172402
A string fixed at both ends vibrates in 5 loops as shown in the figure. The total number of nodes and antinodes respectively are-
1 $6 \& 5$
2 $6 \& 10$
3 $2 \& 5$
4 $10 \& 5$
Explanation:
A Nodes are the points that have no displacement from the equilibrium position. While antinodes are points of maximum displacement from the mean position. Hence, There are 6 nodes and 5 Antinodes.
AP EAMCET-20.08.2021
WAVES
172411
Two strings of the same material and same length are given equal tension. If they are vibrating with fundamental frequencies 1600 $\mathrm{Hz}$ and $900 \mathrm{~Hz}$, then the ratio of their respective diameters is
1 $16: 9$
2 $4: 3$
3 $81: 256$
4 $3: 4$
5 $9: 16$
Explanation:
E Given, Fundamental frequencies $\left(f_{1}\right)=1600 \mathrm{~Hz}$ and $\left(f_{2}\right)=900$ $\mathrm{Hz}$ Fundamental frequencies for vibration is given by, $\mathrm{f}=\frac{\mathrm{v}}{2 l}=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mu}}=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\rho \mathrm{A}}}$ Where, $\mathrm{A} \rightarrow$ cross sectional area of string $\mathrm{A}=\pi \mathrm{r}^{2}$ $\mathrm{f} \propto \frac{1}{\sqrt{\mathrm{A}}} \Rightarrow \mathrm{f} \propto \frac{1}{\sqrt{\pi \mathrm{r}^{2}}} \Rightarrow \mathrm{f} \propto \frac{1}{\mathrm{r}}$ $\frac{\mathrm{d}_{1}}{\mathrm{~d}_{2}}=\frac{\mathrm{f}_{2}}{\mathrm{f}_{1}}=\frac{900}{1600}=\frac{9}{16}$ $\mathrm{~d}_{1}: \mathrm{d}_{2}=9: 16$
Kerala CEE 2020
WAVES
172412
A transverse wave is travelling on a string with velocity ' $V$ '. The extension in the string is ' $x$ '. If the string is extended by $50 \%$, the speed of the wave along the string will be nearly (Hooke's law is obeyed)
172413
The fundamental frequency of a wire stretched by $2 \mathrm{~kg} \mathrm{wt}$. is $100 \mathrm{~Hz}$. The weight required to produce its octave will be
1 $12 \mathrm{~kg} \mathrm{wt}$
2 $16 \mathrm{~kg} \mathrm{wt}$
3 $4 \mathrm{~kg} \mathrm{wt}$
4 $8 \mathrm{~kg} \mathrm{wt}$
Explanation:
D Given, $\mathrm{T}_{1}=2 \mathrm{~kg} \mathrm{wt}$ $\mathrm{n}_{1}=100 \mathrm{~Hz}$ The ratio between the frequencies of successive octaves is $1: 2$ $\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}=\frac{1}{2}$ $\mathrm{n}_{2}=2 \mathrm{n}_{1}=200 \mathrm{~Hz}$ We know, $\mathrm{n}=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}$ Here, $l$ and $\mathrm{m}$ are constant. So, $\quad \mathrm{n} \propto \sqrt{\mathrm{T}}$ $\therefore \quad \frac{\mathrm{n}_{1}}{\mathrm{n}_{2}} =\sqrt{\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}}$ $\frac{100}{200} =\sqrt{\frac{2}{\mathrm{~T}_{2}}}$ $\frac{1}{2} =\sqrt{\frac{2}{\mathrm{~T}_{2}}}$ $\frac{1}{4} =\frac{2}{\mathrm{~T}_{2}}$ Hence, $\mathrm{T}_{2}=8 \mathrm{~kg}$ wt
MHT-CET 2020
WAVES
172414
A string of mass $0.1 \mathrm{~kg}$ is under a tension $1.6 \mathrm{~N}$. The length of the string is $1 \mathrm{~m}$. A transverse wave starts from one end of the string. The time taken by the wave to reach the other end is
1 $0.50 \mathrm{~s}$
2 $0.30 \mathrm{~s}$
3 $0.25 \mathrm{~s}$
4 $0.75 \mathrm{~s}$
Explanation:
C Given that, $\mathrm{M}=0.1 \mathrm{~kg}, \mathrm{~T}=1.6 \mathrm{~N}, \mathrm{~L}=1 \mathrm{~m}$ Mass per unit length, $\mu=\frac{M}{L}=\frac{0.1}{1}=0.1 \mathrm{~kg} / \mathrm{m}$ $\text { Velocity, } v=\sqrt{\frac{T}{\mu}}=\sqrt{\frac{1.6}{0.1}}$ $v=4 \mathrm{~m} / \mathrm{s}$ Time taken by wave, $\mathrm{t}=\frac{\mathrm{L}}{\mathrm{v}}=\frac{1}{4}$ $\therefore \quad \mathrm{t} =0.25 \mathrm{sec}$
172402
A string fixed at both ends vibrates in 5 loops as shown in the figure. The total number of nodes and antinodes respectively are-
1 $6 \& 5$
2 $6 \& 10$
3 $2 \& 5$
4 $10 \& 5$
Explanation:
A Nodes are the points that have no displacement from the equilibrium position. While antinodes are points of maximum displacement from the mean position. Hence, There are 6 nodes and 5 Antinodes.
