A fireman wants to slide down a rope. The rope can bear a tension $\dfrac{3}{4}$ of the weight of the man. With what minimum acceleration should the fireman slide down:
(A) $\dfrac{g}{3}$
(B) $\dfrac{g}{6}$
(C) $\dfrac{g}{4}$
(D) $\dfrac{g}{2}$

Hint: Newton’s second law explains that the force needed to accelerate a body is the product of mass of the body and the acceleration of the body. For a large mass the acceleration will be smaller. So for the body to accelerate greater force must be applied. When the fireman slides down the rope the force is transmitted through the rope. The force acting on the rope is called the tension.

Complete step by step solution
 Given the rope can bear a tension $\dfrac{3}{4}$ of the weight of the man. Let’s take the weight of man as $mg$, where, $m$is the mass of the body and $g$ is the acceleration due to gravity. Therefore, the rope can bear a weight of $\dfrac{3}{4}mg$.
The minimum tension at the rope when the fireman slide down is given as,
$mg - \dfrac{3}{4}mg$.
The expression for force is given as
$F = ma$
where $a$ is the acceleration.
This force equating with the minimum tension at the rope, we get
$
  ma = mg - \dfrac{3}{4}mg \\
  \Rightarrow a = g - \dfrac{3}{4}g \\
  \Rightarrow a = g\left( {1 - \dfrac{3}{4}} \right) \\
  \Rightarrow a = \dfrac{g}{4} \\
 $
The fireman should slide down with a minimum acceleration of $\dfrac{1}{4}\;$ of the acceleration due to gravity.

Thus the answer is option C.

Note: The formula for tension has to be noticed. If the fireman slides up the rope the tension on the rope will be $T = mg + ma$. But when the fireman slides down the rope, then tension on the rope will be $T = mg - ma$.