With what minimum acceleration can a fireman slide down a rope whose breaking strength is two third of his weight?
A. $\dfrac{{\text{g}}}{{\text{2}}}$​
B. $\dfrac{{{\text{2g}}}}{{\text{3}}}$
C. $\dfrac{{{\text{3g}}}}{{\text{2}}}$
D. \[\dfrac{{\text{g}}}{{\text{3}}}\]​

- Hint: In order to solve this problem, you need to get the value of force after you get the value of the tension, and then you can get the value of the acceleration from the equation we've gotten.

Complete step by step answer:
On the basis of the question,
Net downward force in the chain, ma = mg – T .... (1)
It is given that the rope breaks when its tension is two-thirds of the man's weight.
So, the tension that the rope can bear is ${{\text{T}}_{{\text{min}}}}{\text{ = }}\dfrac{{\text{2}}}{{\text{3}}}{\text{mg}}$= T
So, the tension that the rope can bear is,
$ \Rightarrow $ma = mg - $\dfrac{{\text{2}}}{{\text{3}}}{\text{mg}}$
From the equation above we can say that acceleration a = g - $\dfrac{{\text{2}}}{{\text{3}}}{\text{g}}$=\[\dfrac{{\text{g}}}{{\text{3}}}\].
Hence, the correct option is D.

Note – Whenever you need to solve such questions always try to obtain the equation which will take you to the term asked in the problem. In these problems sometimes you have to leave some of the unusual information which has nothing to do in the way which will take you to the right answer. So, sometimes you have to ignore some information and have to focus on the important one.