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The radius of curvature of a railway tra...

The radius of curvature of a railway track at a place where the train is moving at a speed of `72 kmh^(-1)` is `625m`. The distance between the rails is 1.5 m. Find the angle and the elevation of the outer rail so that there may be no side pressure on the rails, Take `g = 9.8 ms^(-2)`.

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The correct Answer is:
`3^(@)44', 0.09775 m`
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The radius of curvature of railway track at a place where the train is moving at a speed of 72km h^(-1) is 625m The distance between the rails is 1.5m Find the angle and the elevation of the outer rail so that there may be no side pressure on the rails Take g = 9.8ms^(-2) .

The radius of curvature of a railway line at a place when the train is moving with a speed of 36 kmh^(-1) is 1000 m, the distance between the two rails being 1.5 metre. Calculate the elevation of the outer rail above the inner rail so that there may be no side pressure on the rails.

The radius of curvature of a metre gauge railway line at a place where the train is moving at 36km //h is 50 m .If there is no side thrust on the rails , then the elevation of the outer rail above the inner rail will be (g=10m//s^(2))

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The radius of curvature of railway line at a place is 200 m. If the distance between the rail is 1.6 m and the outer rail is raised by 0.08 m above the inner rail .The speed of the train for which there is no side pressure on the rails (g=10 m//s^(2))

Attempt any THREE : A meter gauge train is moving at 72 km/hr along a curved rail-way of radius of curvature 500m at a certain place. Find the elevation of the outer rail above the inner rail so that there is no side pressure on the rail. (g = 9.8 m//s^(2))

A train is moving at the rate of 60 km per hour around a curve of radius 500 m and the distance between the rails is 170 cm. By how much the outer rail be raised above the inner one so that there may be no lateral thrust on the outer rails ? ( g= 9.8 ms^(-2))

A train is running at 20 m/s on a railway line with radius of curvature 40,000 meters. The distance between the two rails is 1.5 meters. For safe running of train the elevation of outer rail over the inner rail is (g=10m//s^(2))

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