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Manipal Medical Manipal Medical Solved Paper-2012

  • question_answer
    Two block of masses of 1 kg and 2 kg are connected by a metal wire going over a smooth pulley. The breaking stress of metal is\[\frac{40}{3\pi }\times {{10}^{6}}N/{{m}^{2}}\]. What should be the minimum radius of wire used if it should not break? \[(g=10m/{{s}^{2}})\].

    A)  0.5mm           

    B)  1 mm

    C)  1.5 mm           

    D)  2 mm

    Correct Answer: C

    Solution :

     Breaking stress\[=\frac{F}{A}\] \[=\frac{40}{3\pi }\times {{10}^{6}}N/{{m}^{2}}\] According to question \[\frac{40}{3\pi }\times {{10}^{6}}=\frac{3\times 10}{\pi {{r}^{2}}}\] \[{{r}^{2}}=\frac{9}{4\times {{10}^{6}}}\] \[r=1.5\times {{10}^{-3}}m\] \[=1.5\text{ }mm\]


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