Question

The radius of a space is(5.3+-0.1)cm. Calculate the percentage error in volume and surface area with error limits.

Answer 1

solution:-

given by:-
$the \: volume \: is \: given \: as \: V = (4/3)πr3 \\ so, V = (4/3) x 3.14 x 5.33 \\ thus, \\ V = 623.30 cm3 \\ now, \: the \: error \: equation \\ ΔV/V \: = \: 3(Δr/r) \\ or \\ ΔV = V x 3(Δr/r) thus, \\ ΔV = 623.30 x 3x(0.1/5.3) \\ or \: error \: i n \: surface \: area \\ ΔV = +/- 35.28 cm3 \\ thus, \\ ΔV/V x 100 = (35.28/623.30) x 100 \\ so, \\ ΔV/V x 100 = 5.66 %$
hence pecentage error = 5.66%

■I HOPE ITS HELP■

Answer 2

Radius of sphere is given and it is equal to r = (5.3 ± 0.1) cm
We know, volume of sphere = 4/3πr³
E.g., V = 4/3 πr³
difference V with respect to r
dV/dr = 4πr²
dV = 4πr²dr
dividing V both sides,
dV/V = 4πr²dr/{4/3πr³}
dV/V = 3dr/r ⇒∆V/V = 3∆r/r
∴ error in V = 300 × ∆r/r

Here given, ∆r = 0.1 and r = 5.3
∴ % error in V = 300 × 0.1/5.3 = 30/5.3 = 5.66 %

Similarly, you can find relation between surface area and radius of sphere.
e g., % error in A = 200∆r/r
= 200 × 0.1/5.3
= 20/5.3 = 3.773 %