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10 months ago

Class 11th System of Particles and Rotational Motion

7.2 In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 Å (1 Å = 10^-10 m). Find the approximate location of the CM of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.

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Vishal Baghel

Contributor-Level 10

Let us assume that H atom and Cl atom are at a distance of x and y respectively from the CM (Centre of Mass).

If mass of the H atom = m, mass of the Cl atom = 35.5m

Given x + y = 1,27 À

Let us assume that the centre of mass of the given molecule lies at the origin. Therefore,

We can have, : (my+35.5mx)/ (m+35.5m) = 0

mx + 35.5my = 0

x = 35.5 (1.27 – x)

x = 1.24 À

So the centre of mass lies 1.24 À from H atom

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Class 11th Sets

Class 11 Chapter 1 Set Exercise 1.1 NCERT Solutions

Q1. Which of the following are sets? Justify your answer.

(i) The collection of all the months of a year beginning with the letter J.

(ii) The collection of ten most talented writers of India.

(iii) A team of eleven best-cricket batsmen of the world.

(iv) The collection of all boys in your class.

(v) The collection of all natural numbers less than 100.

(vi) A collection of novels written by the writer Munshi Prem Chand.

(vii) The collection of all even integers.

(viii) The collection of questions in this Chapter.

(ix) A collection of most dangerous animals of the world.

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Class 11th System of Particles and Rotational Motion

7.1 Give the location of the centre of mass of a (i) Sphere (ii) Cylinder (iii) Ring and (iv) Cube each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body?

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Vishal Baghel

Contributor-Level 10

All the structures specified are symmetric bodies with uniform mass density. For all these bodies, their centre of mass will lie in their geometric centres.

Not necessarily, the centre of gravity of a circular ring is at the imaginary centre of the ring.

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Class 11th

How do I get admission in class 11th Science in Pune (Maharashtra)?

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Class 11th Physics Gravitation

8.25 A rocket is fired 'vertically' from the surface of mars with a speed of 2 km s–1. If 20% of its initial energy is lost due to Martian atmospheric resistance, how far will the rocket go from the surface of mars before returning to it? Mass of mars = 6.4*10²³ kg; radius of mars = 3395 km; G = 6.67*10^-11 N m² kg^–2.

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Vishal Baghel

Contributor-Level 10

Initial kinetic energy of the rocket = $\frac{1}{2} {m v}^{2}$

Initial potential energy of the rocket = $\frac{- G M m}{R}$

Total initial energy = $\frac{1}{2} {m v}^{2} - \frac{G M m}{R}$

If 20% of initial kinetic energy is lost due to Martian atmosphere resistance, then only 80% of its kinetic energy helps in reaching a height

Total initial energy available = 0.8 $* \frac{1}{2} {m v}^{2}$ $- \frac{G M m}{R}$

Maximum height reached by the rocket = h

At this height, the velocity and hence the kinetic energy of the rocket becomes zero.

Total energy of the rocket at height h = $- \frac{G M m}{R + h}$

Applying the law of conservation of energy for the rocket, we can write:

0.4 $* {m v}^{2}$ $- \frac{G M m}{R}$ = $- \frac{G M m}{R + h}$

0.4 $v^{2}$ = GM( $\frac{1}{R}$ $- \frac{1}{R + h})$

0.4 $v^{2}$ = GM( $\frac{R + h - R}{R (R + h)}$ )

...more

Initial kinetic energy of the rocket = $\frac{1}{2} {m v}^{2}$

Initial potential energy of the rocket = $\frac{- G M m}{R}$

Total initial energy = $\frac{1}{2} {m v}^{2} - \frac{G M m}{R}$

If 20% of initial kinetic energy is lost due to Martian atmosphere resistance, then only 80% of its kinetic energy helps in reaching a height

Total initial energy available = 0.8 $* \frac{1}{2} {m v}^{2}$ $- \frac{G M m}{R}$

Maximum height reached by the rocket = h

At this height, the velocity and hence the kinetic energy of the rocket becomes zero.

Total energy of the rocket at height h = $- \frac{G M m}{R + h}$

Applying the law of conservation of energy for the rocket, we can write:

0.4 $* {m v}^{2}$ $- \frac{G M m}{R}$ = $- \frac{G M m}{R + h}$

0.4 $v^{2}$ = GM( $\frac{1}{R}$ $- \frac{1}{R + h})$

0.4 $v^{2}$ = GM( $\frac{R + h - R}{R (R + h)}$ ) = $\frac{G M h}{R (R + h)}$

$\frac{R + h}{h}$ = $\frac{G M}{{0.4 R v}^{2}}$

$\frac{R}{h}$ + 1 = $\frac{G M}{{0.4 R v}^{2}}$ = $\frac{6.4 * 10^{23} * 6.67 * 10^{- 11}}{0.4 * 3395 * 10^{3} * 2000^{2}}$

h = 495 km

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Class 11th Physics Gravitation

8.24 A spaceship is stationed on Mars. How much energy must be expended on the spaceship to launch it out of the solar system? Mass of the space ship = 1000 kg; mass of the sun = 2*10³⁰ kg; mass of mars = 6.4*10²³ kg; radius of mars = 3395 km; radius of the orbit of mars = 2.28 *10⁸ km; G = 6.67*10^-11 N m2 kg^–2.

