If the length of second's pendulum is decreased by $2\%$, how many seconds it will lose per day ...... $\sec$
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(d) $T \propto \sqrt l $
==> $\frac{{\Delta T}}{T} = \frac{1}{2}\frac{{\Delta l}}{l} = \frac{{0.02}}{2} = 0.01$
==>$\Delta T = 0.01\,T$
Loss of time per day $ = 0.01 \times 24 \times 60 \times 60$$ = 864\sec $
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$\omega =$ frequency of oscillation
Which one of the following graphs shows correctly the variation $a$ with $t$ ?
Here $a=$ acceleration at time $t$
$T=$ time period