MHTCET 20 Apr 2026 Shift 2 Paper

Concept:Since the surface is frictionless and the string is inextensible, all blocks move together with the same common acceleration under the applied force $F$.Explanation:Let the common acceleration of the system be $a$.The total mass of all three blocks is:$$2\ \mathrm{kg} + 8\ \mathrm{kg} + 12\ \mathrm{kg} = 22\ \mathrm{kg}$$Using Newton's second law for the entire system:$$a = \frac{F}{22}$$Now, tension $T_1$ pulls the two blocks on its left, whose combined mass is $2 + 8 = 10\ \mathrm{kg}$.Applying Newton's second law to these two blocks:$$T_1 = 10a$$Next, tension $T_2$ pulls only the $2\ \mathrm{kg}$ block.Applying Newton's second law to the $2\ \mathrm{kg}$ block:$$T_2 = 2a$$Dividing the two equations:$$\frac{T_1}{T_2} = \frac{10a}{2a} = \frac{5}{1}$$Therefore, the required ratio is $T_1 : T_2 = 5 : 1$.Answer:$T_1 : T_2 = 5 : 1$, so the correct option is B.