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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

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 $Hz$ and $900 Hz$ , 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 $(f_{1}) = 1600 Hz$ and $(f_{2}) = 900$ $Hz$
Fundamental frequencies for vibration is given by,
$f = \frac{v}{2 l} = \frac{1}{2 l} \sqrt{\frac{T}{μ}} = \frac{1}{2 l} \sqrt{\frac{T}{ρ A}}$
Where, $A \to$ cross sectional area of string
$A = π r^{2}$
$f \propto \frac{1}{\sqrt{A}} \Rightarrow f \propto \frac{1}{\sqrt{π r^{2}}} \Rightarrow f \propto \frac{1}{r}$
$\frac{d_{1}}{d_{2}} = \frac{f_{2}}{f_{1}} = \frac{900}{1600} = \frac{9}{16}$
$d_{1} : d_{2} = 9 : 16$

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)

1

(0.9) v

2

(1.1) v

3

(0.7) v

4

(1.22) v

Explanation:

D We know that,
$v = \sqrt{\frac{T}{m}}$
$∵ v \propto T$
$\frac{v_{2}}{v_{1}} = \sqrt{\frac{T_{2}}{T_{1}}}$
$\frac{v_{2}}{v_{1}} = \sqrt{\frac{1.5 T_{1}}{T_{1}}} = \sqrt{1.5} [∵ T_{2} = 1.5 T_{1}]$
$v_{2} = (1.22) v$

MHT-CET 2020

WAVES

172413 The fundamental frequency of a wire stretched by $2 kg wt$ . is $100 Hz$ . The weight required to produce its octave will be

1

12 kg wt

2

16 kg wt

3

4 kg wt

4

8 kg wt

Explanation:

D Given,
$T_{1} = 2 kg wt$
$n_{1} = 100 Hz$
The ratio between the frequencies of successive octaves is $1 : 2$
$\frac{n_{1}}{n_{2}} = \frac{1}{2}$
$n_{2} = 2 n_{1} = 200 Hz$
We know,
$n = \frac{1}{2 l} \sqrt{\frac{T}{m}}$
Here, $l$ and $m$ are constant.
So, $n \propto \sqrt{T}$
$∴ \frac{n_{1}}{n_{2}} = \sqrt{\frac{T_{1}}{T_{2}}}$
$\frac{100}{200} = \sqrt{\frac{2}{T_{2}}}$
$\frac{1}{2} = \sqrt{\frac{2}{T_{2}}}$
$\frac{1}{4} = \frac{2}{T_{2}}$
Hence, $T_{2} = 8 kg$ wt

MHT-CET 2020

WAVES

172414 A string of mass $0.1 kg$ is under a tension $1.6 N$ . The length of the string is $1 m$ . 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 s

2

0.30 s

3

0.25 s

4

0.75 s

Explanation:

C Given that, $M = 0.1 kg, T = 1.6 N, L = 1 m$ Mass per unit length,
$μ = \frac{M}{L} = \frac{0.1}{1} = 0.1 kg / m$
$Velocity, v = \sqrt{\frac{T}{μ}} = \sqrt{\frac{1.6}{0.1}}$
$v = 4 m / s$
Time taken by wave,
$t = \frac{L}{v} = \frac{1}{4}$
$∴ t = 0.25 \sec$

MHT-CET 2020

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

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 $Hz$ and $900 Hz$ , 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 $(f_{1}) = 1600 Hz$ and $(f_{2}) = 900$ $Hz$
Fundamental frequencies for vibration is given by,
$f = \frac{v}{2 l} = \frac{1}{2 l} \sqrt{\frac{T}{μ}} = \frac{1}{2 l} \sqrt{\frac{T}{ρ A}}$
Where, $A \to$ cross sectional area of string
$A = π r^{2}$
$f \propto \frac{1}{\sqrt{A}} \Rightarrow f \propto \frac{1}{\sqrt{π r^{2}}} \Rightarrow f \propto \frac{1}{r}$
$\frac{d_{1}}{d_{2}} = \frac{f_{2}}{f_{1}} = \frac{900}{1600} = \frac{9}{16}$
$d_{1} : d_{2} = 9 : 16$

