if the errors involved in the measurements of a side and mass of a cube are 3% and 4% respectively ,what is the maximum permissible error in the density of the material?

we know that

density = mass / volume

or

ρ = M/V = M / L3

here,

L  =  length of each side

so, in terms of percentage errors

(Δρ/ρ) x 100 = [(ΔM/M) x 100] + 3[(ΔL/L) x 100] 

given,

percentage error in M = 4%

percentage error in L = 3%

so,

(Δρ/ρ) x 100 = 4% + 3x(3%) = 4% + 9%

thus,

percentage error in density will be

(Δρ/ρ) x 100 = 13%

  • 112

Percentage error in the measurement of the side of the cube= 3%

Percentage error int he measurement of the mass of the cube= 4%

Density

= mass/volume

= Kg1/m3

therefore

dD/D = dm/m + 3*dm/m

dD/D*100% = (dm/m + 3*dm/m)*100% {calculating the percentage, 100 on LHS and RH cancel out each other}

therfore

dD/D = dm/m + 3*dm/m

= 4%+3*4%

= 4%+12%

= 16%

Thumbs up pls!!!!!!!!

  • -41
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