2. Solved Papers
  3. JEE Main & Advanced

JEE Main & Advanced JEE Main Paper (Held On 25 April 2013)

  • question_answer
                    Two blocks of masses m and M are connected by means of a metal wire of cross-sectional area A Passing over a frictionless fixed pulley as shown in the figure. The system is then released. If  \[\operatorname{M}=2\] m, then the stress produced in the wire is:     JEE Main Online Paper ( Held On 25  April 2013 )

    A)                 \[\frac{2\operatorname{mg}}{3A}\]                      

    B)                                        \[\frac{4\operatorname{mg}}{3A}\]

    C)                                        \[\frac{\operatorname{mg}}{A}\]                                           

    D)                                        \[\frac{3\operatorname{mg}}{4A}\]

    Correct Answer: B

    Solution :

                     Tension in the wire, \[T=\left( \frac{2mM}{m+M} \right)g\] Stress\[=\frac{Force/Tension}{Area}=\frac{2mM}{A(m+M)}g\] \[=\frac{2(m\times 2m)g}{A(m+2m)}=(M=2m\,\text{given})\] \[=\frac{4{{m}^{2}}}{3mA}g=\frac{4mg}{3A}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner