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physics ncert solutions class 11th

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5 months ago

physics ncert solutions class 11th Laws of Motion

5.11 A truck starts from rest and accelerates uniformly at 2.0 m s-2. At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What are the (a) velocity, and (b) acceleration of the stone at t = 11s? (Neglect air resistance.)

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Vishal Baghel

Contributor-Level 10

The acceleration of the truck, a = 2 m/s², t = 10 s, Initial velocity, u = 0, from the equation v = u + at, we get v = 2 $* 10$ = 20 m/s

At time t = 11 s, the horizontal component of the velocity Vx remains unchanged due to no air resistance. Hence Vx = 20 m/s

The vertical component of the velocity Vy is expressed as

Vy = u + $a_{y}$ t

Here t =11-10 = 1 s, $a_{y} = a = g =$ 10 m/s², u = 0

Vy = 10 m/s

The resultant velocity V is given by V = ( $V_{x}^{2}$ + $V_{y}^{2}$ )1/2 = 22.36 m/s

$\tan ? ?$ =Vy/Vx=10/20, $? =$ 26.57 $°$ w.r.t. horizontal

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physics ncert solutions class 11th Laws of Motion

5.10 A body of mass 0.40 kg moving initially with a constant speed of 10 m s-1 to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be t = 0, the position of the body at that time to be x = 0, and predict its position at t = –5 s, 25 s, 100 s.

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Vishal Baghel

Contributor-Level 10

Given, the mass of the body = 0.4 kg, Initial speed = 10m/s, Initial force, Retarding force, F = - 8 N

(a) At t = -5s, Due to constant speed, the acceleration of the body = 0

From the relation, s = ut + (1/2)at², we get

s = 10 $* (- 5)$ + 0 $*$ ${(\frac{1}{2}) (- 5)}^{2}$ = -50 m

At t = 25s, The acceleration of the body due to retarding force, from the relation F = ma, a = -8/0.4 m/s2 = - 20 m/s2

From the relation, s = ut + at², , we get

s = 10 $* (25)$ + ${(\frac{1}{2}) (- 20) (- 25)}^{2}$ = -6000m

At t = 100s

For the first 30 s, while the body moves towards North, the opposite force of 8 N acts on it and from the relation s = ut + (1/2)at², , we get

S₃₀ = 10 $*$ 30 +&nb

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Given, the mass of the body = 0.4 kg, Initial speed = 10m/s, Initial force, Retarding force, F = - 8 N

(a) At t = -5s, Due to constant speed, the acceleration of the body = 0

From the relation, s = ut + (1/2)at², we get

s = 10 $* (- 5)$ + 0 $*$ ${(\frac{1}{2}) (- 5)}^{2}$ = -50 m

At t = 25s, The acceleration of the body due to retarding force, from the relation F = ma, a = -8/0.4 m/s2 = - 20 m/s2

From the relation, s = ut + at², , we get

s = 10 $* (25)$ + ${(\frac{1}{2}) (- 20) (- 25)}^{2}$ = -6000m

At t = 100s

For the first 30 s, while the body moves towards North, the opposite force of 8 N acts on it and from the relation s = ut + (1/2)at², , we get

S₃₀ = 10 $*$ 30 + $(\frac{1}{2}) (- 20) 30^{2}$ = -8700m

The speed after 30 sec, v = u+at

v = 10 – (20 $* 30)$ = -590 m/s

The distance covered in next 70 s,

S₇₀ = -590 $* 70 + (\frac{1}{2}) (* (0) () (70)) * (70)$ = -41300m

The position after 100s = S₃₀+S₇₀ = -8700 -41300 = -50 km

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5 months ago

physics ncert solutions class 11th Laws of Motion

5.9 A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of 5.0 m s^-2. Calculate the initial thrust (force) of the blast.

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Vishal Baghel

Contributor-Level 10

The mass of the rocket, m = 20000 kg

When the rocket is fired, gravitational acceleration tries to pull it down. Hence the effective acceleration on the rocket = rocket acceleration + gravitational acceleration

The acceleration, a = 5 m/s², gravitational acceleration =10 m/s²

Total acceleration = 5 +10 = 15 m/s²

Thrust force = 20000 $*$ 15 N = 3 .0 $*$ $10^{5}$ N

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physics ncert solutions class 11th Laws of Motion

5.8 The driver of a three-wheeler moving with a speed of 36 km/h sees a child standing in the middle of the road and brings his vehicle to rest in 4.0 s just in time to save the child. What is the average retarding force on the vehicle? The mass of the three-wheeler is 400 kg and the mass of the driver is 65 kg.

