NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI15:WAVES
355195
Two stretched strings of same material are vibrating under same tension in fundamental mode. The ratio of their frequencies is \(1: 2\) and ratio of the length of the vibrating segments is \(1: 4\). Then, the ratio of the radii of the string is:
1 In the region between two antinodes vibrate in same phase
2 Of the medium vibrate in same phase
3 On either side of a node vibrate in same phase
4 In the region between two nodes vibrate in same phase
Explanation:
In stationary wave all the particles in one particular segment (i.e., between two nodes) vibrates in the same phase.
PHXI15:WAVES
355197
Two travelling waves \({y_1} = A\sin \left[ {k\left( {x - ct} \right)} \right]\) and \({y_2} = A\sin \left[ {k\left( {x + ct} \right)} \right]\) are superimposed on string. The distance between adjacent nodes is
1 \(c t / 2 \pi\)
2 \(c t / \pi\)
3 \(\pi / k\)
4 \(\pi / 2 k\)
Explanation:
\(y=A \sin (k x-k c t)+A \sin (k x+k c t)\) \(\begin{aligned}& =2 A \sin \left(\dfrac{k x-k c t+k x+k c t}{2}\right) \cos \left(\dfrac{k x-k c t-k x-k c t}{2}\right) \\& =2 A \sin (k x) \cos (k c t)\end{aligned}\) Thus \(\dfrac{2 \pi}{\lambda}=k, \therefore \quad \lambda=\dfrac{2 \pi}{k}\) The distance between adjacent nodes is \(\dfrac{\pi}{k}\)
PHXI15:WAVES
355198
A string of length \(240\;cm\) is connected to two light rings which can freely move along two vertical rods. The maximum three wavelengths of standing waves that can be formed in the string are \(cm\)
1 \(120,60,30\)
2 \(480,240,160\)
3 \(120,80,60\)
4 \(240,180,90\)
Explanation:
Since both ends are free, (the free ends are antinodes. Hence, \(240 = \frac{\lambda }{2}\;\;\;{\mkern 1mu} {\kern 1pt} \therefore \lambda = 480\;cm\) \(240 = \lambda \;\;\;{\mkern 1mu} {\kern 1pt} \therefore \lambda = 240\;cm\) \(240 = \frac{{3\lambda }}{2}\;\;\;{\mkern 1mu} {\kern 1pt} \therefore \lambda = 160\;cm\)
355195
Two stretched strings of same material are vibrating under same tension in fundamental mode. The ratio of their frequencies is \(1: 2\) and ratio of the length of the vibrating segments is \(1: 4\). Then, the ratio of the radii of the string is:
1 In the region between two antinodes vibrate in same phase
2 Of the medium vibrate in same phase
3 On either side of a node vibrate in same phase
4 In the region between two nodes vibrate in same phase
Explanation:
In stationary wave all the particles in one particular segment (i.e., between two nodes) vibrates in the same phase.
PHXI15:WAVES
355197
Two travelling waves \({y_1} = A\sin \left[ {k\left( {x - ct} \right)} \right]\) and \({y_2} = A\sin \left[ {k\left( {x + ct} \right)} \right]\) are superimposed on string. The distance between adjacent nodes is
1 \(c t / 2 \pi\)
2 \(c t / \pi\)
3 \(\pi / k\)
4 \(\pi / 2 k\)
Explanation:
\(y=A \sin (k x-k c t)+A \sin (k x+k c t)\) \(\begin{aligned}& =2 A \sin \left(\dfrac{k x-k c t+k x+k c t}{2}\right) \cos \left(\dfrac{k x-k c t-k x-k c t}{2}\right) \\& =2 A \sin (k x) \cos (k c t)\end{aligned}\) Thus \(\dfrac{2 \pi}{\lambda}=k, \therefore \quad \lambda=\dfrac{2 \pi}{k}\) The distance between adjacent nodes is \(\dfrac{\pi}{k}\)
PHXI15:WAVES
355198
A string of length \(240\;cm\) is connected to two light rings which can freely move along two vertical rods. The maximum three wavelengths of standing waves that can be formed in the string are \(cm\)
1 \(120,60,30\)
2 \(480,240,160\)
3 \(120,80,60\)
4 \(240,180,90\)
Explanation:
Since both ends are free, (the free ends are antinodes. Hence, \(240 = \frac{\lambda }{2}\;\;\;{\mkern 1mu} {\kern 1pt} \therefore \lambda = 480\;cm\) \(240 = \lambda \;\;\;{\mkern 1mu} {\kern 1pt} \therefore \lambda = 240\;cm\) \(240 = \frac{{3\lambda }}{2}\;\;\;{\mkern 1mu} {\kern 1pt} \therefore \lambda = 160\;cm\)
355195
Two stretched strings of same material are vibrating under same tension in fundamental mode. The ratio of their frequencies is \(1: 2\) and ratio of the length of the vibrating segments is \(1: 4\). Then, the ratio of the radii of the string is:
1 In the region between two antinodes vibrate in same phase
2 Of the medium vibrate in same phase
3 On either side of a node vibrate in same phase
4 In the region between two nodes vibrate in same phase
Explanation:
In stationary wave all the particles in one particular segment (i.e., between two nodes) vibrates in the same phase.
