Physics, asked by Raja42661, 1 year ago

The tempearture of an isolated black body falls from T₁ to T₂ in time t, then t is (Let c be a constant)
(a) t=c\bigg \lgroup \frac{1}{T_{2}}-\frac{1}{T_{1}}\bigg \rgroup
(b) t=c\bigg \lgroup \frac{1}{T^{2}_{2}}-\frac{1}{T^{2}_{1}}\bigg \rgroup
(c) t=c\bigg \lgroup \frac{1}{T^{3}_{2}}-\frac{1}{T^{3}_{1}}\bigg \rgroup
(d) t=c\bigg \lgroup \frac{1}{T^{4}_{2}}-\frac{1}{T^{4}_{1}}\bigg \rgroup

Answers

Answered by sp50
0

the answer is yes[tex ]t=c \ bigg\/group \frac {1}{T_{2}}-/frac {1)(T_{1}}\bigg\rgroup [/tex]

Answered by Anonymous
0

Hey Mate!

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Good Question

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Option :↓Choosen below↓

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The tempearture of an isolated black body falls from T₁ to T₂ in time t, then t is (Let c be a constant)

(a) t=c\bigg \lgroup \frac{1}{T_{2}}-\frac{1}{T_{1}}\bigg \rgroup

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