NEET Test Series from KOTA - 10 Papers In MS WORD
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WAVES
172370
Two waves are given by $y_{1}=a \sin (\omega t-k x)$ and $y_{2}=a \cos (\omega t-k x)$. The phase difference between the two waves is
1 $\pi / 4$
2 $\pi$
3 $\pi / 8$
4 $\pi / 2$
Explanation:
D Given that, $\mathrm{y}_{1}=\mathrm{a} \sin (\omega \mathrm{t}-\mathrm{kx})$ $\mathrm{y}_{2}=\mathrm{a} \cos (\omega \mathrm{t}-\mathrm{kx})$ Equation (ii) we can write as $\mathrm{y}_{2}=\mathrm{a} \sin \left(\omega \mathrm{t}-\mathrm{kx}+\frac{\pi}{2}\right)$ Hence, phase difference, $\Delta \phi=\frac{\pi}{2}$
CG PET- 2006
WAVES
172371
If two waves represented by $y_{1}=4 \sin \omega t$ and $y_{2}=3 \sin \left(\omega t+\frac{\pi}{3}\right)$ interfere at a point. The amplitude of the resulting wave will be about
1 7
2 6
3 5
4 3.5
Explanation:
B Given that, $\mathrm{y}_{1}=4 \sin \omega \mathrm{t}$ $\mathrm{y}_{2}=3 \sin \left(\omega \mathrm{t}+\frac{\pi}{3}\right)$ From equation (i) and (ii) we get - Phase difference $(\phi)=\frac{\pi}{3}$ And $A_{1}=4, A_{2}=3$ $\text { Now, } \quad \mathrm{A}_{\text {Resultant }} =\sqrt{\mathrm{A}_{1}^{2}+\mathrm{A}_{2}^{2}+2 \mathrm{~A}_{1} \mathrm{~A}_{2} \cos \phi}$ $\mathrm{A}_{\text {Resultant }} =\sqrt{4^{2}+3^{2}+2 \times 4 \times 3 \times \cos \frac{\pi}{3}}$ $\mathrm{~A}_{\text {Resultant }} =\sqrt{25+24 \times \frac{1}{2}}$ $\mathrm{~A}_{\text {Resultant }} =\sqrt{37}$ $\mathrm{~A}_{\text {Resultant }} \approx 6$
Manipal UGET-2010
WAVES
172373
Two periodic waves of intensities $I_{1}, I_{2}$ pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is
172375
Two waves represented by $x_{1}=2 \sin (20 \pi t) \mathrm{m}$ and $x_{2}=2 \sin \left(20 \pi t+\frac{\pi}{3}\right) m$ are superimposed then the amplitude of the resultant wave.
1 $2 \sqrt{3} \mathrm{~m}$
2 $3 \sqrt{3} \mathrm{~m}$
3 $4 \sqrt{3} \mathrm{~m}$
4 $5 \sqrt{3} \mathrm{~m}$
Explanation:
A The two waves are given by $\mathrm{x}_{1}=2 \sin (20 \pi \mathrm{t}) \mathrm{m}$ $x_{2}=2 \sin \left(20 \pi t+\frac{\pi}{3}\right) m$ $\mathrm{A}_{1}=2, \mathrm{~A}_{2}=2, \phi=\frac{\pi}{3}=60^{\circ}$ The resultant wave amplitude $A=\sqrt{A_{1}^{2}+A_{2}^{2}+2 A_{1} A_{2} \cos \phi}$ $=\sqrt{(2)^{2}+(2)^{2}+2 \times 2 \times 2 \cos 60^{\circ}}$ $=\sqrt{4+4+8 \times \frac{1}{2}}$ $=\sqrt{8+4}$ $A=2 \sqrt{3} \mathrm{~m}$
172370
Two waves are given by $y_{1}=a \sin (\omega t-k x)$ and $y_{2}=a \cos (\omega t-k x)$. The phase difference between the two waves is
1 $\pi / 4$
2 $\pi$
3 $\pi / 8$
4 $\pi / 2$
Explanation:
D Given that, $\mathrm{y}_{1}=\mathrm{a} \sin (\omega \mathrm{t}-\mathrm{kx})$ $\mathrm{y}_{2}=\mathrm{a} \cos (\omega \mathrm{t}-\mathrm{kx})$ Equation (ii) we can write as $\mathrm{y}_{2}=\mathrm{a} \sin \left(\omega \mathrm{t}-\mathrm{kx}+\frac{\pi}{2}\right)$ Hence, phase difference, $\Delta \phi=\frac{\pi}{2}$
CG PET- 2006
WAVES
172371
If two waves represented by $y_{1}=4 \sin \omega t$ and $y_{2}=3 \sin \left(\omega t+\frac{\pi}{3}\right)$ interfere at a point. The amplitude of the resulting wave will be about
1 7
2 6
3 5
4 3.5
Explanation:
B Given that, $\mathrm{y}_{1}=4 \sin \omega \mathrm{t}$ $\mathrm{y}_{2}=3 \sin \left(\omega \mathrm{t}+\frac{\pi}{3}\right)$ From equation (i) and (ii) we get - Phase difference $(\phi)=\frac{\pi}{3}$ And $A_{1}=4, A_{2}=3$ $\text { Now, } \quad \mathrm{A}_{\text {Resultant }} =\sqrt{\mathrm{A}_{1}^{2}+\mathrm{A}_{2}^{2}+2 \mathrm{~A}_{1} \mathrm{~A}_{2} \cos \phi}$ $\mathrm{A}_{\text {Resultant }} =\sqrt{4^{2}+3^{2}+2 \times 4 \times 3 \times \cos \frac{\pi}{3}}$ $\mathrm{~A}_{\text {Resultant }} =\sqrt{25+24 \times \frac{1}{2}}$ $\mathrm{~A}_{\text {Resultant }} =\sqrt{37}$ $\mathrm{~A}_{\text {Resultant }} \approx 6$
Manipal UGET-2010
WAVES
172373
Two periodic waves of intensities $I_{1}, I_{2}$ pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is
172375
Two waves represented by $x_{1}=2 \sin (20 \pi t) \mathrm{m}$ and $x_{2}=2 \sin \left(20 \pi t+\frac{\pi}{3}\right) m$ are superimposed then the amplitude of the resultant wave.
