To determine the height by which the outer rail should be raised, we use the concept of railway banking, which involves balancing the centripetal force needed to negotiate a curve with the gravitational force on the inclined plane. The formula for the height \(h\) of the outer rail is given by:
\(h = \frac{v^2b}{gR}\)
where:
Plugging these values into the formula, we have:
\(h = \frac{(12)^2 \cdot 1.5}{10 \cdot 400}\)
Simplifying:
\(h = \frac{144 \cdot 1.5}{4000}\)
\(h = \frac{216}{4000}\)
\(h = 0.054 \, \text{m}\)
Converting this into centimeters, we find:
\(h = 5.4 \, \text{cm}\)
Thus, the height by which the outer rail should be raised is 5.4 cm.
Therefore, the correct answer is 5.4 cm.
For a train moving around a curve, the required banking angle θ is given by:
\[\tan \theta = \frac{v^2}{Rg}\]
where \(v = 12 \, \text{m/s}\), \(R = 400 \, \text{m}\), and \(g = 10 \, \text{m/s}^2\).
Substitute the values:
\[\tan \theta = \frac{12^2}{10 \times 400} = \frac{144}{4000} = \frac{h}{1.5}\]
where \(h\) is the height by which the outer rail should be raised over the inner rail, and the distance between the rails is 1.5 m.
Solving for \(h\):
\[h = \frac{144 \times 1.5}{4000} = 5.4 \, \text{cm}\]
Thus, the required height is 5.4 cm.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,



What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)