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If L(1) = (3.03 +- 0.02)m and L(2) = (2....

If `L_(1) = (3.03 +- 0.02)m` and `L_(2) = (2.01 +- 0.02)m` then `L_(1) + 2L_(2)` in (in `m`)

A

`7.05 +- 0.06`

B

`6.05 +- 0.06`

C

`6.05 +- 0.02`

D

`7.05 +- 0.02`

Text Solution

Verified by Experts

The correct Answer is:
A
`x = L_(1) + 2L_(2) = 7.05, Delta x = DeltaL, + 2Delta L_(2)`
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