Mathematics in Physics

JEE Main level, single correct, medium

73 of 121

Question 73 of 121

A cylindrical wire of mass(0.4±0.01)g(0.4\pm 0.01)g has length (8±0.04)cm(8\pm 0.04)cm and radius (6±0.03)mm(6\pm 0.03)mm. The maximum error in its density will be

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Answer: (D) 44%

The formula to calculate the density of the material of the wire is given by
ρ=MπR2L...(1)\rho =\dfrac{M}{{\pi R}^{2}L}...\left(1\right)
Hence, the formula to calculate the percentage error in calculating the density of the material can be written as
dρρ×100=[dMM+2dRR+dLL]×100...(2)\dfrac{d\rho }{\rho }\times 100=\left[\dfrac{dM}{M}+\dfrac{2dR}{R}+\dfrac{dL}{L}\right]\times 100...\left(2\right)
Substitute the values of the known parameters into equation (2) to calculate the required percentage error.
\begin{matrix} & = & \left[\dfrac{0.01}{0.4}+\dfrac{2\times 0.03}{6}+\dfrac{0.04}{8}\right]\times 100 \\ & = & 4%\end{matrix}

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