AP EAMCET-20.08.2021
WAVES
172411
Two strings of the same material and same length are given equal tension. If they are vibrating with fundamental frequencies 1600 $\mathrm{Hz}$ and $900 \mathrm{~Hz}$, then the ratio of their respective diameters is
1 $16: 9$
2 $4: 3$
3 $81: 256$
4 $3: 4$
5 $9: 16$
Explanation:
E Given, Fundamental frequencies $\left(f_{1}\right)=1600 \mathrm{~Hz}$ and $\left(f_{2}\right)=900$ $\mathrm{Hz}$ Fundamental frequencies for vibration is given by, $\mathrm{f}=\frac{\mathrm{v}}{2 l}=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mu}}=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\rho \mathrm{A}}}$ Where, $\mathrm{A} \rightarrow$ cross sectional area of string $\mathrm{A}=\pi \mathrm{r}^{2}$ $\mathrm{f} \propto \frac{1}{\sqrt{\mathrm{A}}} \Rightarrow \mathrm{f} \propto \frac{1}{\sqrt{\pi \mathrm{r}^{2}}} \Rightarrow \mathrm{f} \propto \frac{1}{\mathrm{r}}$ $\frac{\mathrm{d}_{1}}{\mathrm{~d}_{2}}=\frac{\mathrm{f}_{2}}{\mathrm{f}_{1}}=\frac{900}{1600}=\frac{9}{16}$ $\mathrm{~d}_{1}: \mathrm{d}_{2}=9: 16$
Kerala CEE 2020
WAVES
172412
A transverse wave is travelling on a string with velocity ' $V$ '. The extension in the string is ' $x$ '. If the string is extended by $50 \%$, the speed of the wave along the string will be nearly (Hooke's law is obeyed)
172413
The fundamental frequency of a wire stretched by $2 \mathrm{~kg} \mathrm{wt}$. is $100 \mathrm{~Hz}$. The weight required to produce its octave will be
1 $12 \mathrm{~kg} \mathrm{wt}$
2 $16 \mathrm{~kg} \mathrm{wt}$
3 $4 \mathrm{~kg} \mathrm{wt}$
4 $8 \mathrm{~kg} \mathrm{wt}$
Explanation:
D Given, $\mathrm{T}_{1}=2 \mathrm{~kg} \mathrm{wt}$ $\mathrm{n}_{1}=100 \mathrm{~Hz}$ The ratio between the frequencies of successive octaves is $1: 2$ $\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}=\frac{1}{2}$ $\mathrm{n}_{2}=2 \mathrm{n}_{1}=200 \mathrm{~Hz}$ We know, $\mathrm{n}=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}$ Here, $l$ and $\mathrm{m}$ are constant. So, $\quad \mathrm{n} \propto \sqrt{\mathrm{T}}$ $\therefore \quad \frac{\mathrm{n}_{1}}{\mathrm{n}_{2}} =\sqrt{\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}}$ $\frac{100}{200} =\sqrt{\frac{2}{\mathrm{~T}_{2}}}$ $\frac{1}{2} =\sqrt{\frac{2}{\mathrm{~T}_{2}}}$ $\frac{1}{4} =\frac{2}{\mathrm{~T}_{2}}$ Hence, $\mathrm{T}_{2}=8 \mathrm{~kg}$ wt
MHT-CET 2020
WAVES
