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WAVES

172404 An elastic string of length $2 m$ is fixed at one end. The string starts to vibrate in third overtone with a frequency $1200 Hz$ . The ratio of frequency of first overtone and fundamental is

1

1 : 1

2

2 : 1

3

3 : 1

4

4 : 1

Explanation:

C String fixed at one end so,
Third overtone $= 5^{th}$ harmonic
First overtone $= 3^{rd}$ harmonic
Fundamental $= 1^{st}$ harmonic
Now, frequency $f = \frac{nV}{4 L} \Rightarrow f \propto n$
Ratio of frequency of first overtone and fundamental,
$\frac{f_{3}}{f_{1}} = \frac{n_{3}}{n_{1}} = \frac{3}{L}$
$f_{3} : f_{1} = 3 : 1$

AP EAMCET-05.10.2021

WAVES

172405 A string is divided into three segments, so that the segment possesses fundamental frequencies in the ratio $1 : 2 : 3$ . Then, the segments are in the ratio

1

6 : 3 : 2

2

4 : 3 : 2

3

4 : 2 : 1

4

3 : 2 : 1

Explanation:

A Given, $f_{1} : f_{2} : f_{3} = 1 : 2 : 3$
A string is divided into three segments. So, that the segment process fundamental frequencies in the ratio $1 : 2 : 3$
Let, $l$ be the length of the string
$∴ l = l_{1} + l_{2} + l_{3}$
We know that
Fundamental frequency $(f) = \frac{1}{2 l} \sqrt{\frac{T}{m}}$
$∵ f = f_{1} + f_{2} + f_{3}$
or $f \propto \frac{1}{l} [∵ T$ and $M$ are constants $]$
While,
$l_{1} : l_{2} : l_{3} = \frac{1}{f_{1}} : \frac{1}{f_{2}} : \frac{1}{f_{3}}$
$l_{1} : l_{2} : l_{3} = \frac{1}{1} : \frac{1}{2} : \frac{1}{3}$
$l_{1} : l_{2} : l_{3} = 6 : 3 : 2$

AP EAMCET-23.08.2021

WAVES

172406 When the area of cross-section of a stretched wire is halved and tension is doubled, the speed of propagation of transverse waves along it becomes $k$ times the initial speed. Then, $k$

1 1

2 4

3 2

4 8

Explanation:

C Given, $A_{2} = \frac{A_{1}}{2}, T_{2} = 2 T_{1}$
Speed of transverse wave in stretched wire is given as
$v = \sqrt{\frac{T}{μ}}$
Where, $T =$ Tension in the wire
$μ =$ Mass per unit length of wire.
So, $v = \sqrt{\frac{T}{M / l}} [∵ μ = \frac{M}{l}]$
$v = \sqrt{\frac{T l}{M}} = \sqrt{\frac{T l}{V \cdot ρ}}$
$v = \sqrt{\frac{T l}{A \cdot l \cdot ρ}}$
$v = \sqrt{\frac{T}{A ρ}} \Rightarrow v \propto \sqrt{\frac{T}{A}}$
$\frac{v_{2}}{v_{1}} = \sqrt{\frac{T_{2}}{T_{1}} \times \frac{A_{1}}{A_{2}}}$
from equation (i), we get-
$\frac{v_{2}}{v_{1}} = \sqrt{\frac{2 T_{1}}{T_{1}} \times \frac{A_{1}}{\frac{A_{1}}{2}}} = \sqrt{4}$
$v_{2} = 2 v_{1}$
So, $k = 2$

AP EAMCET (22.09.2020) Shift-II

WAVES

172407 A uniform rope of length $12 m$ and mass $6 kg$ hangs vertically from a rigid support. A block of mass $2 kg$ is attached to the free end of the rope. A transverse pulse of wavelength $0.06 m$ is produced at the lower end of the rope. Wavelength of the pulse when it reaches the top of the rope is .......... .

