Mathematics in Physics

JEE Main level, numerical answer, medium

96 of 121

Question 96 of 121

The density of a solid metal sphere is diameter. The maximum error in the density of the sphere is (x100)\left(\dfrac{x}{100}\right)%. If the relative errors in measuring the mass and the diameter are 6.06.0% and 1.51.5% respectively, the value of xx is −-

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Answer: 1050

ρ=m43π(d2)3\rho =\dfrac{m}{\dfrac{4}{3}\pi {\left(\dfrac{d}{2}\right)}^{3}}ρ=6mπd3\rho =\dfrac{6m}{\pi {d}^{3}}
taking log
ℓnρ=ℓnm−3ℓnd\ell n\rho =\ell nm-3\ell nd
Maximum error by differentiation
dρρ=dmm+3ddd\dfrac{d\rho }{\rho }=\dfrac{dm}{m}+3\dfrac{dd}{d}
dpρ=6+3×1.5\dfrac{dp}{\rho }=6+3\times 1.5
=10.5=10.5%=\dfrac{1050}{100}%
Ans. x=1050x=1050

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