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PHXI15:WAVES

354652 An earthquake generates both transverse ( $s$ ) and longitudinal ( $p$ ) sound waves in the earth. The speed of ( $s$ ) waves is about $4.5 k m / s$ and that of ( $p$ ) waves is about $8.0 k m / s$ . A seismograph records $P$ and $S$ wave in a gap of 4 minutes. The epicentre of the earthquake is located at a distance of about:

1

250 k m

2

25 k m

3

5000 k m

4

2500 k m

Explanation:

Distance travelled by both the waves is same.
Let the time taken by the $S$ and $P$ waves to reach the seismograph be $t_{1}$ and $t_{2}$ then
$t_{1} - t_{2} = 4 min$
$∴ t_{1} - t_{2} = 240 s (1)$
Let distance of epicentre be $s$ . Then
$s = v_{1} t_{1} = v_{2} t_{2}$
$\Rightarrow 4.5 \times t_{1} = 8 t_{2} (2)$
From eq's (1) and (2)
$\Rightarrow t_{1} = \frac{240 \times 8}{3.5} = 548.58 s$
$∴ s = v_{1} t_{1} = 4.5 \times 548.5 = 2500 k m$

PHXI15:WAVES

354653 Assertion :
The error in Newton's formula of velocity of sound in air was $16 %$ .
Reason :
The experimental value of velocity of sound in air was accurate.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

"The experimental velocity of sound in air is measured at $332 m / s$ , while the value calculated using Newton's formula is $280 m / s$ . The discrepancy between the experimental and theoretical values is $52 m / s$ , resulting in a percentage error". So correct option is (2).

PHXI15:WAVES

354654 Speed of sound in a solid is directly proportional to ( $Y$ - Young's modulus, $ρ$ - density)

1

\sqrt{Y}

2

\frac{1}{\sqrt{ρ}}

3 Both (1) and (2)

4 None of the above

PHXI15:WAVES

354655 The ratio of speed of sound in hydrogen gas to the speed of sound in oxygen gas at the same temperature is

1

1 : 4

2

1 : 2

3

1 : 1

4

4 : 1

Explanation:

Molecular mass of oxygen $(M_{o}) = 32$
Molecular mass of hydrogen $(M_{H}) = 2$
The speed of sound in hydrogen is
Where $η = \frac{C_{p}}{C_{v}} =$ Adiabatic index of diatomic gas.
$v_{H} = \sqrt{\frac{η R T}{M_{H}}} (1)$
The speed of sound in oxygen is
$v_{O} = \sqrt{\frac{η R T}{M_{O_{2}}}} (2)$
On dividing equation (1) by (2), we get
$\frac{v_{H}}{v_{O}} = \sqrt{\frac{η R T / M_{H}}{η R T / M_{O_{2}}}} = \sqrt{\frac{M_{O_{2}}}{M_{H}}} = \sqrt{\frac{32}{2}} = \frac{4}{1}$

JEE - 2023

PHXI15:WAVES

354652 An earthquake generates both transverse ( $s$ ) and longitudinal ( $p$ ) sound waves in the earth. The speed of ( $s$ ) waves is about $4.5 k m / s$ and that of ( $p$ ) waves is about $8.0 k m / s$ . A seismograph records $P$ and $S$ wave in a gap of 4 minutes. The epicentre of the earthquake is located at a distance of about:

1

250 k m

2

25 k m

3

5000 k m

4

2500 k m

Explanation:

Distance travelled by both the waves is same.
Let the time taken by the $S$ and $P$ waves to reach the seismograph be $t_{1}$ and $t_{2}$ then
$t_{1} - t_{2} = 4 min$
$∴ t_{1} - t_{2} = 240 s (1)$
Let distance of epicentre be $s$ . Then
$s = v_{1} t_{1} = v_{2} t_{2}$
$\Rightarrow 4.5 \times t_{1} = 8 t_{2} (2)$
From eq's (1) and (2)
$\Rightarrow t_{1} = \frac{240 \times 8}{3.5} = 548.58 s$
$∴ s = v_{1} t_{1} = 4.5 \times 548.5 = 2500 k m$

