Question

a parallel piped has length a breadth 2a and height 3a length a mesured as (2+-0.02 ) m the maximum percentage error in its volume is ​

Answer 1

Given,

For a parallel pipe,

length= a = (2 ± 0.02)m

breadth= 2a

height= 3a

To find,

Maximum percentage error of volume

Solution,

We know that,

The relative error in the measurement of length = $\frac{error in a}{a}$ (Δa/a)

                                                                                = ± $\frac{0.02}{2}$

                                                                                = ± 0.01

Percentage error in measurement of length = ± 0.01 ×100

                                                                         =± 1 %

Now,

Volume= length× breadth× height

           = a×2a×3a

           = 6a³

The relative error in volume= ΔV/V

                                              = ± 3 (Δa/a)

                                              = ± 3 × 0.01

Percentage error in volume = relative error in volume ×100%

                                              = ± 3 ×0.01 ×100%

                                              = ± 3%

Thus,

The maximum percentage error in the volume is ±3%.