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Vishal Baghel

Contributor-Level 10

Mass of the spaceship, $m_{s}$ = 1000 kg

Mass of the Sun, M = 2 * 10³⁰ kg

Mass of Mars, $m_{m}$ = 6.4*10²³ kg

Orbital radius of Mars, R = 2.28 *10⁸ km = 2.28 *10¹¹ m

Radius of Mars, r = 3395 km = 3.395 $* 10^{6} m$

Universal Gravitational constant, G = 6.67*10^-11 N m2 kg–2

Potential energy of the spaceship due to the gravitational attraction of the Sun = $\frac{- G M m_{s}}{R}$

Potential energy of the spaceship due to the gravitational attraction of Mars= $\frac{- G m_{s} m_{m}}{r}$

Since the spaceship is stationed on Mars, its velocity and hence its kinetic energy will be zero

Total energy of the spaceship = $\frac{- G M m_{s}}{R}$ $- \frac{- G m_{s} m_{m}}{r}$ = $- G m_{s} (\frac{M}{R}$ + $\frac{m_{m}}{r}$ )

The negative sign indicates that th

...more

Mass of the spaceship, $m_{s}$ = 1000 kg

Mass of the Sun, M = 2 * 10³⁰ kg

Mass of Mars, $m_{m}$ = 6.4*10²³ kg

Orbital radius of Mars, R = 2.28 *10⁸ km = 2.28 *10¹¹ m

Radius of Mars, r = 3395 km = 3.395 $* 10^{6} m$

Universal Gravitational constant, G = 6.67*10^-11 N m2 kg–2

Potential energy of the spaceship due to the gravitational attraction of the Sun = $\frac{- G M m_{s}}{R}$

Potential energy of the spaceship due to the gravitational attraction of Mars= $\frac{- G m_{s} m_{m}}{r}$

Since the spaceship is stationed on Mars, its velocity and hence its kinetic energy will be zero

Total energy of the spaceship = $\frac{- G M m_{s}}{R}$ $- \frac{- G m_{s} m_{m}}{r}$ = $- G m_{s} (\frac{M}{R}$ + $\frac{m_{m}}{r}$ )

The negative sign indicates that the system is in bound state.

Energy required for launching the spaceship out of the solar system = - (Total energy of the spaceship)

= $G m_{s} (\frac{M}{R}$ + $\frac{m_{m}}{r}$ )

= 6.67 $* 10^{- 11} * 10^{3} * (\frac{2 * 10^{30}}{2.28 * 10^{11}}$ + $\frac{6.4 * 10^{23}}{3.395 * 10^{6}})$

= 6.67 $* 10^{- 8} * (8.77 * 10^{18}$ + 1.88 $* 10^{17}$ )

= 6.67 $* 10^{- 8} * 8.958 * 10^{18}$

= 5.97 $* 10^{11}$ J

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10 months ago

Class 11th Physics Gravitation

8.23 A star 2.5 times the mass of the sun and collapsed to a size of 12 km rotates with a speed of 1.2 rev. per second. (Extremely compact stars of this kind are known as neutron stars. Certain stellar objects called pulsars belong to this category). Will an object placed on its equator remain stuck to its surface due to gravity? (mass of the sun = 2*10³⁰ kg).

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Vishal Baghel

Contributor-Level 10

Yes, a body gets stuck to the surface of a star if the inward gravitational force is greater than the outward centrifugal force caused by the rotation of the star.

Gravitational force, $f_{g}$ = $\frac{G M m}{R^{2}}$ , where M = mass of the star = 2.5 $* 2 * 10^{30}$ = 5 $* 10^{30}$ kg

M = mass of the body, R = radius of the star = 12km = 1.2 $* 10^{4} m$

$f_{g}$ = $\frac{6.67 * 10^{- 11} * 5 * 10^{30} * m}{{(1.2 * 10^{4})}^{2}^{}} =$ 2.31 $* 10^{12} m N$

Centrifugal force $f_{c}$ = mr $ω^{2}$ where $ω =$ angular speed = 2 $π γ$ and angular frequency $γ$ = 1.2 rev/s

$f_{c} =$ mR( ${2 π γ)}^{2}$ = m $*$ (1.2 $* 10^{4}) * 2 * 3.14 * 1.2 * (2 * 3.14 * 1.2)$ = 6.81 $* 10^{5} m N$

Since $f_{g} > f_{c}$ , the body will remain stuck to the surface of the star.

