Question

The radius of a sphere is expressed as(5.3+ - 0.1) cm. Find the percentage error in the volume of the sphere

Answer 1

Assuming the measurements is of a sphere.

 

Volume of sphere is calculated by formula

 

V=43πr3

 

Log on both sides

 

Log V=log (43π) + 3logr

 

If ΔV and Δr are respective errors in volume and radius, from theory of errors we see that

 

ΔVV=3Δrr

Putting all values

 

ΔVV=3×±0.15.3=±0.0566

 

= ±5.7, rounding up to one decimal point

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Answer 2

Solution :-

Volume of sphere = 4/3πr³

For radius = 5.3 cm

Volume = 4/3*22/7*5.3*5.3*5.3

⇒ 13101.176/21

= 623.8655 cm³

For radius = 5.3 - 0.1 = 5.2 cm

Volume (0)= 4/3*22/7*5.2*5.2*5.2

⇒ 12373.504/21

= 589.2144 cm³

For radius = 5.3 + 0.1 = 5.4 cm

Volume (1) = 4/3*22/7*5.4*5.4*5.4

⇒ 13856.832/21

= 659.8491 cm³
 
The percentage error in volume is -

[V - V (0)]/V = (623.8655 - 589.2144)*100/623.8655

⇒ (34.6511*100)/623.8655

⇒ 5.55 %

[V (1) - V]/V = (659.8491 - 623.8655)*100/623.8655

⇒ (35.9836*100)/623.8655

⇒ 5.76 %

So, the percentage error in volume is in the interval of (5.55 % to 5.76 %)

Answer.