Tags physics ncert solutions class 11th

physics ncert solutions class 11th

Get insights from 952 questions on physics ncert solutions class 11th, answered by students, alumni, and experts. You may also ask and answer any question you like about physics ncert solutions class 11th

Follow Ask Question

952

Questions

Discussions

Active Users

Followers

New answer posted

5 months ago

physics ncert solutions class 11th Motion in a Straight Line

3.18 A police van moving on a highway with a speed of 30 km h^–1 fires a bullet at a thief's car speeding away in the same direction with a speed of 192 km h^–1. If the muzzle speed of the bullet is 150 m s^–1, with what speed does the bullet hit the thief's car?

(Note: Obtain that speed which is relevant for damaging the thief's car).

0 Follower 1.6k Views

Vishal Baghel

Contributor-Level 10

3.18 Speed of the police van = 30 kmph = 8.33 m/s

Speed of the thief's car = 192 kmph = 53.33 m/s

The muzzle speed of the bullet = 150 m/s

Speed of the bullet = speed of the police van + muzzle speed of the bullet = 8.33 + 150m/s = 158.33 m/s

The relative velocity of the bullet = speed of the bullet – speed of the thief's car = 158.33 – 53.33 m/s = 105 m/s

0 0

New answer posted

5 months ago

physics ncert solutions class 11th Motion in a Straight Line

3.17 Figure 3.21 shows the x-t plot of one-dimensional motion of a particle. Is it correct to say from the graph that the particle moves in a straight line for t < 0 and on a parabolic path for t >0? If not, suggest a suitable physical context for this graph.

0 Follower 39 Views

Vishal Baghel

Contributor-Level 10

3.17 For t<0, we cannot say that the particle moved in a straight line and for t>0 on a parabolic path as the x-t graph does not indicate. Instead, this x-t graph denotes a particle is dropped from a height at t=0

0 0

New answer posted

5 months ago

physics ncert solutions class 11th Motion in a Straight Line

3.16 Look at the graphs (a) to (d) (Fig. 3.20) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.

0 Follower 370 Views

Vishal Baghel

Contributor-Level 10

3.16

(a) Figure (a) shows two positions at the same time, which is not possible for a one-dimensional motion

(b) A particle cannot have velocity in two directions at the same time in one-dimensional motion

(c) The graph shows negative speed, which is not possible because speed is always positive

(d) Total path length decreases, which is not possible for a one – dimensional motion

0 0

New answer posted

5 months ago

physics ncert solutions class 11th Motion in a Straight Line

3.15 In Exercises 3.13 and 3.14, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why?

0 Follower 14 Views

Vishal Baghel

Contributor-Level 10

3.15 Instantaneous speed and velocity are applicable for a small interval of time because the magnitude of the displacement is effectively equal to the distance travelled by the particle.

0 0

New answer posted

5 months ago

physics ncert solutions class 11th Motion in a Straight Line

3.14 A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h^–1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h^–1.

What is the

(a) Magnitude of average velocity, and

(b) Average speed of the man over the interval of time

(i) 0 to 30 min,

(ii) 0 to 50 min,

(iii) 0 to 40 min?

[Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero !]

0 Follower 995 Views

Vishal Baghel

Contributor-Level 10

3.14

The distance from home to market = 2.5 km = 2500 m

The speed of the walking while going = 5 kmph = 1.388 m/s

The speed of walking back from market to home = 7.5 kmph = 2.083 m/s

(a) Magnitude of the average velocity = Displacement / time = 0, since the total displacement is zero.

(b)

i. Time taken to reach the market = distance / onward speed = 2500/1.388 = 1801 s = 30 minutes. So the average speed over 0 – 30 min is 5 kmph

ii. Time taken to reach back home = 2500 / 2.083 = 20 minutes. So the average speed = total distance covered / total time taken = (2500 +2500)/ (30 +20) m/min = 100 m/min = 6 kmph

iii. Average speed ove

...more

0 0

New answer posted

5 months ago

physics ncert solutions class 11th Motion in a Straight Line

3.13 Explain clearly, with examples, the distinction between:

(a) Magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval

(b) Magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval]

Show in both (a) and (b) that the second quantity is either greater than or equal to the first

When is the equality sign true? [For simplicity, consider one-dimensional motion

...more

0 Follower 147 Views

Vishal Baghel

Contributor-Level 10

3.13

(a) Let us take the example of football world cup 2022. During Argentina match, Di Mario (A) passed the ball to Messi (B). Messi instantly passed the ball to Di Mario. Now the magnitude of displacement of the ball is 0 as the ball returns to Di Mario at the initial position. But the total length covered by the ball is AB + BA = 2AB. Hence total length covered by the particle (the ball) is more than the magnitude.

(b) If t is the time taken to cover the entire distance, the magnitude of the average velocity of the ball over time interval t = Magnitude of displacement / t = 0 / t = 0

The average speed of the ball = Total lengt

...more

0 0

New answer posted

5 months ago

physics ncert solutions class 11th Motion in a Straight Line

3.12 A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed-time graph of its motion between t = 0 to 12 s.

0 Follower 514 Views

Vishal Baghel

Contributor-Level 10

Given data:

Initial height, h = 90 m

Acceleration due to gravity, g = 9.8 ms2

Speed lost at each collision = 1/10 of its speed

time interval to be considered: 0 to 12 secs.

