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NEET PHYSICSThermal Properties of MatterMedium

Question

A spherical black body with a radius of 12 cm12\text{ cm} radiates 450 W450\text{ W} power at 500 K500\text{ K}. If the radius were halved and the temperature doubled, the power radiated in watts would be:

A

225225

B

450450

C

10001000

D

18001800

Step-by-Step Solution

According to the Stefan-Boltzmann law, the power (PP) radiated by a spherical black body is given by P=σAT4=σ(4πr2)T4P = \sigma A T^4 = \sigma (4\pi r^2) T^4, where rr is the radius and TT is the absolute temperature. Let the initial power be P1=450 WP_1 = 450\text{ W} with radius r1=12 cmr_1 = 12\text{ cm} and temperature T1=500 KT_1 = 500\text{ K}. When the radius is halved (r2=r1/2r_2 = r_1 / 2) and the temperature is doubled (T2=2T1T_2 = 2T_1), the new power P2P_2 will be: P2=σ4π(r1/2)2(2T1)4P_2 = \sigma 4\pi (r_1 / 2)^2 (2T_1)^4 P2=σ4π(r124)(16T14)P_2 = \sigma 4\pi \left(\frac{r_1^2}{4}\right) (16T_1^4) P2=4×[σ(4πr12)T14]P_2 = 4 \times [\sigma (4\pi r_1^2) T_1^4] P2=4×P1P_2 = 4 \times P_1 P2=4×450 W=1800 WP_2 = 4 \times 450\text{ W} = 1800\text{ W}. Therefore, the power radiated would be 1800 W1800\text{ W}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Thermal Properties of Matter. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSThermal Properties of Mattersphericalradiusradiatesradiushalved

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