1

0.02 m

2

0.12 m

3

0.01 m

4

0.03 m

Explanation:

B
Given $T_{1} = 2 g$
$T_{2} = 8 g$
$λ_{1} = 0.06 m$
We know that
$v = f λ$
$v = \sqrt{\frac{T}{μ}}$
So, $v \propto λ \propto \sqrt{T}$
Then,
$v \propto λ \propto \sqrt{T}$
$\frac{\sqrt{T_{1}}}{\sqrt{T_{2}}} = \frac{λ_{1}}{λ_{2}}$
Then $λ_{2} = λ_{1} \sqrt{\frac{T_{2}}{T_{1}}} = 0.06 \sqrt{\frac{8 g}{2 g}}$
$λ_{2} = 0.12 m$

AP EAMCET-23.09.2020

WAVES

172408 A transverse wave is travelling with velocity ' $V$ ' through a metal wire of length ' $L$ ' and density ' $ρ$ '. The tensile stress on the wire is

1

V^{2} ρ

2

V ρ^{2}

3

V ρ

4

\frac{V}{ρ}

Explanation:

A We know that,
$V = \sqrt{\frac{T}{μ}}$ , where $T$ is the tension $& μ$ is mass per unit length.
$∴$ Velocity, $V = \sqrt{\frac{T}{A ρ}} (∵ μ = \frac{Mass}{Length} = \frac{ρ AL}{L} = ρ A)$
$∴ V^{2} = \frac{T}{A ρ}$
$\frac{T}{A} = V^{2} ρ$
$∴$ Tensile stress $(σ) = \frac{T}{A} = V^{2} ρ$

MHT-CET 2020

WAVES

172404 An elastic string of length $2 m$ is fixed at one end. The string starts to vibrate in third overtone with a frequency $1200 Hz$ . The ratio of frequency of first overtone and fundamental is

1

1 : 1

2

2 : 1

3

3 : 1

4

4 : 1

Explanation:

C String fixed at one end so,
Third overtone $= 5^{th}$ harmonic
First overtone $= 3^{rd}$ harmonic
Fundamental $= 1^{st}$ harmonic
Now, frequency $f = \frac{nV}{4 L} \Rightarrow f \propto n$
Ratio of frequency of first overtone and fundamental,
$\frac{f_{3}}{f_{1}} = \frac{n_{3}}{n_{1}} = \frac{3}{L}$
$f_{3} : f_{1} = 3 : 1$

AP EAMCET-05.10.2021

WAVES

172405 A string is divided into three segments, so that the segment possesses fundamental frequencies in the ratio $1 : 2 : 3$ . Then, the segments are in the ratio

1

6 : 3 : 2

2

4 : 3 : 2

3

4 : 2 : 1

4

3 : 2 : 1

Explanation:

A Given, $f_{1} : f_{2} : f_{3} = 1 : 2 : 3$
A string is divided into three segments. So, that the segment process fundamental frequencies in the ratio $1 : 2 : 3$
Let, $l$ be the length of the string
$∴ l = l_{1} + l_{2} + l_{3}$
We know that
Fundamental frequency $(f) = \frac{1}{2 l} \sqrt{\frac{T}{m}}$
$∵ f = f_{1} + f_{2} + f_{3}$
or $f \propto \frac{1}{l} [∵ T$ and $M$ are constants $]$
While,
$l_{1} : l_{2} : l_{3} = \frac{1}{f_{1}} : \frac{1}{f_{2}} : \frac{1}{f_{3}}$
$l_{1} : l_{2} : l_{3} = \frac{1}{1} : \frac{1}{2} : \frac{1}{3}$
$l_{1} : l_{2} : l_{3} = 6 : 3 : 2$

AP EAMCET-23.08.2021

WAVES

172406 When the area of cross-section of a stretched wire is halved and tension is doubled, the speed of propagation of transverse waves along it becomes $k$ times the initial speed. Then, $k$

1 1

2 4

3 2

4 8

Explanation:

C Given, $A_{2} = \frac{A_{1}}{2}, T_{2} = 2 T_{1}$
Speed of transverse wave in stretched wire is given as
$v = \sqrt{\frac{T}{μ}}$
Where, $T =$ Tension in the wire
$μ =$ Mass per unit length of wire.
So, $v = \sqrt{\frac{T}{M / l}} [∵ μ = \frac{M}{l}]$
$v = \sqrt{\frac{T l}{M}} = \sqrt{\frac{T l}{V \cdot ρ}}$
$v = \sqrt{\frac{T l}{A \cdot l \cdot ρ}}$
$v = \sqrt{\frac{T}{A ρ}} \Rightarrow v \propto \sqrt{\frac{T}{A}}$
$\frac{v_{2}}{v_{1}} = \sqrt{\frac{T_{2}}{T_{1}} \times \frac{A_{1}}{A_{2}}}$
from equation (i), we get-
$\frac{v_{2}}{v_{1}} = \sqrt{\frac{2 T_{1}}{T_{1}} \times \frac{A_{1}}{\frac{A_{1}}{2}}} = \sqrt{4}$
$v_{2} = 2 v_{1}$
So, $k = 2$

AP EAMCET (22.09.2020) Shift-II

WAVES

172407 A uniform rope of length $12 m$ and mass $6 kg$ hangs vertically from a rigid support. A block of mass $2 kg$ is attached to the free end of the rope. A transverse pulse of wavelength $0.06 m$ is produced at the lower end of the rope. Wavelength of the pulse when it reaches the top of the rope is .......... .

1

0.02 m

2

0.12 m

3

0.01 m

4

0.03 m

Explanation:

B
Given $T_{1} = 2 g$
$T_{2} = 8 g$
$λ_{1} = 0.06 m$
We know that
$v = f λ$
$v = \sqrt{\frac{T}{μ}}$
So, $v \propto λ \propto \sqrt{T}$
Then,
$v \propto λ \propto \sqrt{T}$
$\frac{\sqrt{T_{1}}}{\sqrt{T_{2}}} = \frac{λ_{1}}{λ_{2}}$
Then $λ_{2} = λ_{1} \sqrt{\frac{T_{2}}{T_{1}}} = 0.06 \sqrt{\frac{8 g}{2 g}}$
$λ_{2} = 0.12 m$

AP EAMCET-23.09.2020

WAVES

172408 A transverse wave is travelling with velocity ' $V$ ' through a metal wire of length ' $L$ ' and density ' $ρ$ '. The tensile stress on the wire is

1

V^{2} ρ

2

V ρ^{2}

3

V ρ

4

\frac{V}{ρ}

Explanation:

A We know that,
$V = \sqrt{\frac{T}{μ}}$ , where $T$ is the tension $& μ$ is mass per unit length.
$∴$ Velocity, $V = \sqrt{\frac{T}{A ρ}} (∵ μ = \frac{Mass}{Length} = \frac{ρ AL}{L} = ρ A)$
$∴ V^{2} = \frac{T}{A ρ}$
$\frac{T}{A} = V^{2} ρ$
$∴$ Tensile stress $(σ) = \frac{T}{A} = V^{2} ρ$

MHT-CET 2020

WAVES

172404 An elastic string of length $2 m$ is fixed at one end. The string starts to vibrate in third overtone with a frequency $1200 Hz$ . The ratio of frequency of first overtone and fundamental is

1

1 : 1

2

2 : 1

3

3 : 1

4

4 : 1

Explanation:

C String fixed at one end so,
Third overtone $= 5^{th}$ harmonic
First overtone $= 3^{rd}$ harmonic
Fundamental $= 1^{st}$ harmonic
Now, frequency $f = \frac{nV}{4 L} \Rightarrow f \propto n$
Ratio of frequency of first overtone and fundamental,
$\frac{f_{3}}{f_{1}} = \frac{n_{3}}{n_{1}} = \frac{3}{L}$
$f_{3} : f_{1} = 3 : 1$

