If the rate of gain of radial acceleration is 0.3 m per sec³ and full centrifugal ratio is developed. On the curve the ratio of the length of the transition curve of same radius on road and railway, is

ANS:A - 2.828

In question, it is asking "length of transition curve of road (L1) to railways (L2) ".
The simplified formula is.
L1/L2 = ( (C.Rfor road) ^3/2) /( (C.R for railway) ^3/2).
Put values in the above formula.
L1/L2 = ( (1/4) ^3/2)/( (1/8) ^3/2) ) = 2.828. Centrifugal ratio for road = 1/4,
Centrifugal ratio for railway = 1/8,

Centrifugal ratio(c.r.)=V^2/Rg where R is the radius and g is acc. due to gravity ---> Eq. 1,
and also V^2/R=Acc.rate x t where t is the time ---> Eq. 2.

V^2 is directly proportional to C.R. since R and g is constant.R is constant since the same radius is given for road and railway.

Therefore V is directly proportional to (C.R.)^(1/2).

Now from Eq. 2.
t=V^2/acc. rate*R ---> Eq. 3.
And we also know that L=V*t, where L is length v, is vel. and t is time.
Putting the value of t from Eq. 3 in the above equation we will get,
L=V^3/acc. rate * R.

Now, L is directly proportional to V^3 since acc. rate and radius are constant for both road and railway.

And V is directly proportional to C.R.^(1/2),
Hence L is directly proportional to C.R.^(3/2),
Now L1(road)=k*(1/4)^(3/2).
L2(railway)=k*(1/8)^(3/2).
Dividing L1 by L2.
We get L1/L2=(√8)=2*(√2) = 2*1.414 = 2.828. C.R.=v^2/Rg ---> eqn1
From this formula, v is proportional to under root of C.R.
L=V^3/CR ---> eqn2.

Now from enq 1 we can say that.
L1=(C.R.1)^(3/2),
L2=(C.R.2)^(3/2),
C.R.1=1/4,
C.R.2=1/8,
THEN L1/L2=(1/4)^(3/2)÷(1/8)^(3/2)=2^(3/2)=2.828. Centrifugal ratio for road = 1/4,
Centrifugal ratio for railway = 1/8,

Centrifugal ratio(c.r.)=V^2/Rg where R is the radius and g is acc. due to gravity ---> Eq. 1,
and also V^2/R=Acc.rate x t where t is the time ---> Eq. 2.

V^2 is directly proportional to C.R. since R and g is constant.R is constant since the same radius is given for road and railway.

Therefore V is directly proportional to (C.R.)^(1/2).

Now from Eq. 2.
t=V^2/acc. rate*R ---> Eq. 3.
And we also know that L=V*t, where L is length v, is vel. and t is time.
Putting the value of t from Eq. 3 in the above equation we will get,
L=V^3/acc. rate*R.

Now, L is directly proportional to V^3 since acc. rate and radius are constant for both road and railway.

And V is directly proportional to C.R.^(1/2),
Hence L is directly proportional to C.R.^(3/2),
Now L1(road)=k*(1/4)^(3/2).
L2(railway)=k*(1/8)^(3/2).
Dividing L1 by L2.
We get L1/L2=( √8)=2*(√2) = 2*1.414 = 2.828. For road centrifugal ratio = 1/4.
For railway = 1/8.
Centrifugal ratio = v^2/Rg.
R = radius

(v^2/r) = Acceleration rate * t.
t = time

t = L/v.
L = length.
v = velocity.

Here we get L=(v^3)/(acceleration * r).

Acceleration is given and r is same for road and railway.
So, L is directly proportional to v^3 and from centrifugal ratio = v^2/Rg.
v is directly proportional to square root of the centrifugal ratio.

From here ratio of length for road and railway;
L1/L2= (1/2)^3/(1/square root 8)^3.
= 2 * square root 2.
= 2 * 1.414.
= 2.828.