155687
Assertion: FM broadcast is better than AM broadcast. Reason: Noise change is maximum in amplitude of $A M$ waves.
1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
A Amplitude modulation (AM) and frequency modulation (FM) are ways of broadcasting radio signals. Both transmit the information in the form of electromagnetic waves. - AM is more susceptible to noise because noise affects amplitude while FM is less susceptible to noise because information in a FM signal is transmitted through varying the frequency, and not the amplitude. So assertion and reason are correct and reason is the correct explanation of assertion.
AIIMS-25.05.2019(M) Shift-1
Electromagnetic Wave
155706
A wave of wavelength $5900 \AA$ emitted by any atom or molecule must have some finite total length which is known as coherence length for sodium light, this length is $2.4 \mathrm{~cm}$. The number of oscillations in this length will be
1 $4.068 \times 10^{8}$
2 $4.068 \times 10^{4}$
3 $4.068 \times 10^{6}$
4 $4.068 \times 10^{5}$
Explanation:
B Given, length $(\mathrm{L})=2.4 \mathrm{~cm}$ or $0.024 \mathrm{~m}$ wavelength $(\lambda)=5900 \AA=5900 \AA \times 10^{-10} \mathrm{~m}$ For the coherence length, Number of oscillation $(n)=\frac{L}{\lambda}$ $\mathrm{n}=\frac{0.024 \mathrm{~m}}{5900 \times 10^{-10} \mathrm{~m}}$ $\mathrm{n}=4.068 \times 10^{4}$
BCECE-2017
Electromagnetic Wave
155708
If levels 1 and 2 are separated by an energy $E_{2^{-}}$ $E_{1}$, such that the corresponding transition frequency falls in the middle of the visible range, calculate the ratio of the populations of two levels in the thermal equilibrium at room temperature.
1 $1.1577 \times 10^{-38}$
2 $2.9 \times 10^{-35}$
3 $2.168 \times 10^{-36}$
4 $1.96 \times 10^{-20}$
Explanation:
A The middle of the visible range $(\lambda)=550 \mathrm{~nm}$ $E_{2}-E_{1}=\frac{h c}{\lambda}$ $=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{550 \times 10^{-9}}$ $=3.6 \times 10^{-19} \mathrm{~J}$ We know that, At thermal equilibrium, ratio of $\frac{\mathrm{N}_{2}}{\mathrm{~N}_{1}}$ is given as- $\frac{\mathrm{N}_{2}}{\mathrm{~N}_{1}}=\exp \left(\frac{\mathrm{E}_{2}-\mathrm{E}_{1}}{\mathrm{KT}}\right)\left(\text { Here we take } \mathrm{T}=27^{\circ}\right.$ room temperature so, in Kelvin $27+273=300$ ) $\frac{\mathrm{N}_{2}}{\mathrm{~N}_{1}}=\exp \left[\frac{-3.6 \times 10^{-19}}{\left(1.38 \times 10^{-23}\right)(300)}\right]$ $\frac{\mathrm{N}_{2}}{\mathrm{~N}_{1}}=\exp (-86.95)$ $\frac{\mathrm{N}_{2}}{\mathrm{~N}_{1}}=1.72 \times 10^{-38} \approx 1.1577 \times 10^{-38}$
Manipal UGET-2015
Electromagnetic Wave
155714
Which of the following electromagnetic radiations has the smallest wavelength?
1 Ultraviolet rays
2 X-rays
3 $\gamma$-rays
4 Microwaves
Explanation:
C | Name | Spectral range | | :--- | :--- | | $\gamma$-rays | $ \lt 0.03$ | | x-rays | $0.03-300$ | | Ultraviolet | $300-380$ | | Blue | $400-500$ | | Green | $500-600$ | | Red | $600-700$ | | Infrared | $700-1300$ | | micro wave | $1 \mathrm{~mm}-30 \mathrm{~cm}$ |
155687
Assertion: FM broadcast is better than AM broadcast. Reason: Noise change is maximum in amplitude of $A M$ waves.