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Class 11th Physics Gravitation

8.22 As you have learnt in the text, a geostationary satellite orbits the earth at a height of nearly 36,000 km from the surface of the earth. What is the potential due to earth's gravity at the site of this satellite? (Take the potential energy at infinity to be zero). Mass of the earth = 6.0*10²⁴ kg, radius = 6400 km.

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Vishal Baghel

Contributor-Level 10

Mass of the Earth, M = 6.0 $* 10^{24}$ kg

Radius of the Earth, R = 6400 km = 6.4 $* 10^{6}$ m

Height of geostationary satellite from the surface of the Earth, h = 36000 km = 3.6 $* 10^{7}$ m

Gravitational potential energy due to Earth's gravity at height h:

= $\frac{- G M}{(R + h)}$

= $- \frac{6.67 * 10^{- 11} * 6.0 * 10^{24}}{3.6 * 10^{7} + 6.4 * 10^{6}}$ = $- \frac{6.67 * 6.0 * 10^{13}}{4.24 * 10^{7}}$ = $-$ 9.44 $* 10^{6}$ J/kg

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10 months ago

Class 11th Physics Gravitation

8.21 Two heavy spheres each of mass 100 kg and radius 0.10 m are placed 1.0 m apart on a horizontal table. What is the gravitational force and potential at the midpoint of the line joining the centers of the spheres? Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable?

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Vishal Baghel

Contributor-Level 10

Mass of each sphere, M = 100 kg

Separation between the spheres, r = 1 m

X is the midpoint between the spheres.

Gravitational force at point x will be zero. This is because gravitational force exerted by each spheres will act in opposite directions.

Gravitational potential at point x:

= $\frac{- G M}{\frac{r}{2}}$ $\frac{- G M}{\frac{r}{2}}$ = $-$ 4 $\frac{G M}{r}$ = $- \frac{4 * 6.67 * 10^{- 11} * 100}{1}$ = $- 2.668 * 10^{- 8}$ J/kg

Any object placed at point x will be in equilibrium state, but the equilibrium is unstable. This is because any change in the position of the object will change the effective force in that direction.

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10 months ago

Class 11th Physics Gravitation

8.20 Two stars each of one solar mass (= 2*10³⁰ kg) are approaching each other for a head on collision. When they are a distance 10⁹ km, their speeds are negligible. What is the speed with which they collide? The radius of each star is 10⁴ km. Assume the stars to remain undistorted until they collide. (Use the known value of G).

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Vishal Baghel

Contributor-Level 10

Mass of each star, M = 2 * 10³⁰ kg, Radius of each star, R = 10⁴ km = $10^{7}$ m

Distance between stars, r = $10^{9}$ km = $10^{12}$ m

For negligible speed, v = 0

The total energy of two stars separated at a distance r

= $- \frac{G M M}{r} + \frac{1}{2} m v^{2}$ = $- \frac{G M M}{r}$ …….(i)

Now, consider the case when the stars are about to collide. Velocity of the stars = V, distance between the centres of the stars = 2R

Total kinetic energy of both stars = $\frac{1}{2}$ M $V^{2}$ + $\frac{1}{2}$ M $V^{2}$ = M $V^{2}$

Total potential energy of both stars = $- \frac{G M M}{2 R}$

Total energy of two stars = M $V^{2} - \frac{G M M}{2 R}$ …….(ii)

Using the law of conservation of energy, we can write

M $V^{2} - \frac{G M M}{2 R}$ =

...more

Mass of each star, M = 2 * 10³⁰ kg, Radius of each star, R = 10⁴ km = $10^{7}$ m

Distance between stars, r = $10^{9}$ km = $10^{12}$ m

For negligible speed, v = 0

The total energy of two stars separated at a distance r

= $- \frac{G M M}{r} + \frac{1}{2} m v^{2}$ = $- \frac{G M M}{r}$ …….(i)

Now, consider the case when the stars are about to collide. Velocity of the stars = V, distance between the centres of the stars = 2R

Total kinetic energy of both stars = $\frac{1}{2}$ M $V^{2}$ + $\frac{1}{2}$ M $V^{2}$ = M $V^{2}$

Total potential energy of both stars = $- \frac{G M M}{2 R}$

Total energy of two stars = M $V^{2} - \frac{G M M}{2 R}$ …….(ii)

Using the law of conservation of energy, we can write

M $V^{2} - \frac{G M M}{2 R}$ = $- \frac{G M M}{r}$

M $V^{2} = \frac{G M M}{2 R}$ $- \frac{G M M}{r}$

$V^{2}$ = GM ( $\frac{1}{2 R}$ $-$ $\frac{1}{r})$ = 6.67 $* 10^{- 11} * 2 * 10^{30}$ ( $\frac{1}{2 * 10^{7}} - \frac{1}{10^{- 12}}$ )

= 1.334 $* 10^{20} *$ ( $\frac{1}{2 * 10^{7}} - \frac{1}{10^{- 12}}$ )

V = 2.6 $* 10^{6}$ m/s

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