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4 months ago

Laws of Motion Class 11th

5.30 An aircraft executes a horizontal loop at a speed of 720 km/h with its wings banked at 15°. What is the radius of the loop ?

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Vishal Baghel

Contributor-Level 10

The speed of the aircraft, v = 720 km/h = 200 m/s

The angle of banking = 15 $°$

From the relation $\tan ? θ$ = $v^{2}$ /rg we get r = $v^{2}$ / ( g $* \tan ? θ)$ = 14928 m

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4 months ago

Laws of Motion Class 11th

5.29 Ten one-rupee coins are put on top of each other on a table. Each coin has a mass m. Give the magnitude and direction of

(a) The force on the 7th coin (counted from the bottom) due to all the coins on its top

(b) The force on the 7th coin by the eighth coin

(c) The reaction of the 6th coin on the 7th coin

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Vishal Baghel

Contributor-Level 10

(a) The force on the 7th coin is due to the 3 coins kept above it

The weight of the 3 coins = 3m

Force exerted on the 7th coin is (3mg) N and the force is acting vertically downwards

(b) The force on the 7th coin by the 8th coin will be the force exerted by the 3 coins above it = (3mg) N, acting downwards

(c) The 6th coin will experience the force of 4 coins above it, acting downwards = 4mg

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4 months ago

Laws of Motion Class 11th

5.28 A stream of water flowing horizontally with a speed of 15 m s^-1 gushes out of a tube of cross-sectional area 10^-2 m², and hits a vertical wall nearby. What is the force exerted on the wall by the impact of water, assuming it does not rebound ?

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Vishal Baghel

Contributor-Level 10

Speed of water, v = 15 m/s

Cross-sectional area of the tube, A = 10^-2 m²

Density of water $ρ$ = 1000 kg/m3

Mass of the water hitting the wall = $ρ * A * v$ = 1000 $*$ 10^-2 $* 15$ = 150 kg/s

Force exerted on the wall because of impact of water = mass $*$ velocity = 150 $* 15$ N = 2250 N

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4 months ago

Laws of Motion Class 11th

5.27 A helicopter of mass 1000 kg rises with a vertical acceleration of 15 m s^-2. The crew and the passengers weigh 300 kg. Give the magnitude and direction of the

(a) Force on the floor by the crew and passengers

(b) Action of the rotor of the helicopter on the surrounding air

(c) Force on the helicopter due to the surrounding air

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Vishal Baghel

Contributor-Level 10

Given

Mass of the helicopter m₁ = 1000 kg

Mass of the crew and passenger m₂ = 300 kg

The vertical acceleration of the helicopter, a = 15 m/s² and acceleration due to gravity g = 10 m/s²

The total mass of the system, m = m₁ + m2 = 1000 + 300 = 1300 kg

(a) Force on the floor by the crew and passengers :

R – m₂g =m₂a

R = m₂ (g+a) = 300 $*$ ( 10 + 15 ) N = 7500 N

(b) Action of the rotor of the helicopter on the surrounding air

R' – mg = ma

R' = m (g+a) = 1300 $* (10 + 15) N$ = 32500 N

(c) Force on the helicopter due to the surrounding air

It is the reaction of the force applied by the rotor on the air. As action and reaction are

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Given

Mass of the helicopter m₁ = 1000 kg

Mass of the crew and passenger m₂ = 300 kg

The vertical acceleration of the helicopter, a = 15 m/s² and acceleration due to gravity g = 10 m/s²

The total mass of the system, m = m₁ + m2 = 1000 + 300 = 1300 kg

(a) Force on the floor by the crew and passengers :

R – m₂g =m₂a

R = m₂ (g+a) = 300 $*$ ( 10 + 15 ) N = 7500 N

(b) Action of the rotor of the helicopter on the surrounding air

R' – mg = ma

R' = m (g+a) = 1300 $* (10 + 15) N$ = 32500 N

(c) Force on the helicopter due to the surrounding air

It is the reaction of the force applied by the rotor on the air. As action and reaction are equal and opposite (Newton's 3rd law), the force of reaction, F = 32500 N, acting vertically upwards.

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4 months ago

Laws of Motion Class 11th

5.26 A stone of mass m tied to the end of a string revolves in a vertical circle of radius R. The net forces at the lowest and highest points of the circle directed vertically downwards are : [Choose the correct alternative]

Lowest Point Highest Point

(a) mg – T₁ mg + T₂

(b) mg + T₁ &nbs

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Lowest Point Highest Point

(a) mg – T₁ mg + T₂

(b) mg + T₁ mg – T₂

(c) mg +T₁ – (m $v_{1}^{2}$ ) / R mg – T₂+ (m $v_{1}^{2}$ ) / R

(d) mg – T₁ – (m $v_{1}^{2}$ ) / R mg + T₂ + (m $v_{1}^{2}$ ) / R

T₁ and v₁ denote the tension and speed at the lowest point. T₂ and v₂ denote corresponding values at the highest point.

