A wire of length l and Mass  per uniform length 6.0×10−3 kg/m is put under tension of 500N . Two consecutive frequencies it resonates at are 420Hz and 490Hz . Then l in meters is

Suppose the wire vibrates at 420Hz  in its nth   harmonic and  at 490Hz  in its n+1th   harmonic. f=n2lTμ ∴490420=n+1n ⇒n=6  Now, from first equation ∴420=62l5006×10−3 ⇒70=12×l500×1036 ⇒140l=5×1056 ⇒l=288.67140=2.0619