1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
A Amplitude modulation (AM) and frequency modulation (FM) are ways of broadcasting radio signals. Both transmit the information in the form of electromagnetic waves. - AM is more susceptible to noise because noise affects amplitude while FM is less susceptible to noise because information in a FM signal is transmitted through varying the frequency, and not the amplitude. So assertion and reason are correct and reason is the correct explanation of assertion.
AIIMS-25.05.2019(M) Shift-1
Electromagnetic Wave
155706
A wave of wavelength $5900 \AA$ emitted by any atom or molecule must have some finite total length which is known as coherence length for sodium light, this length is $2.4 \mathrm{~cm}$. The number of oscillations in this length will be
1 $4.068 \times 10^{8}$
2 $4.068 \times 10^{4}$
3 $4.068 \times 10^{6}$
4 $4.068 \times 10^{5}$
Explanation:
B Given, length $(\mathrm{L})=2.4 \mathrm{~cm}$ or $0.024 \mathrm{~m}$ wavelength $(\lambda)=5900 \AA=5900 \AA \times 10^{-10} \mathrm{~m}$ For the coherence length, Number of oscillation $(n)=\frac{L}{\lambda}$ $\mathrm{n}=\frac{0.024 \mathrm{~m}}{5900 \times 10^{-10} \mathrm{~m}}$ $\mathrm{n}=4.068 \times 10^{4}$
BCECE-2017
Electromagnetic Wave
155708
If levels 1 and 2 are separated by an energy $E_{2^{-}}$ $E_{1}$, such that the corresponding transition frequency falls in the middle of the visible range, calculate the ratio of the populations of two levels in the thermal equilibrium at room temperature.
1 $1.1577 \times 10^{-38}$
2 $2.9 \times 10^{-35}$
3 $2.168 \times 10^{-36}$
4 $1.96 \times 10^{-20}$
Explanation:
A The middle of the visible range $(\lambda)=550 \mathrm{~nm}$ $E_{2}-E_{1}=\frac{h c}{\lambda}$ $=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{550 \times 10^{-9}}$ $=3.6 \times 10^{-19} \mathrm{~J}$ We know that, At thermal equilibrium, ratio of $\frac{\mathrm{N}_{2}}{\mathrm{~N}_{1}}$ is given as- $\frac{\mathrm{N}_{2}}{\mathrm{~N}_{1}}=\exp \left(\frac{\mathrm{E}_{2}-\mathrm{E}_{1}}{\mathrm{KT}}\right)\left(\text { Here we take } \mathrm{T}=27^{\circ}\right.$ room temperature so, in Kelvin $27+273=300$ ) $\frac{\mathrm{N}_{2}}{\mathrm{~N}_{1}}=\exp \left[\frac{-3.6 \times 10^{-19}}{\left(1.38 \times 10^{-23}\right)(300)}\right]$ $\frac{\mathrm{N}_{2}}{\mathrm{~N}_{1}}=\exp (-86.95)$ $\frac{\mathrm{N}_{2}}{\mathrm{~N}_{1}}=1.72 \times 10^{-38} \approx 1.1577 \times 10^{-38}$
Manipal UGET-2015
Electromagnetic Wave
155714
Which of the following electromagnetic radiations has the smallest wavelength?
1 Ultraviolet rays
2 X-rays
3 $\gamma$-rays
4 Microwaves
Explanation:
C | Name | Spectral range | | :--- | :--- | | $\gamma$-rays | $ \lt 0.03$ | | x-rays | $0.03-300$ | | Ultraviolet | $300-380$ | | Blue | $400-500$ | | Green | $500-600$ | | Red | $600-700$ | | Infrared | $700-1300$ | | micro wave | $1 \mathrm{~mm}-30 \mathrm{~cm}$ |
155687
Assertion: FM broadcast is better than AM broadcast. Reason: Noise change is maximum in amplitude of $A M$ waves.
