Three discs A, B and C having radii 2, 4, and 6 cm respectively are coated with carbon black. Wavelength for maximum intensity for the three discs are 300, 400 and 500 nm respectively. If and are power emitted by A, B and C respectively, then

Question

Three discs A, B and C having radii 2, 4, and 6 cm respectively are coated with carbon black. Wavelength for maximum intensity for the three discs are 300, 400 and 500 nm respectively. If $Q_{A}, Q_{B}$ and $Q_{C}$ are power emitted by A, B and C respectively, then [2004]

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(2)

From Wein's displacement law

$λ_{m} T = Constant$

$∴$ $T = \frac{constant}{λ_{m}}$ ${∴ T_{A} = \frac{C}{3 \times 10^{- 7}}, T_{B} = \frac{C}{4 \times 10^{- 7}}, T_{C} = \frac{C}{5 \times 10^{- 7}}}$

Again from Stefan's law

$Q = σ A T^{4}$ $∴$ $Q_{A} = σ . π {(2 \times 10^{- 2})}^{2} \times \frac{C^{4}}{81 \times 10^{- 28}}$

$Q_{B} = σ . π {(4 \times 10^{- 2})}^{2} \times \frac{C^{4}}{256 \times 10^{- 28}}$

$and Q_{C} = σ . π {(6 \times 10^{- 2})}^{2} \times \frac{C^{4}}{625 \times 10^{- 28}}$

$Hence Q_{B} > Q_{C} > Q_{A}$

Solution

Q.
Three discs A, B and C having radii 2, 4, and 6 cm respectively are coated with carbon black. Wavelength for maximum intensity for the three discs are 300, 400 and 500 nm respectively. If $Q_{A}, Q_{B}$ and $Q_{C}$ are power emitted by A, B and C respectively, then [2004]

1 $Q_{A}$ will be maximum

2 $Q_{B}$ will be maximum

3 $Q_{C}$ will be maximum

4 $Q_{A} = Q_{B} = Q_{C}$

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Solution

Q. Three discs A, B and C having radii 2, 4, and 6 cm respectively are coated with carbon black. Wavelength for maximum intensity for the three discs are 300, 400 and 500 nm respectively. If QA, QB and QC are power emitted by A, B and C respectively, then [2004]

1 QA will be maximum

2 QB will be maximum

3 QC will be maximum

4 QA=QB=QC