A metal string A is suspended from a rigid support and its free end is attached to a block of mass M . Second block having mass 2M is suspended at the bottom of the first block using a string B . The area of cross sections of strings A and B are same. The ratio of lengths of strings of A to B is 2 and the ratio of their Young's moduli (Y_A/Y_B) is 0.5 . The ratio of elongations in A to B is ______.

Step-by-step solution

Tension in string B is due to the block of mass 2M : T_B = 2Mg Tension in string A is due to both blocks of mass M and 2M : T_A = Mg + 2Mg = 3Mg The elongation in a string is given by Hooke's law as L = TL AY . The ratio of elongations in string A to string B is: L_A L_B = ( T_A T_B ) ( L_A L_B ) ( A_B A_A ) ( Y_B Y_A ) Given that L_A L_B = 2 , A_B A_A = 1 , and Y_A Y_B = 0.5 Y_B Y_A = 2 . Substituting the values into the ratio equation: L_A L_B = ( 3Mg 2Mg ) (2) (1) (2) L_A L_B = 3 2 4 = 6 Answer: 6