Veer

NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids

These Solutions are part of NCERT Solutions for Class 11 Physics. Here we have given NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids

Question 1.
A steel wire of length 4.7 m and cross-section 3.0 x 10^-5 m² stretches by the same amount as a copper wire of length 3.5 m and cross-section 4.0 x 10^-5 m² under a given load. What is the ratio of Young’s modulus of steel to that of copper?
Answer:
For steel, length of wire, L_s = 4.7 m
Area of cross-section, A_s = 3 x 10^-5 m²
For copper, length of wire, L_c = 3.5 m
Area of cross-section, A_c = 4.0 x 10^-5 m²
Also, extension
l_s = l_c = l

Question 2.
Fig. shows the stress-strain curve for a given material. What are
(a) Young’s modulus and
(b) approximate yield strength for this material?

Answer:
(a) From graph, for stress = 150 x 10⁶ Nm^-2
the corresponding strain = 0.002

Question 3.
The stress-strain graphs for materials A and B are shown in Figure.

The graphs are drawn to the same scale.
(a) Which of the material has greater Young’s modulus?
(b) Which material is more ductile?
(c) Which is more brittle?
(d) Which of the two is stronger material?
Answer:
(a) material A has a greater Young’s modulus because the slope of the graph of material A is more.
(b) Material A is more ductile than material B because it has a large plastic deformation between the elastic limit and the breaking point.
(c) Material B is more brittle than material A because it has a small plastic deformation between the elastic limit and the breaking point.
(d) Material A is stronger than B because more stress is required to break it.

Question 4.
Read each of the statements below carefully and state, with reasons, if it is true or false.
(a) The modulus of elasticity of rubber is greater than that of steel.
(b) The stretching of a coil is determined by its shear modulus.
Answer:
(a) False. Young’s modulus for steel is more than rubber.
(b) True

Question 5.
Calculate the elongation of the steel and brass wire in the Fig. Unloaded length of steel wire is 1.5 m and of brass wire = 1.0.m, the diameter of each wire = 0.25 cm. Young’s modulus of steel is 2.0 x 10¹¹ Pa and that of brass is 0.91 x 10¹¹ Pa.

Answer:

Question 6.
The edges of an aluminium cube are 10 cm long. One face of the cube is firmly fixed to a vertical wall. A mass of 100 kg is then attached to the opposite face of the cube. The shear modulus of aluminium is 25 GPa. What is the vertical deflection of this face? (1 Pa = 1 Nm^-2)
Answer:

Question 7.
Four identical hollow cylindrical columns of steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column. The Young’s modulus of steel is 2.0 x 10¹¹ Pa (1 Pa = 1 N m^-2).
Answer:
Here, outer radius of each column, r₀ = 60 cm = 60 x 10“² m
Inner radius, r,= 30 cm = 30 x 10^-2 m

Question 8.
A piece of copper having a rectangular cross-section of 15.2 mm x 19.1 mm is pulled in tension with 44,500 N force-producing only elastic deformation. Calculate the resulting strain. The shear modulus of elasticity of copper is 42 x 10⁹ Nm-².
Answer:
Here, A = 15.2 x 19.1 x 10-⁶m² . F = 44,500 N ; G = 42 X 10⁹ Nm-².

Question 9.
A steel cable with a radius of 1.5 cm supports a chairlift at a ski area. If the maximum stress is not to exceed 10⁸ Nm^-2, what is the maximum load the cable cab support?
Answer:
r = radius of cable = 1.5 cm.
A = π(0.015)² m²
max stress = 10⁸ Nm^-2
Let F be the maximum force.
stress = \(\frac{F}{A}\)
∴ F = max stress × area.
= 10⁸ × π × (0.015)²
= 7.065 × 10⁴N
= 70,650 N
∴ The maximum load is 70,650 N

Question 10.
A rigid bar of mass 15 kg is supported symmetrically by three wires each 2 m long. These at each end are of copper and middle one is of iron. Determine the ratio of their diameters if each is to have the same tension. Young’s modulus of elasticity for copper and steel are 120 x 10⁹ Nm^-2 and 190 x 10⁹ Nm^-2 respectively.
Answer:
As each wire has same tension F, so each wire has same extension due to mass of rigid bar. As each wire is of same length, hence each wire has same strain. If D is the diameter of wire, then


