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Question Number 47377 by rahul 19 last updated on 09/Nov/18

Commented by rahul 19 last updated on 09/Nov/18

$$Ans:{10}^{5}Pa.$$

Answered by ajfour last updated on 09/Nov/18

$$P=\frac{F}{A}=\frac{△p/△t}{A}=\frac{△p}{Al}\times \frac{l}{△t}$$$$=\frac{v△p}{Al}=\frac{v\left[n\left(2m_{0}v\cos 30°\right)\right]}{Al}$$$$\mathrm{\&}with\frac{n}{Al}=n_0$$$$P=\sqrt{3}{m}_0{n}_0{v}^2$$$$=\sqrt{3}\left(\frac{28\times {10}^{-3}}{6.02\times {10}^{23}}\right)(0.9\times {10}^{25})\left(16\times {10}^4\right)$$$$=\frac{\sqrt{3}\times 0.028\times 9\times 16}{6}\times {10}^{5}Pa$$$$=1.4\times 0.48\sqrt{3}\times {10}^{5}Pa$$$$\approx 1.164\times {10}^{5}Pa.$$

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