$\nu_{n}^{\prime}=\frac{(2 n-1) v}{4\left(L+0.6 r_{1}\right)} \quad$ where $0.6 r_{1}$ is the end correction
Now for fundamental mode i.e $n=1, \quad \nu_{1}^{\prime}=\frac{v}{4\left(L+0.6 r_{1}\right)}$
For open organ pipe with end correction: Frequency of $m^{t h}$ mode,
$\nu_{m}=\frac{m v}{2\left(L+2 \times 0.6 r_{2}\right)} \quad$ where $0.6 r_{2}$ is the end correction on one side.
Now for first overtonei.e $m=2, \quad \nu_{2}=\frac{2 v}{2\left(L+1.2 r_{2}\right)}$
But $\quad \nu_{1}^{\prime}=\nu_{2}$
$\frac{v}{4\left(L+0.6 r_{1}\right)}=\frac{2 v}{2\left(L+1.2 r_{2}\right)}$
$L+1.2 r_{2}=4 L+2.4 r_{1} \Longrightarrow r_{2}-2 r_{1}=2.5 L$
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($M$ is the mass of earth, $R$ is the radius of earth, $G$ is the gravitational constant)
$(A)$ a constant velocity of revolution.
$(B)$ has the least velocity when it is nearest to the sun.
$(C)$ its areal velocity is directly proportional to its velocity.
$(D)$ areal velocity is inversely proportional to its velocity.
$(E)$ to follow a trajectory such that the areal velocity is constant.
Choose the correct answer from the options given below
(density of water $=1000\; \mathrm{kgm}^{-3}$ )