Heating of sphere through resistor!

Let thermal energy stored in the sphere be given by $E$ , and the temperature of the sphere be $T$ (which is also the temperature relative to the surroundings provided we remain in degrees Celsius). $\tau$ is the time required by the question.

The power input (energy per unit time) from resistor to sphere is given by $I^{2}R$ and the power loss from sphere to surroundings by $kT$

Hence the differential equation describing the situation is:

$\frac{dE}{dt} = I^{2}R - kT$

Now, using the heat capacity equation, $E = CT \Rightarrow \frac{dE}{dt} = C\frac{dT}{dt}$

Then substituting into the differential equation:

$C\frac{dT}{dt} = I^{2}R - kT$

The maximum attainable temperature is achieved when $\frac{dE}{dt} = 0$

$\Rightarrow I^{2}R = kT_{max} \Rightarrow T_{max} = \frac{I^{2}R}{k}$

By separation of variables in the differential equation:

$\int_{0}^{\frac{T_{max}}{2}}\frac{C}{I^{2}R-kT}dT = \tau$

$\Rightarrow \tau = -\frac{c}{k}\ln(I^{2}R-kT) \Big|_0^\frac{T_{max}}{2}$

$\Rightarrow \tau = \frac{c}{k}\ln2$

$\therefore \boxed{a=1}$

@NISHANT RAI pls make it O KELVIN ..

Curve proven related to exponential change from 0 towards arrival is a saturation to certain level as:

$M$ (1 - $\displaystyle e^{- \frac {k}{c} t}$ ) = $\displaystyle \frac {M}{2}$ of M (1 - 1) to M (1 - 0) of full time changes.

$\Rightarrow$ (1 - $\displaystyle e^{- \frac {k}{c} t}$ ) = $\displaystyle \frac 12$

$\Rightarrow$ $\displaystyle 2^{- 1}$ = $\displaystyle e^{- \frac {k}{c} t}$

$\Rightarrow$ $\displaystyle e^{- Ln 2}$ = $\displaystyle e^{- \frac {k}{c} t}$

$\Rightarrow$ Ln 2 = $\displaystyle \frac {k}{c} t$

$\Rightarrow$ t = $\displaystyle \frac {c}{k}$ Ln 2

Therefore, a = 1

Answer: 1