PHXI15:WAVES
355197
Two travelling waves \({y_1} = A\sin \left[ {k\left( {x - ct} \right)} \right]\) and \({y_2} = A\sin \left[ {k\left( {x + ct} \right)} \right]\) are superimposed on string. The distance between adjacent nodes is
1 \(c t / 2 \pi\)
2 \(c t / \pi\)
3 \(\pi / k\)
4 \(\pi / 2 k\)
Explanation:
\(y=A \sin (k x-k c t)+A \sin (k x+k c t)\) \(\begin{aligned}& =2 A \sin \left(\dfrac{k x-k c t+k x+k c t}{2}\right) \cos \left(\dfrac{k x-k c t-k x-k c t}{2}\right) \\& =2 A \sin (k x) \cos (k c t)\end{aligned}\) Thus \(\dfrac{2 \pi}{\lambda}=k, \therefore \quad \lambda=\dfrac{2 \pi}{k}\) The distance between adjacent nodes is \(\dfrac{\pi}{k}\)
PHXI15:WAVES
355198
A string of length \(240\;cm\) is connected to two light rings which can freely move along two vertical rods. The maximum three wavelengths of standing waves that can be formed in the string are \(cm\)
1 \(120,60,30\)
2 \(480,240,160\)
3 \(120,80,60\)
4 \(240,180,90\)
Explanation:
Since both ends are free, (the free ends are antinodes. Hence, \(240 = \frac{\lambda }{2}\;\;\;{\mkern 1mu} {\kern 1pt} \therefore \lambda = 480\;cm\) \(240 = \lambda \;\;\;{\mkern 1mu} {\kern 1pt} \therefore \lambda = 240\;cm\) \(240 = \frac{{3\lambda }}{2}\;\;\;{\mkern 1mu} {\kern 1pt} \therefore \lambda = 160\;cm\)
355195
Two stretched strings of same material are vibrating under same tension in fundamental mode. The ratio of their frequencies is \(1: 2\) and ratio of the length of the vibrating segments is \(1: 4\). Then, the ratio of the radii of the string is:
1 In the region between two antinodes vibrate in same phase
2 Of the medium vibrate in same phase
3 On either side of a node vibrate in same phase
4 In the region between two nodes vibrate in same phase
Explanation:
In stationary wave all the particles in one particular segment (i.e., between two nodes) vibrates in the same phase.
PHXI15:WAVES
355197
Two travelling waves \({y_1} = A\sin \left[ {k\left( {x - ct} \right)} \right]\) and \({y_2} = A\sin \left[ {k\left( {x + ct} \right)} \right]\) are superimposed on string. The distance between adjacent nodes is
1 \(c t / 2 \pi\)
2 \(c t / \pi\)
3 \(\pi / k\)
4 \(\pi / 2 k\)
Explanation:
\(y=A \sin (k x-k c t)+A \sin (k x+k c t)\) \(\begin{aligned}& =2 A \sin \left(\dfrac{k x-k c t+k x+k c t}{2}\right) \cos \left(\dfrac{k x-k c t-k x-k c t}{2}\right) \\& =2 A \sin (k x) \cos (k c t)\end{aligned}\) Thus \(\dfrac{2 \pi}{\lambda}=k, \therefore \quad \lambda=\dfrac{2 \pi}{k}\) The distance between adjacent nodes is \(\dfrac{\pi}{k}\)
PHXI15:WAVES
355198
A string of length \(240\;cm\) is connected to two light rings which can freely move along two vertical rods. The maximum three wavelengths of standing waves that can be formed in the string are \(cm\)
1 \(120,60,30\)
2 \(480,240,160\)
3 \(120,80,60\)
4 \(240,180,90\)
Explanation:
Since both ends are free, (the free ends are antinodes. Hence, \(240 = \frac{\lambda }{2}\;\;\;{\mkern 1mu} {\kern 1pt} \therefore \lambda = 480\;cm\) \(240 = \lambda \;\;\;{\mkern 1mu} {\kern 1pt} \therefore \lambda = 240\;cm\) \(240 = \frac{{3\lambda }}{2}\;\;\;{\mkern 1mu} {\kern 1pt} \therefore \lambda = 160\;cm\)
