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Required centripetal force be 31.59 N

Correct option is (D) 14 m/s

Correct option is (B) Mass of the moving vehicle

Correct option is (B) 1000 N

Correct Answer is: (c) 70 g

Correct Answer is: (c) initially be in contact with the inner wall and later with the outer wall

Correct Answer is: (d) any point on the system or outside it

Correct Answer is: (b) mω²l / k - mω²

Let x = the increase in the length of spring. Then the particle moves along a circular path of radius (l + x), and the spring force = kx = centripetal force.

∴ mω² ( l + x ) = kx  or  x = mω²l / k-mω².

Answer is (C) direction of linear velocity changes continuously.

Maximum speed of a car is 10.84 m/s

Negative angular acceleration: When an angular acceleration will have a direction opposite to that of angular velocity. It is termed as negative angular acceleration.

eg: When an electric fan is switched off, the angular velocity of the blades of fan decreases with time.

Positive angular acceleration:

When an angular acceleration will have the same direction as angular velocity, it is termed as positive angular acceleration.

eg: When an electric fan is switched on, the angular velocity of the blades of the fan increases with time.

The right answer is (b) mω₀²a

Explanation:

 The frame of reference is rotating with a uniform angular speed ω₀, so it has an acceleration towards the center and it is a non-inertial plane. To apply Newton's Laws in this frame we apply a pseudo force equal to mω₀²a to the particle but opposite to the direction of the acceleration of the frame. This pseudo force is the centrifugal force on the particle.

(a) 10 cm/s, 10 cm/s ² 

Explanation: 

Velocity of the particle in a circular motion is equal to ωr , so if ω is constant the velocity is proportional to the radius. Here the particle is kept at half radius therefore velocity will be half ie = 10 cm/s. 

Similarly acceleration is ω²r , here too, ω² is constant, so acceleration is also proportional to 'r'. In this case too, making radius half will make acceleration half ie = 10 cm/s².

Answer is (A) \(\frac{v}{b}\) sin²θ

Correct option is (C) 1.5 N

Correct option is (C) Kinetic energy

Answer is (C) centrifugal force may be balanced by the horizontal component of the normal reaction of the rail.

Answer is (D) Cyclist has to counteract the centrifugal force while in the case of car, only the passenger is thrown by this force.

Answer is (A) the radius of curvature of outer rail will be greater than that of the inner rail.

Angular acceleration = 6.28 rad/s²

Greatest velocity is 6.429 m/s

Angle of banking and elevation of the outer track above the inner track are 2°12' ,0.061 m respectively

Answer is (A) 10.84 m/s

Using,

v = \(\sqrt{\mu rg}\) = \(\sqrt{0.4\times30\times9.8}\) = 10.84m/s

(c) The magnitude of acceleration is constant.

Explanation: 

Option (a) and (b) are not correct . Since the speed is constant the particle will have only radial acceleration with constant magnitude. Option (c) is correct. Option (d) is incorrect because option (c) is correct and both options are contradictory.

(b) Gravitational attraction of the sun on the earth is the centripetal force.

To suggest the right action, the values of 'v', 'r' and coefficient of friction between the road and the types of the car must be known. Since the wall is found suddenly it is better to turn the car in a circle of radius 'r' as well as apply the brakes simultaneously to avoid hitting the wall.

When the heavy mass is hanging from a string and is in equilibrium without breaking it, the weight of mass is balanced by tension in the string. But when the same mass is set into oscillation it moves on an arc of a circle having radius equal to length of the string ie it has a circular motion. Due to the circular motion the mass has a centripetal acceleration and to maintain this acceleration the string applies a centripetal force on the mass in addition to the tension in equilibrium position. thus the tension in the string increases due to the oscillation and it breaks.

Answer is (D) Assertion is False but Reason is True.

Answer is (D) Kinetic energy

Maximum number of revolutions per minute is 30

Answer is (B) \(\vec{F}=-\frac{mv^2}{r^3}\vec{r}\)

Answer is (B) mass of the vehicle

Maximum speed of  train is 22.16 m/s

i. Force acting on a particle performing circular motion along the radius of circle and directed towards the centre of the circle is called centripetal force.