AP EAMCET-05.10.2021

WAVES

172405 A string is divided into three segments, so that the segment possesses fundamental frequencies in the ratio $1 : 2 : 3$ . Then, the segments are in the ratio

1

6 : 3 : 2

2

4 : 3 : 2

3

4 : 2 : 1

4

3 : 2 : 1

Explanation:

A Given, $f_{1} : f_{2} : f_{3} = 1 : 2 : 3$
A string is divided into three segments. So, that the segment process fundamental frequencies in the ratio $1 : 2 : 3$
Let, $l$ be the length of the string
$∴ l = l_{1} + l_{2} + l_{3}$
We know that
Fundamental frequency $(f) = \frac{1}{2 l} \sqrt{\frac{T}{m}}$
$∵ f = f_{1} + f_{2} + f_{3}$
or $f \propto \frac{1}{l} [∵ T$ and $M$ are constants $]$
While,
$l_{1} : l_{2} : l_{3} = \frac{1}{f_{1}} : \frac{1}{f_{2}} : \frac{1}{f_{3}}$
$l_{1} : l_{2} : l_{3} = \frac{1}{1} : \frac{1}{2} : \frac{1}{3}$
$l_{1} : l_{2} : l_{3} = 6 : 3 : 2$

AP EAMCET-23.08.2021

WAVES

172406 When the area of cross-section of a stretched wire is halved and tension is doubled, the speed of propagation of transverse waves along it becomes $k$ times the initial speed. Then, $k$

1 1

2 4

3 2

4 8

Explanation:

C Given, $A_{2} = \frac{A_{1}}{2}, T_{2} = 2 T_{1}$
Speed of transverse wave in stretched wire is given as
$v = \sqrt{\frac{T}{μ}}$
Where, $T =$ Tension in the wire
$μ =$ Mass per unit length of wire.
So, $v = \sqrt{\frac{T}{M / l}} [∵ μ = \frac{M}{l}]$
$v = \sqrt{\frac{T l}{M}} = \sqrt{\frac{T l}{V \cdot ρ}}$
$v = \sqrt{\frac{T l}{A \cdot l \cdot ρ}}$
$v = \sqrt{\frac{T}{A ρ}} \Rightarrow v \propto \sqrt{\frac{T}{A}}$
$\frac{v_{2}}{v_{1}} = \sqrt{\frac{T_{2}}{T_{1}} \times \frac{A_{1}}{A_{2}}}$
from equation (i), we get-
$\frac{v_{2}}{v_{1}} = \sqrt{\frac{2 T_{1}}{T_{1}} \times \frac{A_{1}}{\frac{A_{1}}{2}}} = \sqrt{4}$
$v_{2} = 2 v_{1}$
So, $k = 2$

AP EAMCET (22.09.2020) Shift-II

WAVES

172407 A uniform rope of length $12 m$ and mass $6 kg$ hangs vertically from a rigid support. A block of mass $2 kg$ is attached to the free end of the rope. A transverse pulse of wavelength $0.06 m$ is produced at the lower end of the rope. Wavelength of the pulse when it reaches the top of the rope is .......... .

1

0.02 m

2

0.12 m

3

0.01 m

4

0.03 m

Explanation:

B
Given $T_{1} = 2 g$
$T_{2} = 8 g$
$λ_{1} = 0.06 m$
We know that
$v = f λ$
$v = \sqrt{\frac{T}{μ}}$
So, $v \propto λ \propto \sqrt{T}$
Then,
$v \propto λ \propto \sqrt{T}$
$\frac{\sqrt{T_{1}}}{\sqrt{T_{2}}} = \frac{λ_{1}}{λ_{2}}$
Then $λ_{2} = λ_{1} \sqrt{\frac{T_{2}}{T_{1}}} = 0.06 \sqrt{\frac{8 g}{2 g}}$
$λ_{2} = 0.12 m$