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Vishal Baghel

Contributor-Level 10

The net force at the lowest point is denoted by (mg – T₁)and the net force at the highest point is denoted by (mg + T₂), hence option (a) is correct. The forces mg and T₁are in mutually opposite direction at the lowest point and mg and T₂are in the same direction at the highest point.

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4 months ago

Laws of Motion Class 11th

5.25 Figure 5.18 shows a man standing stationary with respect to a horizontal conveyor belt that is accelerating with 1 m s^-2. What is the net force on the man? If the coefficient of static friction between the man's shoes and the belt is 0.2, up to what acceleration of the belt can the man continue to be stationary relative to the belt ?(Mass of the man = 65 kg.)

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Vishal Baghel

Contributor-Level 10

The acceleration of the conveyer belt, a = 1 m/s²

Coefficient of static friction, $μ_{s}$ = 0.2

Mass of the man, m = 65 kg

The net force experienced by the man, MA = 1 $* 65$ N = 65 N

This net force is due to the friction between the belt and the man

At maximum static friction

$μ_{s} * m g = m a_{m a x}$

$a_{m a x}$ = $μ_{s} * g$ = 0.2 $* 10 = 2$ m/s²

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4 months ago

Laws of Motion Class 11th

5.24 Figure 5.17 shows the position-time graph of a body of mass 0.04 kg. Suggest a suitable physical context for this motion. What is the time between two consecutive impulses received by the body ? What is the magnitude of each impulse?

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Vishal Baghel

Contributor-Level 10

This graph could be of a ball rebounding between two walls separated by a distance of 2 cm.

The ball rebounds every 2 secs between the walls with uniform velocity.

Velocity of rebounding = displacement / time = ( 2 $* 10^{- 2}$ )/2 = 0.01 m/s

Initial momentum = mu = 0.04 $* 0.01 =$ 4 $* 10^{- 4}$ kgm/s

Final momentum = -mu = -4 $* 10^{- 4}$ kgm/s

Magnitude of Impulse = Initial momentum – final momentum = 8 $* 10^{- 4}$ kgm/s

The time between two consecutive impulses is 2 secs, so the ball receives an impulse every 2 seconds.

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4 months ago

Laws of Motion Class 11th

5.23 Explain why

(a) A horse cannot pull a cart and run in empty space

(b) Passengers are thrown forward from their seats when a speeding bus stops suddenly

(c) It is easier to pull a lawn mower than to push it

(d) A cricketer moves his hands backwards while holding a catch

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Vishal Baghel

Contributor-Level 10

(a) A horse can pull a cart by the reaction force generated by the ground to its feet. In an empty space, no reaction force will be generated and it cannot pull a cart.

(b) During the movement of the bus, passenger's whole body experiences the inertia of motion of the bus. When a bus brakes, the lower part comes to stand still along with the bus but the upper part continues with the same inertia of motion.

(c) While pulling a lawn mower, the vertical component of the applied force acts upwards and it reduces the effective weight of the lawn mower, make it easy to operate. In case of pushing the lawn mower, the vertical com

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4 months ago

Laws of Motion Class 11th

5.22 If, in Exercise 5.21, the speed of the stone is increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks :

(a) The stone moves radially outwards

(b) The stone flies off tangentially from the instant the string breaks

(c) The stone flies off at an angle with the tangent whose magnitude depends on the speed of the particle ?

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Vishal Baghel

Contributor-Level 10

(a) & (c) solution not possible as the stone will fly off tangentially. The answer is (b)

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4 months ago

Laws of Motion Class 11th

5.21 A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev./min in a horizontal plane. What is the tension in the string ? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N ?

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Vishal Baghel

Contributor-Level 10

Mass of the stone, m = 0.25 kg

Radius of the circular path = 1.5 m

Speed = 40 rev/min = 0.67 rev/s

Angular velocity, $?$ = 2 $? n$ = 2 $* \frac{22}{7} * 0.67$ = 4.19 rev/s

The tension of the string, T = m $?^{2}$ r = 0.25 $* {4.19}^{2} * 1.5$ = 6.59 N

Maximum tension of the string, $T_{m a x}$ = 200 N

From the equation $T_{m a x}$ = ${m v}_{m a x}^{2}$ /r, $v_{m a x}^{2}$ = $T_{m a x} * r / m$

$v_{m a x}^{2}$ = 200 $* 1.5$ /0.25

$v_{m a x}$ = 34.64 m/s

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