2. Sample Papers
  3. NEET

NEET Sample Paper NEET Sample Test Paper-76

  • question_answer
    A 5 m aluminum wire \[\left( y=7\times 1{{0}^{10}}N/{{m}^{2}} \right)\] of diameter 3 mm supports a 40 kg mass. In order to have same elongation in a copper wire \[\left( Y=12\times 1{{0}^{10}}N/{{m}^{2}} \right)\] of same length under same weight, the diameter should be, in mm

    A)  2.0                              

    B)  2.3

    C)  1.75                            

    D)  5.0

    Correct Answer: B

    Solution :

    \[Y=\frac{Fl}{A\Delta l}=\frac{4F}{\pi {{D}^{2}}}\times \frac{l}{\Delta l}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,D=\frac{4Fl}{\pi Y\Delta l}\,\,\,\,\Rightarrow \,\,\,\,D\propto \frac{1}{\sqrt{Y}}\] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\frac{{{D}_{Cu}}}{{{D}_{Al}}}=\sqrt{\frac{{{Y}_{Al}}}{{{Y}_{Cu}}}}\] \[{{D}_{Cu}}\,\,=\,\,3\sqrt{\frac{7\times {{10}^{10}}}{12\times {{10}^{10}}}}\,\,=\,\,2.3\,\,mm\]


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