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Question 1
ప్రశ్న 1
On a conventional Indian Railways horizontal curve, what is meant by positive cant or superelevation?
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Explanation:
Positive cant is the vertical difference between the rail levels on a curve. The outer rail is normally raised relative to the inner rail. This inclination helps balance the lateral effect of negotiating the curve at the selected equilibrium speed. Cant is not a substitute for adequate curve radius, gauge control or track support. Excessive cant can be undesirable for slower trains, so the provided value must suit the traffic mix.
వివరణ:
Question 2
ప్రశ్న 2
For broad-gauge equilibrium-cant calculations using the Indian Railways Permanent Way Manual convention, which effective transverse dimension should normally be adopted?
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Explanation:
The nominal broad gauge of Indian Railways is 1676 mm. Equilibrium-cant calculations use a dynamic gauge that accounts for the transverse wheel–rail contact geometry. The adopted broad-gauge value in the permanent-way cant equation is 1750 mm. Substituting nominal gauge in its place changes the calculated cant. A self-contained numerical question should therefore identify which gauge convention it uses.
వివరణ:
Question 3
ప్రశ్న 3
With equilibrium cant C_e in millimetres, dynamic gauge G in millimetres, speed V in km/h and curve radius R in metres, which is the correct Indian Railways equilibrium-cant formula?
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Explanation:
Equilibrium cant balances the lateral effect of curving in the idealized railway-cant model. Centripetal acceleration is proportional to the square of speed and inversely proportional to radius. Conversion of speed from kilometres per hour and the stated length units gives the constant 127 in the denominator. The resulting relationship is C_e = G V^2/(127 R). The same formula can be rearranged to determine a corresponding equilibrium speed or minimum curve radius.
వివరణ:
Question 4
ప్రశ్న 4
If the running speed on a railway curve is doubled while the curve radius and dynamic gauge remain unchanged, what happens to the theoretical equilibrium cant?
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Explanation:
The equilibrium-cant formula contains the square of the train speed. Doubling speed changes V squared by a factor of four. The curve radius and dynamic gauge are held fixed in this comparison. Therefore the equilibrium cant becomes four times its initial value. In practice, permissible cant and cant deficiency constrain the operating speed rather than allowing cant to increase without limit.
వివరణ:
Question 5
ప్రశ్న 5
For an ideal broad-gauge curve of radius 900 m, a train runs at 90 km/h. Using G = 1750 mm and C_e = G V^2/(127 R), what is the equilibrium cant before rounding to the nearest 5 mm?
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Explanation:
The stated relation gives C_e = 1750 times 90 squared divided by 127 times 900. The square of 90 is 8100. The computed cant is approximately 124.02 mm. Thus the closest unrounded result is 124 mm. Where the Indian Railways procedure requires a multiple of 5 mm, the corresponding rounded design figure would be 125 mm.
వివరణ:
Question 6
ప్రశ్న 6
A broad-gauge railway curve has radius 720 m and a train runs at 72 km/h. For G = 1750 mm, what is the equilibrium cant before prescribed design rounding?
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Explanation:
The equilibrium-cant formula is C_e = G V squared divided by 127 R. The speed squared is 72 squared, or 5184. Substituting G = 1750 mm and R = 720 m gives 9072000/91440 mm. This is approximately 99.2 mm. Actual provided cant must be chosen using the applicable rounding, operational and transition requirements.
వివరణ:
Question 7
ప్రశ్న 7
At a fixed curve radius, equilibrium cant is 80 mm for a train running at 60 km/h. What is the theoretical equilibrium cant at 90 km/h, assuming the same dynamic gauge?
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Explanation:
Equilibrium cant at constant radius is proportional to the square of the train speed. The speed ratio is 90/60 = 1.5. Squaring this ratio gives 2.25. Multiplying the initial cant of 80 mm by 2.25 gives 180 mm. The calculation establishes theoretical equilibrium cant and does not imply that providing 180 mm is automatically permissible for the route.
వివరణ:
Question 8
ప్రశ్న 8
Two curves are traversed at the same train speed on track of the same dynamic gauge. If the second curve has twice the radius of the first, how does its equilibrium cant compare?
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Explanation:
Equilibrium cant is inversely proportional to curve radius when gauge and speed are fixed. Doubling the radius therefore divides the calculated cant by two. The centrifugal acceleration also decreases as the radius increases. A flatter curve generally permits a higher speed for the same cant and cant-deficiency allowance. Actual speed approval also depends on transitions and the applicable sectional restrictions.
వివరణ:
Question 9
ప్రశ్న 9
A broad-gauge curve has R = 1000 m, actual cant C_a = 80 mm and train speed V = 90 km/h. With G = 1750 mm, what cant deficiency does the train experience approximately?
