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 (YA/YB ) is 0.5. The ratio of elongations in A to B is .
- 1
- 4
- 8
- 6
Answer
(4) 6
Solution
[figure]
Tension in a string = weight of everything hanging below it:
TA=(M+2M)g=3Mg, TB=2Mg
Elongation L= TLAY (same area A for both strings)
LA LB= TATB LALB YBYA
= 32 2 10.5
=6
Answer: (4)