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Two block of masses 1 kg and 2 kg are connected by a metal wire going over a smooth pulley as shown in figure. The breaking stress of the metal is . What should2 \times 10^{9} N/m^{2} be the minimum radius of the wire used. If it is not to break?
Take g = 10 m/s^{2}

Option: 1

4.6 \times 10^{-5} m
 


Option: 2

4.6 \times 10^{-6} m


Option: 3

46 \times 10^{-6} m

 


Option: 4

4.6 \times 10^{-7} m


Answers (1)

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Answer-1

\text {The stress in the wire}=\frac{\text {Tension}}{\text {Area of cross-section}}

To avoid breaking, this stress should not exceed the breaking stress.

Let the tension in the wire be T.

The equation of motion of the two blocks are

T-10\; N=\left ( 1\; kg \right )a

20\; N-T=\left ( 2\; kg \right )a

Eliminating 'a' from these equations

T=\left ( \frac{40}{3} \right )N

\text {Stress}=\frac{\left ( \frac{40}{3} \right )N}{\pi r^{2}}

If the minimum radius needed to avoid breaking is r,

2 \times 10^{9}=\frac{\frac{40}{3}}{\pi r^{2}}

r=4.6 \times 10^{-5} m

 

 

Posted by

Ritika Jonwal

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