$
4 $\pi$
Explanation:
C In a transverse wave the particles of the medium vibrate about their mean positions in a direction perpendicular to the direction of wave propagation. Here, the particle velocity is given by $\frac{d y}{d t}$ and wave velocity is given by $\frac{\mathrm{dx}}{\mathrm{dt}}$. Hence, the angle between particle velocity and wave velocity in a transverse wave is $\frac{\pi}{2}$.
UPSEE - 2008
WAVES
172228
The equation of a wave is represented by $y=10^{-4} \sin \left(100 t-\frac{x}{10}\right) m$, then the velocity of wave will be
1 $100 \mathrm{~m} / \mathrm{s}$
2 $4 \mathrm{~m} / \mathrm{s}$
3 $1000 \mathrm{~m} / \mathrm{s}$
4 $10 \mathrm{~m} / \mathrm{s}$
Explanation:
C Given equation. $y=10^{-4} \sin \left(100 t-\frac{x}{10}\right) m$ Comparing the given equation with general equation $\mathrm{y}=\mathrm{a} \sin 2 \pi\left(\frac{\mathrm{t}}{\mathrm{T}}-\frac{\mathrm{x}}{\lambda}\right)$ $\begin{aligned} \frac{2 \pi}{\mathrm{T}}=100, \frac{2 \pi}{\lambda} =\frac{1}{10} \\ \text { So, } \quad \mathrm{T}=\frac{2 \pi}{100}, \quad \lambda =20 \pi \\ \text { We know, } \quad \mathrm{v} =\frac{\lambda}{\mathrm{T}} \\ \mathrm{v} =\frac{20 \pi}{2 \pi / 100} \\ \mathrm{v} =1000 \mathrm{~m} / \mathrm{s}\end{aligned}$
BITSAT-2007
WAVES
172231
The phase velocity $\left(v_{P}\right)$ of travelling wave is
A The wave is, $\mathrm{y}=\mathrm{A} \sin (\omega \mathrm{t}-\mathrm{kx})$ Wave will appear to travel at the phase velocity $\left(v_{p}\right)$. It is given in terms of waves angular velocity $\omega$ and wave number is $\mathrm{k}$ $\mathrm{v}_{\mathrm{p}}=\frac{\omega}{\mathrm{k}}$
CG PET- 2009
WAVES
172248
A boat anchor is rocked by waves whose crests are $100 \mathrm{~m}$ apart and velocity is $25 \mathrm{~m} / \mathrm{sec}$. The boat bounces up once in every :
1 $2500 \mathrm{~s}$
2 $75 \mathrm{~s}$
3 $4 \mathrm{~s}$
4 $0.25 \mathrm{~s}$
Explanation:
C Given, $\lambda=100 \mathrm{~m}, \mathrm{v}=25 \mathrm{~m} / \mathrm{s}$ We know that, Time period $(\mathrm{T})=\frac{\lambda}{\mathrm{V}}$ $\mathrm{T}=\frac{100}{25}$ $\mathrm{~T}=4 \mathrm{sec}$
AIIMS-2006
WAVES
172249
A stone thrown into still water, creates a circular wave pattern moving radially outwards. If $r$ is the distance measured from the centre of the pattern. The amplitude of the wave aries as :
1 $\mathrm{r}^{-1 / 2}$
2 $\mathrm{r}^{-1}$
3 $\mathrm{r}^{-2}$
4 $\mathrm{r}^{-3 / 2}$
Explanation:
A For circular wave pattern Wave intensity, $\mathrm{I} \propto \frac{1}{\mathrm{r}}$ or $\mathrm{r}^{-1}$ Amplitude, $\mathrm{A} \propto \sqrt{\mathrm{I}}$ $\mathrm{A} \propto \sqrt{\frac{1}{\mathrm{r}}}$ $\mathrm{A} \propto \mathrm{r}^{-1 / 2}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
WAVES
172226
The angle between particle velocity and wave velocity in a transverse wave is
1 zero
2 $\pi / 4$
3 $\pi / 2$
4 $\pi$
Explanation:
C In a transverse wave the particles of the medium vibrate about their mean positions in a direction perpendicular to the direction of wave propagation. Here, the particle velocity is given by $\frac{d y}{d t}$ and wave velocity is given by $\frac{\mathrm{dx}}{\mathrm{dt}}$. Hence, the angle between particle velocity and wave velocity in a transverse wave is $\frac{\pi}{2}$.