AP EAMCET-23.09.2020

WAVES

172408 A transverse wave is travelling with velocity ' $V$ ' through a metal wire of length ' $L$ ' and density ' $ρ$ '. The tensile stress on the wire is

1

V^{2} ρ

2

V ρ^{2}

3

V ρ

4

\frac{V}{ρ}

Explanation:

A We know that,
$V = \sqrt{\frac{T}{μ}}$ , where $T$ is the tension $& μ$ is mass per unit length.
$∴$ Velocity, $V = \sqrt{\frac{T}{A ρ}} (∵ μ = \frac{Mass}{Length} = \frac{ρ AL}{L} = ρ A)$
$∴ V^{2} = \frac{T}{A ρ}$
$\frac{T}{A} = V^{2} ρ$
$∴$ Tensile stress $(σ) = \frac{T}{A} = V^{2} ρ$

MHT-CET 2020

WAVES

172404 An elastic string of length $2 m$ is fixed at one end. The string starts to vibrate in third overtone with a frequency $1200 Hz$ . The ratio of frequency of first overtone and fundamental is

1

1 : 1

2

2 : 1

3

3 : 1

4

4 : 1

Explanation:

C String fixed at one end so,
Third overtone $= 5^{th}$ harmonic
First overtone $= 3^{rd}$ harmonic
Fundamental $= 1^{st}$ harmonic
Now, frequency $f = \frac{nV}{4 L} \Rightarrow f \propto n$
Ratio of frequency of first overtone and fundamental,
$\frac{f_{3}}{f_{1}} = \frac{n_{3}}{n_{1}} = \frac{3}{L}$
$f_{3} : f_{1} = 3 : 1$

AP EAMCET-05.10.2021

WAVES

172405 A string is divided into three segments, so that the segment possesses fundamental frequencies in the ratio $1 : 2 : 3$ . Then, the segments are in the ratio

1

6 : 3 : 2

2

4 : 3 : 2

3

4 : 2 : 1

4

3 : 2 : 1

Explanation:

A Given, $f_{1} : f_{2} : f_{3} = 1 : 2 : 3$
A string is divided into three segments. So, that the segment process fundamental frequencies in the ratio $1 : 2 : 3$
Let, $l$ be the length of the string
$∴ l = l_{1} + l_{2} + l_{3}$
We know that
Fundamental frequency $(f) = \frac{1}{2 l} \sqrt{\frac{T}{m}}$
$∵ f = f_{1} + f_{2} + f_{3}$
or $f \propto \frac{1}{l} [∵ T$ and $M$ are constants $]$
While,
$l_{1} : l_{2} : l_{3} = \frac{1}{f_{1}} : \frac{1}{f_{2}} : \frac{1}{f_{3}}$
$l_{1} : l_{2} : l_{3} = \frac{1}{1} : \frac{1}{2} : \frac{1}{3}$
$l_{1} : l_{2} : l_{3} = 6 : 3 : 2$

AP EAMCET-23.08.2021

WAVES

172406 When the area of cross-section of a stretched wire is halved and tension is doubled, the speed of propagation of transverse waves along it becomes $k$ times the initial speed. Then, $k$

1 1

2 4

3 2

4 8

Explanation:

C Given, $A_{2} = \frac{A_{1}}{2}, T_{2} = 2 T_{1}$
Speed of transverse wave in stretched wire is given as
$v = \sqrt{\frac{T}{μ}}$
Where, $T =$ Tension in the wire
$μ =$ Mass per unit length of wire.
So, $v = \sqrt{\frac{T}{M / l}} [∵ μ = \frac{M}{l}]$
$v = \sqrt{\frac{T l}{M}} = \sqrt{\frac{T l}{V \cdot ρ}}$
$v = \sqrt{\frac{T l}{A \cdot l \cdot ρ}}$
$v = \sqrt{\frac{T}{A ρ}} \Rightarrow v \propto \sqrt{\frac{T}{A}}$
$\frac{v_{2}}{v_{1}} = \sqrt{\frac{T_{2}}{T_{1}} \times \frac{A_{1}}{A_{2}}}$
from equation (i), we get-
$\frac{v_{2}}{v_{1}} = \sqrt{\frac{2 T_{1}}{T_{1}} \times \frac{A_{1}}{\frac{A_{1}}{2}}} = \sqrt{4}$
$v_{2} = 2 v_{1}$
So, $k = 2$

