Answer

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**Hint:**To find the answer for the given question, first we need to know the power formula and the body is in a spherical black body with a radius and power are given in the question. Note that the radius is halved and the temperature is doubled. We need to calculate the power radiated in watts.

**Formula used:**

The power formula of spherical black body is

$P = A\sigma {T^4}$

If the $r$ is the radius of the spherical body

Then,

$A$ is the area of the body

$\sigma $ is the Stefan’s constant

$T$ is the temperature of the body

**Complete step by step answer:**

Power formula of spherical body to calculate power radiated

$ \Rightarrow $ $P = A\sigma {T^4}$

The area of the sphere $A = 4\pi {r^2}$

If we substitute the area formula in power formula we get,

$ \Rightarrow $ $P = 4\pi {r^2}.\sigma .{T^4}$

Where $\sigma$ is constant

So, $P \propto {r^2}{T^4}$

If the power radiated $P{}_1$ when the radius of the spherical body is $12cm$

The value of the power $450W$ is given in the question,

Then the equation becomes,

$ \Rightarrow $ $P = \left( {4\pi \sigma } \right){r^2}{T^4} = 450$…………………(1)

In the question the radius is halved and the temperature is doubled

So,

$ \Rightarrow $ $r' = \dfrac{r}{2}$ and $T' = 2T$

$ \Rightarrow $ $ P' = \left( {4\pi \sigma } \right){\left( {\dfrac{r}{2}} \right)^2}{\left( {2T} \right)^4}$

$ \therefore $ $P' = \left( {4\pi \sigma {r^2}{T^4}} \right)\dfrac{{{2^4}}}{{{2^2}}}$

Substitute the equation 1 then we get,

$ \Rightarrow $ $P' = 450 \times 4$

$ \therefore $ $P' = 1800$$W$

The power radiated watts is $1800W$.

**Therefore the option (D) is correct.**

**Note:**In the above equations, Stephen's constant is used because if the energy emitted increases rapidly with increase in temperature which is proportional to the temperature raised to the fourth power. The black body which is used for the lighting, heating and the thermal imaging etc. as well as testing and the measurement applications.

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