1 $2 \sqrt{3} \mathrm{~m}$
2 $3 \sqrt{3} \mathrm{~m}$
3 $4 \sqrt{3} \mathrm{~m}$
4 $5 \sqrt{3} \mathrm{~m}$
Explanation:
A The two waves are given by $\mathrm{x}_{1}=2 \sin (20 \pi \mathrm{t}) \mathrm{m}$ $x_{2}=2 \sin \left(20 \pi t+\frac{\pi}{3}\right) m$ $\mathrm{A}_{1}=2, \mathrm{~A}_{2}=2, \phi=\frac{\pi}{3}=60^{\circ}$ The resultant wave amplitude $A=\sqrt{A_{1}^{2}+A_{2}^{2}+2 A_{1} A_{2} \cos \phi}$ $=\sqrt{(2)^{2}+(2)^{2}+2 \times 2 \times 2 \cos 60^{\circ}}$ $=\sqrt{4+4+8 \times \frac{1}{2}}$ $=\sqrt{8+4}$ $A=2 \sqrt{3} \mathrm{~m}$
172370
Two waves are given by $y_{1}=a \sin (\omega t-k x)$ and $y_{2}=a \cos (\omega t-k x)$. The phase difference between the two waves is
1 $\pi / 4$
2 $\pi$
3 $\pi / 8$
4 $\pi / 2$
Explanation:
D Given that, $\mathrm{y}_{1}=\mathrm{a} \sin (\omega \mathrm{t}-\mathrm{kx})$ $\mathrm{y}_{2}=\mathrm{a} \cos (\omega \mathrm{t}-\mathrm{kx})$ Equation (ii) we can write as $\mathrm{y}_{2}=\mathrm{a} \sin \left(\omega \mathrm{t}-\mathrm{kx}+\frac{\pi}{2}\right)$ Hence, phase difference, $\Delta \phi=\frac{\pi}{2}$
CG PET- 2006
WAVES
172371
If two waves represented by $y_{1}=4 \sin \omega t$ and $y_{2}=3 \sin \left(\omega t+\frac{\pi}{3}\right)$ interfere at a point. The amplitude of the resulting wave will be about
1 7
2 6
3 5
4 3.5
Explanation:
B Given that, $\mathrm{y}_{1}=4 \sin \omega \mathrm{t}$ $\mathrm{y}_{2}=3 \sin \left(\omega \mathrm{t}+\frac{\pi}{3}\right)$ From equation (i) and (ii) we get - Phase difference $(\phi)=\frac{\pi}{3}$ And $A_{1}=4, A_{2}=3$ $\text { Now, } \quad \mathrm{A}_{\text {Resultant }} =\sqrt{\mathrm{A}_{1}^{2}+\mathrm{A}_{2}^{2}+2 \mathrm{~A}_{1} \mathrm{~A}_{2} \cos \phi}$ $\mathrm{A}_{\text {Resultant }} =\sqrt{4^{2}+3^{2}+2 \times 4 \times 3 \times \cos \frac{\pi}{3}}$ $\mathrm{~A}_{\text {Resultant }} =\sqrt{25+24 \times \frac{1}{2}}$ $\mathrm{~A}_{\text {Resultant }} =\sqrt{37}$ $\mathrm{~A}_{\text {Resultant }} \approx 6$
Manipal UGET-2010
WAVES
172373
Two periodic waves of intensities $I_{1}, I_{2}$ pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is
172375
Two waves represented by $x_{1}=2 \sin (20 \pi t) \mathrm{m}$ and $x_{2}=2 \sin \left(20 \pi t+\frac{\pi}{3}\right) m$ are superimposed then the amplitude of the resultant wave.
