Two metallic wires P and Q have same volume and are made up of same material. If their area of cross sections are in the ratio 4:1 and force F1 is applied to P, an extension of l is produced. The force which is required to produce same extension in Q is F2. The value of F1F2 is .
Answer
16
Solution
l= FlAY and volume V=Al l= VA
l= FVA2Y
∵ l, V and Y are the same ⇒ F A2
F1F2= ( APAQ )2= ( 41 )2=16
Answer: 16