Explanation
Applying Bernoulli theorem across 1 and 2,
$$\dfrac{2mg}{A\rho }+2\rho gh=\dfrac{1}{2}\rho v^{2}$$
$$v_{2}=\dfrac{2mg}{A\rho }+2gh$$
$$v^{2}=2g\left ( h+\dfrac{m}{\rho A} \right )$$
$$v = \sqrt{2\left ( gh+\dfrac{mg}{\rho A} \right )}$$
$$Q = 500 cm^{3}/s$$Using continuity equation we have $$Q=A_1v_1=A_2v_2$$or$$500=5(v_1)=2(v_2)$$Thus we get$$v_1=100 cm/s=1 m/s$$ and $$v_2=250 cm/s=2.5 m/s$$
Applying Bernoullis therom, In point (1) and (2),$$\Rightarrow P_{1}-P_{2}=\dfrac{1}{2}\rho(v_2^2-v_1^2)=\dfrac{1}{2}1000(2.5^2-1^2)=2625 N/m^{2}$$$$\Rightarrow P_{1}-P_{2}=2625= \rho gh=13.6\times 1000\times 9.8\times h$$$$\Rightarrow h = 1.97 \ cm$$
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