It is given by F_CP = \(\cfrac{m\text v^2}r\)

where, r = radius of circular path.

ii. Example: Electron revolves around the nucleus of an atom. The necessary centripetal force is provided by electrostatic force of attraction between positively charged nucleus and negatively charged electron.

iii. Unit: N in SI system and dyne in CGS system.

iv. Dimensions: [M¹ L¹ T^-2]

and 

i. A stone tied at the end of a string is revolved in a horizontal circle, the tension in the string provides the necessary centripetal force. It is given by equation T = mrω².

ii. The planets move around the sun in elliptical orbits. The necessary centripetal force is provided to the planet by the gravitational force of attraction exerted by the sun on the planet.

iii. A vehicle is moving along a horizontal circular road with uniform speed. The necessary centripetal force is provided by frictional force between the ground and the tyres of wheel.

iv. Satellite revolves round the earth in circular orbit, necessary centripetal force is provided by gravitational force of attraction between the satellite and the earth.

i. A stone tied at the end of a string is revolved in a horizontal circle, the tension in the string provides the necessary centripetal force. It is given by equation T = mrω² .

ii. The planets move around the sun in elliptical orbits. The necessary centripetal force is provided to the planet by the gravitational force of attraction exerted by the sun on the planet.

iii. A vehicle is moving along a horizontal circular road with uniform speed. The necessary centripetal force is provided by frictional force between the ground and the tyres of wheel.

iv. Satellite revolves round the earth in circular orbit, necessary centripetal force is provided by gravitational force of attraction between the satellite and the earth.

i. Force acting on a particle performing circular motion along the radius of circle and directed towards the centre of the circle is called centripetal force.

It is given by F_CP = \(\cfrac{m\text v^2}r\)

where, r = radius of circular path.

ii. Example: Electron revolves around the nucleus of an atom. The necessary centripetal force is provided by electrostatic force of attraction between positively charged nucleus and negatively charged electron.

iii. Unit: N in SI system and dyne in CGS system.

iv. Dimensions: [M¹ L¹ T^-2]

Answer is (C) centrifugal force

Answer is (B) flat at poles and bulged at the equator.

Answer is (C) Centrifugal force exists in inertial frame of reference.

Answer is (D) lack of proper centripetal force.

Answer is (A) increases

(b) There are other forces on the particle. 
(d) The resultant of the other forces varies in magnitude as well as in direction.

Explanation: 

Above situation is possible when resultant of external forces (R) is perpendicular to direction of motion and always directed towards center with a constant magnitude. So the direction of this resultant force is changing continuously. Since the force applied by the person (F) is constant hence there are other forces acting on the particle and the resultant of these other forces (P) is varying in magnitude and direction so that resultant (R) of the force applied by the person (F) and resultant of other forces (P) has a constant magnitude and always directed towards the center. So options (b) and (d) are correct and (a) & (c) are incorrect.

(b) The magnitude of the frictional force on the car is greater than mv²/r 
(c) The friction coefficient between the ground and the car is not less than α/g. 

Explanation: 

Since the circular motion is not uniform it will have both the radial and tangential components of acceleration and the resultant acceleration will be other than towards the center of path. Option (a) is wrong. 

Since the car will experience a force equal to mv²/r radially outward so the friction force must be more than it to keep it on the circular path. Option (b) is correct. 

Normal force on the car by the ground =mg . So frictional force on the car =µmg. It must not be less than outward force ma i.e. µmg≮ ma, 

→µ≮ a/g. Option (c) is correct. 

In limiting case µ=v²/rg, so option (d) is incorrect.

(b) There are other forces on the particle. 
(d) The resultant of the other forces varies in magnitude as well as in direction.

Explanation: 

Above situation is possible when resultant of external forces (R) is perpendicular to direction of motion and always directed towards center with a constant magnitude. So the direction of this resultant force is changing continuously. Since the force applied by the person (F) is constant hence there are other forces acting on the particle and the resultant of these other forces (P) is varying in magnitude and direction so that resultant (R) of the force applied by the person (F) and resultant of other forces (P) has a constant magnitude and always directed towards the center. So options (b) and (d) are correct and (a) & (c) are incorrect.

i. ω = \(\cfrac{\text v}r\)

ii. ω = \(\cfrac{\theta}t\) 

iii. ω = 2πn

iv. ω = \(\cfrac{2\pi}T\)

Correct option is (B) different at different points on a circle.

i. \(\alpha=\cfrac{\omega_2-\omega_1}t\) 

ii. \(\alpha=\cfrac{2\pi}t\)(n₂ - n₁)

Answer is  (D)  \(\sqrt{2Rg(1+sinθ-cosθ)}\)

Answer is (C) 28 m/s

Using, 

v² = μrg = 0.8 x 100 x 9.8 = 784 

∴ v = 28 m/s