InterviewSolution
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InterviewSolution
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A wire of mass `9.8 xx 10^(-3)` kg per metre passes over a frictionless pulley fixed on the top of an inclined frictionless plane which makes an angle of `30^(@)` with the horizontal Masses `M_(1)` and `M_(2)` are tied at the two ends of the wire The mass `M_(1)` rests on the plane and mass `M_(2)` hangs freely vertically downwards . The whole systemis in equilibrium Now a transverse wave propagates along the wire with a velocity of `100 m//s` If `g = 9.8 m//s^(2)` calculate the valuse of masses `M_(1)` and `M_(2)` . |
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Answer» Here, ` m = 9.8 xx 10^(-3) kg//m` `theta = 30^(@) , g = 9.8 m//s^(2) ` ` upsilon = 100 m//s , M_(1) = ? M_(2) = ? ` The various forces acting on the system are shown in As the system of two masses is in equilibrium therefore , ` T = M_(1) g sin theta = M_(1) g sin 30^(@) = (M_(1) g )/(2) ` ` R = M_(1) g cos theta = M_(1) g cos 30^(@) = M_(1) g sqrt3/(2) ` Also ` T = M_(2) g` From (i) and (iii) , ` T = (M_(1) g )/(2) = M_(2) g ` ` M_(1) = 2 M_(2) ` Now the velocity of transverse waves is ` upsilon= sqrt((T)/(m) ` ` T = upsilon^(2) xx m = (100)^(2) xx 9.8 xx 10^(-3) = 9.8 N ` From (iii) `M_2 = (T)/(g) = (98)/(9.8) = 10 kg ` From `M_(1) = 2 M_(2) = 2 xx 10 kg = 20 kg ` |
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