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)

1

(0.9) v

2

(1.1) v

3

(0.7) v

4

(1.22) v

Explanation:

D We know that,
$v = \sqrt{\frac{T}{m}}$
$∵ v \propto T$
$\frac{v_{2}}{v_{1}} = \sqrt{\frac{T_{2}}{T_{1}}}$
$\frac{v_{2}}{v_{1}} = \sqrt{\frac{1.5 T_{1}}{T_{1}}} = \sqrt{1.5} [∵ T_{2} = 1.5 T_{1}]$
$v_{2} = (1.22) v$

MHT-CET 2020

WAVES

172413 The fundamental frequency of a wire stretched by $2 kg wt$ . is $100 Hz$ . The weight required to produce its octave will be

1

12 kg wt

2

16 kg wt

3

4 kg wt

4

8 kg wt

Explanation:

D Given,
$T_{1} = 2 kg wt$
$n_{1} = 100 Hz$
The ratio between the frequencies of successive octaves is $1 : 2$
$\frac{n_{1}}{n_{2}} = \frac{1}{2}$
$n_{2} = 2 n_{1} = 200 Hz$
We know,
$n = \frac{1}{2 l} \sqrt{\frac{T}{m}}$
Here, $l$ and $m$ are constant.
So, $n \propto \sqrt{T}$
$∴ \frac{n_{1}}{n_{2}} = \sqrt{\frac{T_{1}}{T_{2}}}$
$\frac{100}{200} = \sqrt{\frac{2}{T_{2}}}$
$\frac{1}{2} = \sqrt{\frac{2}{T_{2}}}$
$\frac{1}{4} = \frac{2}{T_{2}}$
Hence, $T_{2} = 8 kg$ wt

MHT-CET 2020

WAVES

172414 A string of mass $0.1 kg$ is under a tension $1.6 N$ . The length of the string is $1 m$ . 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 s

2

0.30 s

3

0.25 s

4

0.75 s

Explanation:

C Given that, $M = 0.1 kg, T = 1.6 N, L = 1 m$ Mass per unit length,
$μ = \frac{M}{L} = \frac{0.1}{1} = 0.1 kg / m$
$Velocity, v = \sqrt{\frac{T}{μ}} = \sqrt{\frac{1.6}{0.1}}$
$v = 4 m / s$
Time taken by wave,
$t = \frac{L}{v} = \frac{1}{4}$
$∴ t = 0.25 \sec$

MHT-CET 2020

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

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 $Hz$ and $900 Hz$ , 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 $(f_{1}) = 1600 Hz$ and $(f_{2}) = 900$ $Hz$
Fundamental frequencies for vibration is given by,
$f = \frac{v}{2 l} = \frac{1}{2 l} \sqrt{\frac{T}{μ}} = \frac{1}{2 l} \sqrt{\frac{T}{ρ A}}$
Where, $A \to$ cross sectional area of string
$A = π r^{2}$
$f \propto \frac{1}{\sqrt{A}} \Rightarrow f \propto \frac{1}{\sqrt{π r^{2}}} \Rightarrow f \propto \frac{1}{r}$
$\frac{d_{1}}{d_{2}} = \frac{f_{2}}{f_{1}} = \frac{900}{1600} = \frac{9}{16}$
$d_{1} : d_{2} = 9 : 16$

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)

1

(0.9) v

2

(1.1) v

3

(0.7) v

4

(1.22) v

Explanation:

D We know that,
$v = \sqrt{\frac{T}{m}}$
$∵ v \propto T$
$\frac{v_{2}}{v_{1}} = \sqrt{\frac{T_{2}}{T_{1}}}$
$\frac{v_{2}}{v_{1}} = \sqrt{\frac{1.5 T_{1}}{T_{1}}} = \sqrt{1.5} [∵ T_{2} = 1.5 T_{1}]$
$v_{2} = (1.22) v$

MHT-CET 2020

WAVES

172413 The fundamental frequency of a wire stretched by $2 kg wt$ . is $100 Hz$ . The weight required to produce its octave will be

1

12 kg wt

2

16 kg wt

3

4 kg wt

4

8 kg wt

Explanation:

