QBManager

WAVES

173122 If the length and diameter of a wire are decreased, then for the same tension the natural frequency of stretched wire will

1 decrease

2 become zero

3 increase

4 not changes

Explanation:

C We know
$Frequency (n) = \frac{1}{2 Lr} \sqrt{\frac{T}{π ρ}}$
$∴$ If $L$ and $r$ decrease, the frequency will increase.
Thus, if the length and diameter of a wire are decreased, then for the same tension the natural frequency of stretched wire will increase.

MHT-CET 2020

WAVES

173127 The resonance tube is filled with a liquid of density higher than that of water, then resonating frequency

1 will not change.

2 may increase or decrease.

3 will decrease.

4 will increase.

MHT-CET 2020

WAVES

173128 In resonance tube experiment, the first and second resonating length of air column is $0.2 m$ and $0.62 m$ respectively. The inner diameter of the tube is

1

3.33 cm

2

0.2 m

3

0.4 m

4

0.33 cm

Explanation:

A Given,
$\begin{array}{r} l_{1} = 0.2 m and l_{2} = 0.62 m \\ So, end correction (e) = \frac{l_{2} - 3 l_{1}}{2} \\ = \frac{0.62 - 0.60}{2} = 0.01 m = 1 cm \end{array}$
$∵ e = 0.3 D$
$D = \frac{e}{0.3} = \frac{1}{0.3} = \frac{10}{3}$
The inner diameter of the tube, $D = 3.33 cm$

MHT-CET 2020

WAVES

173137 In a resonance pipe the first and second resonances are obtained at depths $22.7 cm$ and $70.2 cm$ respectively. What will be the end correction ?

1

1.05 cm

2

115.5 cm

3

92.5 cm

4

113.5 cm

Explanation:

A Let, $l_{1}$ be the depth of first resonance of the pipe $(l_{1}) = 22.7 cm$ and $l_{2}$ be the depth of second resonance of the pipe $(l_{2}) = 70.2 cm$
Now for end correction $x$ ,
$\frac{l_{2} + x}{l_{1} + x} = \frac{3 λ / 4}{λ / 4} = 3$
$x = \frac{l_{2} - 3 l_{1}}{2} = \frac{70.2 - 3 \times 22.7}{2}$
$x = 1.05 cm$

MHT-CET 2009

WAVES

173141 A car moving with a velocity of $36 km / hr$ crosses a siren of frequency $500 Hz$ . The apparent frequency of siren after passing it will be

1

520 Hz

2

485 Hz

3

540 Hz

4

460 Hz

Explanation:

B Given, velocity of observer $(v_{o}) = 36 km / hr$ $= 36 \times \frac{5}{18} = 10 m / \sec, f = 500 Hz$
Velocity of sound frequency $(v_{s}) = 330 m / \sec$
$∴$ Apparent frequency
$v^{'} = \frac{v_{s} - v_{o}}{v_{s}} \times f$
$= \frac{330 - 10}{330} \times 500 = 484.84 = 485 Hz$

WB JEE 2008

WAVES

173122 If the length and diameter of a wire are decreased, then for the same tension the natural frequency of stretched wire will

1 decrease

2 become zero

3 increase

4 not changes

Explanation:

C We know
$Frequency (n) = \frac{1}{2 Lr} \sqrt{\frac{T}{π ρ}}$
$∴$ If $L$ and $r$ decrease, the frequency will increase.
Thus, if the length and diameter of a wire are decreased, then for the same tension the natural frequency of stretched wire will increase.

MHT-CET 2020

WAVES

173127 The resonance tube is filled with a liquid of density higher than that of water, then resonating frequency

1 will not change.

2 may increase or decrease.

3 will decrease.

4 will increase.

MHT-CET 2020

WAVES

173128 In resonance tube experiment, the first and second resonating length of air column is $0.2 m$ and $0.62 m$ respectively. The inner diameter of the tube is

1

3.33 cm

2

0.2 m

3

0.4 m

4

0.33 cm

Explanation:

A Given,
$\begin{array}{r} l_{1} = 0.2 m and l_{2} = 0.62 m \\ So, end correction (e) = \frac{l_{2} - 3 l_{1}}{2} \\ = \frac{0.62 - 0.60}{2} = 0.01 m = 1 cm \end{array}$
$∵ e = 0.3 D$
$D = \frac{e}{0.3} = \frac{1}{0.3} = \frac{10}{3}$
The inner diameter of the tube, $D = 3.33 cm$

MHT-CET 2020

WAVES

173137 In a resonance pipe the first and second resonances are obtained at depths $22.7 cm$ and $70.2 cm$ respectively. What will be the end correction ?

