Question

In the given figure, if the inclined plane and the pulley are frictionless, then the tension \(T\) in the string is

• A. \(\dfrac{Mg}{2}(1-\sin\theta)\)
• B. \(\dfrac{Mg}{2}(1+\sin\theta)\)
• C. \(Mg(1-\sin\theta)\)
• D. \(Mg(1+\sin\theta)\)

Hint:

For two equal masses connected by a string: \[ \boxed{ a=\frac{g(1-\sin\theta)}{2}, \qquad T=\frac{Mg}{2}(1+\sin\theta) } \]

Answer 1

The Correct Option is B

Step 1: Apply Newton's second law to the block on the incline. Taking motion of the block up the incline as positive, \[ T-Mg\sin\theta=Ma \] \[ \boxed{T=Ma+Mg\sin\theta} \]

Step 2: Apply Newton's second law to the hanging block. For the hanging mass, \[ Mg-T=Ma \] \[ \boxed{T=Mg-Ma} \]

Step 3: Solve for acceleration. Adding the two equations, \[ Mg-Mg\sin\theta=2Ma \] \[ a=\frac{g(1-\sin\theta)}{2} \] Substituting into \[ T=Mg-Ma, \] \[ T =Mg-\frac{Mg}{2}(1-\sin\theta) =\frac{Mg}{2}(1+\sin\theta). \] Hence, \[ \boxed{T=\frac{Mg}{2}(1+\sin\theta)} \] Therefore, \[ \boxed{(B)} \] is the correct answer.