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If L(1) = (2.02 +- 0.01)m and L(2) = (1....

If `L_(1) = (2.02 +- 0.01)m` and `L_(2) = (1.02 +- 0.01)m` then `L_(1) + 2L_(2)` is (in m)

A

`4.06 pm 0.02`

B

`4.06 pm 0.03`

C

`4.06 pm 0.0054`

D

`4.06 pm 0.01`

Text Solution

Verified by Experts

The correct Answer is:
B
`L_1 + 2L_2 = 2.02 + 2 xx 1.02 = 4.06`
`Delta L_1 + 2 DeltaL_2 = 0.01 + 2 xx 0.01 = 0.03`
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NARAYNA-UNITS AND MEASUREMENTS-EXERCISE -I (C.W)
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