Problem 14 - A Complete Resource Book in Physics for JEE Main

Question

A string of length \(L\) and mass \(M\) are lying on a horizontal table. A force \(F\) is applied at one of its ends. Tension in the string at a distance \(x\) from the ends at which force is applied is (A) Zero (B) \(F\) (C) \(F(L-x) / L\) (D) \(F(L-x) / M\)

Step-by-Step Solution

Answer

The tension \(T\) in the string at a distance \(x\) from the end where the force is applied is given by: \(T = \frac{F(L-x)}{L}\). The correct answer is (C) \(\frac{F(L-x)}{L}\).

Step 1: Draw a free-body diagram of the segment

First, let's draw a free-body diagram (FBD) of the segment of the string at distance \(x\) from the end where the force is applied. Consider a point P at distance x from the end. The segment of the string from the end where the force is applied to P will experience force F in the horizontal direction and tension T in the opposite direction at the point P.

Step 2: Apply Newton's Second Law

Newton's Second Law states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. In this case, the net force acting on the segment is the difference between the applied force F and tension T. The mass of the segment is \(m = \frac{Mx}{L}\), and the acceleration of this segment is \(a\). Using Newton's Second Law, we get: \(F - T = m \cdot a\)

Step 3: Find the acceleration of the segment

To find the acceleration of this segment, we can use the relationship between force and mass for the entire string: \(F = M \cdot a\) We can solve for acceleration: \(a = \frac{F}{M}\)

Step 4: Substitute the mass of the segment and acceleration

Now, substitute the mass of the segment \(m = \frac{Mx}{L}\) and acceleration \(a = \frac{F}{M}\) into the equation from Step 2: \(F - T = (\frac{Mx}{L}) \cdot (\frac{F}{M})\)

Step 5: Solve for tension T

Solve the equation for T: \(T = F - \frac{Mx}{L} \cdot \frac{F}{M}\) \(T = F - \frac{Fx}{L}\) Now, we can see that the tension T in the string at a distance \(x\) from the end where the force is applied is given by: \(T = \frac{F(L-x)}{L}\) This matches answer choice (C). So, the correct answer is: (C) \(\frac{F(L-x)}{L}\)