Question 11.
A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1 m, is Whirled in a vertical circle with an angular velocity of 2 rad/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm². Calculate the elongation of the wire when the mass is at the lowest point of its path.
Y_steel = 2 x 10¹¹ Nm^-2.
Answer:
Here, m = 14.5 kg ;l = r = lm;v = 2rps;A = 0065 x 10^-4 m²
Total pulling force on mass, when it is at the lowest position of the vertical circle is
F = mg + mr ω²
= 14 .5 x 9.8 + 14.5 x 1 x 4
= 200.1 N

Question 12.
Compute the bulk modulus of water from the following data : Initial volume = 100.0 litre, Pressure Increase =100.0 atm (1 atm = 1.013 x 10⁵Pa), Final volume = 100.5 liter. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.
(1 Pa = 1 Nm^-2 ).
Answer:
Given, V = 100 litre; Vf = 100.5 litre
:. Change in volume, ΔV = V_f – V_i= 100.5 – 100 = 0.5 liter

Question 13.
What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03 x 10³ kg m^-3? (Compressibility of water is 45.8 x 10^-11 Pa^-1 ; 1 Pa = 1 N m^-2)
Answer:
The increase in density is given by dρ = CPρ
where C is compressibility, P is pressure and p is the density at the surface.
Here, C = 45.8 x 10^-11 Pa^-1, P = 80 x 1 x 10⁵ Nm^-2, p = 1.03 x 10³kg m³
∴   dρ= 45.8 x 10^-11 x 80 x 10⁵ x 1.03 x 10³ = 3.774 kg m³
∴Total density at the given depth = ρ+ dp = 1.03 x 10³ + 3.774
= 1.03 x 1000 + 3.774 = 1033.774 kg m^-3 ≈1 .034 X 10³ kg m^-3

Question 14.
Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atmospheres. Bulk modulus of elasticity of glass = 37 x 10⁹ Nm^-2 and 1 atm = 1.103 x 10^s Pa
Answer:
Here ρ= 10 atm = 10 x 1.013 x 10⁵ Pa ; K = 37 x 10⁹ Nm^-2

Question 15.
Determine the volume contraction of a solid copper cube, 10 cm on an edge when subjected to a hydraulic pressure of 7 x 10⁶ Pa. Bulk modulus of copper = 140 G Pal
Answer:
Here,  L = 10 cm = 0.10 m ; p = 7 x 10⁶ Pa ;
K = 140 G Pa = 140 x 10⁹ Pa

Question 16.
How much should be pressure on a liter of water be changed to compress it by 0.10% ? Bulk modulus of elasticity of water = 2.2 x 10⁹ Nm^-2.
Answer:
V = 1 liter = 10^-3 m³ ; ΔV/V = 0.10/100 = 10^-3

Question 17.
Anvils made of single crystals of diamond, with the shape as shown in Fig. are used to investigate behavior of materials under very high pressures. Flat faces at the narrow end of the anvil have a diameter of 0.5 mm, and the wide ends are subjected to a compressional force of 50,000 N. What is the pressure at the tip of the anvil?

Answer:
Here, compressional force, F = 50000 N
Diameter, D = 0.5 mm = 0.5 x 10^-3 m

Question 18.
A rod of length 1.05 m having negligible mass is supported at its ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths as shown in Fig. The cross-sectional area of wires A and B are 1 mm² and 2 mm² respectively. At what point along the rod should a mass m be suspended in order to produce equal stresses and equal strains in both steel and aluminium wires? Given,

Answer:

Question 19.
A mild steel wire of length 1 m and cross-sectional area 0.5 x 10^-2 cm² is stretched, well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the midpoint of the wire. Calculate the depression at the midpoint, g = 10 ms^-2 ; Y = 2 x 10¹¹ Nm^-2.
Answer:


Question 20.
Two strips of metal are riveted together at their ends by four rivets, each of diameter 6 mm. What is the maximum tension that can be exerted by the riveted strip if the shearing stress on the rivet is not to
Answer:
Here, r = 6/2 = 3 mm = 3 x 10^-3 m; Max. stress = 2.9 x 10⁷ Pa ;
Max. load on a rivet = Max. stress x area of cross-section
= 2.9 x 10⁷ x (22/7) x (3 x 10^-3)²

Question 21.
The Marina Trench is located in the Pacific Ocean and at one place it is nearly eleven km beneath the surface of the water. The water pressure at the bottom of the trench is about 1.1 x 10⁸ Pa. A steel ball of initial volume 0.32 m³ is dropped into the ocean and falls to the bottom of the Trench. What is the change in the volume of the ball when it reaches the bottom?
Answer:
p = 1.1 x 10⁸ Pa ; V = 0.32 m³ ; K = 1.6 x 10¹¹ Pa

