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PHXI15:WAVES

354648 The temperature at which the speed of sound in air becomes doubled of its value at $0^{\circ} C$ is

1

1092^{\circ} C

2

819 K

3

819^{\circ} C

4

546^{\circ} C

Explanation:

Speed, $v \propto \sqrt{T}$
$∴ \frac{2 v}{v} = \sqrt{\frac{T_{2}}{T_{1}}}$
$\Rightarrow 2 = \sqrt{\frac{T_{2}}{T_{1}}}$
$\Rightarrow 4 = \frac{T_{2}}{T_{1}}$
$\Rightarrow T_{2} = 4 T_{1} = 4 \times 273 = 1092 K$
$∴ T_{2} = 1092 - 273 = 819^{\circ} C$

PHXI15:WAVES

354649 Velocity of sound in air at the given temperature

1 Decreases with increase in pressure

2 May increase on decrease with pressure

3 Is independent of the variation in pressure

4 Varies directly as the square root of pressure

PHXI15:WAVES

354650 A pressure of $100 k P a$ causes a decrease in volume of water by $5 \times 10^{- 3}$ percent. The speed of sound in water is

1 1414

2 1000

3 2000

4 3000

Explanation:

$v = \sqrt{\frac{B}{δ}} = \sqrt{\frac{d P}{(\frac{d v}{v}) δ}}$
$= \sqrt{\frac{100 \times 10^{3}}{5 \times 10^{- 5} \times 10^{3}}}$
$= \sqrt{\frac{1}{5} \times 10^{7}} = \sqrt{2} \times 10^{3} = 1414 m / s$

PHXI15:WAVES

354651 A steel rod $100 c m$ long is clamped at the middle. The fundamental frequency of longitudinal vibrations of the rod are given to be $2.53 k H z$ . What is the speed of sound in steel?

1

6.2 k m s^{- 1}

2

5.06 k m s^{- 1}

3

7.23 k m s^{- 1}

4

7.45 k m s^{- 1}

Explanation:

In fundamental mode,
$l = 2 (\frac{λ}{4}) = \frac{λ}{2}$
supporting img
$\Rightarrow λ = 2 l (1)$
Given, $l = 100 c m$ ,
$v = 2.53 k H z = 2.53 \times 10^{3} H z$
We know that,
$v = v λ$
$= v \times 2 l [$ From Eq. (1) ]
$= 2.53 \times 10^{3} \times 2 \times 100 \times 10^{- 2}$
$= 5.06 \times 10^{3} m s^{- 1}$
$= 5.06 k m s^{- 1}$

AIIMS - 2018

PHXI15:WAVES

354648 The temperature at which the speed of sound in air becomes doubled of its value at $0^{\circ} C$ is

1

1092^{\circ} C

2

819 K

3

819^{\circ} C

4

546^{\circ} C

Explanation:

Speed, $v \propto \sqrt{T}$
$∴ \frac{2 v}{v} = \sqrt{\frac{T_{2}}{T_{1}}}$
$\Rightarrow 2 = \sqrt{\frac{T_{2}}{T_{1}}}$
$\Rightarrow 4 = \frac{T_{2}}{T_{1}}$
$\Rightarrow T_{2} = 4 T_{1} = 4 \times 273 = 1092 K$
$∴ T_{2} = 1092 - 273 = 819^{\circ} C$

PHXI15:WAVES

354649 Velocity of sound in air at the given temperature

1 Decreases with increase in pressure

2 May increase on decrease with pressure

3 Is independent of the variation in pressure

4 Varies directly as the square root of pressure

PHXI15:WAVES

354650 A pressure of $100 k P a$ causes a decrease in volume of water by $5 \times 10^{- 3}$ percent. The speed of sound in water is

1 1414

2 1000

3 2000

4 3000

Explanation:

$v = \sqrt{\frac{B}{δ}} = \sqrt{\frac{d P}{(\frac{d v}{v}) δ}}$
$= \sqrt{\frac{100 \times 10^{3}}{5 \times 10^{- 5} \times 10^{3}}}$
$= \sqrt{\frac{1}{5} \times 10^{7}} = \sqrt{2} \times 10^{3} = 1414 m / s$

PHXI15:WAVES

354651 A steel rod $100 c m$ long is clamped at the middle. The fundamental frequency of longitudinal vibrations of the rod are given to be $2.53 k H z$ . What is the speed of sound in steel?

