358865
Light with an energy flux of \(20\;\,W/c{m^2}\) falls on a non-reflecting surface at normal incidence. If the surface has an area of \(30\;c{m^2},\) the total momentum delivered (for complete absorption) during \(30\;\min \) is
1 \(1.08 \times {10^7}\;kg - m/s\)
2 \(108 \times {10^4}\;kg - m/s\)
3 \(36 \times {10^{ - 4}}\;kg - m/s\)
4 \(36 \times {10^{ - 5}}\;kg - m/s\)
Explanation:
\(F=P_{\text {rad }} \times\) Area \( = \frac{I}{c} \times {\rm{ Area }}\) \( = \frac{{20}}{{3 \times {{10}^8}}} \times 30 = 2 \times {10^{ - 6}}\;N\) \(=\) Change in momentum in \(1 \mathrm{sec}\) \(\therefore\) Momentum delivered to the surface in \(30 \mathrm{~min}\) \( = 2 \times {10^{ - 6}} \times 30 \times 60 = 36 \times {10^{ - 4}}\;kg\;m\;{s^{ - 1}}\)
NCERT Exempler
PHXI15:WAVES
358866
A small object at rest, absorbs a light pulse of power \(20\;mW\) and duration \(300\;ns\). Assuming speed of light as \(3 \times {10^8}\;m/s\), the momentum of the object becomes equal to
1 \(2 \times {10^{ - 17}}\;kg\;m/s\)
2 \(3 \times {10^{ - 17}}\;kg\;m/s\)
3 \(0.5 \times {10^{ - 17}}\;kg\;m/s\)
4 \(1 \times {10^{ - 17}}\;kg\;m/s\)
Explanation:
\(U\) ( energy \()=\) Power \(\times\) time \( = \left( {20 \times {{10}^{ - 3}}} \right) \times \left( {300 \times {{10}^{ - 9}}\;s} \right)\) \( = 6 \times {10^{ - 9}}{\rm{ joules}}\) Momentum gained by the object is, \(p=\dfrac{U}{c}=\dfrac{6 \times 10^{-9}}{3 \times 10^{8}}\)\(=2 \times 10^{-17} {~kg} {~ms}^{-1}\)
JEE - 2023
PHXI15:WAVES
358867
A plate of mass 10 \(g\) is in equilibrium in air due to the force exerted by a light beam on the plate. Calculate power of the beam (in \(MW\)). Assume that the plate is perfectly absorbing.
1 \(30\,MW\)
2 \(50\,MW\)
3 \(10\,MW\)
4 \(40\,MW\)
Explanation:
For equilibrium, the force exerted by the light beam should balance the weight of plate. \({F_{{\rm{photon }}}} = mg\) \(\left. {\left( {{F_{{\text{photon}}}} = \frac{{IA}}{c} = \frac{P}{c}} \right.,{\text{where}}\,\,{\text{power}}\,\,P = IA} \right)\) \( \Rightarrow \frac{P}{c} = 10 \times {10^{ - 3}} \times 10\) \( \Rightarrow P = 30 \times {10^6}\;W = 30\,\,MW\)
PHXI15:WAVES
358868
A radiation of \(200\;W\) is incident on a surface which is \(60 \%\) reflecting and \(40 \%\) absorbing. The total force on the surface is
1 \(1.07 \times {10^{ - 6}}\;N\)
2 \(1.07 \times {10^{ - 7}}\;N\)
3 \(1.03 \times {10^{ - 7}}\;N\)
4 \(1.3 \times {10^{ - 6}}\;N\)
Explanation:
\(P_{r a d}=(1+\rho) \dfrac{I}{c}\) where \(\rho=\) coefficient of reflection \(F=(1+\rho) \dfrac{I}{C} \times A\) \(\therefore F = (1 + 0.6)\frac{{200}}{{3 \times {{10}^8} \times A}} \times A = 1.07 \times {10^{ - 6}}\;N\)
358865
