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

Correct answer: B. (2.0 ± 0.2) × 10^ 11 ~N / m^2 aligned Y & =(F L)/(A l)=(4 F L)/(π d^2 l) & = (4)(1.0 × 9.8)(2) π(0.4 × 10^ -3 )^2(0.8 × 10^ -3 ) =20 × 10^ 11 N / m^2 aligned Further (Δ Y)/(Y)=2 ((Δ d)/(d))+((Δ l)/(l)) aligned Δ Y & = \ 2((Δ d)/(d))+((Δ l)/(l))\ Y= \ 2 × (0.01)/(0.4)+(0.05)/(0.8)\ ×2.0 ×10^ 11 & =0.2 ×10^ 11 N/ m^2 or (Y+Δ Y)=(2+0.2) ×10^ 11 N/ m^2 aligned The correct option is (b).

JEE Advanced 2007Paper 2

Exam: JEE Advanced
Year: 2007
Paper: Paper 2
Subject: Physics
Chapter: Mathematics in Physics

JEE Advanced 2007 · Paper 2 · Physics · Mathematics in Physics

A student performs an experiment to determine the Young's modulus...

A student performs an experiment to determine the Young's modulus of a wire, exactly

2 \mathrm{~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 \mathrm{~mm}

with an uncertainty of

\pm 0.05 \mathrm{~mm}

at a load of exactly

1.0 \mathrm{~kg}

. The student also measures the diameter of the wire to be

0.4 \mathrm{~mm}

with an uncertainty of

\pm 0.01 \mathrm{~mm}

. Take

g=9.8 \mathrm{~m} / \mathrm{s}^2

(exact). The Young's modulus obtained from the reading is

Options

A $(2.0 \pm 0.3) 10^{11} \mathrm{~N} / \mathrm{m}^2$
B $(2.0 \pm 0.2) \times 10^{11} \mathrm{~N} / \mathrm{m}^2$
C $(2.0 \pm 0.1) \times 10^{11} \mathrm{~N} / \mathrm{m}^2$
D $(2.0 \pm 0.05) \times 10^{11} \mathrm{~N} / \mathrm{m}^2$

Correct answer

B. $(2.0 \pm 0.2) \times 10^{11} \mathrm{~N} / \mathrm{m}^2$

Step-by-step solution

\begin{aligned} Y & =\frac{F L}{A l}=\frac{4 F L}{\pi d^2 l} \\ & =\frac{(4)(1.0 \times 9.8)(2)}{\pi\left(0.4 \times 10^{-3}\right)^2\left(0.8 \times 10^{-3}\right)}=20 \times 10^{11} \mathrm{ N} / \mathrm{m}^2\end{aligned}

Further

\quad \frac{\Delta Y}{Y}=2 \left(\frac{\Delta d}{d}\right)+\left(\frac{\Delta l}{l}\right)

\begin{aligned} \therefore \quad \Delta Y & = \left\{2\left(\frac{\Delta d}{d}\right)+\left(\frac{\Delta l}{l}\right)\right \}Y= \left\{2 \times \frac{0.01}{0.4}+\frac{0.05}{0.8}\right\} \times2.0 \times10^{11} \\ & =0.2 \times10^{11} \mathrm{ N}/ \mathrm{m}^2 \text{\text{ or } }}(Y+\Delta Y)=(2+0.2) \times10^{11} \mathrm{ N}/ \mathrm{m}^2 \end{aligned}

$$ The correct option is (b).

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