D Given,
$T_{1} = 2 kg wt$
$n_{1} = 100 Hz$
The ratio between the frequencies of successive octaves is $1 : 2$
$\frac{n_{1}}{n_{2}} = \frac{1}{2}$
$n_{2} = 2 n_{1} = 200 Hz$
We know,
$n = \frac{1}{2 l} \sqrt{\frac{T}{m}}$
Here, $l$ and $m$ are constant.
So, $n \propto \sqrt{T}$
$∴ \frac{n_{1}}{n_{2}} = \sqrt{\frac{T_{1}}{T_{2}}}$
$\frac{100}{200} = \sqrt{\frac{2}{T_{2}}}$
$\frac{1}{2} = \sqrt{\frac{2}{T_{2}}}$
$\frac{1}{4} = \frac{2}{T_{2}}$
Hence, $T_{2} = 8 kg$ wt

MHT-CET 2020

WAVES

172414 A string of mass $0.1 kg$ is under a tension $1.6 N$ . The length of the string is $1 m$ . 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 s

2

0.30 s

3

0.25 s

4

0.75 s

Explanation:

C Given that, $M = 0.1 kg, T = 1.6 N, L = 1 m$ Mass per unit length,
$μ = \frac{M}{L} = \frac{0.1}{1} = 0.1 kg / m$
$Velocity, v = \sqrt{\frac{T}{μ}} = \sqrt{\frac{1.6}{0.1}}$
$v = 4 m / s$
Time taken by wave,
$t = \frac{L}{v} = \frac{1}{4}$
$∴ t = 0.25 \sec$

MHT-CET 2020

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

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 $Hz$ and $900 Hz$ , 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 $(f_{1}) = 1600 Hz$ and $(f_{2}) = 900$ $Hz$
Fundamental frequencies for vibration is given by,
$f = \frac{v}{2 l} = \frac{1}{2 l} \sqrt{\frac{T}{μ}} = \frac{1}{2 l} \sqrt{\frac{T}{ρ A}}$
Where, $A \to$ cross sectional area of string
$A = π r^{2}$
$f \propto \frac{1}{\sqrt{A}} \Rightarrow f \propto \frac{1}{\sqrt{π r^{2}}} \Rightarrow f \propto \frac{1}{r}$
$\frac{d_{1}}{d_{2}} = \frac{f_{2}}{f_{1}} = \frac{900}{1600} = \frac{9}{16}$
$d_{1} : d_{2} = 9 : 16$

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)

1

(0.9) v

2

(1.1) v

3

(0.7) v

4

(1.22) v

Explanation:

D We know that,
$v = \sqrt{\frac{T}{m}}$
$∵ v \propto T$
$\frac{v_{2}}{v_{1}} = \sqrt{\frac{T_{2}}{T_{1}}}$
$\frac{v_{2}}{v_{1}} = \sqrt{\frac{1.5 T_{1}}{T_{1}}} = \sqrt{1.5} [∵ T_{2} = 1.5 T_{1}]$
$v_{2} = (1.22) v$

MHT-CET 2020

WAVES

172413 The fundamental frequency of a wire stretched by $2 kg wt$ . is $100 Hz$ . The weight required to produce its octave will be

1

12 kg wt

2

16 kg wt

3

4 kg wt

4

8 kg wt

Explanation:

D Given,
$T_{1} = 2 kg wt$
$n_{1} = 100 Hz$
The ratio between the frequencies of successive octaves is $1 : 2$
$\frac{n_{1}}{n_{2}} = \frac{1}{2}$
$n_{2} = 2 n_{1} = 200 Hz$
We know,
$n = \frac{1}{2 l} \sqrt{\frac{T}{m}}$
Here, $l$ and $m$ are constant.
So, $n \propto \sqrt{T}$
$∴ \frac{n_{1}}{n_{2}} = \sqrt{\frac{T_{1}}{T_{2}}}$
$\frac{100}{200} = \sqrt{\frac{2}{T_{2}}}$
$\frac{1}{2} = \sqrt{\frac{2}{T_{2}}}$
$\frac{1}{4} = \frac{2}{T_{2}}$
Hence, $T_{2} = 8 kg$ wt