We hope the NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids, help you. If you have any query regarding NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 11 Physics Chapter 8 Gravitation

These Solutions are part of NCERT Solutions for Class 11 Physics. Here we have given NCERT Solutions for Class 11 Physics Chapter 8 Gravitation

Question 1.
Answer the following:
(a) You can shield a charge from electrical forces by putting it inside a hollow conductor. Can you shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means?
(b) An astronaut inside a small space ship orbiting around the earth cannot detect gravity. If the space station orbiting around the Earth has a large size, can he hope to detect gravity?
(c) If you compare the gravitational force on the Earth due to the Sun to that due to the moon, you would find that the Sun’s pull is greater than the moon’s pull (you can check this yourself using the data available in the succeeding exercises). However, the tidal effect of the moon’s pull is greater than the tidal effect of the Sun. Why?
Answer:
(a) Gravitational force on a body inside a hollow sphere is zero. However, a gravitational force acts on a body inside a hollow sphere due to bodies lying outside the hollow sphere. Hence a body cannot be shielded from the gravitational influence of nearby matter.
(b) Yes, with a larger mass, the value of gravity will be significant inside the spaceship.
(c) Tidal effect

Although Sun’s pull is more than the moon, yet tidal effect due to the moon is more because the moon is nearer to the earth.

Question 2.
Choose the correct alternative:
(a) Acceleration due to gravity increases/decreases with increasing altitude.
(b) Acceleration due to gravity increases/decreases with increasing depth (assume the Earth to be a sphere of uniform density).
(c) Acceleration due to gravity is independent of the mass of the Earth/mass of the body.
(d) The formula -G Mm (1/r₂ – 1/r₁) is more/less accurate than the formula mg (r₂ – r₁) for the difference of potential energy between two points r₂ and r₁ distance away from the center of the Earth.
Answer:

Question 3.
Suppose there existed a planet that went around the sun twice as fast as the Earth. What would be its orbital size as compared to that of the Earth?
Answer:

Question 4.
One of the satellites of Jupiter has an orbital period of 1.769 days and the radius of the orbit is 4.22 x 10⁸ m. Show that the mass of Jupiter is about one thousand times that of the sun. (Take 1 year = 365.25 mean solar day).
Answer:

Question 5.
Let us consider that our galaxy consists of 2.5 x 10¹¹ stars each of one solar mass. How long will this star at a distance of 50,000 ly from the galactic center take to complete one revolution? Take the diameter of the Milky Way to be 10⁵ ly. G = 6.67 x 10¹¹ Nm² kg².
Answer:
r = 50,000 ly = 50,000 x 9.46 x 10¹⁵m = 4.73 x 10²⁰m
M = 2.5 x 10¹¹ solar mass = 2.5 x 10ⁿ x (2 x 10³⁰) kg
= 5.0 x 10⁴¹ kg

Question 6.
Choose the correct alternative:
(1) If the zero of the potential energy is at infinity, the total energy of an orbiting satellite is negative of its kinetic/potential energy.
(2) The energy required to rocket an orbiting satellite out of earth’s gravitational influence is more/less than the energy required to project a stationary object at the same height (as the satellite) out of Earth’s influence.
Answer:
1. Kinetic energy
The potential energy of a satellite rotating in its orbit is zero. The total energy of a system is the sum of its kinetic energy (+ve) and potential energy. Since the earth satellite system is a bound system. The satellite has negative total energy. So, the energy of the satellite is the negative of its kinetic energy.

2. Less
An orbiting satellite has more energy than a stationary object at the same height. This additional energy is provided by the orbit. It requires lesser energy to make it move out of the earth’s influence than a stationary object.

Question 7.
Does the escape speed of a body from the Earth depend on
(a) the mass of the body,
(b) the location from where it is projected,
(c) the direction of projection,
(d) the height of the location from where the body is launched? Explain your answer.
Answer:
(a) Escape speed of a body is independent of the mass of a body

where M is the mass of early
(b) Escape speed of a body depends upon the location.
(c) Escape speed of a body is independent of the direction of projection.
(d) Escape speed of a body depends upon the height of the location from where the body is projected because the escape velocity depends upon the gravitational potential at the point from which it is projected and this potential depends upon height also

Question 8.
A comet orbits the sun in a highly elliptical orbit. Does the comet have a constant
(a) linear speed,
(b) angular speed,
(c) angular momentum,
(d) kinetic energy,
(e) potential energy,
(f) total energy throughout its orbit?
Neglect any mass loss of the comet when it comes very close to the Sun.
Answer:
Angular momentum and total energy do not vary throughout the orbit whereas the rest all the quantities vary in the orbit.

