JEE MAIN - Physics (2019 - 10th April Evening Slot - No. 6)
Light is incident normally on a completely absorbing surface with an energy flux of 25 W cm–2. If the surface
has an area of 25 cm2, the momentum transferred to the surface in 40 min time duration will be :
6.3 × 10–4 Ns
5.0 × 10–3 Ns
1.4 × 10–6 Ns
3.5 × 10–6 Ns
Explanation
$$P = {{\Delta E} \over C}$$
$$ = {{\left( {25 \times 25} \right) \times 40 \times 60} \over {3 \times {{10}^8}}}N - s$$
$$ = 5 \times {10^{ - 2}}N - s$$
$$ = {{\left( {25 \times 25} \right) \times 40 \times 60} \over {3 \times {{10}^8}}}N - s$$
$$ = 5 \times {10^{ - 2}}N - s$$
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