Answer
$3\;cm$
Work Step by Step
The volume flow rate is the same for the two holes
$A_1v_1=A_2v_2$
or, $\frac{A_2}{2}v_1=A_2v_2$
or, $\frac{v_2}{v_1}=\frac{1}{2}$
Using the the Bernoulli’s equation, we obtain
$v_1=\sqrt {2gh_1}$
and $v_2=\sqrt {2gh_2}$
Therefore, the ratio becomes
$\sqrt {\frac{h_2}{h_1}}=\frac{1}{2}$
or, $h_2=h_1\times\frac{1}{4}$
or, $h_2=12\times\frac{1}{4}\;cm$
or, $\boxed{h_2=3\;cm}$
Therefore, the height above the hole in tank 2 is $3\;cm$
