Ncert Exercises & Solutions

Light with an energy flux of \(20~\text{W/cm}^2\) falls on a non-reflecting surface at normal incidence. If the surface has an area of \(30~\text{cm}^2\), the momentum delivered (for complete absorption) during \(30\) minutes is:
1. \(36\times10^{-5}~\text{kg-m/s}\)
2. \(36\times10^{-4}~\text{kg-m/s}\)
3. \(108\times10^{4}~\text{kg-m/s}\)
4. \(1.08\times10^{7}~\text{kg-m/s}\)

(b) Hint: The omentum transferred depends on the energy transferred to the wall.

Step 1: Given, energy flux,

ϕ

= 20 W/

{cm}^{2}

Area, A= 30

{cm}^{2}

Time, t = 30 min = 30 x 60 sec
Now, the total energy falling on the surface in time is, U =

ϕ

At =20 x 30 x (30 x 60) J

Step 2: The momentum of the incident light =

\frac{U}{c}

= \frac{20 \times 30 \times (30 \times 60)}{3 \times 10^{8}} \Rightarrow 36 \times 10^{- 4} kg - {ms}^{- 1}

Step 3: Momentum of the reflected light = 0

∴

Momentum delivered to the surface

= 36 \times 10^{- 4} - 0 = 36 \times 10^{- 4} kg - {ms}^{- 1}

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions