[amp_mcq option1=”1 Pa” option2=”91 Pa” option3=”981 Pa” option4=”9810 Pa” correct=”option3″]<\/p>\n
The correct answer is C. 981 Pa.<\/p>\n
The pressure at a depth of 1 m below the free surface of a body of water is equal to the weight of the water column above it. The density of water is 1000 kg\/m3, and the acceleration due to gravity is 9.8 m\/s2. Therefore, the pressure is:<\/p>\n
$P = \\rho g h = 1000 \\frac{kg}{m^3} \\times 9.8 \\frac{m}{s^2} \\times 1 m = 981 Pa$<\/p>\n
Option A is incorrect because it is the pressure at a depth of 1 cm below the free surface of water. Option B is incorrect because it is the pressure at a depth of 10 cm below the free surface of water. Option D is incorrect because it is the pressure at a depth of 100 cm below the free surface of water.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”1 Pa” option2=”91 Pa” option3=”981 Pa” option4=”9810 Pa” correct=”option3″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[648],"tags":[],"class_list":["post-7154","post","type-post","status-publish","format-standard","hentry","category-hydraulics-and-fluid-mechanics","no-featured-image-padding"],"yoast_head":"\n