Question 9.
Which of the following symptoms is likely to afflict an astronaut in space :
(a) swollen feet,
(b) swollen face,
(c) headache,
(d) orientational problem?
Answer:
The astronaut in space will suffer from
(b) swollen face,
(c) headache and
(d) orientational problem.

Question 10.
The gravitation intensity at the center of the drum head defined by a hemispherical shell has the direction indicated by the arrow (see Fig.), (i) a, (ii) b, (iii) c, (iv) zero.

Answer:
Intensity inside a shell is zero, so it will be zero at P and Q also (Potential is constant).

Question 11.
For the above problem, the direction of the gravitational intensity at an arbitrary point P is indicated by the arrow (i) d, (ii)y, (iii) f, (iv) g.
Answer:
As in the fig, upper portion of the shell is missing, so gravitational intensity at P and Q should act along e and c respectively.

Question 12.
A rocket is fired from the earth towards the Sun. At what distance from the Earth’s center is the gravitational force on the rocket zero? Mass of the Sun = 2 x 10³⁰ kg, mass of the Earth = 6 x 10²⁴ kg. Neglect the effect of other planets etc. (orbital radius = 1.5 x 10¹¹ m).
Answer:

Mass of Sun, M = 2 x 10³⁰ kg
Mass of Earth, m = 6 x 10²⁴ kg
Distance between Sun and Earth, r = 1.5 x 10¹¹ m
Let at the point P,
the gravitational force on the rocket due to Earth = gravitational force on the rocket due to Sun
x = distance of the point P from the Earth

Question 13.
How will you weigh the Sun, that is estimate its mass? You will need to know the period of one of its planets and the radius of the planetary orbit. The mean orbital radius of the Earth around the Sun is 1.5 x 10⁸ km. Estimate the mass of the Sun.
Answer:
Here radius of earth’s orbit (R + x) = 1.5 x 10⁸ x 10³ = 1.5 x 10¹¹ m
Time period of sun, T = 365 days = 365 x 24 x 60 x 60 s

Question 14.
A Saturn year is 29.5 times the Earth year. How far is Saturn from the Sun if the Earth is 1.50 x 10⁸ km away from the Sun?
Answer:
As we know the Kepler’s third law

Question 15.
A body weighs 63 N on the surface of the Earth. What is the gravitational force on it due to the Earth at a height equal to half the radius of the Earth?
Answer:

Question 16.
Assuming the Earth to be a sphere of uniform mass density, how much would a body weigh halfway down to the center of the Earth if it weighed 250 N on the surface?
Answer:

Question 17.
A rocket is fired vertically with a speed of 5 km s^-1 from the Earth’s surface. How far from the Earth does the rocket go before returning to the Earth? Mass of the earth = 6.0 x 10²⁴ kg, mean radius of the Earth = 6.4 x 10⁶ m ; G = 6.67 x 10^–11 N m² kg^-2.
Answer:

Question 18.
The escape speed of a projectile on the Earth’s surface is 11.2 km s^–1. A body is projected out with thrice this speed. What is the speed of the body far away from the Earth? Ignore the presence of the Sun and the other planets.
Answer:

Question 19.
A satellite orbits the Earth at a height of 400 km above the surface. How much energy must be expended to rocket the satellite out of the Earth’s gravitational influence? Mass of the satellite = 200 kg ; mass of the Earth = 6.0 x 10²⁴ kg ; radius of the Earth = 6.4 x 10⁶ m ; G = 6.67 x 10^–11 N m² kg^–2
Answer:
As we know, the total energy of a satellite in orbit,

Question 20.
Two stars each of one solar mass (= 2 x 10³⁰ kg) are approaching each other for a head-on collision. When they are at a distance of 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).
Answer:
Here, M = 2 x 10³⁰ kg ; R = 10⁴ km = 10⁷ m ; r = 10⁹ km = 10¹² m

This decrease in the potential energy of the system of two stars will appear as an increase in their K.E.
When the stars are at a distance of 10⁹ km, their speeds are negligible and hence their initial K.E.s are also negligible. If υ is the speed with which the two stars collide, then increase in K.E. of the system of the two stars

Question 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 Held 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?
Answer:

The gravitational field at P due to sphere A is equal and opposite to the gravitational field at P due to sphere B.
Hence, the net gravitational field at P is zero.
Gravitational potential at P, V = V_A + V_B

