Explanation:
Using Bernoulli's theorem,
\({p_1} + \frac{1}{2}\rho v_1^2 = {p_2} + \frac{1}{2}\rho v_2^2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\)
Here, \(p_{1}=\rho_{m} g h_{1}=13600 \times 9.8 \times 10^{-2}\)
\(p_{2}=13600 \times 9.8 \times h\)
\(\rho = 1000\,kg{m^{ - 3}}\)
\({v_1} = 35 \times {10^{ - 2}}\;m{s^{ - 1}},{v_2} = 65 \times {10^{ - 2}}\;m{s^{ - 1}}\)
\(\therefore\) From Eq.(1), we get
\(\Rightarrow 13600 \times 9.8 \times 10^{-2}+\dfrac{1}{2} \times 1000 \times(0.35)^{2}\)
\(=13600 \times 9.8 \times h+\dfrac{1}{2} \times 1000 \times(0.65)^{2}\)
\(\Rightarrow 13600 \times 9.8 \times h\)
\(=13600 \times 9.8 \times 10^{-2}+\dfrac{1}{2} 10^{3}\left[(0.35)^{2}-(0.65)^{2}\right]\)
\(\Rightarrow 13600 \times 9.8 \times h\)
\(=13600 \times 9.8 \times 10^{-2}-500(0.3)\)
After solving, \(h = 0.89\;cm\) of \(Hg\).