JEE Mains · Physics · STD 11 - Units,dimensions and measurement
- JEE Mains 2020
- Hard
- Numerical
The density of a solid metal sphere is determined by measuring its mass and its diameter. The maximum error in the density of the sphere is \(\left(\frac{x}{100}\right) \% .\) If the relative errors in measuring the mass and the diameter are \(6.0 \%\) and \(1.5 \%\) respectively, the value of \(x\) is
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(1050\)
Step-by-step Solution
Detailed explanation
\(\rho=\frac{M}{V}=\frac{M}{\frac{4}{3} \pi\left(\frac{D}{2}\right)^{3}}\)
\(\rho=\frac{6}{\pi} M D ^{-3}\)
taking log
\(\ell n \rho=\ell n \left(\frac{6}{\pi}\right)+\ell n M -3 \ell m D\)
Differentiates
\(\frac{ d p}{\rho}=0+\frac{ d M }{ M }-3 \frac{ d ( D )}{ D }\)
for maximum error
\(100 \times \frac{ d \rho}{\rho}=\frac{ dM }{ M } \times 100+\frac{3 d D }{ D } \times 100\)
\(=6+3 \times 1.5\)
\(=10.5 \%\)
\(=\frac{1050}{100} \%\)
\(x=1050.00\)
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