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Standard XII

Physics

Stefan-Boltzman's Law

The temperatu...

Question

The temperature of an isolated black body falls from $T_{1}$ to $T_{2}$ in time $t$ . The value of $t$ is (Let $c$ be a constant) :

t = c (\frac{1}{T_{2}} - \frac{1}{T_{2}})

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t = c (\frac{1}{T_{2}^{2}} - \frac{1}{T_{1}^{2}})

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t = c (\frac{1}{T_{2}^{3}} - \frac{1}{T_{1}^{3}})

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t = c (\frac{1}{T_{2}^{4}} - \frac{1}{T_{1}^{4}})

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Solution

The correct option is C $t = c (\frac{1}{T_{2}^{3}} - \frac{1}{T_{1}^{3}})$
By stefan-boltzmann law, rate of energy radiated by black body is given as:

$q = A σ (T^{4}) \Rightarrow m s d T / d t = A σ (T^{4})$ , where $T$ is the temperature of the body.

So, $d t = C (d T / T^{4})$ , where $C$ is the constant of proportionality. Integrating above we get,

$\int_{0}^{t} d t = C \int_{T_{1}}^{T_{2}} d T / T^{4}$

$\Rightarrow t = (- C / 3) (1 / T^{3}) |_{T_{1}}^{T_{2}}$ $\Rightarrow t = c (\frac{1}{T_{2}^{3}} - \frac{1}{T_{1}^{3}})$ where, $c = - C / 3$

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$l o g_{e} (n + 1) - l o g_{e} (n - 1) = 4 a [(\frac{1}{n}) + (\frac{1}{3 n^{3}}) + (\frac{1}{5 n^{5}}) + . . . \infty]$ Find $8 a$ .

Q. Choose the correct alternative for each of the following. If

P (E_{1}) = \frac{1}{6}

P (E_{2}) = \frac{1}{3}

P (E_{3}) = \frac{1}{6}

, where

E_{1}

E_{2}

E_{3}

E_{4}

are elementary events of a random experiment, then P(

E_{4}

) is equal to

The product of the following series $(1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + . . .)$ $\times$ $(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + . . .)$ is

Q. Assertion :

Δ = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{(1 + a_{1} b_{1}) (1 + a_{1}^{2} b_{1}^{2} - a_{1} b_{1})}{1 + a_{1} b_{1}} & \frac{(1 + a_{1} b_{2}) (1 + a_{1}^{2} b_{2}^{2} - a_{1} b_{2})}{1 + a_{1} b_{2}} & \frac{(1 + a_{1} b_{3}) (1 + a_{1}^{2} b_{3}^{2} - a_{1} b_{3})}{1 + a_{1} b_{3}} \frac{(1 + a_{2} b_{1}) (1 + a_{2}^{2} b_{1}^{2} - a_{2} b_{1})}{1 + a_{2} b_{1}} & \frac{(1 + a_{2} b_{2}) (1 + a_{2}^{2} b_{2}^{2} - a_{2} b_{2})}{1 + a_{2} b_{2}} & \frac{(1 + a_{2} b_{3}) (1 + a_{2}^{2} b_{3}^{2} - a_{2} b_{3})}{1 + a_{2} b_{3}} \frac{(1 + a_{3} b_{1}) (1 + a_{3}^{2} b_{1}^{2} - a_{3} b_{1})}{1 + a_{3} b_{1}} & \frac{(1 + a_{3} b_{2}) (1 + a_{2}^{2} b_{2}^{2} - a_{3} b_{2})}{1 + a_{3} b_{2}} & \frac{(1 + a_{3} b_{3}) (1 + a_{3}^{2} b_{3}^{2} - a_{3} b_{3})}{1 + a_{3} b_{3}} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

Δ = 0

Reason:

Δ

can be written as product of two determinants.

Q. lf

α

lies in the third quadrant, then

\sqrt{\frac{1 - sin α}{1 + sin α}} + \sqrt{\frac{1 + sin α}{1 - sin α}} =

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The temperature of an isolated black body falls from T1 to T2 in time t. The value of t is (Let c be a constant) :

The temperature of an isolated black body falls from $T_{1}$ to $T_{2}$ in time $t$ . The value of $t$ is (Let $c$ be a constant) :