00:01
Dear student, so we have been given atmospheric pressure that equals 10 kbascal.
00:16
And pressure due to the lid of 4kgs is given by, let's say, the pressure due to the lid equals force per unit area.
00:32
So its force should be equal to its weight that is mass times gravitational acceleration divided by area, area equals pi times the square of the radius.
00:45
So let's plug in the values.
00:47
Mass of the lid is given there is 4 kgs.
00:50
The gravitational acceleration is 9 .8 meters per second square.
00:55
I'm not writing the units, dear student, because all the units are in si system.
01:00
Let's take the value of pi equals 3 .14 times radius equals 10 centimeters.
01:10
10 centi is equal to 10 raised to the power minus 2 and its whole squared.
01:17
So from here the pressure exerted by the lid equals 1 .248 kilo pascal.
01:36
All idea student.
01:38
Now the total pressure should be equal to the pressure due to the lead plus the pressure due to the atmospheric or atmospheric pressure let's plug in the values the pressure due to the lid is 1 .248 kilo pascal plus the pressure due to atmosphere is given there is 101 kiosk so your sum is equal to 102 .248 kilo pascal so this is the total pressure to your student now using the standard pressure temperature table, we can write date...