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Explanation:
The theoretical equilibrium cant at 90 km/h is 1750 times 8100 divided by 127000. This equals approximately 111.61 mm. Actual cant is 80 mm, which is lower than the theoretical requirement. Cant deficiency is therefore 111.61 minus 80 = 31.61 mm. The closest listed answer is 32 mm, and its admissibility must be assessed under the relevant rolling-stock provisions.
వివరణ:
Question 10
ప్రశ్న 10
A slow goods train negotiates a broad-gauge curve of radius 800 m at 50 km/h. Actual cant is 100 mm and G = 1750 mm. What cant excess does it experience approximately?
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Explanation:
The theoretical equilibrium cant at 50 km/h is 1750 times 2500 divided by 127 times 800. This gives approximately 43.06 mm. The actual cant of 100 mm exceeds the theoretical requirement at that speed. Cant excess is therefore 100 minus 43.06 = 56.94 mm. The answer is approximately 57 mm, and excessive values can increase loading and wear on the inner rail.
వివరణ:
Question 11
ప్రశ్న 11
On a broad-gauge curve of radius 700 m with actual cant 100 mm, a passenger train runs at 100 km/h while a goods train runs at 50 km/h. With G = 1750 mm, which pair is approximately correct?
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Explanation:
The passenger train has equilibrium cant of approximately 196.85 mm at 100 km/h. Relative to 100 mm actual cant, its deficiency is approximately 96.85 mm. The goods train has equilibrium cant of approximately 49.21 mm at 50 km/h. Relative to the same 100 mm actual cant, its cant excess is approximately 50.79 mm. This example shows why actual cant must be selected by considering both fast and slow traffic and the applicable limits for each.
వివరణ:
Question 12
ప్రశ్న 12
For a fully transitioned broad-gauge curve, assume R = 600 m, actual cant C_a = 90 mm and the applicable permissible cant deficiency C_d = 75 mm. Using V = 0.27 sqrt[R(C_a + C_d)], what curve speed is obtained approximately, before other speed constraints?
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Explanation:
The permitted speed expression includes the sum of actual cant and allowed deficiency. Here that sum is 90 + 75 = 165 mm. Multiplication by the radius gives 600 times 165 = 99000. Hence V = 0.27 times the square root of 99000, or approximately 84.95 km/h. The resulting 85 km/h is a cant-based value and remains subject to transition, sectional-speed and other restrictions.
వివరణ:
Question 13
ప్రశ్న 13
A fully transitioned broad-gauge curve has radius 800 m, actual cant 80 mm and an explicitly permitted cant deficiency of 75 mm. Using V = 0.27 sqrt[R(C_a + C_d)], what is the approximate cant-based limiting speed?
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Explanation:
The available total equivalent cant is 80 + 75 = 155 mm. The product of radius and equivalent cant is 800 times 155 = 124000. Taking the square root and multiplying by 0.27 gives approximately 95.08 km/h. Thus the closest answer is 95 km/h. A final sanctioned speed must also respect transition suitability and all other operating restrictions.
వివరణ:
Question 14
ప్రశ్న 14
For a broad-gauge train at 100 km/h, suppose the maximum combined value of actual cant plus permissible cant deficiency is 175 mm. With G = 1750 mm and C_e = G V^2/(127 R), what minimum curve radius follows from the cant criterion alone?
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Explanation:
At the stated speed the theoretical equilibrium cant must not exceed the permitted sum of 175 mm. Rearranging the equilibrium-cant equation gives R = G V squared divided by 127 times C_e. The substitution is 1750 times 10000 divided by 127 times 175. This produces approximately 787.4 m, which rounds to about 790 m. The value is the minimum arising from the specified cant criterion alone and does not independently establish full track design compliance.
వివరణ:
Question 15
ప్రశ్న 15
A broad-gauge train travels at 90 km/h on a curve of radius 800 m. If G = 1750 mm, the maximum permitted cant deficiency is 75 mm, and actual cant must be selected in multiples of 5 mm, what is the smallest acceptable actual cant under this criterion?
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Explanation:
At 90 km/h and 800 m radius, the equilibrium cant is approximately 139.52 mm. The actual cant must be at least 139.52 minus 75 = 64.52 mm. The required value is rounded upward to the next available multiple of 5 mm. Therefore 65 mm is the smallest specified cant that satisfies the deficiency criterion. This answer addresses only cant deficiency and assumes other design requirements are separately satisfied.
