A wire of `9.8xx10^(-3) kg/m` passes over a frictionless light pulley fixed on the top of a frictionless inclined plane which makes an angle of `30^(@)` with the horizontal. Masses m and M are tied at the two ends of wire such that m rests on the plane and M hangs freely vertically downwards. the entire system is in equilibrium and a transverse wave propagates along the wire with a velocities of `100 m//s`.

Options

A. m = 20 kg
B. m = 5 kg
C. m = 2 kg
D. m = 7 kg

Answer

A

Explanation

Speed `(v)=sqrt((T)/(mu))`
At equilibrium
Mg = `mgsin30^(@)=T`
`therefore M=(m)/(2)`
`implies 100=sqrt((Mg)/(9.8xx10^(-3)))`
`implies M=10kg and m=20kg`.