A block of mass $1 μ g$ is connected with an elastic string of stiffness constant $K = 10^{- 5} N / m$ . Now a light pulse of intensity $I = 20 W / m^{2}$ strikes the block at its vertical surface of surface area $10 c m^{2}$ at an angle $53 °$ as shown. If surface of block is 100% absorbing the light and the duration of light pulse is 6ms then the max displacement of block from it’s mean position is $N \times 10^{- 5}$ mm. find value of N.

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answer is 144.

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Detailed Solution

Impulse on block = $(\frac{I A}{C}) \cos^{2} 53^{°} \times (Δ t)$
$= \frac{(20) (10 \times 10^{- 4})}{3 \times 10^{8}} \times {(0.6)}^{2} \times 6 \times 10^{- 3} = \frac{72}{5} \times 10^{- 14} k g m / s$
Now we have
Impulse =mv
$\frac{72}{5} \times 10^{- 14} = 1 \times 10^{- 9} v v = \frac{72}{5} \times 10^{- 5} m / s$
Now we have
$\frac{1}{2} k x^{2} = \frac{1}{2} m v^{2} 10^{- 5} x^{2} = 10^{- 9} \times 24 \times 10^{- 5} \times 24 \times 10^{- 5} x = \frac{72}{5} \times 10^{- 7} m N = \frac{7.2}{5} = 1.44$