Question

calculate the pressure due to watercolumn of height of 10cm ( g = ms)

Answer 1

$\rho = density = {10}^{3} \: kg / {m}^{3} \\ g = 10 \: m/ {s}^{2} \\ h = 10 \: cm \\ \\ \implies \: pressure = \rho \times g \times h \\ \\ \implies \: P = {10}^{3} \times 10 \times 10 \\ \\ \implies \: P = {10}^{4} \times \frac{10}{100} \: \: \: \because1m = 100cm \\ \\ \implies \: p = {10}^{3} \: n/ {m}^{2} \\ \\ \implies \: p = {10}^{3} Pa \\ \\ \implies\: p = 1 \: kPa$