2. Questions Bank
  3. JEE Main & Advanced
  4. Physics
  5. Elasticity

JEE Main & Advanced Physics Elasticity Question Bank Youngs Modulus and Breaking Stress

  • question_answer
    A 5 m long aluminium wire (\[Y=7\times {{10}^{10}}N/{{m}^{2}})\] of diameter 3 mm supports a 40 kg mass. In order to have the same elongation in a copper wire \[(Y=12\times {{10}^{10}}N/{{m}^{2}})\] of the same length under the same weight, the diameter should now be, in mm.                         [AMU 2000]

    A) 1.75    

    B)             1.5

    C)             2.5      

    D)             5.0

    Correct Answer: C

    Solution :

                    \[l=\frac{FL}{\pi {{r}^{2}}Y}\Rightarrow {{r}^{2}}\propto \frac{1}{Y}\]       (F,L and l are constant)             \[\frac{{{r}_{2}}}{{{r}_{1}}}={{\left( \frac{{{Y}_{1}}}{{{Y}_{2}}} \right)}^{1/2}}={{\left( \frac{7\times {{10}^{10}}}{12\times {{10}^{10}}} \right)}^{1/2}}\] Þ \[{{r}_{2}}=1.5\times {{\left( \frac{7}{12} \right)}^{1/2}}\]= 1.145 mm \ dia = 2.29 mm


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