{"id":1644,"date":"2018-03-28T16:07:50","date_gmt":"2018-03-28T08:07:50","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=1644"},"modified":"2023-08-10T00:33:49","modified_gmt":"2023-08-10T00:33:49","slug":"pascals-principle-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/pascals-principle-problems-and-solutions.htm","title":{"rendered":"Pascals principle &#8211; problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">Pascals principle &#8211; problems and solutions<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">1. <u>Known :<\/u><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">The area of A<sub>1 <\/sub>= 10 cm<sup>2<\/sup><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">The area of A<sub>2<\/sub> = 100 cm<sup>2<\/sup><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">Force 2 (F<sub>2<\/sub>) = 100 Newton<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><u>Wanted :<\/u> Force 1 (F<sub>1<\/sub>)<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>Solution :<\/u><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"color: #000000; font-size: 12pt; font-family: 'times new roman', times, serif;\">P = F \/ A<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><em><span style=\"color: #000000;\"><i>P = <\/i><\/span><\/em><em><span style=\"color: #000000;\"><i>pressure<\/i><\/span><\/em><em><span style=\"color: #000000;\"><i>, F = <\/i><\/span><\/em><em><span style=\"color: #000000;\"><i>force<\/i><\/span><\/em><em><span style=\"color: #000000;\"><i>, A = <\/i><\/span><\/em><em><span style=\"color: #000000;\"><i>area<\/i><\/span><\/em><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><span style=\"color: #000000;\">P<\/span><sub><span style=\"color: #000000;\">1 <\/span><\/sub><span style=\"color: #000000;\">= F<\/span><sub><span style=\"color: #000000;\">1<\/span><\/sub><span style=\"color: #000000;\"> \/ A<\/span><sub><span style=\"color: #000000;\">1 <\/span><\/sub><span style=\"color: #000000;\"><br \/>\nP<\/span><sub><span style=\"color: #000000;\">2<\/span><\/sub><span style=\"color: #000000;\"> = F<\/span><sub><span style=\"color: #000000;\">2<\/span><\/sub><span style=\"color: #000000;\">\u00a0\/ A<\/span><sub><span style=\"color: #000000;\">2<\/span><\/sub><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><span style=\"color: #000000;\">P<\/span><sub><span style=\"color: #000000;\">1<\/span><\/sub><span style=\"color: #000000;\"> = P<\/span><sub><span style=\"color: #000000;\">2<\/span><\/sub><span style=\"color: #000000;\"><br \/>\nF<\/span><sub><span style=\"color: #000000;\">1<\/span><\/sub><span style=\"color: #000000;\">\u00a0\/ A<\/span><span style=\"color: #000000;\"><sub>1\u00a0<\/sub><\/span><span style=\"color: #000000;\">= F<\/span><span style=\"color: #000000;\"><sub>2\u00a0<\/sub><\/span><span style=\"color: #000000;\">\/ A<\/span><sub><span style=\"color: #000000;\">2<\/span><\/sub><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><span style=\"color: #000000;\">F<\/span><sub><span style=\"color: #000000;\">1<\/span><\/sub><span style=\"color: #000000;\">\u00a0\/ <\/span><span style=\"color: #000000;\">10 cm<\/span><span style=\"color: #000000;\"><sup>2 <\/sup><\/span><span style=\"color: #000000;\"><sub>\u00a0<\/sub><\/span><span style=\"color: #000000;\">= 100<\/span><span style=\"color: #000000;\"> N <\/span><span style=\"color: #000000;\">\/ 100 cm<\/span><span style=\"color: #000000;\"><sup>2<\/sup><\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><span style=\"color: #000000;\">F<\/span><sub><span style=\"color: #000000;\">1<\/span><\/sub><span style=\"color: #000000;\">\u00a0\/ <\/span><span style=\"color: #000000;\">10<\/span><span style=\"color: #000000;\"><sub>\u00a0<\/sub><\/span><span style=\"color: #000000;\">= 1<\/span><span style=\"color: #000000;\"> N<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><span style=\"color: #000000;\">F<\/span><sub><span style=\"color: #000000;\">1<\/span><\/sub><span style=\"color: #000000;\">\u00a0= (10)(1 N)<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><span style=\"color: #000000;\">F<\/span><sub><span style=\"color: #000000;\">1<\/span><\/sub><span style=\"color: #000000;\">\u00a0= 10 Newton<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">2. If the area of A<sub>1<\/sub> = 0.001 m<sup>2<\/sup> and the area of A<sub>2<\/sub> = 0.1 m<sup>2<\/sup> , external input force F<sub>1<\/sub> = 100 N, then the external output force F<sub>2<\/sub> ? <\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>Known :<img loading=\"lazy\" decoding=\"async\" class=\"alignright size-full wp-image-1647\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/03\/Pascal\u2019s-principle-problems-and-solutions-2.png\" alt=\"Pascal\u2019s principle - problems and solutions 2\" width=\"156\" height=\"91\" \/><\/u><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">The area of A<sub>1<\/sub> = 0.001 m<sup>2<\/sup><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">The