A rod of length L and mass M is acted on by two unequal forces $F_1$ and $F_2\left(<F\right)$as shown in the figure.

The tension in the rod at a distance y from the end A is given bya. $F_1\left(1-\frac{y}{L}\right)+F_2\left(\frac{y}{L}\right)$ b. $F_2\left(1-\frac{y}{L}\right)+F_1\left(\frac{Y}{L}\right)$ c. $\left(F_1-F_2\right)\frac{y}{L}$ d. None of these

Sure, here are the step-by-step solutions: 1. Net force on the rod = F1 - F2 2. As mass of the rod is M, acceleration of the rod is given by a = (F1 - F2) / M 3. Considering the motion of the rod, whose mass is (m/L)y, the equation of motion is: F1 - T = (M/L)y * a 4. Simplifying the equation, we get: T = F1(1 - y/L) + F2(y/L) 5. Alternatively, we can consider the motion of the other end of the rod (BC) and use the same equation of motion to get the same result: T - F2 = (M/L)(l - y) * a T = F2(M/L)(L - y) * (F1 - F2) / M Simplifying the equation, we get: T = F1(1 - y/L) + F2(y/L)