Light with an energy flux of 18 W/cm$$2\text{cm}$$^2cm2 falls on a non-reflecting surface at normal incidence. If the surface has an area of 20 cm$$2\text{cm}$$^2cm2, find the total momentum delivered (for complete absorption) during a 30 minute time span?

2.16×10−5 kg/s2.16 \times 10^{-5} $$\mathrm{~kg}$$ / $$\mathrm{s}2$$.16×10−5 kg/s

2.16×10−2 kg/s2.16 \times 10^{-2} $$\mathrm{~kg}$$ / $$\mathrm{s}2$$.16×10−2 kg/s

2.16×10−3 kg⋅m/s 2.16 \times 10^{-3}\, $$\text{kg}$$ \cdot $$\text{m/s}2$$.16×10−3kg⋅m/s

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Light with an energy flux of 18 W/cm2\text{cm}^2cm2 falls on a non-reflecting surface at normal incidence. If the surface has an area of 20 cm2\

Question

Light with an energy flux of 18 W/ $\text{cm}^2$ falls on a non-reflecting surface at normal incidence. If the surface has an area of 20 $\text{cm}^2$ , find the total momentum delivered (for complete absorption) during a 30 minute time span?

$2.16 \times 10^{-5} \mathrm{~kg} / \mathrm{s}$

$2.16 \times 10^{-2} \mathrm{~kg} / \mathrm{s}$

2.16 kg m/s

$2.16 \times 10^{-3}\, \text{kg} \cdot \text{m/s}$

Solution

Correct option is D

$\textbf{Given:} \\\quad \bullet\ \text{Energy flux } = 18\, \text{W/cm}^2 \\\quad \bullet\ \text{Area } A = 20\, \text{cm}^2 = 20 \times 10^{-4}\, \text{m}^2 \\\quad \bullet\ \text{Time } t = 30\, \text{minutes} = 1800\, \text{s} \\\quad \bullet\ \text{Speed of light } c = 3 \times 10^8\, \text{m/s} \\\\\quad \text{Power incident: } P = \text{Energy flux} \times \text{Area} \\\quad \text{Momentum delivered: } p = \frac{P \cdot t}{c} \\\\\quad P = 18 \times 10^4 \times 20 \times 10^{-4} = 360\, \text{W} \\\quad p = \frac{360 \times 1800}{3 \times 10^8} = 2.16 \times 10^{-3}\, \text{kg} \cdot \text{m/s}$

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Light with an energy flux of 18 W/ $\text{cm}^2$ falls on a non-reflecting surface at normal incidence. If the surface has an area of 20 $\text{cm}^2$ , find the total momentum delivered (for complete absorption) during a 30 minute time span?

$2.16 \times 10^{-5} \mathrm{~kg} / \mathrm{s}$

$2.16 \times 10^{-2} \mathrm{~kg} / \mathrm{s}$

2.16 kg m/s

$2.16 \times 10^{-3}\, \text{kg} \cdot \text{m/s}$

Correct option is D

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Light with an energy flux of 18 W/cm2\text{cm}^2cm2​ falls on a non-reflecting surface at normal incidence. If the surface has an area of 20 cm2\text{cm}^2cm2​, find the total momentum delivered (for complete absorption) during a 30 minute time span?

​2.16×10−5 kg/s2.16 \times 10^{-5} \mathrm{~kg} / \mathrm{s}2.16×10−5 kg/s​​

​2.16×10−2 kg/s2.16 \times 10^{-2} \mathrm{~kg} / \mathrm{s}2.16×10−2 kg/s​​

2.16 kg m/s

​​​2.16×10−3 kg⋅m/s 2.16 \times 10^{-3}\, \text{kg} \cdot \text{m/s}2.16×10−3kg⋅m/s​​

Correct option is D

Similar Questions

Access ‘AAI JE ATC’ Mock Tests with

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Light with an energy flux of 18 W/ $\text{cm}^2$ falls on a non-reflecting surface at normal incidence. If the surface has an area of 20 $\text{cm}^2$ , find the total momentum delivered (for complete absorption) during a 30 minute time span?

$2.16 \times 10^{-5} \mathrm{~kg} / \mathrm{s}$

$2.16 \times 10^{-2} \mathrm{~kg} / \mathrm{s}$

$2.16 \times 10^{-3}\, \text{kg} \cdot \text{m/s}$