MHT-CET 2020

WAVES

172414 A string of mass $0.1 kg$ is under a tension $1.6 N$ . The length of the string is $1 m$ . 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 s

2

0.30 s

3

0.25 s

4

0.75 s

Explanation:

C Given that, $M = 0.1 kg, T = 1.6 N, L = 1 m$ Mass per unit length,
$μ = \frac{M}{L} = \frac{0.1}{1} = 0.1 kg / m$
$Velocity, v = \sqrt{\frac{T}{μ}} = \sqrt{\frac{1.6}{0.1}}$
$v = 4 m / s$
Time taken by wave,
$t = \frac{L}{v} = \frac{1}{4}$
$∴ t = 0.25 \sec$

MHT-CET 2020

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

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 $Hz$ and $900 Hz$ , 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 $(f_{1}) = 1600 Hz$ and $(f_{2}) = 900$ $Hz$
Fundamental frequencies for vibration is given by,
$f = \frac{v}{2 l} = \frac{1}{2 l} \sqrt{\frac{T}{μ}} = \frac{1}{2 l} \sqrt{\frac{T}{ρ A}}$
Where, $A \to$ cross sectional area of string
$A = π r^{2}$
$f \propto \frac{1}{\sqrt{A}} \Rightarrow f \propto \frac{1}{\sqrt{π r^{2}}} \Rightarrow f \propto \frac{1}{r}$
$\frac{d_{1}}{d_{2}} = \frac{f_{2}}{f_{1}} = \frac{900}{1600} = \frac{9}{16}$
$d_{1} : d_{2} = 9 : 16$

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)

1

(0.9) v

2

(1.1) v

3

(0.7) v

4

(1.22) v

Explanation:

D We know that,
$v = \sqrt{\frac{T}{m}}$
$∵ v \propto T$
$\frac{v_{2}}{v_{1}} = \sqrt{\frac{T_{2}}{T_{1}}}$
$\frac{v_{2}}{v_{1}} = \sqrt{\frac{1.5 T_{1}}{T_{1}}} = \sqrt{1.5} [∵ T_{2} = 1.5 T_{1}]$
$v_{2} = (1.22) v$

MHT-CET 2020

WAVES

172413 The fundamental frequency of a wire stretched by $2 kg wt$ . is $100 Hz$ . The weight required to produce its octave will be

1

12 kg wt

2

16 kg wt

3

4 kg wt

4

8 kg wt

Explanation:

D Given,
$T_{1} = 2 kg wt$
$n_{1} = 100 Hz$
The ratio between the frequencies of successive octaves is $1 : 2$
$\frac{n_{1}}{n_{2}} = \frac{1}{2}$
$n_{2} = 2 n_{1} = 200 Hz$
We know,
$n = \frac{1}{2 l} \sqrt{\frac{T}{m}}$
Here, $l$ and $m$ are constant.
So, $n \propto \sqrt{T}$
$∴ \frac{n_{1}}{n_{2}} = \sqrt{\frac{T_{1}}{T_{2}}}$
$\frac{100}{200} = \sqrt{\frac{2}{T_{2}}}$
$\frac{1}{2} = \sqrt{\frac{2}{T_{2}}}$
$\frac{1}{4} = \frac{2}{T_{2}}$
Hence, $T_{2} = 8 kg$ wt

MHT-CET 2020

WAVES

172414 A string of mass $0.1 kg$ is under a tension $1.6 N$ . The length of the string is $1 m$ . 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 s

2

0.30 s

3

0.25 s

4

0.75 s

Explanation:

C Given that, $M = 0.1 kg, T = 1.6 N, L = 1 m$ Mass per unit length,
$μ = \frac{M}{L} = \frac{0.1}{1} = 0.1 kg / m$
$Velocity, v = \sqrt{\frac{T}{μ}} = \sqrt{\frac{1.6}{0.1}}$
$v = 4 m / s$
Time taken by wave,
$t = \frac{L}{v} = \frac{1}{4}$
$∴ t = 0.25 \sec$

MHT-CET 2020