Light with an energy flux of \(20\;\,W/c{m^2}\) falls on a non-reflecting surface at normal incidence. If the surface has an area of \(30\;c{m^2},\) the total momentum delivered (for complete absorption) during \(30\;\min \) is
1 \(1.08 \times {10^7}\;kg - m/s\)
2 \(108 \times {10^4}\;kg - m/s\)
3 \(36 \times {10^{ - 4}}\;kg - m/s\)
4 \(36 \times {10^{ - 5}}\;kg - m/s\)
Explanation:
\(F=P_{\text {rad }} \times\) Area \( = \frac{I}{c} \times {\rm{ Area }}\) \( = \frac{{20}}{{3 \times {{10}^8}}} \times 30 = 2 \times {10^{ - 6}}\;N\) \(=\) Change in momentum in \(1 \mathrm{sec}\) \(\therefore\) Momentum delivered to the surface in \(30 \mathrm{~min}\) \( = 2 \times {10^{ - 6}} \times 30 \times 60 = 36 \times {10^{ - 4}}\;kg\;m\;{s^{ - 1}}\)
NCERT Exempler
PHXI15:WAVES
358866
A small object at rest, absorbs a light pulse of power \(20\;mW\) and duration \(300\;ns\). Assuming speed of light as \(3 \times {10^8}\;m/s\), the momentum of the object becomes equal to
1 \(2 \times {10^{ - 17}}\;kg\;m/s\)
2 \(3 \times {10^{ - 17}}\;kg\;m/s\)
3 \(0.5 \times {10^{ - 17}}\;kg\;m/s\)
4 \(1 \times {10^{ - 17}}\;kg\;m/s\)
Explanation:
\(U\) ( energy \()=\) Power \(\times\) time \( = \left( {20 \times {{10}^{ - 3}}} \right) \times \left( {300 \times {{10}^{ - 9}}\;s} \right)\) \( = 6 \times {10^{ - 9}}{\rm{ joules}}\) Momentum gained by the object is, \(p=\dfrac{U}{c}=\dfrac{6 \times 10^{-9}}{3 \times 10^{8}}\)\(=2 \times 10^{-17} {~kg} {~ms}^{-1}\)
JEE - 2023
PHXI15:WAVES
358867
A plate of mass 10 \(g\) is in equilibrium in air due to the force exerted by a light beam on the plate. Calculate power of the beam (in \(MW\)). Assume that the plate is perfectly absorbing.
1 \(30\,MW\)
2 \(50\,MW\)
3 \(10\,MW\)
4 \(40\,MW\)
Explanation:
For equilibrium, the force exerted by the light beam should balance the weight of plate. \({F_{{\rm{photon }}}} = mg\) \(\left. {\left( {{F_{{\text{photon}}}} = \frac{{IA}}{c} = \frac{P}{c}} \right.,{\text{where}}\,\,{\text{power}}\,\,P = IA} \right)\) \( \Rightarrow \frac{P}{c} = 10 \times {10^{ - 3}} \times 10\) \( \Rightarrow P = 30 \times {10^6}\;W = 30\,\,MW\)
PHXI15:WAVES
358868
A radiation of \(200\;W\) is incident on a surface which is \(60 \%\) reflecting and \(40 \%\) absorbing. The total force on the surface is
1 \(1.07 \times {10^{ - 6}}\;N\)
2 \(1.07 \times {10^{ - 7}}\;N\)
3 \(1.03 \times {10^{ - 7}}\;N\)
4 \(1.3 \times {10^{ - 6}}\;N\)
Explanation:
\(P_{r a d}=(1+\rho) \dfrac{I}{c}\) where \(\rho=\) coefficient of reflection \(F=(1+\rho) \dfrac{I}{C} \times A\) \(\therefore F = (1 + 0.6)\frac{{200}}{{3 \times {{10}^8} \times A}} \times A = 1.07 \times {10^{ - 6}}\;N\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI15:WAVES
