A spring balance is attached on both ends to strings; the strings hang over frictionless pulleys and are each connected to 20 N weights as shown in Fig. The reading on the scale will be closest to(A) 0 N(B) 10 N(C) 20 N(D) 40 N

 

 

I am answering according to this image-

 

The magnitude of tension in the string supporting the block is equal to the magnitude of its weight say W, but in a direction opposite to the direction of W, which is downwards. This is necessary to establish vertical equilibrium, or else the blocks would move. Also the magnitude of tension in the entire string must be uniform.Since the tension is available from both the blocks, one could infer that the magnitude of net tension in the string must be twice the amount from the single block, that is 2W.However, to account for uniform tension and the condition of vertical equilibrium, it is clear that the magnitude of tension in the string can never exceed the magnitude of weight W. Therefore, the magnitude of tension in the string is equal to W and not to 2W.The magnitude of this tension in the string is what read by the spring balance, and therefore the scale would read to the value of W which is given to be 20 N.Thus, the correct option is (C) 20 N while the others are ruled out.The correct option is:   RegardsArun (askIITians forum expert)