1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
A Amplitude modulation (AM) and frequency modulation (FM) are ways of broadcasting radio signals. Both transmit the information in the form of electromagnetic waves. - AM is more susceptible to noise because noise affects amplitude while FM is less susceptible to noise because information in a FM signal is transmitted through varying the frequency, and not the amplitude. So assertion and reason are correct and reason is the correct explanation of assertion.
AIIMS-25.05.2019(M) Shift-1
Electromagnetic Wave
155706
A wave of wavelength $5900 \AA$ emitted by any atom or molecule must have some finite total length which is known as coherence length for sodium light, this length is $2.4 \mathrm{~cm}$. The number of oscillations in this length will be
1 $4.068 \times 10^{8}$
2 $4.068 \times 10^{4}$
3 $4.068 \times 10^{6}$
4 $4.068 \times 10^{5}$
Explanation:
B Given, length $(\mathrm{L})=2.4 \mathrm{~cm}$ or $0.024 \mathrm{~m}$ wavelength $(\lambda)=5900 \AA=5900 \AA \times 10^{-10} \mathrm{~m}$ For the coherence length, Number of oscillation $(n)=\frac{L}{\lambda}$ $\mathrm{n}=\frac{0.024 \mathrm{~m}}{5900 \times 10^{-10} \mathrm{~m}}$ $\mathrm{n}=4.068 \times 10^{4}$
BCECE-2017
Electromagnetic Wave
155708
If levels 1 and 2 are separated by an energy $E_{2^{-}}$ $E_{1}$, such that the corresponding transition frequency falls in the middle of the visible range, calculate the ratio of the populations of two levels in the thermal equilibrium at room temperature.
1 $1.1577 \times 10^{-38}$
2 $2.9 \times 10^{-35}$
3 $2.168 \times 10^{-36}$
4 $1.96 \times 10^{-20}$
Explanation:
A The middle of the visible range $(\lambda)=550 \mathrm{~nm}$ $E_{2}-E_{1}=\frac{h c}{\lambda}$ $=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{550 \times 10^{-9}}$ $=3.6 \times 10^{-19} \mathrm{~J}$ We know that, At thermal equilibrium, ratio of $\frac{\mathrm{N}_{2}}{\mathrm{~N}_{1}}$ is given as- $\frac{\mathrm{N}_{2}}{\mathrm{~N}_{1}}=\exp \left(\frac{\mathrm{E}_{2}-\mathrm{E}_{1}}{\mathrm{KT}}\right)\left(\text { Here we take } \mathrm{T}=27^{\circ}\right.$ room temperature so, in Kelvin $27+273=300$ ) $\frac{\mathrm{N}_{2}}{\mathrm{~N}_{1}}=\exp \left[\frac{-3.6 \times 10^{-19}}{\left(1.38 \times 10^{-23}\right)(300)}\right]$ $\frac{\mathrm{N}_{2}}{\mathrm{~N}_{1}}=\exp (-86.95)$ $\frac{\mathrm{N}_{2}}{\mathrm{~N}_{1}}=1.72 \times 10^{-38} \approx 1.1577 \times 10^{-38}$
Manipal UGET-2015
Electromagnetic Wave
155714
Which of the following electromagnetic radiations has the smallest wavelength?
1 Ultraviolet rays
2 X-rays
3 $\gamma$-rays
4 Microwaves
Explanation:
C | Name | Spectral range | | :--- | :--- | | $\gamma$-rays | $ \lt 0.03$ | | x-rays | $0.03-300$ | | Ultraviolet | $300-380$ | | Blue | $400-500$ | | Green | $500-600$ | | Red | $600-700$ | | Infrared | $700-1300$ | | micro wave | $1 \mathrm{~mm}-30 \mathrm{~cm}$ |
NEET Test Series from KOTA - 10 Papers In MS WORD
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Electromagnetic Wave
155687
Assertion: FM broadcast is better than AM broadcast. Reason: Noise change is maximum in amplitude of $A M$ waves.