Question 22.
As you have learned 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 x 10²⁴ kg, radius = 6400 km.
Answer:
Distance of satellite from the center of earth = R + r + x
= 6400 + 36000
= 42400 km
= 4.24 x 10⁷ m

Question 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 observed stellar objects called pulsars are believed to belong to this category). Will an object placed on its equator remain stuck to its surface due to gravity? (mass of the Sun = 2 x 10³⁰ kg).
Answer:
The centripetal acceleration of the object placed at the equator of the star, a_c

Question 24.
A spaceship is stationed on Mars. How much energy must be expended on the spaceship to rocket it out of the solar system ? Mass of the spaceship = 1000 kg ; mass of the Sun = 2 x 10³⁰ kg ; mass of Mars = 6.4 x 10²³ kg ; radius of Mars = 3395 km ; radius of the orbit of Mars = 2.28 x 10⁸ km ; G = 6.67 x 10^-11 Nm² kg^-2.
Answer:
The total energy of the spaceship in the orbit of mars,

Question 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 x 10²³ kg; radius of Mars = 3395 km; G = 6.67 x 10^-11N m² kg^-2.
Answer:
Let m = Mass of the rocket, M = Mass of Mars, R = Radius of Mars

We hope the NCERT Solutions for Class 11 Physics Chapter 8 Gravitation, help you. If you have any query regarding NCERT Solutions for Class 11 Physics Chapter 8 Gravitation, Energy and Power, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion

These Solutions are part of NCERT Solutions for Class 11 Physics. Here we have given NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion

Question 1.
Give the location of the center of mass of a (i) sphere, (ii) cylinder, (iii) ring, and (iv) cube, each of uniform mass density. Does not the center of mass of a body necessarily lie inside the body?
Answer:
The Centre of the mass of sphere cylinder, ring, and cube with homogenous mass distribution lies at its geometric centre. It is not necessary that CM (centre of mass) lies inside the body as in the cases of the ring or hollow hemisphere.

Question 2.
In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 A (1 A = 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.
Answer:

Let us choose the nucleus of the hydrogen atom as the origin for measuring distance.
Mass of hydrogen atom, m₁ = 1 unit (say)
Mass of chlorine atom, m₂ = 35.5 units (say)
Now, x₁= 0 and x₂ = 1.27 A = 1.27 x 10^-10 m
The distance of C.M. of HCl molecule from the origin is given by

Question 3.
A child sits stationary at one end of a long trolley moving uniformly with a speed V on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of C.M. of the (trolley + child) system?
Answer:
There is no external force acting on the child and the trolley system, hence the momentum of the entire system is conserved. Therefore the CM keeps moving with its initial speed of V in the same direction.

Question 4.
Show that the area of the triangle contained between the vectors a and b is one-half of the magnitude of a x b.
Answer:

Question 5.
Show that a.(b x c) is equal in magnitude to the volume of the parallelepiped formed on the three vectors a, b and c.
Answer:

Let a parallelepiped be formed on the three vectors

Question 6.
Find the components along the x, y, z axes of the angular momentum \(\vec { l } \)of a particle, whose position vector is \(\vec { r } \) with components x, y, z and momentum is with components x, y, z and momentum is \(\vec { p } \) with components p_x, p_y and pz. Show that if the particle moves only in the x-y plane, the angular momentum has only a z-component.
Answer:

Question 7.
Two particles, each of mass m and speed u, travel in opposite directions along parallel lines separated by a distance d. Show that the vector angular momentum of the two-particle system is the same whatever be the point about which the angular momentum is taken.
Answer:

Question 8.
A non-uniform bar of weight W is suspended at rest, by two strings of negligible weight as shown in Fig. The angles made by the strings with the vertical are 36.9° and 53.1° respectively. The bar is 2 m long. Calculate the distance d of the center of gravity of the bar from its left end.
Answer:

As is clear from Fig.,
θ₁ = 36.9°,θ₂ = 53.1°.
If T₁, T₂ are the tensions in the two strings, then for equilibrium along the horizontal,
T₁  sin θ1 = T₂ sin θ₂


Question 9.
A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its center of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.
Answer:

Here,     m = 1800 kg
Distance of center of gravity (C) behind the front axle = 1.05 m.
Let R₁. R₂ be the force exerted by the level ground on front wheels and back wheels. As is clear from fig.,
R₁+ R₂ = mg = 1800 x 9.8 = 17640 N
For rotational equilibrium about C,