358865
Light with an energy flux of \(20\;\,W/c{m^2}\) falls on a non-reflecting surface at normal incidence. If the surface has an area of \(30\;c{m^2},\) the total momentum delivered (for complete absorption) during \(30\;\min \) is
1 \(1.08 \times {10^7}\;kg - m/s\)
2 \(108 \times {10^4}\;kg - m/s\)
3 \(36 \times {10^{ - 4}}\;kg - m/s\)
4 \(36 \times {10^{ - 5}}\;kg - m/s\)
Explanation:
\(F=P_{\text {rad }} \times\) Area \( = \frac{I}{c} \times {\rm{ Area }}\) \( = \frac{{20}}{{3 \times {{10}^8}}} \times 30 = 2 \times {10^{ - 6}}\;N\) \(=\) Change in momentum in \(1 \mathrm{sec}\) \(\therefore\) Momentum delivered to the surface in \(30 \mathrm{~min}\) \( = 2 \times {10^{ - 6}} \times 30 \times 60 = 36 \times {10^{ - 4}}\;kg\;m\;{s^{ - 1}}\)
NCERT Exempler
PHXI15:WAVES
358866
A small object at rest, absorbs a light pulse of power \(20\;mW\) and duration \(300\;ns\). Assuming speed of light as \(3 \times {10^8}\;m/s\), the momentum of the object becomes equal to
1 \(2 \times {10^{ - 17}}\;kg\;m/s\)
2 \(3 \times {10^{ - 17}}\;kg\;m/s\)
3 \(0.5 \times {10^{ - 17}}\;kg\;m/s\)
4 \(1 \times {10^{ - 17}}\;kg\;m/s\)
Explanation:
\(U\) ( energy \()=\) Power \(\times\) time \( = \left( {20 \times {{10}^{ - 3}}} \right) \times \left( {300 \times {{10}^{ - 9}}\;s} \right)\) \( = 6 \times {10^{ - 9}}{\rm{ joules}}\) Momentum gained by the object is, \(p=\dfrac{U}{c}=\dfrac{6 \times 10^{-9}}{3 \times 10^{8}}\)\(=2 \times 10^{-17} {~kg} {~ms}^{-1}\)
JEE - 2023
PHXI15:WAVES
358867
A plate of mass 10 \(g\) is in equilibrium in air due to the force exerted by a light beam on the plate. Calculate power of the beam (in \(MW\)). Assume that the plate is perfectly absorbing.
1 \(30\,MW\)
2 \(50\,MW\)
3 \(10\,MW\)
4 \(40\,MW\)
Explanation:
For equilibrium, the force exerted by the light beam should balance the weight of plate. \({F_{{\rm{photon }}}} = mg\) \(\left. {\left( {{F_{{\text{photon}}}} = \frac{{IA}}{c} = \frac{P}{c}} \right.,{\text{where}}\,\,{\text{power}}\,\,P = IA} \right)\) \( \Rightarrow \frac{P}{c} = 10 \times {10^{ - 3}} \times 10\) \( \Rightarrow P = 30 \times {10^6}\;W = 30\,\,MW\)
PHXI15:WAVES
358868
A radiation of \(200\;W\) is incident on a surface which is \(60 \%\) reflecting and \(40 \%\) absorbing. The total force on the surface is
1 \(1.07 \times {10^{ - 6}}\;N\)
2 \(1.07 \times {10^{ - 7}}\;N\)
3 \(1.03 \times {10^{ - 7}}\;N\)
4 \(1.3 \times {10^{ - 6}}\;N\)
Explanation:
\(P_{r a d}=(1+\rho) \dfrac{I}{c}\) where \(\rho=\) coefficient of reflection \(F=(1+\rho) \dfrac{I}{C} \times A\) \(\therefore F = (1 + 0.6)\frac{{200}}{{3 \times {{10}^8} \times A}} \times A = 1.07 \times {10^{ - 6}}\;N\)