1

1.05 cm

2

115.5 cm

3

92.5 cm

4

113.5 cm

Explanation:

A Let, $l_{1}$ be the depth of first resonance of the pipe $(l_{1}) = 22.7 cm$ and $l_{2}$ be the depth of second resonance of the pipe $(l_{2}) = 70.2 cm$
Now for end correction $x$ ,
$\frac{l_{2} + x}{l_{1} + x} = \frac{3 λ / 4}{λ / 4} = 3$
$x = \frac{l_{2} - 3 l_{1}}{2} = \frac{70.2 - 3 \times 22.7}{2}$
$x = 1.05 cm$

MHT-CET 2009

WAVES

173141 A car moving with a velocity of $36 km / hr$ crosses a siren of frequency $500 Hz$ . The apparent frequency of siren after passing it will be

1

520 Hz

2

485 Hz

3

540 Hz

4

460 Hz

Explanation:

B Given, velocity of observer $(v_{o}) = 36 km / hr$ $= 36 \times \frac{5}{18} = 10 m / \sec, f = 500 Hz$
Velocity of sound frequency $(v_{s}) = 330 m / \sec$
$∴$ Apparent frequency
$v^{'} = \frac{v_{s} - v_{o}}{v_{s}} \times f$
$= \frac{330 - 10}{330} \times 500 = 484.84 = 485 Hz$

WB JEE 2008

WAVES

173122 If the length and diameter of a wire are decreased, then for the same tension the natural frequency of stretched wire will

1 decrease

2 become zero

3 increase

4 not changes

Explanation:

C We know
$Frequency (n) = \frac{1}{2 Lr} \sqrt{\frac{T}{π ρ}}$
$∴$ If $L$ and $r$ decrease, the frequency will increase.
Thus, if the length and diameter of a wire are decreased, then for the same tension the natural frequency of stretched wire will increase.

MHT-CET 2020

WAVES

173127 The resonance tube is filled with a liquid of density higher than that of water, then resonating frequency

1 will not change.

2 may increase or decrease.

3 will decrease.

4 will increase.

MHT-CET 2020

WAVES

173128 In resonance tube experiment, the first and second resonating length of air column is $0.2 m$ and $0.62 m$ respectively. The inner diameter of the tube is

1

3.33 cm

2

0.2 m

3

0.4 m

4

0.33 cm

Explanation:

A Given,
$\begin{array}{r} l_{1} = 0.2 m and l_{2} = 0.62 m \\ So, end correction (e) = \frac{l_{2} - 3 l_{1}}{2} \\ = \frac{0.62 - 0.60}{2} = 0.01 m = 1 cm \end{array}$
$∵ e = 0.3 D$
$D = \frac{e}{0.3} = \frac{1}{0.3} = \frac{10}{3}$
The inner diameter of the tube, $D = 3.33 cm$

MHT-CET 2020

WAVES

173137 In a resonance pipe the first and second resonances are obtained at depths $22.7 cm$ and $70.2 cm$ respectively. What will be the end correction ?

1

1.05 cm

2

115.5 cm

3

92.5 cm

4

113.5 cm

Explanation:

A Let, $l_{1}$ be the depth of first resonance of the pipe $(l_{1}) = 22.7 cm$ and $l_{2}$ be the depth of second resonance of the pipe $(l_{2}) = 70.2 cm$
Now for end correction $x$ ,
$\frac{l_{2} + x}{l_{1} + x} = \frac{3 λ / 4}{λ / 4} = 3$
$x = \frac{l_{2} - 3 l_{1}}{2} = \frac{70.2 - 3 \times 22.7}{2}$
$x = 1.05 cm$

MHT-CET 2009

WAVES

173141 A car moving with a velocity of $36 km / hr$ crosses a siren of frequency $500 Hz$ . The apparent frequency of siren after passing it will be

1

520 Hz

2

485 Hz

3

540 Hz

4

460 Hz

Explanation:

B Given, velocity of observer $(v_{o}) = 36 km / hr$ $= 36 \times \frac{5}{18} = 10 m / \sec, f = 500 Hz$
Velocity of sound frequency $(v_{s}) = 330 m / \sec$
$∴$ Apparent frequency
$v^{'} = \frac{v_{s} - v_{o}}{v_{s}} \times f$
$= \frac{330 - 10}{330} \times 500 = 484.84 = 485 Hz$

WB JEE 2008

WAVES

173122 If the length and diameter of a wire are decreased, then for the same tension the natural frequency of stretched wire will

1 decrease

2 become zero

3 increase

4 not changes

Explanation:

C We know
$Frequency (n) = \frac{1}{2 Lr} \sqrt{\frac{T}{π ρ}}$
$∴$ If $L$ and $r$ decrease, the frequency will increase.
Thus, if the length and diameter of a wire are decreased, then for the same tension the natural frequency of stretched wire will increase.

MHT-CET 2020

WAVES

173127 The resonance tube is filled with a liquid of density higher than that of water, then resonating frequency

1 will not change.

2 may increase or decrease.

3 will decrease.

4 will increase.