Question 10.
(a) Find the moment of inertia of a sphere about a tangent to the sphere, given the moment of inertia of the sphere about any of its diameters to be 2 MR²/5, where M is the mass of the sphere and R is the radius of the sphere.
(b) What is the moment of inertia of a uniform disc of radius R and mass M about an axis
(1) passing through its center and normal to the disc
(2) passing through a point on its edge and normal to the disc?
The moment of inertia of the disc about any of its diameters is given to be \(\frac { 1}{ 4 } \) MR².
Answer:

Question 11.
Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its center. Which of the two will acquire a greater angular speed after a given time?
Answer:
M.I. of the cylinder =I₁= MR²
M.I. of the cylinder=I₂=\(\frac {2}{ 5 } \) MR²

ω = ω₀ +αt, therefore sphere acquires a greater speed than a cylinder as α₂ >α₁

Question 12.
A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad s^-1. The radius of the cylinder is 0.25 What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis?
Answer:

Question 13.
(a) A child stands at the center of a turntable with his two arms outstretched. The turntable is set to rotate with an angular speed of 40 rev/min. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to 2/5 times the initial value? Assume that the turntable rotates without friction.
(b) Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy?
Answer:
(a) Suppose, the initial moment of inertia of the child is I₁. Then the final moment of inertia,
I₂ = \(\frac {2}{ 5 } \)
Also, v₁= 40 rev min

Clearly, final (K.E.)_rot becomes more because the child uses his internal energy when he folds his hands to increase the kinetic energy.

Question 14.
A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N? What is the linear acceleration of the rope? Assume that there is no slipping.
Answer:
Here, M = 3 kg, R = 40 cm = 0.4 m
M.I. of the hollow cylinder about its axis = I = MR² = 3 x (0.4)² = 0.48 kg m²
When the force of 30 N is applied over the rope wound on the cylinder, the torque will act on the cylinder. It is given by
τ= FR = 30 x 0.4 = 12 N m
If a is angular acceleration produced, then
τ = Iα

Question 15.
To maintain a rotor at a uniform angular speed of 200 rad s^-1, an engine needs to transmit a torque of 180 Nm. What is the power required by the engine? (Note: uniform angular velocity in the absence of friction implies zero torque. In practice, applied torque is needed to counter frictional torque). Assume that the engine is 100% efficient.
Answer:
Here τ = 18 Nm, ω = 200 rad s^-1
P =τω
∴ V P = 180 x 200 = 36000 W = 36 kW.

Question 16.
From a uniform disk of radius R, a circular section of radius R/2 is cut out. The center of the hole is at R/2 from the center of the original disc. Locate the center of mass of the resulting flat body.
Answer:
Let the mass of disc = M


Mass of the portion removed from the disc is concentrated at [O] and the mass of the remaining disc is supposed to be concentrated at O₂ at a distance x from the center of the disc (O).

Question 17.
A meter stick is balanced on a knife-edge at its center. When two coins, each of mass 5 g are put one on top of the other at the 12*0 cm mark, the stick is found to be balanced at 45-0 cm. What is the mass of the meter stick?
Answer:

Let m be the mass of the stick concentrated at C, the 50 cm mark (Fig.)
According to the principle of moments
Moment of the mass of coins about C’ = moment of the mass of the rod about C’
10 g (45 – 12) = mg (50 – 45)

Question 18.
A solid sphere rolls down two different inclined planes of the same heights but different angles of inclination. Will it reach the bottom with the same speed in each case? Will it take longer to roll down one plane than the other? If so, which ones and why
Answer:

Since sphere rolls down two inclined planes of same height, so velocity of sphere in both the cases is same.

Since θ is different in both cases, so sphere will take longer time in case of inclined plane having a smaller inclination angle (0).

Question 19.
A hoop of radius 2 m weighs 100 kg. It rolls along a horizontal floor so that its center of mass has a speed of 20 cm/s. How much work has to be done to stop it?
Answer:
Here,  R = 2m, 100 kg
u = 20 cm/s = 0.2 m/s

Question 20.
The oxygen molecule has a mass of 5.30 x 10^–26 kg and a moment of inertia of 1.94 x 10^-46 kg m² about an axis through its center perpendicular to the line joining the two atoms. Suppose the mean speed of such a molecule in a gas is 500 m/s and that its kinetic energy of rotation is two-thirds of its kinetic energy of translation. Find the average angular velocity of the molecule.
Answer:

Here, m = 5.30 x 10^–26 kg
I = 1.94 X 10^-46 kg m²
υ = 500 m/s
If m/2 is mass of each atom of oxygen and 2r is distance between the two atoms as shown in Fig., then

Question 21.
A cylinder rolls up an inclined plane of the angle of inclination of 30°. At the bottom of the inclined plane, the center of mass of the cylinder has a speed of 5 m/s.
(a) How far will the cylinder go up the plane?
(b) How long will it take to return to the bottom?
Answer:

Here, θ = 30°, υ= 5 m/s
Let the cylinder go up the plane upto a height h.

