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Momentum of Object Hit by Light

Solution

Momentum received by object = Momentum of the pulse

$\displaystyle \therefore p=\frac{h}{\lambda }=\frac{h\nu }{\lambda \nu }=\frac{Energy}{c}=\frac{Pt}{c}$

$\displaystyle Or,p=\frac{30\times {{10}^{-3}}\times 100\times {{10}^{-9}}}{3\times {{10}^{8}}}=1.0\times {{10}^{-17}}kgm/s$

Hence, (B).

${\scriptscriptstyle 6 + f(x) = 2f( – x) + 3{x^2}\int\limits_{ – 1}^1 {f(t)dt}}$
$\scriptstyle f(x) = {\left( {\frac{x}{\pi }} \right)^x} + {\left( {\frac{\pi }{x}} \right)^x}$
$\scriptstyle \sqrt {7 – \sqrt {7 + \sqrt {7 – \sqrt {7 + …..} } } }$
$\scriptstyle f(x) = \int\limits_0^x {{\pi ^t}(t – e)(t – \pi )dt} $
$\scriptstyle 7f(x) + 5f\left( {\frac{1}{x}} \right) = x + 1$ …

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