172414
A string of mass $0.1 \mathrm{~kg}$ is under a tension $1.6 \mathrm{~N}$. The length of the string is $1 \mathrm{~m}$. A transverse wave starts from one end of the string. The time taken by the wave to reach the other end is
1 $0.50 \mathrm{~s}$
2 $0.30 \mathrm{~s}$
3 $0.25 \mathrm{~s}$
4 $0.75 \mathrm{~s}$
Explanation:
C Given that, $\mathrm{M}=0.1 \mathrm{~kg}, \mathrm{~T}=1.6 \mathrm{~N}, \mathrm{~L}=1 \mathrm{~m}$ Mass per unit length, $\mu=\frac{M}{L}=\frac{0.1}{1}=0.1 \mathrm{~kg} / \mathrm{m}$ $\text { Velocity, } v=\sqrt{\frac{T}{\mu}}=\sqrt{\frac{1.6}{0.1}}$ $v=4 \mathrm{~m} / \mathrm{s}$ Time taken by wave, $\mathrm{t}=\frac{\mathrm{L}}{\mathrm{v}}=\frac{1}{4}$ $\therefore \quad \mathrm{t} =0.25 \mathrm{sec}$
NEET Test Series from KOTA - 10 Papers In MS WORD
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WAVES
172402
A string fixed at both ends vibrates in 5 loops as shown in the figure. The total number of nodes and antinodes respectively are-
1 $6 \& 5$
2 $6 \& 10$
3 $2 \& 5$
4 $10 \& 5$
Explanation:
A Nodes are the points that have no displacement from the equilibrium position. While antinodes are points of maximum displacement from the mean position. Hence, There are 6 nodes and 5 Antinodes.
AP EAMCET-20.08.2021
WAVES
172411
Two strings of the same material and same length are given equal tension. If they are vibrating with fundamental frequencies 1600 $\mathrm{Hz}$ and $900 \mathrm{~Hz}$, then the ratio of their respective diameters is
1 $16: 9$
2 $4: 3$
3 $81: 256$
4 $3: 4$
5 $9: 16$
Explanation:
E Given, Fundamental frequencies $\left(f_{1}\right)=1600 \mathrm{~Hz}$ and $\left(f_{2}\right)=900$ $\mathrm{Hz}$ Fundamental frequencies for vibration is given by, $\mathrm{f}=\frac{\mathrm{v}}{2 l}=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mu}}=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\rho \mathrm{A}}}$ Where, $\mathrm{A} \rightarrow$ cross sectional area of string $\mathrm{A}=\pi \mathrm{r}^{2}$ $\mathrm{f} \propto \frac{1}{\sqrt{\mathrm{A}}} \Rightarrow \mathrm{f} \propto \frac{1}{\sqrt{\pi \mathrm{r}^{2}}} \Rightarrow \mathrm{f} \propto \frac{1}{\mathrm{r}}$ $\frac{\mathrm{d}_{1}}{\mathrm{~d}_{2}}=\frac{\mathrm{f}_{2}}{\mathrm{f}_{1}}=\frac{900}{1600}=\frac{9}{16}$ $\mathrm{~d}_{1}: \mathrm{d}_{2}=9: 16$
Kerala CEE 2020
WAVES
172412