MHT-CET 2020

WAVES

173128 In resonance tube experiment, the first and second resonating length of air column is $0.2 m$ and $0.62 m$ respectively. The inner diameter of the tube is

1

3.33 cm

2

0.2 m

3

0.4 m

4

0.33 cm

Explanation:

A Given,
$\begin{array}{r} l_{1} = 0.2 m and l_{2} = 0.62 m \\ So, end correction (e) = \frac{l_{2} - 3 l_{1}}{2} \\ = \frac{0.62 - 0.60}{2} = 0.01 m = 1 cm \end{array}$
$∵ e = 0.3 D$
$D = \frac{e}{0.3} = \frac{1}{0.3} = \frac{10}{3}$
The inner diameter of the tube, $D = 3.33 cm$

MHT-CET 2020

WAVES

173137 In a resonance pipe the first and second resonances are obtained at depths $22.7 cm$ and $70.2 cm$ respectively. What will be the end correction ?

1

1.05 cm

2

115.5 cm

3

92.5 cm

4

113.5 cm

Explanation:

A Let, $l_{1}$ be the depth of first resonance of the pipe $(l_{1}) = 22.7 cm$ and $l_{2}$ be the depth of second resonance of the pipe $(l_{2}) = 70.2 cm$
Now for end correction $x$ ,
$\frac{l_{2} + x}{l_{1} + x} = \frac{3 λ / 4}{λ / 4} = 3$
$x = \frac{l_{2} - 3 l_{1}}{2} = \frac{70.2 - 3 \times 22.7}{2}$
$x = 1.05 cm$

MHT-CET 2009

WAVES

173141 A car moving with a velocity of $36 km / hr$ crosses a siren of frequency $500 Hz$ . The apparent frequency of siren after passing it will be

1

520 Hz

2

485 Hz

3

540 Hz

4

460 Hz

Explanation:

B Given, velocity of observer $(v_{o}) = 36 km / hr$ $= 36 \times \frac{5}{18} = 10 m / \sec, f = 500 Hz$
Velocity of sound frequency $(v_{s}) = 330 m / \sec$
$∴$ Apparent frequency
$v^{'} = \frac{v_{s} - v_{o}}{v_{s}} \times f$
$= \frac{330 - 10}{330} \times 500 = 484.84 = 485 Hz$

WB JEE 2008

WAVES

173122 If the length and diameter of a wire are decreased, then for the same tension the natural frequency of stretched wire will

1 decrease

2 become zero

3 increase

4 not changes

Explanation:

C We know
$Frequency (n) = \frac{1}{2 Lr} \sqrt{\frac{T}{π ρ}}$
$∴$ If $L$ and $r$ decrease, the frequency will increase.
Thus, if the length and diameter of a wire are decreased, then for the same tension the natural frequency of stretched wire will increase.

MHT-CET 2020

WAVES

173127 The resonance tube is filled with a liquid of density higher than that of water, then resonating frequency

1 will not change.

2 may increase or decrease.

3 will decrease.

4 will increase.

MHT-CET 2020

WAVES

173128 In resonance tube experiment, the first and second resonating length of air column is $0.2 m$ and $0.62 m$ respectively. The inner diameter of the tube is

1

3.33 cm

2

0.2 m

3

0.4 m

4

0.33 cm

Explanation:

A Given,
$\begin{array}{r} l_{1} = 0.2 m and l_{2} = 0.62 m \\ So, end correction (e) = \frac{l_{2} - 3 l_{1}}{2} \\ = \frac{0.62 - 0.60}{2} = 0.01 m = 1 cm \end{array}$
$∵ e = 0.3 D$
$D = \frac{e}{0.3} = \frac{1}{0.3} = \frac{10}{3}$
The inner diameter of the tube, $D = 3.33 cm$

MHT-CET 2020

WAVES

173137 In a resonance pipe the first and second resonances are obtained at depths $22.7 cm$ and $70.2 cm$ respectively. What will be the end correction ?

1

1.05 cm

2

115.5 cm

3

92.5 cm

4

113.5 cm

Explanation:

A Let, $l_{1}$ be the depth of first resonance of the pipe $(l_{1}) = 22.7 cm$ and $l_{2}$ be the depth of second resonance of the pipe $(l_{2}) = 70.2 cm$
Now for end correction $x$ ,
$\frac{l_{2} + x}{l_{1} + x} = \frac{3 λ / 4}{λ / 4} = 3$
$x = \frac{l_{2} - 3 l_{1}}{2} = \frac{70.2 - 3 \times 22.7}{2}$
$x = 1.05 cm$

MHT-CET 2009

WAVES

173141 A car moving with a velocity of $36 km / hr$ crosses a siren of frequency $500 Hz$ . The apparent frequency of siren after passing it will be

1

520 Hz

2

485 Hz

3

540 Hz

4

460 Hz

Explanation:

B Given, velocity of observer $(v_{o}) = 36 km / hr$ $= 36 \times \frac{5}{18} = 10 m / \sec, f = 500 Hz$
Velocity of sound frequency $(v_{s}) = 330 m / \sec$
$∴$ Apparent frequency
$v^{'} = \frac{v_{s} - v_{o}}{v_{s}} \times f$
$= \frac{330 - 10}{330} \times 500 = 484.84 = 485 Hz$

WB JEE 2008