A spherical body of radius r and density falls freely through a viscous liquid having density and viscosity and attains a terminal velocity v_0. Estimated maximum error in the quantity is : (Ignore e…

Question

A spherical body of radius r and density falls freely through a viscous liquid having density and viscosity and attains a terminal velocity v_0. Estimated maximum error in the quantity is : (Ignore errors associated with , and g, gravitational acceleration)

Accepted Answer

Correct answer: A. (a) The terminal velocity v_0 of a spherical body falling through a viscous liquid is given by Stokes' Law: v_0= 2r^2 ( - )g9 Rearranging the formula to express viscosity in terms of other variables: = 2r^2 ( - )g9v_0Since we are told to ignore errors associated with , , and g, these quantities along with the numerical factor 29 are treated as constants. The expression for can be written as =k r^2v_0, where k= 2 ( - )g9 is a constant. Taking the natural logarithm on both sides: ln ⁡ = ln ⁡k+2 ln ⁡r- ln ⁡v_0Differentiating both sides to find the relative error: d =0+2 drr- dv_0v_0For the estimated maximum fractional error, we take the absolute sum of the individual relative errors: =2 rr+ v_0v_0

$A spherical body of radius r and density σ falls freely through a viscous liquid having density ρ and viscosity η and attains a terminal velocity v_{0} . Estimated maximum error in the quantity η is : (Ignore errors associated with σ, ρ and g, gravitational acceleration)$

Solution

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