A transverse wave is travelling on a string with velocity ' $V$ '. The extension in the string is ' $x$ '. If the string is extended by $50 \%$, the speed of the wave along the string will be nearly (Hooke's law is obeyed)
172413
The fundamental frequency of a wire stretched by $2 \mathrm{~kg} \mathrm{wt}$. is $100 \mathrm{~Hz}$. The weight required to produce its octave will be
1 $12 \mathrm{~kg} \mathrm{wt}$
2 $16 \mathrm{~kg} \mathrm{wt}$
3 $4 \mathrm{~kg} \mathrm{wt}$
4 $8 \mathrm{~kg} \mathrm{wt}$
Explanation:
D Given, $\mathrm{T}_{1}=2 \mathrm{~kg} \mathrm{wt}$ $\mathrm{n}_{1}=100 \mathrm{~Hz}$ The ratio between the frequencies of successive octaves is $1: 2$ $\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}=\frac{1}{2}$ $\mathrm{n}_{2}=2 \mathrm{n}_{1}=200 \mathrm{~Hz}$ We know, $\mathrm{n}=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}$ Here, $l$ and $\mathrm{m}$ are constant. So, $\quad \mathrm{n} \propto \sqrt{\mathrm{T}}$ $\therefore \quad \frac{\mathrm{n}_{1}}{\mathrm{n}_{2}} =\sqrt{\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}}$ $\frac{100}{200} =\sqrt{\frac{2}{\mathrm{~T}_{2}}}$ $\frac{1}{2} =\sqrt{\frac{2}{\mathrm{~T}_{2}}}$ $\frac{1}{4} =\frac{2}{\mathrm{~T}_{2}}$ Hence, $\mathrm{T}_{2}=8 \mathrm{~kg}$ wt
MHT-CET 2020
WAVES
172414
A string of mass $0.1 \mathrm{~kg}$ is under a tension $1.6 \mathrm{~N}$. The length of the string is $1 \mathrm{~m}$. A transverse wave starts from one end of the string. The time taken by the wave to reach the other end is
1 $0.50 \mathrm{~s}$
2 $0.30 \mathrm{~s}$
3 $0.25 \mathrm{~s}$
4 $0.75 \mathrm{~s}$
Explanation:
C Given that, $\mathrm{M}=0.1 \mathrm{~kg}, \mathrm{~T}=1.6 \mathrm{~N}, \mathrm{~L}=1 \mathrm{~m}$ Mass per unit length, $\mu=\frac{M}{L}=\frac{0.1}{1}=0.1 \mathrm{~kg} / \mathrm{m}$ $\text { Velocity, } v=\sqrt{\frac{T}{\mu}}=\sqrt{\frac{1.6}{0.1}}$ $v=4 \mathrm{~m} / \mathrm{s}$ Time taken by wave, $\mathrm{t}=\frac{\mathrm{L}}{\mathrm{v}}=\frac{1}{4}$ $\therefore \quad \mathrm{t} =0.25 \mathrm{sec}$
172402
A string fixed at both ends vibrates in 5 loops as shown in the figure. The total number of nodes and antinodes respectively are-
1 $6 \& 5$
2 $6 \& 10$
3 $2 \& 5$
4 $10 \& 5$
Explanation:
A Nodes are the points that have no displacement from the equilibrium position. While antinodes are points of maximum displacement from the mean position. Hence, There are 6 nodes and 5 Antinodes.
