Uniform Circular Motion Numerical on Curved Tracks Banking of Roads and Railway Tracks Class-11 Nootan ISC Physics Solutions Ch-8. Step by step solutions of Kumar and Mittal Physics of Nageen Prakashan as council latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ISC Board Class-11 Physics.
Uniform Circular Motion Numerical on Curved Tracks Banking of Roads and Railway Tracks Class-11 Nootan ISC Physics
| Board | ISC |
| Class | 11 |
| Subject | Physics |
| Writer | Kumar and Mittal |
| Publication | Nageen Prakashan |
| Chapter-8 | Uniform Circular Motion |
| Topics | Numerical on Curved Tracks Banking of Roads and Railway Tracks |
| Academic Session | 2024-2025 |
Numerical on Curved Tracks Banking of Roads and Railway Tracks
Uniform Circular Motion Class-11 ISC Nootan Solutions Ch-8 of Kumar and Mittal Physics of Nageen Prakashan
Que-20: A motor-cyclist driving with a speed of 0.6 km/minute goes over a turn of radius of curvature 50 metre. Calculate his centripetal acceleration and his slope with the vertical (g = 10 m/s²).
Ans- v = 0.6 km/min = 0.6 x 10³/ 60 m/s = 10 m/s
a = v²/r
=> 10 x 10 / 50
=> 2.0 m/s²
and tan θ = v² / r g
=> 10 x 10 / 50 x 10
=> 0.2
θ = tan¯¹ (0.2)
Que-21: For traffic moving at 60 km/h, if the radius of the circular curve of a highway is 0.1 km, what is the correct angle of banking of the road? Take g = 10 m/s². Given tan¯¹ (0.278) = 15.5°.
Ans- v = 60 km/h = 60 x 5/18 = 50/3 m/s
r = 0.1 km = 100 m
now tan θ = v² / r g
=> (50 x 50) / (9 x 100 x 10)
=> 5 / 18
tan θ = 0.278
θ = 15.5°
Que-22: The distance between the rails of a railway line is 1 m. How much should the outer rail be raised relative to the inner one so that a train may cross over a turn of radius 400 m with a speed of 20 m/s without friction? g = 10 m/s².
Ans- tan θ = v² / r g
=> 20 x 20 / 400 x 10
=> 1/ 10
again tan θ = h /1 = 1 /10
h = 0.1 m = 10 cm
Que-23: The radius of the turn of a road is 500 m. The width of the road is 10 m. Its outer edge is 0.15 m higher than the inner edge. For which speed of the vehicle running over the road this banking is ideal? (g = 10 m/s²)
Ans- tan θ = v² / r g = 0.15 / 10
v = √0.15 x r g /10
=> √0.15 x 500 /10
=> √75
=> 8.66 m/s
Que-24: Find the maximum speed at which a car can turn round a curve of 30 m radius on a level road if the coefficient of friction between the tyres and the road is 0.4. Take g = 10 m/s²
Ans- m v² / r = μ m g
v = √μ m g
=> √ 0.4 x 30 x 10
=> √120
=> 10.95 m/s
Que-25: A 1000 kg car is moving with a uniform speed of 30 m/s on a circular path of radius 200 m. How much frictional force is needed to provide the necessary centripetal force to the car? If the coefficient of friction is 0.8, then what can be the maximum speed of the car? (g = 9.8 m/s²).
Ans- frictional force = centripetal force
μ m g = m v² / r
=>1000 x 30 x 30 / 200
=> 4500 N
for μ = 0.8
v = √μ r g
=> √0.8 x 200 x 9.8
=> 39.6 m/s
Que-26: A train is moving with a speed of 72 km/h on a curved railway-line of 400 m radius. A spring balance loaded with a block of weight of 5 kg is suspended from the roof of the train. What would be the reading of the balance? (g = 10 m/s²).
Ans- 5 kg block is subjected to two forces due to gravity, i.e. 50 N (F = mg) and centripetal force
Calculating first the centripetal force;
F = mv² / r , m = 5 kg , r = 400m
(train speed 72 km/h = 20 m/s ) , v = 20m/s
Hence centripetal force is = [5 x (20)²] / 400 = 5N
Now calculating the resultant force ;
T = √((50)²+(5)² )
T = 50.25 N
Now weight on balance
F = m g
m = 50.25 / 10
=> 5.025 kg
Que-27: If, in the above case, the spring makes an angle of 10° with the vertical, then what is the centripetal acceleration of the train? (g = 10 m/s². tan 10° 0.176).
Ans- centripetal acceleration
a = v² / r
=> but tan θ = v² / r g
=> v² / r = tan θ x g
=> tan 10 ° x 10
=> 0.176 x 10
=> 1.76 m/s²
—: end of Uniform Circular Motion Numerical on Curved Tracks Banking of Roads and Railway Tracks Class-11 Nootan ISC :—
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