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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} k g - m / s

2

108 \times 10^{4} k g - m / s

3

36 \times 10^{- 4} k g - m / s

4

36 \times 10^{- 5} k g - m / s

Explanation:

$F = P_{rad} \times$ Area
$= \frac{I}{c} \times A r e a$
$= \frac{20}{3 \times 10^{8}} \times 30 = 2 \times 10^{- 6} N$
$=$ Change in momentum in $1 \sec$
$∴$ Momentum delivered to the surface in $30 \min$
$= 2 \times 10^{- 6} \times 30 \times 60 = 36 \times 10^{- 4} k g m s^{- 1}$

NCERT Exempler

PHXI15:WAVES

358866 A small object at rest, absorbs a light pulse of power $20 m W$ and duration $300 n s$ . Assuming speed of light as $3 \times 10^{8} m / s$ , the momentum of the object becomes equal to

1

2 \times 10^{- 17} k g m / s

2

3 \times 10^{- 17} k g m / s

3

0.5 \times 10^{- 17} k g m / s

4

1 \times 10^{- 17} k g m / s

Explanation:

$U$ ( energy $) =$ Power $\times$ time
$= (20 \times 10^{- 3}) \times (300 \times 10^{- 9} s)$
$= 6 \times 10^{- 9} j o u l e s$
Momentum gained by the object is,
$p = \frac{U}{c} = \frac{6 \times 10^{- 9}}{3 \times 10^{8}}$ $= 2 \times 10^{- 17} k g {m s}^{- 1}$

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 $M W$ ). Assume that the plate is perfectly absorbing.
supporting img

1

30 M W

2

50 M W

3

10 M W

4

40 M W

Explanation:

For equilibrium, the force exerted by the light beam should balance the weight of plate.
$F_{p h o t o n} = m g$
$(F_{photon} = \frac{I A}{c} = \frac{P}{c}, where power P = I A)$
$\Rightarrow \frac{P}{c} = 10 \times 10^{- 3} \times 10$
$\Rightarrow P = 30 \times 10^{6} W = 30 M W$

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:

$P_{r a d} = (1 + ρ) \frac{I}{c}$ where
$ρ =$ coefficient of reflection
$F = (1 + ρ) \frac{I}{C} \times A$
$∴ F = (1 + 0.6) \frac{200}{3 \times 10^{8} \times A} \times A = 1.07 \times 10^{- 6} N$

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} k g - m / s

2

108 \times 10^{4} k g - m / s

3

36 \times 10^{- 4} k g - m / s

4

36 \times 10^{- 5} k g - m / s

Explanation:

$F = P_{rad} \times$ Area
$= \frac{I}{c} \times A r e a$
$= \frac{20}{3 \times 10^{8}} \times 30 = 2 \times 10^{- 6} N$
$=$ Change in momentum in $1 \sec$
$∴$ Momentum delivered to the surface in $30 \min$
$= 2 \times 10^{- 6} \times 30 \times 60 = 36 \times 10^{- 4} k g m s^{- 1}$

NCERT Exempler

PHXI15:WAVES

358866 A small object at rest, absorbs a light pulse of power $20 m W$ and duration $300 n s$ . Assuming speed of light as $3 \times 10^{8} m / s$ , the momentum of the object becomes equal to

1

2 \times 10^{- 17} k g m / s

2

3 \times 10^{- 17} k g m / s

3

0.5 \times 10^{- 17} k g m / s

4

1 \times 10^{- 17} k g m / s

Explanation:

$U$ ( energy $) =$ Power $\times$ time
$= (20 \times 10^{- 3}) \times (300 \times 10^{- 9} s)$
$= 6 \times 10^{- 9} j o u l e s$
Momentum gained by the object is,
$p = \frac{U}{c} = \frac{6 \times 10^{- 9}}{3 \times 10^{8}}$ $= 2 \times 10^{- 17} k g {m s}^{- 1}$

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 $M W$ ). Assume that the plate is perfectly absorbing.
supporting img

1

30 M W

2

50 M W

3

10 M W

4

40 M W

Explanation:

For equilibrium, the force exerted by the light beam should balance the weight of plate.
$F_{p h o t o n} = m g$
$(F_{photon} = \frac{I A}{c} = \frac{P}{c}, where power P = I A)$
$\Rightarrow \frac{P}{c} = 10 \times 10^{- 3} \times 10$
$\Rightarrow P = 30 \times 10^{6} W = 30 M W$

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:

$P_{r a d} = (1 + ρ) \frac{I}{c}$ where
$ρ =$ coefficient of reflection
$F = (1 + ρ) \frac{I}{C} \times A$
$∴ F = (1 + 0.6) \frac{200}{3 \times 10^{8} \times A} \times A = 1.07 \times 10^{- 6} N$