AP EAMCET-20.08.2021
WAVES
172411
Two strings of the same material and same length are given equal tension. If they are vibrating with fundamental frequencies 1600 $\mathrm{Hz}$ and $900 \mathrm{~Hz}$, then the ratio of their respective diameters is
1 $16: 9$
2 $4: 3$
3 $81: 256$
4 $3: 4$
5 $9: 16$
Explanation:
E Given, Fundamental frequencies $\left(f_{1}\right)=1600 \mathrm{~Hz}$ and $\left(f_{2}\right)=900$ $\mathrm{Hz}$ Fundamental frequencies for vibration is given by, $\mathrm{f}=\frac{\mathrm{v}}{2 l}=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mu}}=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\rho \mathrm{A}}}$ Where, $\mathrm{A} \rightarrow$ cross sectional area of string $\mathrm{A}=\pi \mathrm{r}^{2}$ $\mathrm{f} \propto \frac{1}{\sqrt{\mathrm{A}}} \Rightarrow \mathrm{f} \propto \frac{1}{\sqrt{\pi \mathrm{r}^{2}}} \Rightarrow \mathrm{f} \propto \frac{1}{\mathrm{r}}$ $\frac{\mathrm{d}_{1}}{\mathrm{~d}_{2}}=\frac{\mathrm{f}_{2}}{\mathrm{f}_{1}}=\frac{900}{1600}=\frac{9}{16}$ $\mathrm{~d}_{1}: \mathrm{d}_{2}=9: 16$
Kerala CEE 2020
WAVES
172412
A transverse wave is travelling on a string with velocity ' $V$ '. The extension in the string is ' $x$ '. If the string is extended by $50 \%$, the speed of the wave along the string will be nearly (Hooke's law is obeyed)
172413
The fundamental frequency of a wire stretched by $2 \mathrm{~kg} \mathrm{wt}$. is $100 \mathrm{~Hz}$. The weight required to produce its octave will be
1 $12 \mathrm{~kg} \mathrm{wt}$
2 $16 \mathrm{~kg} \mathrm{wt}$
3 $4 \mathrm{~kg} \mathrm{wt}$
4 $8 \mathrm{~kg} \mathrm{wt}$
Explanation:
D Given, $\mathrm{T}_{1}=2 \mathrm{~kg} \mathrm{wt}$ $\mathrm{n}_{1}=100 \mathrm{~Hz}$ The ratio between the frequencies of successive octaves is $1: 2$ $\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}=\frac{1}{2}$ $\mathrm{n}_{2}=2 \mathrm{n}_{1}=200 \mathrm{~Hz}$ We know, $\mathrm{n}=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}$ Here, $l$ and $\mathrm{m}$ are constant. So, $\quad \mathrm{n} \propto \sqrt{\mathrm{T}}$ $\therefore \quad \frac{\mathrm{n}_{1}}{\mathrm{n}_{2}} =\sqrt{\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}}$ $\frac{100}{200} =\sqrt{\frac{2}{\mathrm{~T}_{2}}}$ $\frac{1}{2} =\sqrt{\frac{2}{\mathrm{~T}_{2}}}$ $\frac{1}{4} =\frac{2}{\mathrm{~T}_{2}}$ Hence, $\mathrm{T}_{2}=8 \mathrm{~kg}$ wt
MHT-CET 2020
WAVES
172414
A string of mass $0.1 \mathrm{~kg}$ is under a tension $1.6 \mathrm{~N}$. The length of the string is $1 \mathrm{~m}$. A transverse wave starts from one end of the string. The time taken by the wave to reach the other end is
1 $0.50 \mathrm{~s}$
2 $0.30 \mathrm{~s}$
3 $0.25 \mathrm{~s}$
4 $0.75 \mathrm{~s}$
Explanation:
C Given that, $\mathrm{M}=0.1 \mathrm{~kg}, \mathrm{~T}=1.6 \mathrm{~N}, \mathrm{~L}=1 \mathrm{~m}$ Mass per unit length, $\mu=\frac{M}{L}=\frac{0.1}{1}=0.1 \mathrm{~kg} / \mathrm{m}$ $\text { Velocity, } v=\sqrt{\frac{T}{\mu}}=\sqrt{\frac{1.6}{0.1}}$ $v=4 \mathrm{~m} / \mathrm{s}$ Time taken by wave, $\mathrm{t}=\frac{\mathrm{L}}{\mathrm{v}}=\frac{1}{4}$ $\therefore \quad \mathrm{t} =0.25 \mathrm{sec}$
172402
A string fixed at both ends vibrates in 5 loops as shown in the figure. The total number of nodes and antinodes respectively are-
1 $6 \& 5$
2 $6 \& 10$
3 $2 \& 5$
4 $10 \& 5$
Explanation:
A Nodes are the points that have no displacement from the equilibrium position. While antinodes are points of maximum displacement from the mean position. Hence, There are 6 nodes and 5 Antinodes.