వివరణ:
Question 16
ప్రశ్న 16
A goods train runs at 45 km/h on a broad-gauge curve of radius 600 m. If G = 1750 mm, allowable cant excess is 75 mm and provided cant must be a multiple of 5 mm, what is the largest permitted actual cant under this criterion?
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Explanation:
The equilibrium cant at 45 km/h and 600 m radius is approximately 46.51 mm. Actual cant may not exceed this equilibrium value by more than 75 mm. The resulting upper bound is approximately 121.51 mm. Among multiples of 5 mm, the largest value not exceeding that bound is 120 mm. This limit may need to be reduced further if other track or operating requirements govern.
వివరణ:
Question 17
ప్రశ్న 17
A mixed-traffic broad-gauge curve has radius 1000 m and carries trains at 100 km/h and 50 km/h. If both the fast-train cant deficiency and slow-train cant excess must not exceed 75 mm, which listed actual cant satisfies both requirements using G = 1750 mm?
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Explanation:
Equilibrium cant is approximately 137.80 mm at 100 km/h. For the fast train, actual cant must therefore be at least 137.80 minus 75 = 62.80 mm. Equilibrium cant at 50 km/h is approximately 34.45 mm, so the slow-train limit requires actual cant to be no more than 109.45 mm. Among the listed values, only 90 mm lies within both bounds. Its approximate fast-train deficiency and slow-train excess are 47.80 mm and 55.55 mm respectively.
వివరణ:
Question 18
ప్రశ్న 18
Which statement correctly distinguishes Indian Railways cant-deficiency provisions for ordinary rolling stock and specially nominated rolling stock under the 2024 Permanent Way Manual?
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Explanation:
The manual distinguishes normal rolling stock from specially nominated rolling stock. A cant-deficiency value of 75 mm applies to stock outside the nominated categories. Higher values for nominated stock are conditional on the prescribed satisfactory oscillation trial and speed certificate. Those permissions are not interchangeable with a blanket approval for every train or turnout configuration. Engineering speed calculations must therefore state the applicable rolling-stock category and its authorized deficiency.
వివరణ:
Question 19
ప్రశ్న 19
Under the Indian Railways Permanent Way Manual 2024, which combination correctly states the maximum design cant for the specified broad-gauge route categories?
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Explanation:
The current manual specifies a route-category distinction for maximum design cant. Group A and B broad-gauge routes have a maximum design cant of 185 mm, subject to the relevant provisions. Other broad-gauge routes have a maximum of 165 mm. On track with turnouts, design cant is limited to 140 mm. These route-dependent values should not be replaced by older textbook generalizations that assign one limit to every broad-gauge route.
వివరణ:
Question 20
ప్రశ్న 20
A corridor is used by high-speed passenger trains and slower heavy goods trains. Which method is most appropriate for selecting the actual cant on its railway curves?
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Explanation:
Actual cant is not necessarily the equilibrium cant for the fastest train. Slow goods trains can experience significant cant excess when cant is chosen for high passenger-train speeds. The chief engineering authority therefore considers the speeds actually achieved by different train categories. The final value must respect the allowable deficiency and excess as well as transition and track restrictions. Balanced selection can improve both running safety and long-term maintenance performance.
వివరణ:
Question 21
ప్రశ్న 21
What is the principal purpose of a railway transition curve between a straight track and a circular curve carrying cant?
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Explanation:
A transition curve provides a gradual change from straight-track curvature to the circular-curve curvature. Cant can be introduced over the same appropriate length instead of appearing abruptly. This reduces sudden changes in lateral acceleration and the vehicle roll response. The transition must meet the applicable gradient and rate-of-change requirements. Even if a circular curve satisfies a cant-based speed calculation, an inadequate transition can require a lower operating speed.
వివరణ:
Question 22
ప్రశ్న 22
An illustrative cant run-off raises the outer rail by 100 mm relative to the inner rail. If the longitudinal cant gradient must not be steeper than 1 in 720, what is the minimum run-off length based on this gradient criterion alone?
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Explanation:
A cant gradient of 1 in 720 permits one millimetre of cant change over 720 millimetres of longitudinal track length. For a total change of 100 mm, the minimum required length is 100 times 720 = 72000 mm. This equals 72 m. The value addresses only the stated cant-gradient condition. The actual transition may have to be longer to meet rate-of-change and other design requirements.
వివరణ:
Question 23
ప్రశ్న 23
Cant increases uniformly from 0 to 100 mm along a 100 m railway transition. A train travels at 72 km/h. What is its experienced rate of change of cant in mm/s under the stated uniform assumption?