UPSEE - 2008
WAVES
172228
The equation of a wave is represented by $y=10^{-4} \sin \left(100 t-\frac{x}{10}\right) m$, then the velocity of wave will be
1 $100 \mathrm{~m} / \mathrm{s}$
2 $4 \mathrm{~m} / \mathrm{s}$
3 $1000 \mathrm{~m} / \mathrm{s}$
4 $10 \mathrm{~m} / \mathrm{s}$
Explanation:
C Given equation. $y=10^{-4} \sin \left(100 t-\frac{x}{10}\right) m$ Comparing the given equation with general equation $\mathrm{y}=\mathrm{a} \sin 2 \pi\left(\frac{\mathrm{t}}{\mathrm{T}}-\frac{\mathrm{x}}{\lambda}\right)$ $\begin{aligned} \frac{2 \pi}{\mathrm{T}}=100, \frac{2 \pi}{\lambda} =\frac{1}{10} \\ \text { So, } \quad \mathrm{T}=\frac{2 \pi}{100}, \quad \lambda =20 \pi \\ \text { We know, } \quad \mathrm{v} =\frac{\lambda}{\mathrm{T}} \\ \mathrm{v} =\frac{20 \pi}{2 \pi / 100} \\ \mathrm{v} =1000 \mathrm{~m} / \mathrm{s}\end{aligned}$
BITSAT-2007
WAVES
172231
The phase velocity $\left(v_{P}\right)$ of travelling wave is
A The wave is, $\mathrm{y}=\mathrm{A} \sin (\omega \mathrm{t}-\mathrm{kx})$ Wave will appear to travel at the phase velocity $\left(v_{p}\right)$. It is given in terms of waves angular velocity $\omega$ and wave number is $\mathrm{k}$ $\mathrm{v}_{\mathrm{p}}=\frac{\omega}{\mathrm{k}}$
CG PET- 2009
WAVES
172248
A boat anchor is rocked by waves whose crests are $100 \mathrm{~m}$ apart and velocity is $25 \mathrm{~m} / \mathrm{sec}$. The boat bounces up once in every :
1 $2500 \mathrm{~s}$
2 $75 \mathrm{~s}$
3 $4 \mathrm{~s}$
4 $0.25 \mathrm{~s}$
Explanation:
C Given, $\lambda=100 \mathrm{~m}, \mathrm{v}=25 \mathrm{~m} / \mathrm{s}$ We know that, Time period $(\mathrm{T})=\frac{\lambda}{\mathrm{V}}$ $\mathrm{T}=\frac{100}{25}$ $\mathrm{~T}=4 \mathrm{sec}$
AIIMS-2006
WAVES
172249
A stone thrown into still water, creates a circular wave pattern moving radially outwards. If $r$ is the distance measured from the centre of the pattern. The amplitude of the wave aries as :
1 $\mathrm{r}^{-1 / 2}$
2 $\mathrm{r}^{-1}$
3 $\mathrm{r}^{-2}$
4 $\mathrm{r}^{-3 / 2}$
Explanation:
A For circular wave pattern Wave intensity, $\mathrm{I} \propto \frac{1}{\mathrm{r}}$ or $\mathrm{r}^{-1}$ Amplitude, $\mathrm{A} \propto \sqrt{\mathrm{I}}$ $\mathrm{A} \propto \sqrt{\frac{1}{\mathrm{r}}}$ $\mathrm{A} \propto \mathrm{r}^{-1 / 2}$
172226
The angle between particle velocity and wave velocity in a transverse wave is
1 zero
2 $\pi / 4$
3 $\pi / 2$
4 $\pi$
Explanation:
C In a transverse wave the particles of the medium vibrate about their mean positions in a direction perpendicular to the direction of wave propagation. Here, the particle velocity is given by $\frac{d y}{d t}$ and wave velocity is given by $\frac{\mathrm{dx}}{\mathrm{dt}}$. Hence, the angle between particle velocity and wave velocity in a transverse wave is $\frac{\pi}{2}$.