AP EAMCET-20.08.2021
WAVES
172411
Two strings of the same material and same length are given equal tension. If they are vibrating with fundamental frequencies 1600 $\mathrm{Hz}$ and $900 \mathrm{~Hz}$, then the ratio of their respective diameters is
1 $16: 9$
2 $4: 3$
3 $81: 256$
4 $3: 4$
5 $9: 16$
Explanation:
E Given, Fundamental frequencies $\left(f_{1}\right)=1600 \mathrm{~Hz}$ and $\left(f_{2}\right)=900$ $\mathrm{Hz}$ Fundamental frequencies for vibration is given by, $\mathrm{f}=\frac{\mathrm{v}}{2 l}=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mu}}=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\rho \mathrm{A}}}$ Where, $\mathrm{A} \rightarrow$ cross sectional area of string $\mathrm{A}=\pi \mathrm{r}^{2}$ $\mathrm{f} \propto \frac{1}{\sqrt{\mathrm{A}}} \Rightarrow \mathrm{f} \propto \frac{1}{\sqrt{\pi \mathrm{r}^{2}}} \Rightarrow \mathrm{f} \propto \frac{1}{\mathrm{r}}$ $\frac{\mathrm{d}_{1}}{\mathrm{~d}_{2}}=\frac{\mathrm{f}_{2}}{\mathrm{f}_{1}}=\frac{900}{1600}=\frac{9}{16}$ $\mathrm{~d}_{1}: \mathrm{d}_{2}=9: 16$
Kerala CEE 2020
WAVES
172412
A transverse wave is travelling on a string with velocity ' $V$ '. The extension in the string is ' $x$ '. If the string is extended by $50 \%$, the speed of the wave along the string will be nearly (Hooke's law is obeyed)
172413
The fundamental frequency of a wire stretched by $2 \mathrm{~kg} \mathrm{wt}$. is $100 \mathrm{~Hz}$. The weight required to produce its octave will be
1 $12 \mathrm{~kg} \mathrm{wt}$
2 $16 \mathrm{~kg} \mathrm{wt}$
3 $4 \mathrm{~kg} \mathrm{wt}$
4 $8 \mathrm{~kg} \mathrm{wt}$
Explanation:
D Given, $\mathrm{T}_{1}=2 \mathrm{~kg} \mathrm{wt}$ $\mathrm{n}_{1}=100 \mathrm{~Hz}$ The ratio between the frequencies of successive octaves is $1: 2$ $\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}=\frac{1}{2}$ $\mathrm{n}_{2}=2 \mathrm{n}_{1}=200 \mathrm{~Hz}$ We know, $\mathrm{n}=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}$ Here, $l$ and $\mathrm{m}$ are constant. So, $\quad \mathrm{n} \propto \sqrt{\mathrm{T}}$ $\therefore \quad \frac{\mathrm{n}_{1}}{\mathrm{n}_{2}} =\sqrt{\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}}$ $\frac{100}{200} =\sqrt{\frac{2}{\mathrm{~T}_{2}}}$ $\frac{1}{2} =\sqrt{\frac{2}{\mathrm{~T}_{2}}}$ $\frac{1}{4} =\frac{2}{\mathrm{~T}_{2}}$ Hence, $\mathrm{T}_{2}=8 \mathrm{~kg}$ wt
MHT-CET 2020
WAVES
172414
A string of mass $0.1 \mathrm{~kg}$ is under a tension $1.6 \mathrm{~N}$. The length of the string is $1 \mathrm{~m}$. A transverse wave starts from one end of the string. The time taken by the wave to reach the other end is
1 $0.50 \mathrm{~s}$
2 $0.30 \mathrm{~s}$
3 $0.25 \mathrm{~s}$
4 $0.75 \mathrm{~s}$
Explanation:
C Given that, $\mathrm{M}=0.1 \mathrm{~kg}, \mathrm{~T}=1.6 \mathrm{~N}, \mathrm{~L}=1 \mathrm{~m}$ Mass per unit length, $\mu=\frac{M}{L}=\frac{0.1}{1}=0.1 \mathrm{~kg} / \mathrm{m}$ $\text { Velocity, } v=\sqrt{\frac{T}{\mu}}=\sqrt{\frac{1.6}{0.1}}$ $v=4 \mathrm{~m} / \mathrm{s}$ Time taken by wave, $\mathrm{t}=\frac{\mathrm{L}}{\mathrm{v}}=\frac{1}{4}$ $\therefore \quad \mathrm{t} =0.25 \mathrm{sec}$
