12 questions · timed · auto-graded
$\text{T}=2\pi\sqrt{\frac{\ell}{\text{g}}}$
$\Rightarrow2=2\pi\sqrt{\frac{\ell}{10}}$
$\Rightarrow\frac{\ell}{10}=\frac{1}{\pi^2}$
$\Rightarrow\ell=1\text{cm}$ $(\therefore\pi^2\approx10)$
$\text{T}=2\pi\sqrt{\frac{\text{m}}{\text{k}}}=2\sec$ [Time period of pendulum of a clock = 2 sec]
So, $4\pi^{2+}\Big(\frac{\text{m}}{\text{k}}\Big)=4$
$\therefore\text{m}=\frac{\text{k}}{\pi^2}=\frac{0.1}{10}=0.01\text{kg}\approx10\text{gm}.$
$\theta=\pi\sin[\pi\sec^{-1}\text{t}]$
$\therefore\omega=\pi\sec^{-1}$ (comparing with the equation of SHM)
$\Rightarrow\frac{2\pi}{\text{T}}=\pi$
$\Rightarrow\text{T}=2\sec.$
We know that $\text{T}=2\pi\sqrt{\frac{\ell}{\text{g}}}$$\Rightarrow2=2\sqrt{\frac{\ell}{\text{g}}}$
$\Rightarrow1=\sqrt{\frac{\ell}{\text{g}}}$
$\Rightarrow\ell=1\text{m}.$
$\therefore$ Length of the pendulum is 1m.
$\text{T}_2=\frac{24\times3600}{\Big(\frac{24\times3600-24}{2}\Big)}=2\times\frac{3600}{3599}$
Now, $\frac{\text{g}_2}{\text{g}_1}=\Big(\frac{\text{T}_1}{\text{T}_2}\Big)^2$$\therefore\text{g}_2=(9.8)\Big(\frac{3599}{3600}\Big)^2=9.795\text{m/s}^2$
$\Rightarrow\sqrt{\frac{1}{\text{g}}}=\sqrt{\frac{\text{m}}{\text{k}}}$
$\Rightarrow\frac{1}{\text{g}}=\frac{\text{m}}{\text{k}}$
$\Rightarrow1=\frac{\text{mg}}{\text{k}}=\frac{\text{F}}{\text{k}}=\text{x}.$ (Because, restoring force = weight = F = mg)
$\Rightarrow1=\text{x }(\text{proved})$
$\therefore\text{gd}=\text{g}(1-\frac{\text{d}}{\text{R}})=9.8\Big(1-\frac{1600}{6400}\Big)$
$=9.8\Big(1-\frac{1}{4}\Big)=9.8\times\frac{3}{4}=7.35\text{m/s}^2$
$\therefore$ Time period $\text{T}'=2\pi\sqrt{\frac{\ell}{\text{g}\delta}}$
$=2\pi\sqrt{\frac{0.4}{7.35}}=2\pi\sqrt{0.054}=2\pi\times0.23$
$=2\times3.14\times0.23=1.465\approx1.47\sec.$
$\text{r}^2-\text{y}^2=\text{y}^2$
$\Rightarrow2\text{y}^2=\text{r}^2$
$\Rightarrow\text{y}=\frac{\text{r}}{\sqrt{2}}=\frac{10}{\sqrt{2}}=5\sqrt{2}\text{cm}$ From the the mean position.
$\text{T}=2\pi\sqrt{\frac{\text{l}}{\text{g}}}$ first l increases and then it decreases.