UPSEE - 2008
WAVES
172228
The equation of a wave is represented by $y=10^{-4} \sin \left(100 t-\frac{x}{10}\right) m$, then the velocity of wave will be
1 $100 \mathrm{~m} / \mathrm{s}$
2 $4 \mathrm{~m} / \mathrm{s}$
3 $1000 \mathrm{~m} / \mathrm{s}$
4 $10 \mathrm{~m} / \mathrm{s}$
Explanation:
C Given equation. $y=10^{-4} \sin \left(100 t-\frac{x}{10}\right) m$ Comparing the given equation with general equation $\mathrm{y}=\mathrm{a} \sin 2 \pi\left(\frac{\mathrm{t}}{\mathrm{T}}-\frac{\mathrm{x}}{\lambda}\right)$ $\begin{aligned} \frac{2 \pi}{\mathrm{T}}=100, \frac{2 \pi}{\lambda} =\frac{1}{10} \\ \text { So, } \quad \mathrm{T}=\frac{2 \pi}{100}, \quad \lambda =20 \pi \\ \text { We know, } \quad \mathrm{v} =\frac{\lambda}{\mathrm{T}} \\ \mathrm{v} =\frac{20 \pi}{2 \pi / 100} \\ \mathrm{v} =1000 \mathrm{~m} / \mathrm{s}\end{aligned}$
BITSAT-2007
WAVES
172231
The phase velocity $\left(v_{P}\right)$ of travelling wave is
A The wave is, $\mathrm{y}=\mathrm{A} \sin (\omega \mathrm{t}-\mathrm{kx})$ Wave will appear to travel at the phase velocity $\left(v_{p}\right)$. It is given in terms of waves angular velocity $\omega$ and wave number is $\mathrm{k}$ $\mathrm{v}_{\mathrm{p}}=\frac{\omega}{\mathrm{k}}$
CG PET- 2009
WAVES
172248
A boat anchor is rocked by waves whose crests are $100 \mathrm{~m}$ apart and velocity is $25 \mathrm{~m} / \mathrm{sec}$. The boat bounces up once in every :
1 $2500 \mathrm{~s}$
2 $75 \mathrm{~s}$
3 $4 \mathrm{~s}$
4 $0.25 \mathrm{~s}$
Explanation:
C Given, $\lambda=100 \mathrm{~m}, \mathrm{v}=25 \mathrm{~m} / \mathrm{s}$ We know that, Time period $(\mathrm{T})=\frac{\lambda}{\mathrm{V}}$ $\mathrm{T}=\frac{100}{25}$ $\mathrm{~T}=4 \mathrm{sec}$
AIIMS-2006
WAVES
172249
A stone thrown into still water, creates a circular wave pattern moving radially outwards. If $r$ is the distance measured from the centre of the pattern. The amplitude of the wave aries as :
1 $\mathrm{r}^{-1 / 2}$
2 $\mathrm{r}^{-1}$
3 $\mathrm{r}^{-2}$
4 $\mathrm{r}^{-3 / 2}$
Explanation:
A For circular wave pattern Wave intensity, $\mathrm{I} \propto \frac{1}{\mathrm{r}}$ or $\mathrm{r}^{-1}$ Amplitude, $\mathrm{A} \propto \sqrt{\mathrm{I}}$ $\mathrm{A} \propto \sqrt{\frac{1}{\mathrm{r}}}$ $\mathrm{A} \propto \mathrm{r}^{-1 / 2}$