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Vishal Baghel

Contributor-Level 10

The initial speed of 3 wheeler, u = 36 km/h = 10 m/s

The final velocity of the 3 wheeler, v = 0

The time t = 4 s

From the relation, v = u-at, we get, a = u/t = 10/4 m/s²= 2.5 m/s²

The total mass acting on the 3 wheeler, m = mass of the 3 wheeler + mass of the driver = 400 +65 kg = 465 kg

The average retarding force F is given by F = MA = 465 $*$ 2.5 N = 1162.5 N = 1.2 $*$ $10^{3}$ N

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5 months ago

physics ncert solutions class 11th Laws of Motion

5.7 A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Give the magnitude and direction of the acceleration of the body.

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Vishal Baghel

Contributor-Level 10

Given, the mass m = 5 kg

The resultant force F of the two forces 6N and 8N is given by,

F = $\sqrt{8^{2} + 6^{2}}$ = 10 N

Acceleration is given by F = ma, a = F/m = 10/5 m/s² = 2 m/s²

$\tan ? θ$ = 6/8, $θ$ = 36.86 $°$

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5 months ago

physics ncert solutions class 11th Laws of Motion

5.6 A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 m s^-1 to 3.5 m s^-1 in 25 s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?

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Vishal Baghel

Contributor-Level 10

Initial speed, u = 2.0 m/s, the final speed v = 3.5 m/s, the time t = 25 s

From the relation v = u + at, we get acceleration a = (v-u)/t = (3.5 – 2.0 )/ 25s = 0.06 m/s²

The force F = ma, F = 3 $*$ 0.06 N = 0.18 N

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5 months ago

physics ncert solutions class 11th Laws of Motion

5.5 A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 m s-1. How long does the body take to stop?

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Vishal Baghel

Contributor-Level 10

The retarding force, F = 50N, the mass m = 20 kg

From the equation F = ma, we get acceleration a = 50 / 20 m/s² = 2.5 m/s²

The initial speed, u = 15 m/s, the final speed v =0, from the relation v = u-at, we get

T = u/a = 15/2.5 s = 6s

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physics ncert solutions class 11th Laws of Motion

5.4 One end of a string of length l is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed v the net force on the particle (directed towards the centre) is:

(i) T

(ii) T - $\frac{m v^{2}}{l}$

(iii) T + $\frac{m v^{2}}{l}$

(iv) 0 T is the tension in the string. [Choose the correct alternative]

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5 months ago

physics ncert solutions class 11th Laws of Motion

5.3 Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg,

(a) Just after it is dropped from the window of a stationary train

(b) Just after it is dropped from the window of a train running at a constant velocity of 36 km/h

(c) Just after it is dropped from the window of a train accelerating with 1 m s^-2

(d) Lying on the floor of a train which is accelerating with 1 m s^-2, the stone being at rest relative to the train. (Neglect air resistance throughout.)

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Vishal Baghel

Contributor-Level 10

(a) Mass of the stone, m = 0.1 kg, Acceleration due to gravity = 10 m/s²

Net force = 0.1 $*$ 10 N = 1.0 N. The direction of force is downwards

(b) Since the train is moving at a constant velocity, the acceleration imparted by the train is zero. Only acceleration acting on the stone is gravity. So the force will be 1.0 N, acting downwards

(c) The acceleration of the train = 1 m/s² so the force exerted by the train = 0.1 $*$ 1 N = 0.1 N and it is acting in the horizontal direction. But when the stone is dropped, the force started acting on the stone is only due to gravity, the train moving force has no effect. So the net

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(a) Mass of the stone, m = 0.1 kg, Acceleration due to gravity = 10 m/s²

Net force = 0.1 $*$ 10 N = 1.0 N. The direction of force is downwards

(d) When the stone is lying on the train floor, the only force acting on the stone is due to the acceleration of the train. So the net force on the stone = 0.1 $*$ 1 N. The direction is horizontal along the motion of the train

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physics ncert solutions class 11th Laws of Motion

5.2 A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble,

(a) During its upward motion

(b) During its downward motion

(c) At the highest point where it is momentarily at rest. Do your answers change if the pebble was thrown at an angle of 45° with the horizontal direction?

(Ignore air resistance.)

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Vishal Baghel

Contributor-Level 10

(a) During the upward motion f the pebble, the acceleration due to gravity acts downwards. The magnitude of the force is

F = MA = 0.05 kg $*$ 10 m/s² = 0.5 N and the direction of force is downwards.

(b) During its downward motion – the force is 0.5 N and the direction is downwards.

(c) When the pebble is thrown at an angle of 45 $°$ with the horizontal direction, it will have both horizontal component of force and vertical component of force during its upward journey, When it reaches the maximum height, the vertical component of velocity will becomes zero and the force acting will be 0.5 N and acting downwards.

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