area of A<sub>2<\/sub> = 0.1 m<sup>2<\/sup><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">External input force F<sub>1<\/sub> = 100 Newton<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><u>Wanted <\/u><u>:<\/u> External output force (F<sub>2<\/sub>)<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>Solution :<\/u><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><span style=\"color: #000000;\">P<\/span><sub><span style=\"color: #000000;\">1<\/span><\/sub><span style=\"color: #000000;\"> = P<\/span><sub><span style=\"color: #000000;\">2<\/span><\/sub><span style=\"color: #000000;\"><br \/>\nF<\/span><sub><span style=\"color: #000000;\">1<\/span><\/sub><span style=\"color: #000000;\">\u00a0\/ A<\/span><sub><span style=\"color: #000000;\">1\u00a0<\/span><\/sub><span style=\"color: #000000;\">= F<\/span><span style=\"color: #000000;\"><sub>2\u00a0<\/sub><\/span><span style=\"color: #000000;\">\/ A<\/span><sub><span style=\"color: #000000;\">2<\/span><\/sub><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><span style=\"color: #000000;\">100 N \/ <\/span><span style=\"color: #000000;\">0.001 m<\/span><span style=\"color: #000000;\"><sup>2<\/sup><\/span> <span style=\"color: #000000;\">= <\/span><span style=\"color: #000000;\">F<\/span><sub><span style=\"color: #000000;\">2<\/span><\/sub> <span style=\"color: #000000;\">\/ 0.1 m<\/span><span style=\"color: #000000;\"><sup>2<\/sup><\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><span style=\"color: #000000;\">100 N \/ <\/span><span style=\"color: #000000;\">0.001 <\/span><span style=\"color: #000000;\">= <\/span><span style=\"color: #000000;\">F<\/span><sub><span style=\"color: #000000;\">2<\/span><\/sub> <span style=\"color: #000000;\">\/ 0.1 <\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><span style=\"color: #000000;\">100,000 N<\/span> <span style=\"color: #000000;\">= <\/span><span style=\"color: #000000;\">F<\/span><sub><span style=\"color: #000000;\">2<\/span><\/sub> <span style=\"color: #000000;\">\/ 0.1<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><span style=\"color: #000000;\">F<\/span><sub><span style=\"color: #000000;\">2<\/span><\/sub><span style=\"color: #000000;\"> = (0.1)(100,000 N)<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><span style=\"color: #000000;\">F<\/span><sub><span style=\"color: #000000;\">2<\/span><\/sub><span style=\"color: #000000;\"> = 10,000 N<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><span style=\"color: #000000;\">3. <\/span><span style=\"color: #000000;\">Car&#8217;s weight = <\/span>16,000 N. What is the external input force F&#8230;<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>Known :<img loading=\"lazy\" decoding=\"async\" class=\"alignright size-full wp-image-1648\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/03\/Pascal\u2019s-principle-problems-and-solutions-3.png\" alt=\"Pascal\u2019s principle - problems and solutions 3\" width=\"222\" height=\"123\" \/><\/u><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">Car&#8217;s <a href=\"https:\/\/gurumuda.net\/physics\/gravitational-force-weight-problems-and-solutions.htm\" target=\"_blank\" rel=\"noopener\">weight <\/a>(w) = 16,000 N<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">Area of B (A<sub>B<\/sub>) = 4000 cm<sup>2<\/sup> = 4000 \/ 10,000 m<sup>2<\/sup> = 4 \/ 10 m<sup>2<\/sup> = 0.4 m<sup>2 <\/sup><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">Area of A (A<sub>A<\/sub>) = 50 cm<sup>2<\/sup> = 50 \/ 10,000 m<sup>2<\/sup> = 0.005 m<sup>2 <\/sup><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><u>Wanted :<\/u> Force F<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>Solution :<\/u><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">F \/ A<sub>A<\/sub> = w \/ A<sub>B<\/sub><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">F \/ 0.005 m<sup>2 <\/sup>= 16,000 N \/ 0.4 m<sup>2 <\/sup><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">F \/ 0.005 = 16,000 N \/ 0.4<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">F \/ 0.005 = 40,000 N<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">F = (40,000 N)(0.005)<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">F = 200 Newton<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">4.<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">Area of A is 60 cm<sup>2<\/sup> and area of B is 4,200 cm<sup>2<\/sup>, determine the external input force of F.