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Question

The breaking stress of the metal used for the wire connecting the blocks is 2×109 N/m2. Minimum radius of the wire is:
[Take 4006π=4.60 ,g=10 m/s2 for calculation.]


A
4.6×105 m
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B
9.6×105 m
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C
8.6×105 m
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D
5.6×105 m
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Solution

The correct option is A 4.6×105 m
Tension in the wire will produce stress. At any cross-section of wire, stress will be
σ=FA
Let the tension in the wire be T when the wire is about to break. At that instant, stress in the wire will be breaking stress.
σ=σbreaking
σbreaking=TA ...(1)


Applying Newton's second law F=ma along the direction of acceleration of the blocks,
For left block,
T10=1×a=a ...(2)
For right block,
20T=2a ...(3)

On adding Eq. (2) and (3),
10=3a
a=103 m/s2
On putting value of a in Eq.(2),
T=403 N
Substituting in Eq.(1),
σbreaking=(403)A
A=4032×109
Breaking stress, σbreaking=2×109 N/m2
Breaking stress will correspond to the minimum cross-sectional area of wire.
πr2min=406×109

rmin=406×109×π=4006π×1010

rmin=4.6×105 m
Minimum value of required radius is 4.6×105 m

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