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**Hint:**Heat is a form of energy and it flows from high temperature to low temperature. In the above question, a metallic sphere is cooling. It means that the heat is flowing from the sphere to the surrounding. The amount of heat that is transferred from the sphere to the surrounding is a function of time. We can use newton’s law of cooling.

**Complete step by step solution:**

Step 1: First let us understand what newton’s law of cooling is. Newton's law of cooling states that the rate of heat loss of a body is directly proportional to the difference in the temperatures between the body and its surroundings. This holds equality with the heat transfer coefficient. Express Newton's law of cooling mathematically.

$\therefore \dfrac{{\Delta T}}{{\Delta t}} = k\left( {T(t) - {T_{env}}} \right)$

Where, $\Delta T$ is the temperature change, $T(t)$ is the average temperature change of the body ( it depends on time), ${T_{env}}$ is the temperature of the environment, and $k$ is the heat transfer coefficient.

Step 2: substitute the values

$\therefore \dfrac{{50 - 40}}{5} = k\left( {\dfrac{{50 + 40}}{2} - 20} \right)$

$\Delta t = 5\min $

$ \Rightarrow \dfrac{{10}}{5} = k\left( {\dfrac{{90}}{2} - 20} \right)$

$ \Rightarrow 2 = k(45 - 20) = k25$

$ \Rightarrow k = \dfrac{2}{{25}}$

Step 3: Now after 5min of 40 $^\circ C$, we can again apply newton’s law of cooling. Therefore if the temperature of the sphere after 5 min is $T$ then,

$\therefore \dfrac{{40 - T}}{5} = \dfrac{2}{{25}}\left( {\dfrac{{40 + T}}{2} - 20} \right)$

$ \Rightarrow 40 - T = \dfrac{2}{5}\left( {\dfrac{{40 + T}}{2} - 20} \right)$

$ \Rightarrow 200 - 5T = 40 + T - 40$

Further, simplify for T

$\therefore 6T = 200$

$ \Rightarrow T = \dfrac{{200}}{6}$

$ \Rightarrow T = 33.3 ^\circ $

**Hence the correct option is Option B.**

**Note:**The rate of transfer of heat depends on many factors like material of the object, temperature difference between the object and the surrounding, etc. The heat transfer coefficient is constant for a specific material. In the above question, we are not given the heat transfer coefficient. Therefore, first, calculate the value of the heat transfer coefficient then approach another part of the problem.

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