1 $2 \sqrt{3} \mathrm{~m}$
2 $3 \sqrt{3} \mathrm{~m}$
3 $4 \sqrt{3} \mathrm{~m}$
4 $5 \sqrt{3} \mathrm{~m}$
Explanation:
A The two waves are given by $\mathrm{x}_{1}=2 \sin (20 \pi \mathrm{t}) \mathrm{m}$ $x_{2}=2 \sin \left(20 \pi t+\frac{\pi}{3}\right) m$ $\mathrm{A}_{1}=2, \mathrm{~A}_{2}=2, \phi=\frac{\pi}{3}=60^{\circ}$ The resultant wave amplitude $A=\sqrt{A_{1}^{2}+A_{2}^{2}+2 A_{1} A_{2} \cos \phi}$ $=\sqrt{(2)^{2}+(2)^{2}+2 \times 2 \times 2 \cos 60^{\circ}}$ $=\sqrt{4+4+8 \times \frac{1}{2}}$ $=\sqrt{8+4}$ $A=2 \sqrt{3} \mathrm{~m}$
172370
Two waves are given by $y_{1}=a \sin (\omega t-k x)$ and $y_{2}=a \cos (\omega t-k x)$. The phase difference between the two waves is
1 $\pi / 4$
2 $\pi$
3 $\pi / 8$
4 $\pi / 2$
Explanation:
D Given that, $\mathrm{y}_{1}=\mathrm{a} \sin (\omega \mathrm{t}-\mathrm{kx})$ $\mathrm{y}_{2}=\mathrm{a} \cos (\omega \mathrm{t}-\mathrm{kx})$ Equation (ii) we can write as $\mathrm{y}_{2}=\mathrm{a} \sin \left(\omega \mathrm{t}-\mathrm{kx}+\frac{\pi}{2}\right)$ Hence, phase difference, $\Delta \phi=\frac{\pi}{2}$
CG PET- 2006
WAVES
172371
If two waves represented by $y_{1}=4 \sin \omega t$ and $y_{2}=3 \sin \left(\omega t+\frac{\pi}{3}\right)$ interfere at a point. The amplitude of the resulting wave will be about
1 7
2 6
3 5
4 3.5
Explanation:
B Given that, $\mathrm{y}_{1}=4 \sin \omega \mathrm{t}$ $\mathrm{y}_{2}=3 \sin \left(\omega \mathrm{t}+\frac{\pi}{3}\right)$ From equation (i) and (ii) we get - Phase difference $(\phi)=\frac{\pi}{3}$ And $A_{1}=4, A_{2}=3$ $\text { Now, } \quad \mathrm{A}_{\text {Resultant }} =\sqrt{\mathrm{A}_{1}^{2}+\mathrm{A}_{2}^{2}+2 \mathrm{~A}_{1} \mathrm{~A}_{2} \cos \phi}$ $\mathrm{A}_{\text {Resultant }} =\sqrt{4^{2}+3^{2}+2 \times 4 \times 3 \times \cos \frac{\pi}{3}}$ $\mathrm{~A}_{\text {Resultant }} =\sqrt{25+24 \times \frac{1}{2}}$ $\mathrm{~A}_{\text {Resultant }} =\sqrt{37}$ $\mathrm{~A}_{\text {Resultant }} \approx 6$
Manipal UGET-2010
WAVES
172373
Two periodic waves of intensities $I_{1}, I_{2}$ pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is
172375
Two waves represented by $x_{1}=2 \sin (20 \pi t) \mathrm{m}$ and $x_{2}=2 \sin \left(20 \pi t+\frac{\pi}{3}\right) m$ are superimposed then the amplitude of the resultant wave.
1 $2 \sqrt{3} \mathrm{~m}$
2 $3 \sqrt{3} \mathrm{~m}$
3 $4 \sqrt{3} \mathrm{~m}$
4 $5 \sqrt{3} \mathrm{~m}$
Explanation:
A The two waves are given by $\mathrm{x}_{1}=2 \sin (20 \pi \mathrm{t}) \mathrm{m}$ $x_{2}=2 \sin \left(20 \pi t+\frac{\pi}{3}\right) m$ $\mathrm{A}_{1}=2, \mathrm{~A}_{2}=2, \phi=\frac{\pi}{3}=60^{\circ}$ The resultant wave amplitude $A=\sqrt{A_{1}^{2}+A_{2}^{2}+2 A_{1} A_{2} \cos \phi}$ $=\sqrt{(2)^{2}+(2)^{2}+2 \times 2 \times 2 \cos 60^{\circ}}$ $=\sqrt{4+4+8 \times \frac{1}{2}}$ $=\sqrt{8+4}$ $A=2 \sqrt{3} \mathrm{~m}$