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} k g - m / s

2

108 \times 10^{4} k g - m / s

3

36 \times 10^{- 4} k g - m / s

4

36 \times 10^{- 5} k g - m / s

Explanation:

$F = P_{rad} \times$ Area
$= \frac{I}{c} \times A r e a$
$= \frac{20}{3 \times 10^{8}} \times 30 = 2 \times 10^{- 6} N$
$=$ Change in momentum in $1 \sec$
$∴$ Momentum delivered to the surface in $30 \min$
$= 2 \times 10^{- 6} \times 30 \times 60 = 36 \times 10^{- 4} k g m s^{- 1}$

NCERT Exempler

PHXI15:WAVES

358866 A small object at rest, absorbs a light pulse of power $20 m W$ and duration $300 n s$ . Assuming speed of light as $3 \times 10^{8} m / s$ , the momentum of the object becomes equal to

1

2 \times 10^{- 17} k g m / s

2

3 \times 10^{- 17} k g m / s

3

0.5 \times 10^{- 17} k g m / s

4

1 \times 10^{- 17} k g m / s

Explanation:

$U$ ( energy $) =$ Power $\times$ time
$= (20 \times 10^{- 3}) \times (300 \times 10^{- 9} s)$
$= 6 \times 10^{- 9} j o u l e s$
Momentum gained by the object is,
$p = \frac{U}{c} = \frac{6 \times 10^{- 9}}{3 \times 10^{8}}$ $= 2 \times 10^{- 17} k g {m s}^{- 1}$

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 $M W$ ). Assume that the plate is perfectly absorbing.
supporting img

1

30 M W

2

50 M W

3

10 M W

4

40 M W

Explanation:

For equilibrium, the force exerted by the light beam should balance the weight of plate.
$F_{p h o t o n} = m g$
$(F_{photon} = \frac{I A}{c} = \frac{P}{c}, where power P = I A)$
$\Rightarrow \frac{P}{c} = 10 \times 10^{- 3} \times 10$
$\Rightarrow P = 30 \times 10^{6} W = 30 M W$

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:

$P_{r a d} = (1 + ρ) \frac{I}{c}$ where
$ρ =$ coefficient of reflection
$F = (1 + ρ) \frac{I}{C} \times A$
$∴ F = (1 + 0.6) \frac{200}{3 \times 10^{8} \times A} \times A = 1.07 \times 10^{- 6} N$

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} k g - m / s

2

108 \times 10^{4} k g - m / s

3

36 \times 10^{- 4} k g - m / s

4

36 \times 10^{- 5} k g - m / s

Explanation:

$F = P_{rad} \times$ Area
$= \frac{I}{c} \times A r e a$
$= \frac{20}{3 \times 10^{8}} \times 30 = 2 \times 10^{- 6} N$
$=$ Change in momentum in $1 \sec$
$∴$ Momentum delivered to the surface in $30 \min$
$= 2 \times 10^{- 6} \times 30 \times 60 = 36 \times 10^{- 4} k g m s^{- 1}$

NCERT Exempler

PHXI15:WAVES

358866 A small object at rest, absorbs a light pulse of power $20 m W$ and duration $300 n s$ . Assuming speed of light as $3 \times 10^{8} m / s$ , the momentum of the object becomes equal to

1

2 \times 10^{- 17} k g m / s

2

3 \times 10^{- 17} k g m / s

3

0.5 \times 10^{- 17} k g m / s

4

1 \times 10^{- 17} k g m / s

Explanation:

$U$ ( energy $) =$ Power $\times$ time
$= (20 \times 10^{- 3}) \times (300 \times 10^{- 9} s)$
$= 6 \times 10^{- 9} j o u l e s$
Momentum gained by the object is,
$p = \frac{U}{c} = \frac{6 \times 10^{- 9}}{3 \times 10^{8}}$ $= 2 \times 10^{- 17} k g {m s}^{- 1}$

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 $M W$ ). Assume that the plate is perfectly absorbing.
supporting img

1

30 M W

2

50 M W

3

10 M W

4

40 M W

Explanation:

For equilibrium, the force exerted by the light beam should balance the weight of plate.
$F_{p h o t o n} = m g$
$(F_{photon} = \frac{I A}{c} = \frac{P}{c}, where power P = I A)$
$\Rightarrow \frac{P}{c} = 10 \times 10^{- 3} \times 10$
$\Rightarrow P = 30 \times 10^{6} W = 30 M W$

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:

$P_{r a d} = (1 + ρ) \frac{I}{c}$ where
$ρ =$ coefficient of reflection
$F = (1 + ρ) \frac{I}{C} \times A$
$∴ F = (1 + 0.6) \frac{200}{3 \times 10^{8} \times A} \times A = 1.07 \times 10^{- 6} N$

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