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JEE Mains · Physics · STD 11 - Units,dimensions and measurement
- JEE Mains 2019
- Medium
- MCQ
The diameter and height of a cylinder are measured by a meter scale to be \(12.6 \pm 0.1\, cm\) and \(34.2 \pm 0.1\, cm\), respectively. What will be the value of its volume in appropriate significant figures?
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(4260 \pm 80\,cm^3\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l}
V = \pi {R^2}h = \frac{\pi }{4}{D^2}h\\
\,\,\,\,\,\,\, = 4260\,c{m^2}\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{\Delta V}}{V} = 2\frac{{\Delta D}}{D} + \frac{{\Delta h}}{h}\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {2 \times \frac{{0.1}}{{12.6}} + \frac{{0.1}}{{34.2}}} \right)V\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{2 \times 426}}{{12.6}} + \frac{{426}}{{34.2}}\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 67.61 + 12.459 = 80.075\\
\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,V = 4260 \pm 80\,c{m^3}
\end{array}\)
V = \pi {R^2}h = \frac{\pi }{4}{D^2}h\\
\,\,\,\,\,\,\, = 4260\,c{m^2}\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{\Delta V}}{V} = 2\frac{{\Delta D}}{D} + \frac{{\Delta h}}{h}\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {2 \times \frac{{0.1}}{{12.6}} + \frac{{0.1}}{{34.2}}} \right)V\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{2 \times 426}}{{12.6}} + \frac{{426}}{{34.2}}\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 67.61 + 12.459 = 80.075\\
\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,V = 4260 \pm 80\,c{m^3}
\end{array}\)
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