NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI15:WAVES
354621
Assertion : Sound waves cannot propagate through vacuum but light waves can. Reason : Sound waves cannot be polarised but light waves can be polarised.
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:
"Sound waves cannot travel through a vacuum because they are mechanical waves. Light waves can propagate through a vacuum because they are electromagnetic waves. Since sound waves are longitudinal waves, with particles moving in the direction of propagation, they cannot be polarized." So correct option is (2).
PHXI15:WAVES
354622
Sound waves transfer
1 Only energy not momentum
2 Energy
3 Momentum
4 Both energy and momentum
Explanation:
Conceptual Question
PHXI15:WAVES
354623
Assertion : Two persons on the surface of moon cannot talk to each other. Reason : There is no atmosphere on moon.
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:
Sound waves require material medium to travel. As there is no atmosphere (Vacuum) on the surface of moon, therefore the sound waves cannot reach from one person to another.
PHXI15:WAVES
354624
Velocity of sound waves in air is \(330\;m{s^{ - 1}}.\) For a particular sound in air, a path difference of \(40\;cm\) is equivalent to a phase difference of \(1.6 \pi\). The frequency of the wave is
1 \(165\;Hz\)
2 \(150\;Hz\)
3 \(660\;Hz\)
4 \(330\;Hz\)
Explanation:
Phase difference of \(1.6 \pi\) corresponds to path difference of \(40\;cm.\) Hence, phase difference of \(2 \pi\) will correspond to a path difference of \(50\;cm,\) \(i.\,e.\) \(\lambda = 50\;cm\) or \(0.5\;m.\) \(\therefore \quad n = \frac{v}{\lambda } = \frac{{330}}{{0.5}} = 660\;Hz\)
354621
Assertion : Sound waves cannot propagate through vacuum but light waves can. Reason : Sound waves cannot be polarised but light waves can be polarised.
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:
"Sound waves cannot travel through a vacuum because they are mechanical waves. Light waves can propagate through a vacuum because they are electromagnetic waves. Since sound waves are longitudinal waves, with particles moving in the direction of propagation, they cannot be polarized." So correct option is (2).
PHXI15:WAVES
354622
Sound waves transfer
1 Only energy not momentum
2 Energy
3 Momentum
4 Both energy and momentum
Explanation:
Conceptual Question
PHXI15:WAVES
354623
Assertion : Two persons on the surface of moon cannot talk to each other. Reason : There is no atmosphere on moon.
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:
Sound waves require material medium to travel. As there is no atmosphere (Vacuum) on the surface of moon, therefore the sound waves cannot reach from one person to another.
PHXI15:WAVES
354624
Velocity of sound waves in air is \(330\;m{s^{ - 1}}.\) For a particular sound in air, a path difference of \(40\;cm\) is equivalent to a phase difference of \(1.6 \pi\). The frequency of the wave is
1 \(165\;Hz\)
2 \(150\;Hz\)
3 \(660\;Hz\)
4 \(330\;Hz\)
Explanation:
Phase difference of \(1.6 \pi\) corresponds to path difference of \(40\;cm.\) Hence, phase difference of \(2 \pi\) will correspond to a path difference of \(50\;cm,\) \(i.\,e.\) \(\lambda = 50\;cm\) or \(0.5\;m.\) \(\therefore \quad n = \frac{v}{\lambda } = \frac{{330}}{{0.5}} = 660\;Hz\)
354621
Assertion : Sound waves cannot propagate through vacuum but light waves can. Reason : Sound waves cannot be polarised but light waves can be polarised.
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:
"Sound waves cannot travel through a vacuum because they are mechanical waves. Light waves can propagate through a vacuum because they are electromagnetic waves. Since sound waves are longitudinal waves, with particles moving in the direction of propagation, they cannot be polarized." So correct option is (2).
PHXI15:WAVES
354622
Sound waves transfer
1 Only energy not momentum
2 Energy
3 Momentum
4 Both energy and momentum
Explanation:
Conceptual Question
PHXI15:WAVES
354623
Assertion : Two persons on the surface of moon cannot talk to each other. Reason : There is no atmosphere on moon.
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:
Sound waves require material medium to travel. As there is no atmosphere (Vacuum) on the surface of moon, therefore the sound waves cannot reach from one person to another.
PHXI15:WAVES
354624
Velocity of sound waves in air is \(330\;m{s^{ - 1}}.\) For a particular sound in air, a path difference of \(40\;cm\) is equivalent to a phase difference of \(1.6 \pi\). The frequency of the wave is
1 \(165\;Hz\)
2 \(150\;Hz\)
3 \(660\;Hz\)
4 \(330\;Hz\)
Explanation:
Phase difference of \(1.6 \pi\) corresponds to path difference of \(40\;cm.\) Hence, phase difference of \(2 \pi\) will correspond to a path difference of \(50\;cm,\) \(i.\,e.\) \(\lambda = 50\;cm\) or \(0.5\;m.\) \(\therefore \quad n = \frac{v}{\lambda } = \frac{{330}}{{0.5}} = 660\;Hz\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
PHXI15:WAVES
354621
Assertion : Sound waves cannot propagate through vacuum but light waves can. Reason : Sound waves cannot be polarised but light waves can be polarised.
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:
"Sound waves cannot travel through a vacuum because they are mechanical waves. Light waves can propagate through a vacuum because they are electromagnetic waves. Since sound waves are longitudinal waves, with particles moving in the direction of propagation, they cannot be polarized." So correct option is (2).
PHXI15:WAVES
354622
Sound waves transfer
1 Only energy not momentum
2 Energy
3 Momentum
4 Both energy and momentum
Explanation:
Conceptual Question
PHXI15:WAVES
354623
Assertion : Two persons on the surface of moon cannot talk to each other. Reason : There is no atmosphere on moon.
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:
Sound waves require material medium to travel. As there is no atmosphere (Vacuum) on the surface of moon, therefore the sound waves cannot reach from one person to another.
PHXI15:WAVES
354624
Velocity of sound waves in air is \(330\;m{s^{ - 1}}.\) For a particular sound in air, a path difference of \(40\;cm\) is equivalent to a phase difference of \(1.6 \pi\). The frequency of the wave is
1 \(165\;Hz\)
2 \(150\;Hz\)
3 \(660\;Hz\)
4 \(330\;Hz\)
Explanation:
Phase difference of \(1.6 \pi\) corresponds to path difference of \(40\;cm.\) Hence, phase difference of \(2 \pi\) will correspond to a path difference of \(50\;cm,\) \(i.\,e.\) \(\lambda = 50\;cm\) or \(0.5\;m.\) \(\therefore \quad n = \frac{v}{\lambda } = \frac{{330}}{{0.5}} = 660\;Hz\)