The time period 'T' of a simple pendulum of length 'l' is given by T = `2pisqrt(l/g)`.Find the percentage error in the value of 'T' if the percentage error in the value of 'l' is 2%.

Explanation

Given that `( dl)/l xx 100= 2%`
`"Now "T=2pisqrt(l/g)`
` rArr log T = log (2pi)/sqrtg + 1/2 log l`
`rArr 1/T*(dT)/(dl) = 1/(2l)`
`rArr (dT)/T = 1/2 * (dl)/l`
` rArr (dT)/T xx 100 = 1/2*(dl)/l xx 100 = 1/2 xx 2% = 1%`
Therefore the percentage error in the value of T = 1%.