A metallic sphere cools from ${50}^{\circ }\mathrm{C}\text{ to }{40}^{\circ }\mathrm{C}\text{ in }300\mathrm{s}\text{. }$ If atmospheric temperature around is $20°\mathrm{C}$ , then the sphere’s temperature after the next 5 minutes will be close to :

From Newton’s Law of cooling, 
$\frac{T_1-T_2}{t}=K\left[\frac{T_1+T_2}{2}-T_0\right]$ 
Here, $T_1={50}^{\circ }\mathrm{C},T_2={40}^{\circ }\mathrm{C}$ 
$\begin{array}{l}\text{ an }{T}_0={20}^{\circ }C,\mathrm{t}=600\mathrm{s}=5\text{ minutes }\\ \Rightarrow \frac{50-40}{5}=K\left(\frac{50+40}{2}-20\right) ...\left(i\right)\end{array}$
Let T be the temperature of sphere after next 5 minutes.
Then 
$\frac{40-T}{5}=K\left(\frac{40+T}{2}-20\right)$ ...(ii)
Dividing equation, (ii) by (i), we get
$\begin{array}{l}\frac{40-T}{10}=\frac{40+T-40}{50+40-40}=\frac{T}{50}\\ \Rightarrow 40-T=\frac{T}{5}\Rightarrow 200-5T=T\\ \therefore T=\frac{200}{6}={33.3}^{\circ }\mathrm{C}\end{array}$