1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
A Amplitude modulation (AM) and frequency modulation (FM) are ways of broadcasting radio signals. Both transmit the information in the form of electromagnetic waves. - AM is more susceptible to noise because noise affects amplitude while FM is less susceptible to noise because information in a FM signal is transmitted through varying the frequency, and not the amplitude. So assertion and reason are correct and reason is the correct explanation of assertion.
AIIMS-25.05.2019(M) Shift-1
Electromagnetic Wave
155706
A wave of wavelength $5900 \AA$ emitted by any atom or molecule must have some finite total length which is known as coherence length for sodium light, this length is $2.4 \mathrm{~cm}$. The number of oscillations in this length will be
1 $4.068 \times 10^{8}$
2 $4.068 \times 10^{4}$
3 $4.068 \times 10^{6}$
4 $4.068 \times 10^{5}$
Explanation:
B Given, length $(\mathrm{L})=2.4 \mathrm{~cm}$ or $0.024 \mathrm{~m}$ wavelength $(\lambda)=5900 \AA=5900 \AA \times 10^{-10} \mathrm{~m}$ For the coherence length, Number of oscillation $(n)=\frac{L}{\lambda}$ $\mathrm{n}=\frac{0.024 \mathrm{~m}}{5900 \times 10^{-10} \mathrm{~m}}$ $\mathrm{n}=4.068 \times 10^{4}$
BCECE-2017
Electromagnetic Wave
155708
If levels 1 and 2 are separated by an energy $E_{2^{-}}$ $E_{1}$, such that the corresponding transition frequency falls in the middle of the visible range, calculate the ratio of the populations of two levels in the thermal equilibrium at room temperature.
1 $1.1577 \times 10^{-38}$
2 $2.9 \times 10^{-35}$
3 $2.168 \times 10^{-36}$
4 $1.96 \times 10^{-20}$
Explanation:
A The middle of the visible range $(\lambda)=550 \mathrm{~nm}$ $E_{2}-E_{1}=\frac{h c}{\lambda}$ $=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{550 \times 10^{-9}}$ $=3.6 \times 10^{-19} \mathrm{~J}$ We know that, At thermal equilibrium, ratio of $\frac{\mathrm{N}_{2}}{\mathrm{~N}_{1}}$ is given as- $\frac{\mathrm{N}_{2}}{\mathrm{~N}_{1}}=\exp \left(\frac{\mathrm{E}_{2}-\mathrm{E}_{1}}{\mathrm{KT}}\right)\left(\text { Here we take } \mathrm{T}=27^{\circ}\right.$ room temperature so, in Kelvin $27+273=300$ ) $\frac{\mathrm{N}_{2}}{\mathrm{~N}_{1}}=\exp \left[\frac{-3.6 \times 10^{-19}}{\left(1.38 \times 10^{-23}\right)(300)}\right]$ $\frac{\mathrm{N}_{2}}{\mathrm{~N}_{1}}=\exp (-86.95)$ $\frac{\mathrm{N}_{2}}{\mathrm{~N}_{1}}=1.72 \times 10^{-38} \approx 1.1577 \times 10^{-38}$
Manipal UGET-2015
Electromagnetic Wave
155714
Which of the following electromagnetic radiations has the smallest wavelength?
1 Ultraviolet rays
2 X-rays
3 $\gamma$-rays
4 Microwaves
Explanation:
C | Name | Spectral range | | :--- | :--- | | $\gamma$-rays | $ \lt 0.03$ | | x-rays | $0.03-300$ | | Ultraviolet | $300-380$ | | Blue | $400-500$ | | Green | $500-600$ | | Red | $600-700$ | | Infrared | $700-1300$ | | micro wave | $1 \mathrm{~mm}-30 \mathrm{~cm}$ |
