A fighter plane moves in a vertical circle of radius 2500 m . The mass of the plane is 15000 kg and its speed at the olowest point of its motion is 900 km/ hour . What is the force exerted by air on the plane at the lowest point `(g=10 m//s^(2))`
A
`3xx10^(5)N`
B
`4.1xx10^(5)N`
C
`5.25xx10^(5)N`
D
`6.5xx10^(5)N`
Text Solution
Verified by Experts
The correct Answer is:
C
At the lowest point , the weight mg acts vertically downwards while the force exerted by air is vertically upwards ` therefore ` The resultant upward force `=F - mg ` this gives the C.P force `=(mv^2)/(r ) ` ` therefore (mv^2)/(r) = F - mg or F= (mv^(2)) /(r ) + mg ` in this case , `V= 900 xx(5)/(18) = 250 m//s , r= 2500m` `therefore F = (1500xx250xx250)/(2500)+15000xx10` `=15000xx25+15000xx10` `= 15000xx35=525 xx10^(3) = 5.25 xx10^(5) N`
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