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NEET Sample Paper NEET Sample Test Paper-64

  • question_answer
    A student performs an experiment to determine the Young?s modulus of a wire, exactly 2 m long, by Searle?s method. In a particular reading, the student measures the extension in the length of the wire to be 0.8 mm with an uncertainty of 0.05 mm, at a load of exactly kg. The student also measures the diameter of the wire to be 0.4 mm with an uncertainty of 0.01 mm. Take \[\operatorname{g}= 9.8 m/{{s}^{2}}\] (exact). The Young?s modulus obtained from the reading is

    A) \[(2.0\,\,\pm \,\,0.3)\,\,\times \,\,{{10}^{11}}\,N/{{m}^{2}}\]

    B) \[(2.0\,\,\pm \,\,0.2)\,\,\times \,\,{{10}^{11}}\,N/{{m}^{2}}\]

    C) \[(2.0\,\,\pm \,\,0.1)\,\,\times \,\,{{10}^{11}}\,N/{{m}^{2}}\]

    D) \[(2.0\,\,\pm \,\,0.05)\,\,\times \,\,{{10}^{11}}\,N/{{m}^{2}}\]

    Correct Answer: B

    Solution :

    \[Y=\frac{Fl}{xA}=\frac{1\times 9.8\times 2}{(0.8\times {{10}^{-3}})\times \pi {{\left( \frac{0.4\times {{10}^{-3}}}{2} \right)}^{2}}}\] \[= 1.94 \times  1{{0}^{11}}N/{{m}^{2}}\approx  2.0 \times  1{{0}^{11}}\,N/{{m}^{2}}\] \[\frac{\Delta Y}{Y}=\frac{\Delta x}{x}+\frac{\Delta A}{A}=\frac{\Delta x}{x}+2\frac{\Delta d}{d}\] \[=\,\,\frac{0.05}{0.8}+\frac{2(0.01)}{0.4}=0.1125\] \[\Rightarrow \,\,\Delta Y=\left( 0.1125 \right)Y\] \[=0.1125\times 1.94\times {{10}^{11}}\] \[=\text{ }0.2185\times {{10}^{11}}\approx 0.2\times {{10}^{11}}\]


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