The breaking stress of the metal used for the wire connecting the blocks is 2×109N/m2. Minimum radius of the wire is: [Take √4006π=4.60,g=10m/s2 for calculation.]
A
4.6×10−5m
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B
9.6×10−5m
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C
8.6×10−5m
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D
5.6×10−5m
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Solution
The correct option is A4.6×10−5m 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, T−10=1×a=a...(2) For right block, 20−T=2a...(3)
On adding Eq. (2) and (3), 10=3a ⇒a=103m/s2 On putting value of a in Eq.(2), T=403N Substituting in Eq.(1), σbreaking=(403)A ⇒A=4032×109 Breaking stress, σbreaking=2×109N/m2 Breaking stress will correspond to the minimum cross-sectional area of wire. ⇒πr2min=406×109
⇒rmin=√406×109×π=√4006π×1010
⇒rmin=4.6×10−5m Minimum value of required radius is 4.6×10−5m