UPSEE Solved Model Paper 6

Given that the length of pendulum increases by 1%. Now we need to find out how many seconds will a pendulum lose or gain per day Equation of time period of pendulum is that : T = $2\pi\sqrt{\frac{L}{g}}$ = $2\pi\left(\frac{L}{g}\right)^{1/2}$ where : T is the time period , L is length of pendulum and g is constant Differentiate this equation with respect to L : $\frac{d}{dL}(T)$ = $\frac{d}{dL}\left[2\pi\sqrt{\frac{L}{g}}\right]$ = $\frac{d}{dL}\left[2\pi\left(\frac{L}{g}\right)^{1/2}\right]$ = $2\pi\times\frac{d}{dL}\left(\frac{L}{g}\right)^{1/2}$ = $2\pi\times\frac{1}{2}\times\left(\frac{L}{g}\right)^{-1/2}\times\frac{1}{g}$ = $\frac{\pi}{g}\times\frac{1}{\sqrt{\frac{L}{g}}}$ $\left(\frac{dT}{dL}\right)^2$ = $\frac{\pi^2}{g^2}\times\frac{1}{\frac{L}{g}}$ = $\frac{\pi^2}{gL}$ $\frac{dT}{dL}$ = $\frac{\pi}{\sqrt{gL}}$ dT = $\frac{\pi}{\sqrt{gL}}\times dL$ Given that the length of pendulum increases by 1% $\frac{dL}{L}$ = 1% = $\frac{1}{100}$ = 0.01 dL = 0.01 L ∴ dT = $\frac{\pi}{\sqrt{gL}}$ × 0.01 L dT = $\frac{0.01\pi}{\sqrt{g}}\times\frac{L}{\sqrt{L}}$ = $\frac{0.01\pi}{\sqrt{g}}\times\frac{\sqrt{L}\times\sqrt{L}}{\sqrt{L}}$ = $\frac{0.01\pi\sqrt{L}}{\sqrt{g}}$ = $0.01\pi\sqrt{\frac{L}{g}}$ = 0.01 × $\frac{1}{2}\times2\pi\sqrt{\frac{L}{g}}$ = 0.005 × T Total time per day T = 24 × 60 × 60 = 86400 dT = 0.005 × 86400 = 432 s ∴ Time lost per day , dT = 432 s