1

6.2 k m s^{- 1}

2

5.06 k m s^{- 1}

3

7.23 k m s^{- 1}

4

7.45 k m s^{- 1}

Explanation:

In fundamental mode,
$l = 2 (\frac{λ}{4}) = \frac{λ}{2}$
supporting img
$\Rightarrow λ = 2 l (1)$
Given, $l = 100 c m$ ,
$v = 2.53 k H z = 2.53 \times 10^{3} H z$
We know that,
$v = v λ$
$= v \times 2 l [$ From Eq. (1) ]
$= 2.53 \times 10^{3} \times 2 \times 100 \times 10^{- 2}$
$= 5.06 \times 10^{3} m s^{- 1}$
$= 5.06 k m s^{- 1}$

AIIMS - 2018

PHXI15:WAVES

354648 The temperature at which the speed of sound in air becomes doubled of its value at $0^{\circ} C$ is

1

1092^{\circ} C

2

819 K

3

819^{\circ} C

4

546^{\circ} C

Explanation:

Speed, $v \propto \sqrt{T}$
$∴ \frac{2 v}{v} = \sqrt{\frac{T_{2}}{T_{1}}}$
$\Rightarrow 2 = \sqrt{\frac{T_{2}}{T_{1}}}$
$\Rightarrow 4 = \frac{T_{2}}{T_{1}}$
$\Rightarrow T_{2} = 4 T_{1} = 4 \times 273 = 1092 K$
$∴ T_{2} = 1092 - 273 = 819^{\circ} C$

PHXI15:WAVES

354649 Velocity of sound in air at the given temperature

1 Decreases with increase in pressure

2 May increase on decrease with pressure

3 Is independent of the variation in pressure

4 Varies directly as the square root of pressure

PHXI15:WAVES

354650 A pressure of $100 k P a$ causes a decrease in volume of water by $5 \times 10^{- 3}$ percent. The speed of sound in water is

1 1414

2 1000

3 2000

4 3000

Explanation:

$v = \sqrt{\frac{B}{δ}} = \sqrt{\frac{d P}{(\frac{d v}{v}) δ}}$
$= \sqrt{\frac{100 \times 10^{3}}{5 \times 10^{- 5} \times 10^{3}}}$
$= \sqrt{\frac{1}{5} \times 10^{7}} = \sqrt{2} \times 10^{3} = 1414 m / s$

PHXI15:WAVES

354651 A steel rod $100 c m$ long is clamped at the middle. The fundamental frequency of longitudinal vibrations of the rod are given to be $2.53 k H z$ . What is the speed of sound in steel?

1

6.2 k m s^{- 1}

2

5.06 k m s^{- 1}

3

7.23 k m s^{- 1}

4

7.45 k m s^{- 1}

Explanation:

In fundamental mode,
$l = 2 (\frac{λ}{4}) = \frac{λ}{2}$
supporting img
$\Rightarrow λ = 2 l (1)$
Given, $l = 100 c m$ ,
$v = 2.53 k H z = 2.53 \times 10^{3} H z$
We know that,
$v = v λ$
$= v \times 2 l [$ From Eq. (1) ]
$= 2.53 \times 10^{3} \times 2 \times 100 \times 10^{- 2}$
$= 5.06 \times 10^{3} m s^{- 1}$
$= 5.06 k m s^{- 1}$

AIIMS - 2018

PHXI15:WAVES

354648 The temperature at which the speed of sound in air becomes doubled of its value at $0^{\circ} C$ is

1

1092^{\circ} C

2

819 K

3

819^{\circ} C

4

546^{\circ} C

Explanation:

Speed, $v \propto \sqrt{T}$
$∴ \frac{2 v}{v} = \sqrt{\frac{T_{2}}{T_{1}}}$
$\Rightarrow 2 = \sqrt{\frac{T_{2}}{T_{1}}}$
$\Rightarrow 4 = \frac{T_{2}}{T_{1}}$
$\Rightarrow T_{2} = 4 T_{1} = 4 \times 273 = 1092 K$
$∴ T_{2} = 1092 - 273 = 819^{\circ} C$

PHXI15:WAVES

354649 Velocity of sound in air at the given temperature

1 Decreases with increase in pressure

2 May increase on decrease with pressure

3 Is independent of the variation in pressure

4 Varies directly as the square root of pressure

PHXI15:WAVES

354650 A pressure of $100 k P a$ causes a decrease in volume of water by $5 \times 10^{- 3}$ percent. The speed of sound in water is

1 1414

2 1000

3 2000

4 3000

Explanation:

$v = \sqrt{\frac{B}{δ}} = \sqrt{\frac{d P}{(\frac{d v}{v}) δ}}$
$= \sqrt{\frac{100 \times 10^{3}}{5 \times 10^{- 5} \times 10^{3}}}$
$= \sqrt{\frac{1}{5} \times 10^{7}} = \sqrt{2} \times 10^{3} = 1414 m / s$

PHXI15:WAVES

354651 A steel rod $100 c m$ long is clamped at the middle. The fundamental frequency of longitudinal vibrations of the rod are given to be $2.53 k H z$ . What is the speed of sound in steel?

1

6.2 k m s^{- 1}

2

5.06 k m s^{- 1}

3

7.23 k m s^{- 1}

4

7.45 k m s^{- 1}

Explanation:

In fundamental mode,
$l = 2 (\frac{λ}{4}) = \frac{λ}{2}$
supporting img
$\Rightarrow λ = 2 l (1)$
Given, $l = 100 c m$ ,
$v = 2.53 k H z = 2.53 \times 10^{3} H z$
We know that,
$v = v λ$
$= v \times 2 l [$ From Eq. (1) ]
$= 2.53 \times 10^{3} \times 2 \times 100 \times 10^{- 2}$
$= 5.06 \times 10^{3} m s^{- 1}$
$= 5.06 k m s^{- 1}$

AIIMS - 2018

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