The monkey B shown in figure above is holding on to the tail of the monkey A who is climbing up an ideal rope. The masses of the monkeys A and B are 5 kg and 2 Kg respectively. If A can tolerate a tension of 30 N in its tail, what force should it apply on the rope in order to carry the monkey B with it? Take g = $10m/{s^2}$.

(A) Between 50 N and 80
(B) Between 30 N and 80 N
(C) Between 70 N and 105 N

 (D) Between 60 N and 90 N

Hint To answer this question, we need to form two expressions for the tension. After the evaluation of the expressions we will be left with the value of the tension and a. This will give us the required answer for this question.

Complete step by step answer
We know that from the question it can be given that:
Mass of the first monkey is ${M_1}$ which is 5 kg and the mass of the second monkey is ${M_2}$ which is 2 kg. The force is given as 30 N.
So, the first expression that can be formed from the diagram is given as:
$T - {F_g} - 30 - 5a = 0.....................(i)$
So, the second expression that can be formed from the diagram is given as:
$30 - 2g - 2a = 0.....................(ii)$
Now we have to multiply by 2 in the equation (i) and then multiply by 5 in the equation (ii) and then solve them to get the following results:
T = 105 N and $a = 5m/{s^2}$
Thus, the maximum force that is applied by the monkey A to carry monkey B is 105 N.

Hence the correct answer is option C.

Note It should be known to us that the tension force is defined as the force that is transmitted through a cable, rope or even a wire or a string when it is pulled tight by the forces acting from the opposite ends. The tension force is directed along the length of the cable and then pulls equally on the objects on the opposite ends of the respective wire.