<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>Known :<img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-2827 alignright\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/03\/Pascal\u2019s-principle-problems-and-solutions-3-1.png\" alt=\"Pascal\u2019s principle - problems and solutions 3\" width=\"195\" height=\"178\" \/><\/u><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">Area of A (A<sub>A<\/sub>) = 60 cm<sup>2<\/sup><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">Area of B (A<sub>B<\/sub>) = 4200 cm<sup>2<\/sup><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">Weight w (w) = 3500 Newton<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><u>Wanted :<\/u> F<sub>1<\/sub><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>Solution :<\/u><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">Force of F calculated using the equation of <a href=\"https:\/\/gurumuda.net\/physics\/pascals-principle.htm\" target=\"_blank\" rel=\"noopener\">Pascal&#8217;s principle<\/a> :<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">F<sub>1<\/sub> \/ A<sub>1<\/sub> = F<sub>2<\/sub> \/ A<sub>2<\/sub><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">F<sub>1<\/sub> \/ 60 cm<sup>2 <\/sup>= 3500 N \/ 4200 cm<sup>2<\/sup><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">F<sub>1<\/sub> \/ 60 = 35 N \/ 42<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">F<sub>1<\/sub> = (60)(35) \/ 42<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">F<sub>1<\/sub> = 2100 \/ 42<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">F<sub>1<\/sub> = 50 Newton<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><span lang=\"en-US\">5. The hydraulic lift has a large cross section and a small cross section. Large cross-sectional area is 20 times the small cross-sectional area. If on the small cross section is given an input force of 25 N, then determine the output force.<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>Known :<\/u><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><span lang=\"en-US\">Small cross section area <\/span>(A<sub>1<\/sub>) = A<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><span lang=\"en-US\">Large cross-sectional area <\/span>(A<sub>2<\/sub>) = 20A<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">Input force (F<sub>1<\/sub>) = 25 N<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><u>Wanted :<\/u> Output force (F<sub>2<\/sub>)<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>Solution :<\/u><\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-3266\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/03\/Pascal\u2019s-principle-problems-and-solutions-7.png\" alt=\"Pascal\u2019s principle - problems and solutions 7\" width=\"125\" height=\"154\" \/><\/span><\/p>\n<ol style=\"text-align: justify;\">\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>What is Pascal&#8217;s Principle?<\/strong><\/span>\n<ul>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer<\/strong>: Pascal&#8217;s Principle states that any change in pressure applied at any point in a fluid is transmitted undiminished throughout the fluid in all directions.<\/span><\/li>\n<\/ul>\n<\/li>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>How does a hydraulic lift work based on Pascal&#8217;s Principle?<\/strong><\/span>\n<ul>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer<\/strong>: In a hydraulic lift, a small force applied to a small-area piston creates a pressure that is transmitted undiminished to a larger-area piston. This results in a larger force on the larger piston, enabling it to lift heavy objects.<\/span><\/li>\n<\/ul>\n<\/li>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Why is the pressure change undiminished in Pascal&#8217;s Principle?<\/strong><\/span>\n<ul>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer<\/strong>: Because fluids are incompressible, when pressure is applied to one part of the fluid, it cannot compress, so it transfers the pressure change uniformly to every other part of the fluid.<\/span><\/li>\n<\/ul>\n<\/li>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>How does an increase in depth affect fluid pressure, considering Pascal\u2019s principle?<\/strong><\/span>\n<ul>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer<\/strong>: As depth increases in a fluid, the pressure also increases due to the weight of the fluid above. This is consistent with Pascal\u2019s principle as the pressure increase is experienced uniformly at a particular depth.<\/span><\/li>\n<\/ul>\n<\/li>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Why do our ears pop when ascending or descending quickly in an airplane or diving deep underwater?<\/strong><\/span>\n<ul>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer<\/strong>: Rapid altitude or depth changes lead to rapid pressure changes in the surrounding medium (air or water). Our ears pop as a way to equalize the internal pressure with the external pressure, in line with Pascal&#8217;s Principle.<\/span><\/li>\n<\/ul>\n<\/li>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>If a sealed syringe with no air bubbles is depressed, why is it hard to compress the fluid inside?<\/strong><\/span>\n<ul>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer<\/strong>: Fluids are virtually incompressible. When you try to decrease the volume of the fluid inside the syringe, the pressure increases, resisting the compression, according to Pascal&#8217;s Principle.