If the radius of a sphere is mentioned as 7 m with an error of 0.02 m then the approximate error in calculating its volume is

Question

If the radius of a sphere is mentioned as 7 m with an error of 0.02 m then the approximate error in calculating its volume is (A) 1.83 π m 3 (B) 2.25 π m3 (C) 4.39 π m 3 (D) 3.92 π m 3

Tardigrade Admin · Accepted Answer

Radius (r)=7 m Error in radius (d r)=0.02 m Volume of sphere (v)=(4/3) π r3 differentiate w.r. to ' r ' on both sides, (d v/d r)=(4 π/3)(3 r2) d v=4 π(r2) ⋅ d r=4 π(49) ⋅(0.02) d v=3.92 π m3

Q. If the radius of a sphere is mentioned as $7 m$ with an error of $0.02 m$ then the approximate error in calculating its volume is

$1.83 π m^{3}$

$2.25 π m^{3}$

$4.39 π m^{3}$

$3.92 π m^{3}$

Radius $(r) = 7 m$
Error in radius $(d r) = 0.02 m$
Volume of sphere $(v) = \frac{4}{3} π r^{3}$
differentiate w.r. to ' $r$ ' on both sides,
$\frac{d v}{d r} = \frac{4 π}{3} (3 r^{2})$
$d v = 4 π (r^{2}) \cdot d r = 4 π (49) \cdot (0.02)$
$d v = 3.92 π m^{3}$

Q. If the radius of a sphere is mentioned as 7m with an error of 0.02m then the approximate error in calculating its volume is

1.83πm3

2.25πm3

4.39πm3

3.92πm3