# The total length of a sonometer wire between fixed ends is 110 m .Two bridges are placed to divide the length of wire in ratio 6 :3 :2 .The tension in the wire is 400 N and the mass per unit length is 0.01 kg/m . What is the minimum common frequency with which three parts can vibrate ?

$(a)\;1100\;Hz\qquad(b)\;1000\;Hz\qquad(c)\;166\;Hz\qquad(d)\;100\;Hz$

As,wire divided in ratio 6:3:2 which gives the length 60:30:20 so HCF will give the λ/2, which is 10 so. λ=20 , Then we know for sonometer , frequency (f) =p/2L(T/µ )^1/2. P- no. of antinode L- length of sonometer wire in meter T-tension µ -mass per unit length And the value, 2L/p=λ So, p=11 L=1.1m ,T=400N, µ=0.01kg/m Putting in formula, f=11/2*1.1(400/0.01)^1/2 f=1000Hz