CBSE Class 12-science Answered
pls solve sir
Asked by mufeedatvp2000 | 16 Apr, 2020, 07:35: PM
Expert Answer
If a metallic wire is stretched by a force F, we have , Y = (F/A) / ( Δl / l ) ...............(1)
where Y is Youngs modulus, A is area of crosssection, Δl is elongation of wire due to applied force and l is its original length
From the above eqn.(1), we get workdone W or Stored energy as, W = F×Δl = Y × [ (Δl)2 / l ] × A ...................(1)
In the given system of two masses connected by wire and rotated about the free end of wire,
Centrifugal force Fm acting on mass m which is a stretching force of wire of length 2l is, Fm = m ω2 ( 2 l ) = 2mω2l
Centrifugal force F2m acting on mass 2m which is a stretching force of wire of length 3l is, F2m = 2m ω2 ( 3 l ) = 6mω2l
Ratio of stretching forces , Fm : F2m = 2mω2l : 6mω2l = 1 : 3
Let Δl be the total elongation . Due to stretching force ratio, elongation ratio of wires connecting masses is given by (1/4)Δl : (3/4)Δl
Ratio of stored energy is obtained from eqn.(1) as, { Y × [ (1/16) (Δl)2 / l ] × A } : { Y × [ (9/16)(Δl)2 / l ] × A } or 1 : 9
Answered by Thiyagarajan K | 17 Apr, 2020, 08:27: AM
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