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JEE Main 20206 SeptemberMorning Shift
Exam
JEE Main
Year
2020
Date
6 September
Shift
Morning Shift
Subject
Physics
Chapter
Capacitance

JEE Main 2020 · 6 September · Morning Shift · Physics · Capacitance

Suppose that intensity of a laser is ((315)/(π))W m ^ - 2 . The rms...

Suppose that intensity of a laser is (315π)Wm−2.\left(\dfrac{315}{\pi}\right)W{m}^{- 2} . The rms electric field, in units of Vm−1V{m}^{- 1} associated with this source is close to the nearest integer is - \left({ε}_{0} = 8 . 86 \times {10}^{- 12}{C}^{2}N{m}^{- 2}; c = 3 \times {10}^{8}m{s}^{- 1}\right)

Correct answer

0

Step-by-step solution

12C∈0Ems2=315π\dfrac{1}{2}C{\in}_{0}{E}_{ms}^{2} = \dfrac{315}{\pi}

Erms=315×2π×3×108×8.86×10−12{E}_{rms} = \sqrt{\dfrac{315 \times 2}{\pi \times 3 \times {10}^{8} \times 8 . 86 \times {10}^{- 12}}}

=63083.4612×10−4= \sqrt{\dfrac{630}{83 . 4612 \times {10}^{- 4}}}

=7.5484×104= \sqrt{7 . 5484 \times {10}^{4}}

=2.75×102=2.75×102=275= 2 . 75 \times {10}^{2} = 2 . 75 \times {10}^{2} = 275

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