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NEET PHYSICSEasy

The energy that will be ideally radiated by a 100 kW100\text{ kW} transmitter in 1 hr1\text{ hr} is:

A

1×105 J1 \times 10^5\text{ J}

B

36×107 J36 \times 10^7\text{ J}

C

36×104 J36 \times 10^4\text{ J}

D

36×105 J36 \times 10^5\text{ J}

Step-by-Step Solution

  1. Identify the Formula: Power (PP) is the rate at which energy (EE) is consumed or radiated. The relationship is given by E=P×tE = P \times t [Class 11 Physics, Ch 6, Sec 6.10].
  2. Convert Units to SI:
  • Power P=100 kW=100×103 W=105 J/sP = 100\text{ kW} = 100 \times 10^3\text{ W} = 10^5\text{ J/s}.
  • Time t=1 hr=60×60 s=3600 st = 1\text{ hr} = 60 \times 60\text{ s} = 3600\text{ s}.
  1. Calculate Energy: E=P×t=(105 J/s)×(3600 s)E = P \times t = (10^5\text{ J/s}) \times (3600\text{ s}) E=3600×105 JE = 3600 \times 10^5\text{ J} E=36×102×105 JE = 36 \times 10^2 \times 10^5\text{ J} E=36×107 JE = 36 \times 10^7\text{ J}
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