IIT-JEE 2007 · Paper 2 · Physics Q2

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PhysicsSingle correct+3 / -1Question 2 of 66

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 $\pm0.05$ mm at a load of exactly 1.0 kg. The student also measures the diameter of the wire to be 0.4 mm with an uncertainty of $\pm0.01$ mm. Take $g=9.8\ \mathrm{m/s^2}$ (exact). The Young's modulus obtained from the reading is

(2.0\pm0.3)\times10^{11}\ \mathrm{N/m^2}

(2.0\pm0.2)\times10^{11}\ \mathrm{N/m^2}

(2.0\pm0.1)\times10^{11}\ \mathrm{N/m^2}

(2.0\pm0.05)\times10^{11}\ \mathrm{N/m^2}

Answer and worked solution

Official answer: A

Young's modulus.
$Y=\frac{4MgL}{\pi d^2\,\Delta l}=\frac{4(1.0)(9.8)(2)}{\pi(0.4\times10^{-3})^2(0.8\times10^{-3})}=1.95\times10^{11}\approx2.0\times10^{11}\ \mathrm{N/m^2}.$

Relative error. $M$ , $g$ and $L$ are exact:
$\frac{\Delta Y}Y=\frac{\Delta(\Delta l)}{\Delta l}+2\frac{\Delta d}d=\frac{0.05}{0.8}+2\cdot\frac{0.01}{0.4}=0.0625+0.05=0.1125.$

Absolute error. $\Delta Y=0.1125\times2.0\times10^{11}\approx0.225\times10^{11}$ , which rounds to 0.2 or 0.3; the key accepts both.

(A) (key: A or B)

Topics:Young's modulus by Searle's method Errors in measurement

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IIT-JEE 2007 · Paper 2, Physics question 2