<\/span><\/li>\n<\/ul>\n<\/li>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>How does Pascal&#8217;s Principle explain the phenomenon of a dam bursting if a small hole is created at its base?<\/strong><\/span>\n<ul>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer<\/strong>: Due to the weight of the water, the pressure at the base of the dam is high. Even a small opening allows this high pressure to be exerted on the surrounding structures, which can lead to catastrophic failure of the dam.<\/span><\/li>\n<\/ul>\n<\/li>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Why is it easier to squirt liquid from a nearly full squeeze bottle than a nearly empty one?<\/strong><\/span>\n<ul>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer<\/strong>: A full bottle has a larger amount of liquid which, when squeezed, transmits the applied pressure more effectively throughout the fluid due to Pascal&#8217;s Principle. An almost empty bottle has more air, which is compressible, so the same pressure doesn&#8217;t produce as forceful a squirt of liquid.<\/span><\/li>\n<\/ul>\n<\/li>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>In hydraulic systems, is the output force always greater than the input force?<\/strong><\/span>\n<ul>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer<\/strong>: Not necessarily. The ratio of the output force to the input force is determined by the ratio of the areas of the output and input pistons. The system is designed to amplify force by having a larger output piston area than input, but it&#8217;s the area ratio that determines force amplification.<\/span><\/li>\n<\/ul>\n<\/li>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Considering Pascal\u2019s Principle, why do deep-sea creatures have sturdy and compact bodies?<\/strong><\/span><\/li>\n<\/ol>\n<ul>\n<li style=\"text-align: justify;\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer<\/strong>: Deep-sea creatures experience enormous pressures due to the weight of the overlying water column. To survive, these creatures have evolved to have sturdy and compact bodies that can withstand these high pressures.<\/span><\/li>\n<\/ul>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"<p>Pascals principle &#8211; problems and solutions 1. Known : The area of A1 = 10 cm2 The area of A2 = 100 cm2 Force 2 (F2) = 100 Newton Wanted : Force 1 (F1) Solution : P = F \/ A P = pressure, F = force, A = area P1 = F1 \/ A1 &#8230; <a title=\"Pascals principle &#8211; problems and solutions\" class=\"read-more\" href=\"https:\/\/gurumuda.net\/physics\/pascals-principle-problems-and-solutions.htm\" aria-label=\"Read more about Pascals principle &#8211; problems and solutions\">Read more<\/a><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_seopress_titles_title":"","_seopress_titles_desc":"","_seopress_robots_index":"","_seopress_robots_follow":"","_seopress_robots_imageindex":"","_seopress_robots_snippet":"","_seopress_robots_primary_cat":"","_seopress_robots_breadcrumbs":"","_seopress_robots_freeze_modified_date":"","_seopress_robots_custom_modified_date":"","_seopress_robots_canonical":"","_seopress_social_fb_title":"","_seopress_social_fb_desc":"","_seopress_social_fb_img":"","_seopress_social_fb_img_attachment_id":0,"_seopress_social_fb_img_width":0,"_seopress_social_fb_img_height":0,"_seopress_social_twitter_title":"","_seopress_social_twitter_desc":"","_seopress_social_twitter_img":"","_seopress_social_twitter_img_attachment_id":0,"_seopress_social_twitter_img_width":0,"_seopress_social_twitter_img_height":0,"_seopress_redirections_value":"","_seopress_redirections_enabled":"","_seopress_redirections_enabled_regex":"","_seopress_redirections_logged_status":"","_seopress_redirections_param":"","_seopress_redirections_type":0,"_seopress_analysis_target_kw":"Pascals principle - problems and solutions","_seopress_news_disabled":"","_seopress_video_disabled":"","_seopress_video":[],"_seopress_pro_schemas_manual":[],"_seopress_pro_rich_snippets_disable_all":"","_seopress_pro_rich_snippets_disable":[],"_seopress_pro_schemas":[],"footnotes":""},"categories":[3],"tags":[],"class_list":["post-1644","post","type-post","status-publish","format-standard","hentry","category-solved-problems-in-basic-physics"],"gt_translate_keys":[{"key":"link","format":"url"}],"_links":{"self":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/1644","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/comments?post=1644"}],"version-history":[{"count":2,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/1644\/revisions"}],"predecessor-version":[{"id":8735,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/1644\/revisions\/8735"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=1644"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=1644"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=1644"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}