354830
For a sound source of intensity \(I\;W/{m^2}\), corresponding sound level is \(B_{0}\) decibel. If the intensity is increased to \(4I\), new sound level becomes:
1 \(\left(B_{0}+3\right) d B\)
2 \(2 B_{0} d B\)
3 \(4 B_{0} d B\)
4 \(\left(B_{0}+6\right) d B\)
Explanation:
Given that \(B_{o}=10 \log \left(\dfrac{I}{I_{o}}\right)\) \(B=10 \log \left(\dfrac{4 I}{I_{o}}\right)=10\left[\log \left(\dfrac{I}{I_{o}}\right)+\log 4\right]\) \(B=B_{o}+10 \log (4)=B_{o}+20 \log 2=\left(B_{o}+6\right) d B\)
PHXI15:WAVES
354831
Two sound waves travel in the same direction in a medium. The amplitude of each wave is \(A\) and the phase difference between the two waves is \(120^{\circ}\). The resultant amplitude will be
1 \(2\;A\)
2 \(\sqrt{2} A\)
3 \(4\;A\)
4 \(A\)
Explanation:
Here, \(A_{1}=A, A_{2}=A, \phi=120^{\circ}\) The amplitude of the resultant wave is \(\begin{aligned}& A_{R}=\sqrt{A_{1}^{2}+A_{2}^{2}+2 A_{1} A_{2} \cos \phi} \\& =\sqrt{A^{2}+A^{2}+2 A A \cos 120^{\circ}} \\& \therefore A_{R}=A .\end{aligned}\)
PHXI15:WAVES
354832
A stationary point source of sound emits sound uniformly in all directions in a non absorbing medium. Two points \(P\) and \(Q\) are at a distance of \(4\;m\) and \(9\;m\) respectively from the source. The ratio of amplitudes of the waves at \(P\) and \(Q\) is
1 \(3 / 2\)
2 \(4 / 9\)
3 \(2 / 3\)
4 \(9 / 4\)
Explanation:
For an isotropic point source of power \(P\), intensity \(I\) at a distance \(r\) from it will be \(I=\dfrac{P_{0}}{4 \pi r^{2}}\) Since power \(\left( {{P_0}} \right)\) remains the same, \(\therefore \dfrac{I_{1}}{I_{2}}=\left(\dfrac{r_{2}}{r_{1}}\right)^{2}=\left(\dfrac{9}{4}\right)^{2}\) \(\because I \propto A^{2}\), where \(A\) is the amplitude of a wave \(\therefore \dfrac{A_{1}}{A_{2}}=\sqrt{\dfrac{I_{1}}{I_{2}}}=\dfrac{9}{4}\)
354830
For a sound source of intensity \(I\;W/{m^2}\), corresponding sound level is \(B_{0}\) decibel. If the intensity is increased to \(4I\), new sound level becomes:
1 \(\left(B_{0}+3\right) d B\)
2 \(2 B_{0} d B\)
3 \(4 B_{0} d B\)
4 \(\left(B_{0}+6\right) d B\)
Explanation:
Given that \(B_{o}=10 \log \left(\dfrac{I}{I_{o}}\right)\) \(B=10 \log \left(\dfrac{4 I}{I_{o}}\right)=10\left[\log \left(\dfrac{I}{I_{o}}\right)+\log 4\right]\) \(B=B_{o}+10 \log (4)=B_{o}+20 \log 2=\left(B_{o}+6\right) d B\)
PHXI15:WAVES
354831
Two sound waves travel in the same direction in a medium. The amplitude of each wave is \(A\) and the phase difference between the two waves is \(120^{\circ}\). The resultant amplitude will be
1 \(2\;A\)
2 \(\sqrt{2} A\)
3 \(4\;A\)
4 \(A\)
Explanation:
Here, \(A_{1}=A, A_{2}=A, \phi=120^{\circ}\) The amplitude of the resultant wave is \(\begin{aligned}& A_{R}=\sqrt{A_{1}^{2}+A_{2}^{2}+2 A_{1} A_{2} \cos \phi} \\& =\sqrt{A^{2}+A^{2}+2 A A \cos 120^{\circ}} \\& \therefore A_{R}=A .\end{aligned}\)
PHXI15:WAVES
354832
A stationary point source of sound emits sound uniformly in all directions in a non absorbing medium. Two points \(P\) and \(Q\) are at a distance of \(4\;m\) and \(9\;m\) respectively from the source. The ratio of amplitudes of the waves at \(P\) and \(Q\) is
1 \(3 / 2\)
2 \(4 / 9\)
3 \(2 / 3\)
4 \(9 / 4\)
Explanation:
For an isotropic point source of power \(P\), intensity \(I\) at a distance \(r\) from it will be \(I=\dfrac{P_{0}}{4 \pi r^{2}}\) Since power \(\left( {{P_0}} \right)\) remains the same, \(\therefore \dfrac{I_{1}}{I_{2}}=\left(\dfrac{r_{2}}{r_{1}}\right)^{2}=\left(\dfrac{9}{4}\right)^{2}\) \(\because I \propto A^{2}\), where \(A\) is the amplitude of a wave \(\therefore \dfrac{A_{1}}{A_{2}}=\sqrt{\dfrac{I_{1}}{I_{2}}}=\dfrac{9}{4}\)