Question 22.
As shown in Fig., the two sides of a step ladder BA and CA are 1.6 m long and hinged at A. A rope DE, 0.5 m is tied halfway up. A weight of 40 kg is suspended from a point F, 1.2 m from B along with the ladder BA. Assuming the floor to be frictionless and neglecting the weight of the ladder, find the tension in the rope and forces exerted by the floor on the ladder. (Take g=9.8 m/s²)
(Hint. Consider the equilibrium of each side of the ladder separately.)

Answer:

Let T be the tension in the rope DE. R_B and R_c are the normal reactions of the floor at B and C respectively.
Since the ladder is in translational equilibrium, therefore, R_B + R_c = W = mg = 40 x 9.8 = 392 N …(i)
A ladder is also in rotational equilibrium, therefore, net torque on arms AB and AC is zero.
For arm AB, R_B x BG – W x IG = T x AJ

Question 23.
A man stands on a rotating platform, with his arms stretched horizontally holding a 5 kg weight in each hand. The angular speed of the platform is 30 revolutions per minute. The man then brings his arms close to his body with the distance of each weight from the axis changing from 90 cm to 20 cm. The moment of inertia of the man together with the platform may be taken to be constant and equal to 7.6 kg m². What is his new angular speed? (Neglect friction.)Is kinetic energy conserved in the process? If not, from where does the change come about?
Answer:

No, kinetic energy is not conserved in the process. In fact, as a moment of inertia decreases, the K.E. of rotation increases. This change comes about as work is done by the man in bringing his arms closer to his body.

Question 24.
A bullet of mass 10 g and speed 500 m/s is fired into a door and gets embedded exactly at the center of the door. The door is 1.0 m wide and weighs 12 kg. It is hinged at one end and rotates about a vertical axis practically without friction. Find the angular speed of the door just after the bullet embeds into it.(Hint. The moment of inertia of the door about the vertical axis at one end is ML²/3.)
Answer:

Question 25.
Two discs of moments of inertia I, and I₂ about their respective axes (normal to the disc and passing through the center), and rotating with angular speeds ω₁ and ω₂ are brought into contact face to face with their axes of rotation coincident,
(a) What is the angular speed of the two-disc system?
(b) Show that the kinetic energy of the combined system is less than the sum of the initial kinetic energies of the two discs. How do you account for this loss in energy? Take ω_1 ≠ω₂
Answer:
Initial angular moment of the discs = I₁ω₁ + I₂ω₂
M.I. of two discs combined as a system = I₁+ I₂
Final angular moment of the combination = (I+ I₂)ω
By using the law of conservation of angular momentum,
we get I₁ω₁ + I₂ω₂ = ( I₁+ I₂ )ω

As the above term comes out to be positive, thus, the rotational kinetic energy of the combined disc is less than the total initial energy.

Question 26.
(a) Prove the theorem of perpendicular axes (Hint. Square of the distance of a point (x, y) in the x-y plane from an axis perpendicular to the plane through the origin is x² + y²).
(b) Prove the theorem of parallel axes (Hint. If the center of mass is chosen to be the origin εm_ir_i= 0).
Answer:
(a) The theorem of perpendicular axes: According to this theorem, the moment of inertia of a plane lamina (i.e., a two-dimensional body of any shape/size) about any axis OZ perpendicular to the plane of the lamina is equal to sum of the moments of inertia of the lamina about any two mutually perpendicular axes OX and OY in the plane of lamina, meeting at a point where the given axis OZ passes through the lamina. Suppose at the point ‘R’ m{ particle is situated moment of inertia about Z-axis of lamina
= moment of inertia of the body about r-axis
= moment of inertia of the body about the y-axis.