PHXI15:WAVES

354653 Assertion :
The error in Newton's formula of velocity of sound in air was $16 %$ .
Reason :
The experimental value of velocity of sound in air was accurate.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

"The experimental velocity of sound in air is measured at $332 m / s$ , while the value calculated using Newton's formula is $280 m / s$ . The discrepancy between the experimental and theoretical values is $52 m / s$ , resulting in a percentage error". So correct option is (2).

PHXI15:WAVES

354654 Speed of sound in a solid is directly proportional to ( $Y$ - Young's modulus, $ρ$ - density)

1

\sqrt{Y}

2

\frac{1}{\sqrt{ρ}}

3 Both (1) and (2)

4 None of the above

PHXI15:WAVES

354655 The ratio of speed of sound in hydrogen gas to the speed of sound in oxygen gas at the same temperature is

1

1 : 4

2

1 : 2

3

1 : 1

4

4 : 1

Explanation:

Molecular mass of oxygen $(M_{o}) = 32$
Molecular mass of hydrogen $(M_{H}) = 2$
The speed of sound in hydrogen is
Where $η = \frac{C_{p}}{C_{v}} =$ Adiabatic index of diatomic gas.
$v_{H} = \sqrt{\frac{η R T}{M_{H}}} (1)$
The speed of sound in oxygen is
$v_{O} = \sqrt{\frac{η R T}{M_{O_{2}}}} (2)$
On dividing equation (1) by (2), we get
$\frac{v_{H}}{v_{O}} = \sqrt{\frac{η R T / M_{H}}{η R T / M_{O_{2}}}} = \sqrt{\frac{M_{O_{2}}}{M_{H}}} = \sqrt{\frac{32}{2}} = \frac{4}{1}$

JEE - 2023

PHXI15:WAVES

354652 An earthquake generates both transverse ( $s$ ) and longitudinal ( $p$ ) sound waves in the earth. The speed of ( $s$ ) waves is about $4.5 k m / s$ and that of ( $p$ ) waves is about $8.0 k m / s$ . A seismograph records $P$ and $S$ wave in a gap of 4 minutes. The epicentre of the earthquake is located at a distance of about:

1

250 k m

2

25 k m

3

5000 k m

4

2500 k m

Explanation:

Distance travelled by both the waves is same.
Let the time taken by the $S$ and $P$ waves to reach the seismograph be $t_{1}$ and $t_{2}$ then
$t_{1} - t_{2} = 4 min$
$∴ t_{1} - t_{2} = 240 s (1)$
Let distance of epicentre be $s$ . Then
$s = v_{1} t_{1} = v_{2} t_{2}$
$\Rightarrow 4.5 \times t_{1} = 8 t_{2} (2)$
From eq's (1) and (2)
$\Rightarrow t_{1} = \frac{240 \times 8}{3.5} = 548.58 s$
$∴ s = v_{1} t_{1} = 4.5 \times 548.5 = 2500 k m$

PHXI15:WAVES

354653 Assertion :
The error in Newton's formula of velocity of sound in air was $16 %$ .
Reason :
The experimental value of velocity of sound in air was accurate.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

"The experimental velocity of sound in air is measured at $332 m / s$ , while the value calculated using Newton's formula is $280 m / s$ . The discrepancy between the experimental and theoretical values is $52 m / s$ , resulting in a percentage error". So correct option is (2).

