1. 1 × 10⁵ J 2. 36 × 10⁷ J 3. 36 × 10⁴ J 4. 36 × 10⁵ J

Answer: 2. 36 × 10⁷ J Energy radiated equals power times time. E=Pt=100×10³\ W×3600\ s=3.6×10⁸\ J=36×10⁷\ J Option (2) is correct.

The energy that will be ideally radiated by a 100 kW transmitter in 1 hour is :

1. \(1 \times 10^{5}\) J

2. \(36 \times 10^{7}\) J

3. \(36 \times 10^{4}\) J

4. \(36 \times 10^{5}\) J

Doubtly Asked 30 Oct 2026 NEET UG 2022 Practice test neet Save WhatsApp Report

Best answer

Answer: 2. \(36 \times 10^{7}\) J

Energy radiated equals power times time.

\(E=Pt=100\times10^{3}\ \text{W}\times3600\ \text{s}=3.6\times10^{8}\ \text{J}=36\times10^{7}\ \text{J}\)

Option (2) is correct.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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