358865
Light with an energy flux of \(20\;\,W/c{m^2}\) falls on a non-reflecting surface at normal incidence. If the surface has an area of \(30\;c{m^2},\) the total momentum delivered (for complete absorption) during \(30\;\min \) is
1 \(1.08 \times {10^7}\;kg - m/s\)
2 \(108 \times {10^4}\;kg - m/s\)
3 \(36 \times {10^{ - 4}}\;kg - m/s\)
4 \(36 \times {10^{ - 5}}\;kg - m/s\)
Explanation:
\(F=P_{\text {rad }} \times\) Area \( = \frac{I}{c} \times {\rm{ Area }}\) \( = \frac{{20}}{{3 \times {{10}^8}}} \times 30 = 2 \times {10^{ - 6}}\;N\) \(=\) Change in momentum in \(1 \mathrm{sec}\) \(\therefore\) Momentum delivered to the surface in \(30 \mathrm{~min}\) \( = 2 \times {10^{ - 6}} \times 30 \times 60 = 36 \times {10^{ - 4}}\;kg\;m\;{s^{ - 1}}\)
NCERT Exempler
PHXI15:WAVES
358866
A small object at rest, absorbs a light pulse of power \(20\;mW\) and duration \(300\;ns\). Assuming speed of light as \(3 \times {10^8}\;m/s\), the momentum of the object becomes equal to
1 \(2 \times {10^{ - 17}}\;kg\;m/s\)
2 \(3 \times {10^{ - 17}}\;kg\;m/s\)
3 \(0.5 \times {10^{ - 17}}\;kg\;m/s\)
4 \(1 \times {10^{ - 17}}\;kg\;m/s\)
Explanation:
\(U\) ( energy \()=\) Power \(\times\) time \( = \left( {20 \times {{10}^{ - 3}}} \right) \times \left( {300 \times {{10}^{ - 9}}\;s} \right)\) \( = 6 \times {10^{ - 9}}{\rm{ joules}}\) Momentum gained by the object is, \(p=\dfrac{U}{c}=\dfrac{6 \times 10^{-9}}{3 \times 10^{8}}\)\(=2 \times 10^{-17} {~kg} {~ms}^{-1}\)
JEE - 2023
PHXI15:WAVES
358867
A plate of mass 10 \(g\) is in equilibrium in air due to the force exerted by a light beam on the plate. Calculate power of the beam (in \(MW\)). Assume that the plate is perfectly absorbing.
1 \(30\,MW\)
2 \(50\,MW\)
3 \(10\,MW\)
4 \(40\,MW\)
Explanation:
For equilibrium, the force exerted by the light beam should balance the weight of plate. \({F_{{\rm{photon }}}} = mg\) \(\left. {\left( {{F_{{\text{photon}}}} = \frac{{IA}}{c} = \frac{P}{c}} \right.,{\text{where}}\,\,{\text{power}}\,\,P = IA} \right)\) \( \Rightarrow \frac{P}{c} = 10 \times {10^{ - 3}} \times 10\) \( \Rightarrow P = 30 \times {10^6}\;W = 30\,\,MW\)
PHXI15:WAVES
358868
A radiation of \(200\;W\) is incident on a surface which is \(60 \%\) reflecting and \(40 \%\) absorbing. The total force on the surface is
1 \(1.07 \times {10^{ - 6}}\;N\)
2 \(1.07 \times {10^{ - 7}}\;N\)
3 \(1.03 \times {10^{ - 7}}\;N\)
4 \(1.3 \times {10^{ - 6}}\;N\)
Explanation:
\(P_{r a d}=(1+\rho) \dfrac{I}{c}\) where \(\rho=\) coefficient of reflection \(F=(1+\rho) \dfrac{I}{C} \times A\) \(\therefore F = (1 + 0.6)\frac{{200}}{{3 \times {{10}^8} \times A}} \times A = 1.07 \times {10^{ - 6}}\;N\)
