Calculate percentage error in determination of time period of a pendulumT=2πlgwhere l and g are measured with ±1% and ±2%.

T=2πlgT=(2π)(l)+1/2(g)-1/2Taking logarithm of both sides, we haveln(T)=ln(2π)+12(lnl)-12ln(g)  …… (i)Here, 2π is a constant, therefore ln(2π) is also a constant.Differentiating Eq. (i), we have1TdT=0+121l(dl)-121g(dg)dTTmax= maximum value of ±12dll∓12dgg=12dll+12dggThis can also be written asΔTT×100max=12Δll×100+12Δgg×100or percentage error in time period=±12( percentage error in l)+12( percentage error in g)=±12×1+12×2=±1.5%