1.

A one metre steel wire of negligible mass and area of cross-section 0.01 cm^(-2) is kept on a smooth horizontal table with one end fixed.A ball of mass 1 kg is attached to the other end. The ball and the wire are rotating with an angular velocity of omega. If the elongation of the wire is 2 mm, then omega is (Young's modulus of steel = 2 xx 10^(11) Nm^(-2) )

Answer»

5 rad `s^(-1)`
10 rad `s^(-1)`
15 rad `s^(-1)`
20 rad `s^(-1)`

SOLUTION :Given, elongation of the wire, `Delta`l= 2mm
= `2 xx 10^(-3)` m
Mass of the ball, m=1 kg
Length of wire, l = 1 m
area of cross - sectional of wire, A= 0.01`cm^(2) = 0.01 xx 10^(-4)` m
Young.s modulus of steel, R = `2 xx 10^(11) Nm^(-2)`
`therefore` TENSION force in wire , T = m `OMEGA^(2)`l
`therefore ` Stress = `("Tension")/("Area") = (m omega^(2)l)/(A )`
Strain = `( Delta l)/(l ) = ("stress")/("Young.s modulus")`
` Delta l=(m omega^(2) l^(2))/(YA)`
or`omega = sqrt((YA Delta)/(ML^(2))`
putting the given values, we get
`=sqrt((2 x 10^(11) xx 0.01 xx 10^(-4) xx 2 xx 10^(-3))/(1 xx (1)^2))`
`omega = 20 rad/"sec"^(-1)`


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