23P3Q01

1a)

\displaystyle x=\frac{{{L}^{3}}gM}{4w{{t}^{3}}E}

Units of x \displaystyle =\text{m}

\displaystyle \begin{aligned} \text{Units of }\frac{{{L}^{3}}gM}{4w{{t}^{3}}E}   & =\frac{\text{(}{{\text{m}}^{3}}\text{)(m }{{\text{s}}^{-2}}\text{)(kg)}}{\text{(m)(}{{\text{m}}^{3}})(\text{N }{{\text{m}}^{-2}})} \\ & =\frac{{{\text{s}}^{-2}}\text{.kg}}{\text{(kg m }{{\text{s}}^{-2}})({{\text{m}}^{-2}})} \\ & =\text{m} \end{aligned}

1b)

\displaystyle \begin{aligned} E&=\frac{{{L}^{3}}gM}{4w{{t}^{3}}x} \\ & =\frac{{{(0.800)}^{3}}(9.81)(300\times {{10}^{-3}})}{4(2.10\times {{10}^{-2}}){{(4.56\times {10}^{-3})}^{3}}(1.00\times {{10}^{-2}})} \\ &=1.8918\times {{10}^{10}}\text{ Pa} \end{aligned}

\displaystyle \begin{aligned} \frac{\Delta E}{E} &=3\frac{\Delta L}{L}+\frac{\Delta g}{g}+\frac{\Delta M}{M}+\frac{\Delta w}{w}+3\frac{\Delta t}{t}+\frac{\Delta x}{x} \\ &=3\frac{0.005}{0.800}+\frac{0.01}{9.81}+\frac{2}{300}+\frac{0.02}{2.10}+3\frac{0.01}{4.56}+\frac{0.01}{1.00} \\ &=5.2539\% \end{aligned}

\displaystyle E=1.89\times {{10}^{10}}\text{ Pa }\pm 5\%

COMMENT: This question is rather unusual, in the sense that . In all the past year questions, the requirement was to present the final answer using the “1 s.f. matching d.p.” rule. And the answer would have been as follow:

\displaystyle \begin{aligned} \Delta E&=0.052539\times 1.8918\times {{10}^{10}} \\ & =9.940\times {{10}^{8}} \\ & =1\times {{10}^{9}}\text{ (1 sf)} \end{aligned}

\displaystyle E=(1.9\pm 0.1)\times {{10}^{10}}

COMMENT: It is debatable whether the uncertainty for g should be incorporated. Since we are using 9.81 m s-2 as the value for g, one may argue that g has a small percentage uncertainty of about \displaystyle \frac{\Delta g}{g}=\frac{0.01}{9.81}=0.102\%. However, regardless, it does not affect our final answer.

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