Step 1: Time to fall from 90 m

h = ½ gt2 ⇒ 90 = 21 * 9.8 * t2 ⇒ t2 = 9.8180 ≈ 18.37 ⇒ t ≈ 4.29 seconds

Step 2: Speed before first impact

v = gt = 9.8 * 4.29 ≈ 42 m/s

After hitting the floor with 42 m/s, ball rebound with;

vrebound = 0.9*42 = 37.8m/s

Step 3:

Max height after rebound:

Use:

t = g / v = 37.8 / 9.8 ≈ 3.86s

Again, the b

...more

0 0

New answer posted

5 months ago

physics ncert solutions class 11th Motion in a Straight Line

3.11 Read each statement below carefully and state with reasons and examples, if it is true or false;

A particle in one-dimensional motion

(a) With zero speed at an instant may have non-zero acceleration at that instant

(b) With zero speed may have non-zero velocity

(c) With constant speed must have zero acceleration

(d) With positive value of acceleration must be speeding up

0 Follower 159 Views

Vishal Baghel

Contributor-Level 10

3.11

(a) True – zero speed may have a non zero instantaneous non-zero acceleration

(b) False – zero speed will always make zero velocity

(c) True – constant speed will have zero acceleration

(d) False – If the positive value of acceleration is in the direction of its motion then it can speed up

0 0

New answer posted

5 months ago

physics ncert solutions class 11th Motion in a Straight Line

3.10 A player throws a ball upwards with an initial speed of 29.4 m s^–1.

(a) What is the direction of acceleration during the upward motion of the ball?

(b) What are the velocity and acceleration of the ball at the highest point of its motion?

(c) Choose the x = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of x axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion

(d) To what height does the ball rise and after how long does the ball return to the player's hands

...more

0 Follower 468 Views

Vishal Baghel

Contributor-Level 10

3.10

(a) The direction of the acceleration is downward during the upward motion of the ball because the acceleration is due to gravity. Gravity always pulls the object toward the centre of the earth.

(b) Velocity: At the peak the velocity is zero because the ball momentarily stops before falling downwards.

Acceleration: g = 9.8 ms^–2

(c) If we consider the highest point x = 0 m, t= 0 s,

During upward motion of the ball before it reaches the maximum height, x = +ve, velocity = -ve, acceleration = +ve. During downward motion, x = +ve, velocity = +ve, acceleration = +ve

(d) At the highest point, final velocity v = 0, initial vel

...more

0 0

New answer posted

5 months ago

physics ncert solutions class 11th Motion in a Straight Line

3.9 Two towns A and B are connected by a regular bus service with a bus leaving in either direction every T minutes. A man cycling with a speed of 20 km h^–1 in the direction A to B notices that a bus goes past him every 18 min in the direction of his motion, and every 6 min in the opposite direction. What is the period T of the bus service and with what speed (assumed constant) do the buses ply on the road?

0 Follower 78 Views

Vishal Baghel

Contributor-Level 10

3.9 Cycling speed $V_{c}$ = 20 kmph

Let as assume speed of the bus is $V_{b}$

The relative velocity of the buses in the direction of motion of the cyclist = $V_{b}$ - $V_{c}$

Buses go past him every 18 mins = 18/60h = 0.3h

Distance covered = ( $V_{b} - V_{c}$ ) * 0.3

Since the buses leave every T minutes, ( $V_{b} - V_{c}$ ) * 0.3 = $V_{b}$ * (T/60) ……(1)

The relative velocity of the buses in the opposite direction of the cyclist = $V_{b}$ + $V_{c}$

Buses go past him in the opposite direction = every 6 m = 6/60 h

Distance covered = ( $V_{b}$ + $V_{c}$ ) * 1/10

Since the buses leave every T minute, we get ( $V_{b}$ + $V_{c}$ ) * 1/10

...more

3.9 Cycling speed $V_{c}$ = 20 kmph

Let as assume speed of the bus is $V_{b}$

The relative velocity of the buses in the direction of motion of the cyclist = $V_{b}$ - $V_{c}$

Buses go past him every 18 mins = 18/60h = 0.3h

Distance covered = ( $V_{b} - V_{c}$ ) * 0.3

Since the buses leave every T minutes, ( $V_{b} - V_{c}$ ) * 0.3 = $V_{b}$ * (T/60) ……(1)

The relative velocity of the buses in the opposite direction of the cyclist = $V_{b}$ + $V_{c}$

Buses go past him in the opposite direction = every 6 m = 6/60 h

Distance covered = ( $V_{b}$ + $V_{c}$ ) * 1/10

Since the buses leave every T minute, we get ( $V_{b}$ + $V_{c}$ ) * 1/10 = $V_{b}$ * T/60 …….(2)

From equations (1) & (2), we get

( $V_{b}$ + $V_{c}$ ) / 10 = ( $V_{b} - V_{c}$ ) * 0.3

$V_{b}$ + $V_{c}$ = 3( $V_{b}$ - 3 $V_{c}$ )

2 $V_{b}$ = 4 $V_{c}$ , $V_{b}$ = 2 $V_{c}$ So $V_{b}$ = 2 * 20 = 40 kmph

From the equation (1), we get

(40-20) * 0.3 = 40 * T/60, T = 9 min

less

0 0

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
682k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Need guidance on career and education? Ask our experts

Your Question

physics ncert solutions class 11th

Questions

Discussions

Active Users

Followers

All

Discussions

Unanswered Questions

Share Your College Life Experience