Question

the volume of two bodies are measured to be v1=(10.2±0.02)cm^3 and v2=(6.4±0.01)cm^3. calculate the sum and difference of percentage error

Answer 1

v1 =  ( A ± ΔA ) = ( 10.2 ± 0.02 )cm^3
v2 = ( B ± ΔB )  = ( 6.4 ± 0.01 )cm^3

let v3 be the sum of the given volumes
 therefore, v3= ( Z ± ΔZ )cm^3
   where Z is the true value of the sum and ± ΔZ is the error limit

ATQ-

              Z = A + B
                 =  10.2 + 6.4
                 =  16.6 cm^3

and        ± ΔZ = ±  (ΔA + ΔB )
                       =  ±  (0.02 +0.01 )    
                       =  ±  0.03  

therefore, v3 =  ( Z ± ΔZ )cm^3  = ( 16.6 ±  0.03 )  cm^3  


now, let v4 be the difference of the given volumes.
 therefore, v4 = ( X ± ΔX )cm^3

hence,    
              X = A - B
                 =  10.2 - 6.4
                 =  3.8 cm^3

and        ± ΔX = ±  (ΔA + ΔB )
                       =  ±  (0.02 +0.01 )    
                       =  ±  0.03   

therefore, v4 =  ( X ± ΔX )cm^3  = ( 3.8 ±  0.03 )  cm^3  

Answer 2

Answer:

• The sum value = ( 16.6 ± 0.03 ) cm³
• The difference value =  ( 3.8 ± 0.03 ) cm³
• The sum of percentage error = 16.6 ± 0.18%
• The difference of Percentage error = 3.8 ± 0.78%

Explanation:

Given:

V₁ = ( 10.2 ± 0.02 ) cm³

V₂ = ( 6.4 ± 0.01 ) cm³

To find:

Sum and difference of percentage error.

Solution:

ΔV = ± ( ΔV₁ + ΔV₂)      ..............................(1)

     = ± ( 0.02 + 0.01 ) cm³

ΔV = ± 0.03 cm³

The sum value is given by:

V₁ + V₂ = ( 10.2 + 6.4) cm³

V₁ + V₂ =  16.6 cm³

The sum of value = ( 16.6 ± 0.03 ) cm³

Therefore, the difference value is given by:

V₁ - V₂ = ( 10.2 - 6.4) cm³

V₁ - V₂ =  3.8 cm³

The difference value =  ( 3.8 ± 0.03 ) cm³

To find the sum of and difference of error percentage:

The final expression is given in the form of a = $a_{m}$±  Δ $a_{mean}$

Therefore,

$Relative error = \frac{a_{mean} }{a_{m} }\\ \\Percentage error = \frac{a_{mean} }{a_{m} } * 100$

The Percentage error = $\frac{0.03}{16.6} * 100$

                             = 0.18%

The sum of percentage error = 16.6 ± 0.18%

The percentage error = $\frac{0.03}{3.8} * 100$

                                    = 0.78%

The difference of Percentage error = 3.8 ± 0.78%

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