Explanation

(b) Here, $p_{1} = 2\ cm\ of\ Hg$
$\ = 2 \times 13.6 \times 980$
$= 2.666 \times 10^{4}dyne\ {cm}^{- 2}$
$\ \ \ v_{1} = 32\ {cms}^{- 1},\ \ \ \ \ \ v_{2} = 65{cms}^{- 1}$
For a horizontal pipe, according to Bernoulli’s theorem,
$\ \ \ \ \frac{p_{1}}{\rho} + \frac{1}{2}v_{1}^{2} = \frac{p_{2}}{\rho} + \frac{1}{2}v_{2}^{2}$
$or\ \ p_{2} = p_{1} + \frac{1}{2}\rho\left( v_{1}^{2} - v_{2}^{2} \right)$
$or\ \ p_{2} = 2.666 \times 10^{4} + \frac{1}{2} \times \lbrack(32)^{2} - {(65}^{2})\rbrack$
$\ \ \ = 2.666 \times 10^{4} - 0.16 \times 10^{4}$
$\ \ \ = \ \frac{2.506 \times 10^{4}}{13.6 \times 980} = 1.88\ cm\ of\ Hg$