354830
For a sound source of intensity \(I\;W/{m^2}\), corresponding sound level is \(B_{0}\) decibel. If the intensity is increased to \(4I\), new sound level becomes:
1 \(\left(B_{0}+3\right) d B\)
2 \(2 B_{0} d B\)
3 \(4 B_{0} d B\)
4 \(\left(B_{0}+6\right) d B\)
Explanation:
Given that \(B_{o}=10 \log \left(\dfrac{I}{I_{o}}\right)\) \(B=10 \log \left(\dfrac{4 I}{I_{o}}\right)=10\left[\log \left(\dfrac{I}{I_{o}}\right)+\log 4\right]\) \(B=B_{o}+10 \log (4)=B_{o}+20 \log 2=\left(B_{o}+6\right) d B\)
PHXI15:WAVES
354831
Two sound waves travel in the same direction in a medium. The amplitude of each wave is \(A\) and the phase difference between the two waves is \(120^{\circ}\). The resultant amplitude will be
1 \(2\;A\)
2 \(\sqrt{2} A\)
3 \(4\;A\)
4 \(A\)
Explanation:
Here, \(A_{1}=A, A_{2}=A, \phi=120^{\circ}\) The amplitude of the resultant wave is \(\begin{aligned}& A_{R}=\sqrt{A_{1}^{2}+A_{2}^{2}+2 A_{1} A_{2} \cos \phi} \\& =\sqrt{A^{2}+A^{2}+2 A A \cos 120^{\circ}} \\& \therefore A_{R}=A .\end{aligned}\)
PHXI15:WAVES
354832
A stationary point source of sound emits sound uniformly in all directions in a non absorbing medium. Two points \(P\) and \(Q\) are at a distance of \(4\;m\) and \(9\;m\) respectively from the source. The ratio of amplitudes of the waves at \(P\) and \(Q\) is
1 \(3 / 2\)
2 \(4 / 9\)
3 \(2 / 3\)
4 \(9 / 4\)
Explanation:
For an isotropic point source of power \(P\), intensity \(I\) at a distance \(r\) from it will be \(I=\dfrac{P_{0}}{4 \pi r^{2}}\) Since power \(\left( {{P_0}} \right)\) remains the same, \(\therefore \dfrac{I_{1}}{I_{2}}=\left(\dfrac{r_{2}}{r_{1}}\right)^{2}=\left(\dfrac{9}{4}\right)^{2}\) \(\because I \propto A^{2}\), where \(A\) is the amplitude of a wave \(\therefore \dfrac{A_{1}}{A_{2}}=\sqrt{\dfrac{I_{1}}{I_{2}}}=\dfrac{9}{4}\)
354830
For a sound source of intensity \(I\;W/{m^2}\), corresponding sound level is \(B_{0}\) decibel. If the intensity is increased to \(4I\), new sound level becomes:
1 \(\left(B_{0}+3\right) d B\)
2 \(2 B_{0} d B\)
3 \(4 B_{0} d B\)
4 \(\left(B_{0}+6\right) d B\)
Explanation:
Given that \(B_{o}=10 \log \left(\dfrac{I}{I_{o}}\right)\) \(B=10 \log \left(\dfrac{4 I}{I_{o}}\right)=10\left[\log \left(\dfrac{I}{I_{o}}\right)+\log 4\right]\) \(B=B_{o}+10 \log (4)=B_{o}+20 \log 2=\left(B_{o}+6\right) d B\)
PHXI15:WAVES
354831
Two sound waves travel in the same direction in a medium. The amplitude of each wave is \(A\) and the phase difference between the two waves is \(120^{\circ}\). The resultant amplitude will be
1 \(2\;A\)
2 \(\sqrt{2} A\)
3 \(4\;A\)
4 \(A\)
Explanation:
Here, \(A_{1}=A, A_{2}=A, \phi=120^{\circ}\) The amplitude of the resultant wave is \(\begin{aligned}& A_{R}=\sqrt{A_{1}^{2}+A_{2}^{2}+2 A_{1} A_{2} \cos \phi} \\& =\sqrt{A^{2}+A^{2}+2 A A \cos 120^{\circ}} \\& \therefore A_{R}=A .\end{aligned}\)
PHXI15:WAVES
354832
A stationary point source of sound emits sound uniformly in all directions in a non absorbing medium. Two points \(P\) and \(Q\) are at a distance of \(4\;m\) and \(9\;m\) respectively from the source. The ratio of amplitudes of the waves at \(P\) and \(Q\) is
1 \(3 / 2\)
2 \(4 / 9\)
3 \(2 / 3\)
4 \(9 / 4\)
Explanation:
For an isotropic point source of power \(P\), intensity \(I\) at a distance \(r\) from it will be \(I=\dfrac{P_{0}}{4 \pi r^{2}}\) Since power \(\left( {{P_0}} \right)\) remains the same, \(\therefore \dfrac{I_{1}}{I_{2}}=\left(\dfrac{r_{2}}{r_{1}}\right)^{2}=\left(\dfrac{9}{4}\right)^{2}\) \(\because I \propto A^{2}\), where \(A\) is the amplitude of a wave \(\therefore \dfrac{A_{1}}{A_{2}}=\sqrt{\dfrac{I_{1}}{I_{2}}}=\dfrac{9}{4}\)
