TS EAMCET 10 May 2024 Shift 2 Solved Paper

Section: Physics

Question : 113 of 160

Marks: +1, -0

The efficiency of a bulb of power 60 W is 16%. The peak value of the electric field produced by the electromagnetic radiation from the bulb at a distance of 2 m from the bulb is
(‌

4πE0

=9×109Nm2C−2)

[TG EAPCET 10 May 2024 Shift 2]

24Vm−1
16Vm−1
9Vm−1
12Vm−1

Validate

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