PHXI15:WAVES

354654 Speed of sound in a solid is directly proportional to ( $Y$ - Young's modulus, $ρ$ - density)

1

\sqrt{Y}

2

\frac{1}{\sqrt{ρ}}

3 Both (1) and (2)

4 None of the above

PHXI15:WAVES

354655 The ratio of speed of sound in hydrogen gas to the speed of sound in oxygen gas at the same temperature is

1

1 : 4

2

1 : 2

3

1 : 1

4

4 : 1

Explanation:

Molecular mass of oxygen $(M_{o}) = 32$
Molecular mass of hydrogen $(M_{H}) = 2$
The speed of sound in hydrogen is
Where $η = \frac{C_{p}}{C_{v}} =$ Adiabatic index of diatomic gas.
$v_{H} = \sqrt{\frac{η R T}{M_{H}}} (1)$
The speed of sound in oxygen is
$v_{O} = \sqrt{\frac{η R T}{M_{O_{2}}}} (2)$
On dividing equation (1) by (2), we get
$\frac{v_{H}}{v_{O}} = \sqrt{\frac{η R T / M_{H}}{η R T / M_{O_{2}}}} = \sqrt{\frac{M_{O_{2}}}{M_{H}}} = \sqrt{\frac{32}{2}} = \frac{4}{1}$

JEE - 2023

PHXI15:WAVES

354652 An earthquake generates both transverse ( $s$ ) and longitudinal ( $p$ ) sound waves in the earth. The speed of ( $s$ ) waves is about $4.5 k m / s$ and that of ( $p$ ) waves is about $8.0 k m / s$ . A seismograph records $P$ and $S$ wave in a gap of 4 minutes. The epicentre of the earthquake is located at a distance of about:

1

250 k m

2

25 k m

3

5000 k m

4

2500 k m

Explanation:

Distance travelled by both the waves is same.
Let the time taken by the $S$ and $P$ waves to reach the seismograph be $t_{1}$ and $t_{2}$ then
$t_{1} - t_{2} = 4 min$
$∴ t_{1} - t_{2} = 240 s (1)$
Let distance of epicentre be $s$ . Then
$s = v_{1} t_{1} = v_{2} t_{2}$
$\Rightarrow 4.5 \times t_{1} = 8 t_{2} (2)$
From eq's (1) and (2)
$\Rightarrow t_{1} = \frac{240 \times 8}{3.5} = 548.58 s$
$∴ s = v_{1} t_{1} = 4.5 \times 548.5 = 2500 k m$

PHXI15:WAVES

354653 Assertion :
The error in Newton's formula of velocity of sound in air was $16 %$ .
Reason :
The experimental value of velocity of sound in air was accurate.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

"The experimental velocity of sound in air is measured at $332 m / s$ , while the value calculated using Newton's formula is $280 m / s$ . The discrepancy between the experimental and theoretical values is $52 m / s$ , resulting in a percentage error". So correct option is (2).

PHXI15:WAVES

354654 Speed of sound in a solid is directly proportional to ( $Y$ - Young's modulus, $ρ$ - density)

1

\sqrt{Y}

2

\frac{1}{\sqrt{ρ}}

3 Both (1) and (2)

4 None of the above

PHXI15:WAVES

354655 The ratio of speed of sound in hydrogen gas to the speed of sound in oxygen gas at the same temperature is

1

1 : 4

2

1 : 2

3

1 : 1

4

4 : 1

Explanation:

Molecular mass of oxygen $(M_{o}) = 32$
Molecular mass of hydrogen $(M_{H}) = 2$
The speed of sound in hydrogen is
Where $η = \frac{C_{p}}{C_{v}} =$ Adiabatic index of diatomic gas.
$v_{H} = \sqrt{\frac{η R T}{M_{H}}} (1)$
The speed of sound in oxygen is
$v_{O} = \sqrt{\frac{η R T}{M_{O_{2}}}} (2)$
On dividing equation (1) by (2), we get
$\frac{v_{H}}{v_{O}} = \sqrt{\frac{η R T / M_{H}}{η R T / M_{O_{2}}}} = \sqrt{\frac{M_{O_{2}}}{M_{H}}} = \sqrt{\frac{32}{2}} = \frac{4}{1}$

JEE - 2023