Question

$\huge\underline\bold\red{Ouestion:-}$

➠ Temperature of a tungsten filament of 90watt electric bulb is 2000K Fond surface area of filament.

Er= 0.3​

Answer 1

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➠ Temperature of a tungsten filament of 90watt electric bulb is 2000K Fond surface area of filament.

Er= 0.3

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Temperature = T = 2000K

Power = 60 watt

er = 0.3

$\large{ \pink{\mathfrak{ \underline{\overline{\mid \:\: To\: find\:\: \mid}}}}}$

Surface area of filament = ?

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E = $\dfrac{Q}{AT}$

E= $\sigma \: {T}^{4} \times Er$

P = $\dfrac{Q}{T}$

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$\large{ \pink{\mathfrak{ \underline{\overline{\mid \:\: given \:\: \mid}}}}}$

We know,

E = $\dfrac{Q}{AT}$

Also,

E= $\sigma \: {T}^{4} \times Er$

Thus,

$\dfrac{Q}{AT} = \sigma \: {T}^{4} \times 0.3$

$\implies \dfrac{P}{A} = 5.67 \times {10}^{ - 8} \times 16 \times {10}^{12} \times 0.3$

$\implies A = \dfrac{600}{5.67 \times 16 \times {10}^{4} \times 0.3}$

$\implies A = \dfrac{150}{22.68} \times {10}^{ - 4}$

$\red{\bold{\boxed{\large{\implies A = 15 \times {10}^{ - 4}}}}}$

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