AP EAMCET (22.09.2020) Shift-II

WAVES

172407 A uniform rope of length $12 m$ and mass $6 kg$ hangs vertically from a rigid support. A block of mass $2 kg$ is attached to the free end of the rope. A transverse pulse of wavelength $0.06 m$ is produced at the lower end of the rope. Wavelength of the pulse when it reaches the top of the rope is .......... .

1

0.02 m

2

0.12 m

3

0.01 m

4

0.03 m

Explanation:

B
Given $T_{1} = 2 g$
$T_{2} = 8 g$
$λ_{1} = 0.06 m$
We know that
$v = f λ$
$v = \sqrt{\frac{T}{μ}}$
So, $v \propto λ \propto \sqrt{T}$
Then,
$v \propto λ \propto \sqrt{T}$
$\frac{\sqrt{T_{1}}}{\sqrt{T_{2}}} = \frac{λ_{1}}{λ_{2}}$
Then $λ_{2} = λ_{1} \sqrt{\frac{T_{2}}{T_{1}}} = 0.06 \sqrt{\frac{8 g}{2 g}}$
$λ_{2} = 0.12 m$

AP EAMCET-23.09.2020

WAVES

172408 A transverse wave is travelling with velocity ' $V$ ' through a metal wire of length ' $L$ ' and density ' $ρ$ '. The tensile stress on the wire is

1

V^{2} ρ

2

V ρ^{2}

3

V ρ

4

\frac{V}{ρ}

Explanation:

A We know that,
$V = \sqrt{\frac{T}{μ}}$ , where $T$ is the tension $& μ$ is mass per unit length.
$∴$ Velocity, $V = \sqrt{\frac{T}{A ρ}} (∵ μ = \frac{Mass}{Length} = \frac{ρ AL}{L} = ρ A)$
$∴ V^{2} = \frac{T}{A ρ}$
$\frac{T}{A} = V^{2} ρ$
$∴$ Tensile stress $(σ) = \frac{T}{A} = V^{2} ρ$

MHT-CET 2020

WAVES

172404 An elastic string of length $2 m$ is fixed at one end. The string starts to vibrate in third overtone with a frequency $1200 Hz$ . The ratio of frequency of first overtone and fundamental is

1

1 : 1

2

2 : 1

3

3 : 1

4

4 : 1

Explanation:

C String fixed at one end so,
Third overtone $= 5^{th}$ harmonic
First overtone $= 3^{rd}$ harmonic
Fundamental $= 1^{st}$ harmonic
Now, frequency $f = \frac{nV}{4 L} \Rightarrow f \propto n$
Ratio of frequency of first overtone and fundamental,
$\frac{f_{3}}{f_{1}} = \frac{n_{3}}{n_{1}} = \frac{3}{L}$
$f_{3} : f_{1} = 3 : 1$

AP EAMCET-05.10.2021

WAVES

172405 A string is divided into three segments, so that the segment possesses fundamental frequencies in the ratio $1 : 2 : 3$ . Then, the segments are in the ratio

1

6 : 3 : 2

2

4 : 3 : 2

3

4 : 2 : 1

4

3 : 2 : 1

Explanation:

A Given, $f_{1} : f_{2} : f_{3} = 1 : 2 : 3$
A string is divided into three segments. So, that the segment process fundamental frequencies in the ratio $1 : 2 : 3$
Let, $l$ be the length of the string
$∴ l = l_{1} + l_{2} + l_{3}$
We know that
Fundamental frequency $(f) = \frac{1}{2 l} \sqrt{\frac{T}{m}}$
$∵ f = f_{1} + f_{2} + f_{3}$
or $f \propto \frac{1}{l} [∵ T$ and $M$ are constants $]$
While,
$l_{1} : l_{2} : l_{3} = \frac{1}{f_{1}} : \frac{1}{f_{2}} : \frac{1}{f_{3}}$
$l_{1} : l_{2} : l_{3} = \frac{1}{1} : \frac{1}{2} : \frac{1}{3}$
$l_{1} : l_{2} : l_{3} = 6 : 3 : 2$

AP EAMCET-23.08.2021

WAVES

172406 When the area of cross-section of a stretched wire is halved and tension is doubled, the speed of propagation of transverse waves along it becomes $k$ times the initial speed. Then, $k$

1 1

2 4

3 2

4 8

Explanation:

C Given, $A_{2} = \frac{A_{1}}{2}, T_{2} = 2 T_{1}$
Speed of transverse wave in stretched wire is given as
$v = \sqrt{\frac{T}{μ}}$
Where, $T =$ Tension in the wire
$μ =$ Mass per unit length of wire.
So, $v = \sqrt{\frac{T}{M / l}} [∵ μ = \frac{M}{l}]$
$v = \sqrt{\frac{T l}{M}} = \sqrt{\frac{T l}{V \cdot ρ}}$
$v = \sqrt{\frac{T l}{A \cdot l \cdot ρ}}$
$v = \sqrt{\frac{T}{A ρ}} \Rightarrow v \propto \sqrt{\frac{T}{A}}$
$\frac{v_{2}}{v_{1}} = \sqrt{\frac{T_{2}}{T_{1}} \times \frac{A_{1}}{A_{2}}}$
from equation (i), we get-
$\frac{v_{2}}{v_{1}} = \sqrt{\frac{2 T_{1}}{T_{1}} \times \frac{A_{1}}{\frac{A_{1}}{2}}} = \sqrt{4}$
$v_{2} = 2 v_{1}$
So, $k = 2$

AP EAMCET (22.09.2020) Shift-II

WAVES

172407 A uniform rope of length $12 m$ and mass $6 kg$ hangs vertically from a rigid support. A block of mass $2 kg$ is attached to the free end of the rope. A transverse pulse of wavelength $0.06 m$ is produced at the lower end of the rope. Wavelength of the pulse when it reaches the top of the rope is .......... .

1

0.02 m

2

0.12 m

3

0.01 m

4

0.03 m

Explanation:

B
Given $T_{1} = 2 g$
$T_{2} = 8 g$
$λ_{1} = 0.06 m$
We know that
$v = f λ$
$v = \sqrt{\frac{T}{μ}}$
So, $v \propto λ \propto \sqrt{T}$
Then,
$v \propto λ \propto \sqrt{T}$
$\frac{\sqrt{T_{1}}}{\sqrt{T_{2}}} = \frac{λ_{1}}{λ_{2}}$
Then $λ_{2} = λ_{1} \sqrt{\frac{T_{2}}{T_{1}}} = 0.06 \sqrt{\frac{8 g}{2 g}}$
$λ_{2} = 0.12 m$

AP EAMCET-23.09.2020

WAVES

172408 A transverse wave is travelling with velocity ' $V$ ' through a metal wire of length ' $L$ ' and density ' $ρ$ '. The tensile stress on the wire is

1

V^{2} ρ

2

V ρ^{2}

3

V ρ

4

\frac{V}{ρ}

Explanation:

A We know that,
$V = \sqrt{\frac{T}{μ}}$ , where $T$ is the tension $& μ$ is mass per unit length.
$∴$ Velocity, $V = \sqrt{\frac{T}{A ρ}} (∵ μ = \frac{Mass}{Length} = \frac{ρ AL}{L} = ρ A)$
$∴ V^{2} = \frac{T}{A ρ}$
$\frac{T}{A} = V^{2} ρ$
$∴$ Tensile stress $(σ) = \frac{T}{A} = V^{2} ρ$

MHT-CET 2020