Question

answer this question​

Answer 1

Answer:

Given,

radius of curvature, \sf{r=1\ km=1000\ m}r=1 km=1000 m

distance between the rails, \sf{d=1\ m.}d=1 m.

speed of the train, \sf{v=36\ km\,h^{-1}=10\ m\,s^{-1}}v=36 kmh

−1

=10 ms

−1

Let \sf{g=10\ m\,s^{-2}.}g=10 ms

−2

.

There would be no side pressure against the rails when the train moves along the rail with maximum safe speed, which is given by,

\sf{\longrightarrow v^2=rg\tan\theta}⟶v

2

=rgtanθ

Then the angle of elevation will be,

\sf{\longrightarrow \theta=\tan^{-1}\left(\dfrac{v^2}{rg}\right)}⟶θ=tan

−1

(

rg

v

2

)

\sf{\longrightarrow \theta=\tan^{-1}\left(\dfrac{10^2}{1000\times10}\right)}⟶θ=tan

−1

(

1000×10

10

2

)

\sf{\longrightarrow \theta=\tan^{-1}\left(\dfrac{1}{100}\right)}⟶θ=tan

−1

(

100

1

)

\sf{\longrightarrow \theta=0.01\ rad}⟶θ=0.01 rad

Then the elevation of outer rail will be given by,

\sf{\longrightarrow h=d\sin\theta}⟶h=dsinθ

\sf{\longrightarrow h=1\times\sin(0.01)}⟶h=1×sin(0.01)

\sf{\longrightarrow\underline{\underline{h=0.01\ m}}}⟶

h=0.01 m