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CBSE Class 9 Answered

A glass full of water has a bottom of area 20 cm2 , top of area 20 cm2 , height 20 cm and volume half a litre.

(a) Find the force exerted by the water on  the bottom

(b) Considering the equilibrium of the water, find the resultant force exerted by the sides of the glass on the warer.

Atmospheric pressure = 1.0 * 105 N m-2 Density of water = 1000 kg m-3 and g = 10 m s-2. .Take all numbers to be exact

Asked by Paresh | 30 Dec, 2015, 07:45: PM
answered-by-expert Expert Answer
begin mathsize 12px style left parenthesis straight a right parenthesis space Given colon straight h space equals space 0.2 space straight m straight g space equals space 10 space straight m divided by straight s squared straight rho space equals space 1000 space kg divided by straight m cubed straight A space equals space 20 space cross times space 10 to the power of negative 4 end exponent space straight m squared Total space pressure space at space the space bottom space equals space atmospheric space pressure space plus space pressure space of space the space liquid Pressure space of space the space liquid comma space straight P space equals space hρg space straight P space equals space 0.2 space cross times space 10 space cross times space 1000 space equals space 0.02 space cross times space 10 to the power of 5 space pa therefore space Total space pressure space equals space 1.01 space cross times space 10 to the power of 5 space cross times space 0.02 space cross times space 10 to the power of 5 space space equals space 1.03 space cross times 10 to the power of 5 space pa Force space equals space straight P space cross times space straight A space equals 1.03 space cross times 10 to the power of 5 space cross times space 20 space cross times space 10 to the power of negative 4 end exponent space equals space 206 space straight N left parenthesis straight b right parenthesis space Consider space the space curved space surface space as space sphere Let space straight r space be space the space radius space of space the space sphere comma space straight r space equals space 10 space cm space equals space 0.1 space straight m Area space of space sphere comma space straight A space equals space 4 πr squared space equals space 4 space cross times space 3.14 space cross times space 0.1 squared space equals space 0.123 space straight m squared The space average space liquid space pressure space along space the space depth straight P space equals space 1 over straight h integral subscript 0 superscript straight H space straight rho space straight g space straight h space dh therefore space straight P space equals space straight h ρg over straight h straight H squared over 2 space equals space ρgh over 2 Therefore comma space average space water space pressure comma space straight P space equals fraction numerator 1000 space cross times space 10 space cross times space 0.2 over denominator 2 end fraction space equals space 1000 Total space pressure space space equals space atmospheric space pressure space plus Avg space water space space pressure space straight P space equals space 1.01 space cross times space 10 to the power of 5 space plus space 0.01 space cross times space 10 to the power of 5 space equals space 1.02 space cross times space 10 to the power of 5 space pa straight F space equals space PA space equals space 1.02 space cross times space 10 to the power of 5 space cross times space 0.126 space equals space 1.3 space cross times space 10 to the power of 4 space straight N end style
Answered by Priyanka Kumbhar | 31 Dec, 2015, 09:48: AM

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