Correct Answer: $\frac{3}{4}$

Solution : Given: If $\alpha$ is an acute angle and $2\sin \alpha+15\cos^2 \alpha=7$.
$2\sin \alpha+15\cos^2\alpha=7$
⇒ $2\sin \alpha+15(1–\sin^2 \alpha)=7$
Let $\sin \alpha$ be $x$.
⇒ $2x+15(1–x^2)=7$
⇒ $2x+15–15x^2=7$
⇒ $15x^2–2x–8=0$
⇒ $15x^2–12x+10x–8=0$
⇒ $3x(5x–4)+2(5x–4)=0$
⇒ $(5x–4)(3x+2)=0$
⇒ $x=\frac{4}{5},\frac{–2}{3}$
$\sin \alpha=\frac{4}{5},\frac{–2}{3}$
Since $\alpha$ is an acute angle, the value of $\sin\alpha=\frac{4}{5}$
So, the $\text{Perpendicular}$ is 4 and the $\text{Hypotenuse}$ is 5.
Using Pythagoras theorem, $(\text{Hypotenuse})^2=(\text{Base})^2+(\text{Perpendicular})^2$.
⇒ $(5)^2=(\text{Base})^2+4^2$
⇒ $25=(\text{Base})^2+16$
⇒ $(\text{Base})^2=9$
⇒ $\text{Base}=3$
The value of $\cot\alpha$ is given as $\frac{\text{Base}}{\text{Perpendicular}}$ i.e., $\cot \alpha=\frac{3}{4}$
Hence, the correct answer is $\frac{3}{4}$.