If the error in measuring the radius of the sphere is 2 % and that in measuring its mass is 3 % , then the error in measuring the density of material of the sphere is

Question

If the error in measuring the radius of the sphere is 2 % and that in measuring its mass is 3 % , then the error in measuring the density of material of the sphere is (A)  5 %  (B)  7 %  (C)  9 %  (D)  11 %

Tardigrade Admin · Accepted Answer

As density =( textmass/ textVolume) ∴ ρ=(M/(4)3π R3)=(3/4)(M/π R3) ∴ The percentage error in density is (Δρ/ρ)×100 %=((Δ M/M)+3(Δ R/R))×100 % =3 %+3(2 %)=3 %+6 %=9 %

Q. If the error in measuring the radius of the sphere is $2%$ and that in measuring its mass is $3%$ , then the error in measuring the density of material of the sphere is

$5%$

$7%$

$9%$

$11%$

As density $= \frac{mass}{Volume}$
$∴ ρ = \frac{M}{\frac{4}{3} π R ^{3}} = \frac{3}{4} \frac{M}{π R ^{3}}$
$∴$ The percentage error in density is
$\frac{Δ ρ}{ρ} \times 100% = (\frac{Δ M}{M} + 3 \frac{Δ R}{R}) \times 100%$
$= 3% + 3 (2%) = 3% + 6% = 9%$

Q. If the error in measuring the radius of the sphere is 2% and that in measuring its mass is 3% , then the error in measuring the density of material of the sphere is

5%

7%

9%

11%