Question

A train has to negotiate a curve of 400m by how much the outer rail be raised with respect to inner rail for speed 48km/hr. The distance between rails is 1m?

Answer 1

 radius of curved path is r. width of path is b m. the outer edge is raised by h w.r.t inner edge so that  a car with speed v can pass safe over it. the value of h is
tanθ = v²/rg=h/b
h=v ²b/rg
substituting the   values according to given question,
v=48km/h =(48x 1000)/60x60m/s⇒40/3 m/s
b=1m
r=400m
g=9.8m/s2
h=(40/3)²x1/400x9.8
⇒0.0045m
⇒4.54cm

Answer 2

Answer:

2 / 45 m

Explanation:

Given,

A train has to negotiate a curve of 400m.

The distance between rails is 1m.

To find: by how much the outer rail be raised with respect to inner rail for speed 48km/hr.

Solution:

velocity = 48 km/hr = 40 / 3 m/s

For safe turn without friction  

tanФ = v^2/ rg = h /x

where h = outer edge

here, x = 1 m , g = 10 m/s² , r = 400 m

then, h = v² / rg = (40/3)² / 400*10 = 2 / 45 m

Therefore, the outer rail has be raised 2/45m.