The period of oscillation T of a pendulum of length l at a place of acceleleration due to gravity g is given by `T=2pisqrt(l/g`. If the calculated length is 0.992 times the actual length and if the value assumed for g is 1.002times its actal value, the relative error in the computed value of T, is

Options

A. `0.005`
B. `-0.005`
C. `0.003`
D. `-0.003`

Answer

B

Explanation

we have, `l+ Deltal= 0.992 l and g+ Delta g =1.002 g`
`rArr" "=-0.008 l and Delta g =0.002 g` `rArr" " (Delta l)/l =-0.008 and (Delta g)/g= 0.002`
Now, `T=2pisqrt(l/g)`
`rArr" "log T = log 2 pi + 1/2 (log l-log g)`
`rArr" "d(log T) = d(log" " 2pi)+1/2d(log l-log g)`
`rArr (dT)/T=1/2((dl)/l-(dg)/g)`
`rArr" "(Delta T)/T=1/2((Deltal)/l-(Deltag)/g)" "[:' dT cong Deltat,dlcongDeltal and dg cong Deltag]`
`rArr" "(DeltaT)/T=1/2 (-0.008-0.002) = -0.005`