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Explanation:
The train speed of 72 km/h is converted to 20 m/s by dividing by 3.6. The time taken to traverse the 100 m transition is 100/20 = 5 seconds. The cant change during that period is 100 mm. Therefore the rate of change of cant is 100/5 = 20 mm/s. This is an average and constant rate because the question assumes a uniform cant increase and constant speed.
వివరణ:
Question 24
ప్రశ్న 24
Using the ordinary fully transitioned-curve design expressions in the Indian Railways Permanent Way Manual 2024, take C_a = 100 mm, C_d = 75 mm and V_m = 100 km/h. If the desirable transition length is the largest of 0.0056 C_a V_m, 0.0056 C_d V_m and 0.72 C_a, rounded upward to the next multiple of 10 m, what length results?
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Explanation:
The criterion based on cant rate gives 0.0056 times 100 times 100 = 56 m. The cant-deficiency-rate criterion gives 0.0056 times 75 times 100 = 42 m. The cant-gradient criterion gives 0.72 times 100 = 72 m. The largest of these values is 72 m. Rounding upward to the next multiple of 10 m yields a desirable transition length of 80 m.
వివరణ:
Question 25
ప్రశ్న 25
An existing broad-gauge curve has adequate cant deficiency margin for the proposed higher speed, but its available transition is too short to satisfy the applicable cant and deficiency rate-of-change criteria. Which decision is most defensible?
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Explanation:
A curve speed cannot be justified solely from equilibrium cant and allowable deficiency. The transition introduces curvature and cant at rates relevant to vehicle dynamics and passenger comfort. An insufficient transition can govern the permissible speed even if the circular arc itself passes a cant calculation. Appropriate options include improving the geometry or revising the selected cant, deficiency and operating speed in accordance with the rules. The final permitted speed must also remain within the authorized sectional and rolling-stock limits.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: B. The outer rail is raised above the inner rail to help counterbalance the outward lateral effect of curving. సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: D. The adopted dynamic gauge of 1750 mm, representing the nominal gauge plus the rail-head-width allowance. సమాధానం: D.
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Question 3 ప్రశ్న 3Answer: A. C_e = G V^2/(127 R). సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: C. It becomes four times its former value. సమాధానం: C.
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Question 5 ప్రశ్న 5Answer: B. Approximately 124 mm. సమాధానం: B.
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Question 6 ప్రశ్న 6Answer: D. Approximately 99.2 mm. సమాధానం: D.
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Question 7 ప్రశ్న 7Answer: A. 180 mm. సమాధానం: A.
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Question 8 ప్రశ్న 8Answer: C. It is half as large. సమాధానం: C.
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Question 9 ప్రశ్న 9Answer: B. 32 mm. సమాధానం: B.
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Question 10 ప్రశ్న 10Answer: D. 57 mm. సమాధానం: D.
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Question 11 ప్రశ్న 11Answer: C. Passenger cant deficiency 97 mm; goods cant excess 51 mm. సమాధానం: C.
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Question 12 ప్రశ్న 12Answer: A. 85 km/h. సమాధానం: A.
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Question 13 ప్రశ్న 13Answer: B. 95 km/h. సమాధానం: B.
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Question 14 ప్రశ్న 14Answer: D. Approximately 790 m. సమాధానం: D.
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Question 15 ప్రశ్న 15Answer: A. 65 mm. సమాధానం: A.
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Question 16 ప్రశ్న 16Answer: C. 120 mm. సమాధానం: C.
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Question 17 ప్రశ్న 17Answer: B. 90 mm. సమాధానం: B.
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Question 18 ప్రశ్న 18Answer: D. For rolling stock outside the nominated categories the stated maximum is 75 mm, while higher nominated values require the prescribed successful trials and RDSO speed certification. సమాధానం: D.
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Question 19 ప్రశ్న 19Answer: A. 185 mm on Group A and B routes, 165 mm on other routes, and not more than 140 mm on track with turnouts. సమాధానం: A.
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Question 20 ప్రశ్న 20Answer: C. Select an approved nominated equilibrium speed and assess fast-train cant deficiency, slow-train cant excess, transitions and operating constraints together. సమాధానం: C.
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Question 21 ప్రశ్న 21Answer: B. To increase curvature and cant progressively, limiting abrupt changes in lateral acceleration and wheel-load transfer. సమాధానం: B.
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Question 22 ప్రశ్న 22Answer: A. 72 m. సమాధానం: A.
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Question 23 ప్రశ్న 23Answer: D. 20 mm/s. సమాధానం: D.
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Question 24 ప్రశ్న 24Answer: C. 80 m. సమాధానం: C.
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Question 25 ప్రశ్న 25Answer: A. The proposed speed must still satisfy the transition and sectional constraints; cant or deficiency, geometry or speed may need revision. సమాధానం: A.