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Grade: 12th pass
        
 A 0.1kg mass is suspended from a wire of negligible mass. The length of the wire is 1m and its cross-sectional area is 4.9*10-7m2. If the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency 140 rad s -1. If the Young’s modules of the material of the wire is n*109Nm-2, the value of n is.
5 years ago

Answers : (2)

Aditi Chauhan
askIITians Faculty
396 Points
							

We know that 8 = vK/m ……..(i)

Here Y = FL/Al ? F = (YA/L)l

Comparing the above equations with F = kl we get

K= (YA/L) ………….(ii)

From (i) & (ii), ? = v YA/ml

? 140 = vn*109 * 4.9 * 10-7/0.1 * 1

? n =4

5 years ago
Shaana Shankar pandu
13 Points
							
W= k/m
K = wm
K=1960
Now we have F=kx, here x is elongation i.e ∆L 
F=k∆L
Force acting on the mass is it's own weight
Mg=k∆L
∆L= mg/k
∆L=0.0005 m
Now youngs modulas 
Y=FL/A∆L
Y=mgL/A∆L
If you solve this by putting all value you get-
Y=4×109
Now Y=n×109
Therefore n=4
4 months ago
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