CBSE 11th Standard Physics Subject Waves Ncert Exemplar 1 Mark Questions With Solution 2021
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CBSE 11th Standard Physics Subject Waves Ncert Exemplar 1 Mark Questions With Solution 2021
11th Standard CBSE
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Reg.No. :
Physics
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How speed of sound waves in air varies with humidity?
(a)Speed of sound waves in air increases with increases with increase in humidity. This is because presence of moisture decreases the density of air.
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Show that when a string fixed at its two ends vibrates in 1 loop, 2 loops, 3loops and 4loops, the frequencies are in the ratio 1:2:3:4.
(a)In case of a string fixed at two ends, when the string vibrates in n loops
\({ v }_{ n }=\frac { n }{ 2l } \sqrt { \frac { T }{ \mu } } \quad \Rightarrow { v }_{ n }\propto n\)
Hence, when the string vibrates in 1 loop, 2 loops, 3 loops, 4 loops, the frequencies are in the ratio 1:2:3:4. -
A train standing at the outer signal of a railway station blows a whistle of frequency 400 Hz still air. The train begins to move with a speed of 10 ms-1 towards the platform. What is the frequency of the sound for an observer standing on the platform? (Sound velocity in air =330 ms-1).
(a)Given, v=400 Hz, Vs = 10m/s, v = 330m/s
As the source is moving towards stationary observer, apparent frequency will be more than the original.
V'=apparent frequency
\(=\frac { v\times V }{ v-{ v }_{ s } } =\frac { 330\times 400 }{ 330-10 } \)
\(=412.5\ Hz\) -
When two waves of almost equal frequencies n1 and n2 reach at a point, simultaneously. What is the time interval between successive maxima?
(a)Number of beats/s = (n1 - n2)
Hence, time interval between two successive beats = time interval between two successive maxima = \(\frac { 1 }{ { n }_{ 1 }-{ n }_{ 2 } } \) -
A sonometer wire is vibrating in resonance with a tuning fork. Keeping the tension applied same, the length of the wire is doubled. Under what conditions would the tuning fork still be in resonance with the wire?
(a)The sonometer frequency is given by
\(v=\frac { n }{ 2L } \sqrt { \frac { T }{ \mu } } \)
Now, as it vibrates with length L, we assume v=v1
n = n1
\(\therefore { v }_{ 1 }=\frac { { n }_{ 1 } }{ 2L } \sqrt { \frac { T }{ \mu } } \)
When length is doubled, then
\({ v }_{ 2 }=\frac { { n }_{ 2 } }{ 2\times 2L } \sqrt { \frac { T }{ \mu } } \)
Dividing Eq.(i) by Eq. (ii), we get
\(\frac { { v }_{ 1 } }{ { v }_{ 2 } } =\frac { { n }_{ 1 } }{ { n }_{ 2 } } \times 2\)
To keep the resonance,
\( \frac { { v }_{ 1 } }{ { v }_{ 2 } } =1=\frac { { n }_{ 1 } }{ { n }_{ 2 } } \times 2\)
\(\\ \Rightarrow { n }_{ 2 }=2{ n }_{ 1 }\)
Hence, when the wire is doubled, the number of loops also get doubled to produce the resonance. That is, it resonates in second harmonic.
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CBSE 11th Standard Physics Subject Waves Ncert Exemplar 1 Mark Questions With Solution 2021 Answer Keys
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Speed of sound waves in air increases with increases with increase in humidity. This is because presence of moisture decreases the density of air.
Speed of sound waves in air increases with increases with increase in humidity. This is because presence of moisture decreases the density of air.
-
In case of a string fixed at two ends, when the string vibrates in n loops
vn= \(\frac{n}{2l}\sqrt{\frac{T}{μ}}\) ⇒vn ∝ n
Hence, when the string vibrates in 1 loop, 2 loops, 3 loops, 4 loops, the frequencies are in the ratio 1:2:3:4.In case of a string fixed at two ends, when the string vibrates in n loops
vn= \(\frac{n}{2l}\sqrt{\frac{T}{μ}}\) ⇒vn ∝ n
Hence, when the string vibrates in 1 loop, 2 loops, 3 loops, 4 loops, the frequencies are in the ratio 1:2:3:4. -
Given, v=400 Hz, Vs = 10m/s, v = 330m/s
As the source is moving towards stationary observer, apparent frequency will be more than the original.
V'=apparent frequency
\(=\frac{v×V}{v−v_s}=\frac{330×400}{330−10}\)
=412.5 HzGiven, v=400 Hz, Vs = 10m/s, v = 330m/s
As the source is moving towards stationary observer, apparent frequency will be more than the original.
V'=apparent frequency
\(=\frac{v×V}{v−v_s}=\frac{330×400}{330−10}\)
=412.5 Hz -
Number of beats/s = (n1 - n2)
Hence, time interval between two successive beats = time interval between two successive maxima = \(\frac{1}{n_1−n_2}\)Number of beats/s = (n1 - n2)
Hence, time interval between two successive beats = time interval between two successive maxima = \(\frac{1}{n_1−n_2}\) -
The sonometer frequency is given by
v=\(\frac{n}{2L}\sqrt{\frac{T}{μ}}\)
Now, as it vibrates with length L, we assume v=v1
n = n1
∴v1=\(\frac{n_1}{2L}\sqrt{\frac{T}{μ}}\)
When length is doubled, then
v2=\(\frac{n_2}{2 \times2L}\sqrt{\frac{T}{μ}}\)
Dividing Eq.(i) by Eq. (ii), we get
\(\frac{v_1}{v_2}=\frac{n_1}{n_2}×2\)
To keep the resonance,
\(\frac{v_1}{v_2}=1=\frac{n_1}{n_2}×2\)
⇒n2=2n1
Hence, when the wire is doubled, the number of loops also get doubled to produce the resonance. That is, it resonates in second harmonic.The sonometer frequency is given by
v=\(\frac{n}{2L}\sqrt{\frac{T}{μ}}\)
Now, as it vibrates with length L, we assume v=v1
n = n1
∴v1=\(\frac{n_1}{2L}\sqrt{\frac{T}{μ}}\)
When length is doubled, then
v2=\(\frac{n_2}{2 \times2L}\sqrt{\frac{T}{μ}}\)
Dividing Eq.(i) by Eq. (ii), we get
\(\frac{v_1}{v_2}=\frac{n_1}{n_2}×2\)
To keep the resonance,
\(\frac{v_1}{v_2}=1=\frac{n_1}{n_2}×2\)
⇒n2=2n1
Hence, when the wire is doubled, the number of loops also get doubled to produce the resonance. That is, it resonates in second harmonic.