(b) Theorem of parallel axes: According to this theorem, a moment of inertia of a rigid body about any axis AB is equal to the moment of inertia of the body about another axis KL passing through centre of mass C of the body in a direction parallel to AB, plus the product of total mass M of the body and square of the perpendicular distance between the two parallel axes. If h is the perpendicular distance between the axes AB and KL, then Suppose the rigid body is made up of n particles m1, m2, …. mn, mn at perpendicular distances r₁, r₂, r_i…. r_n. respectively from the axis KL passing through centre of mass C of the body.
If h is the perpendicular distance of the particle of mass m{ from KL, then


Question 27.
Prove the result that the velocity of translation of a rolling body (like a ring, disc, cylinder, or sphere) at the bottom of an inclined plane of a height h is given by using dynamical consideration (i.e., by consideration of forces and torques). Note k is the radius of gyration of the body about its symmetry axis, and R is the radius of the body. The body starts from rest at the top of the plane.

Answer:
When a body rolls down an incline of height h, we apply the principle of conservation of energy.
K.E. of translation + K.E. of rotation = P.E. at the top.

Question 28.
A disc rotating about its axis with angular speed ω₀ is placed lightly (without any translational push) on a perfectly frictionless table. The radius of the disc is R. What are the linear velocities of the points A, B, and C on the disc shown in Fig.? Will the disc roll in the direction indicated?
Answer:


The disc will not roll in the given direction because friction is necessary for the same.

Question 29.
Explain why friction is necessary to make the disc in Fig. shown in Q. 28, roll in the direction indicated.
(a) Give the direction of frictional force at B, and the sense of frictional torque, before perfect rolling begins.
(b) What is the force of friction after perfect rolling begins?
Answer:
To make the disc roll, torque is required. This torque will be provided by the frictional force.
(a) At point B, the frictional force supports the angular motion of this point, so frictional force is in the direction of the arrow itself. The direction of frictional torque is normal to the paper in an outward direction.
(b) Frictional force tries to decrease the velocity of point B. When this velocity becomes zero, perfect rolling beings. For zero velocity, the force of friction also becomes zero.

Question 30.
A solid disc and a ring, both of radius 10 cm are placed on a horizontal table simultaneously, with an initial angular speed equal to 10 π rad s^-2. Which of the two will start to roll earlier? The coefficient of kinetic friction is μ_k = 0.2.
Answer:
Force of friction (μ_k  mg) produces an acceleration a in the center of mass (moving with υ = R ω)

Question 31.
A cylinder of mass 10 kg and radius 15 cm is rolling perfectly on a plane of inclination 30°. The coefficient of the static friction μ_s = 0.25.
(a) How much is the force of friction acting on the cylinder?
(b) What is the work done against friction during rolling?
(c) If the inclination 6 of the plane is increased, at what value of 6 does the cylinder begin to skid, and not roll perfectly?
Answer:
Here M = 10 kg; R = 15 cm = 0.15 m; μ_s= 0.25; θ = 30°

Question 32.
Read each statement below carefully, and state, with reason, it is true or false:
(a) During rolling, the force of friction acts in the same direction as the direction of motion of the CM of the body.
(b) The instantaneous speed of the point of contact during rolling is zero.
(c) The instantaneous acceleration of the point of contact during rolling is zero.
(d) For perfect rolling motion, work done against friction is zero.
(e) A wheel moving down a perfectly frictionless inclined plane will undergo a slipping (not rolling) motion.
Answer:
(a) True (The force of friction helps in rolling a body).
(b) True (A rolling body is considered as a rotating body about an axis passing through the point of contact).
(c) False (Since the body is rotating, so its instantaneous acceleration cannot be zero).
(d) True (Since the point of contact is at rest, so work done is zero).
(e) True (In the case of the frictionless inclined plane, there is no tangential force of friction (or torque) and hence wheel cannot roll).

We hope the NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion, help you. If you have any query regarding NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion, drop a comment below and we will get back to you at the earliest

NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory

These Solutions are part of NCERT Solutions for Class 11 Physics. Here we have given NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory

Question 1.
Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the diameter of an oxygen moleculer to be roughly 3 A.
Answer:

Question 2.
Molar volqme is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP : 1 atmospheric pressure, 0°C). Show that it is 22.4 litres.
Answer:
For one mole of an ideal gas, we have

Question 3.
Following figure shows plot of PV/T versus P for 1.00 x 10^-3 kg of oxygen gas at two different temperatures.
Answer:

(a) What does the dotted plot signify ?
(b) What is true : T_x > T₂ or T_x < T₂ ? ‘
(c) What is the value of PV/T where the curves meet on the y-axis ?
(d) If we obtained similar plots for 1.00 X 10^-3 kg of hydrogen, would we get the same value of \(\frac { PV}{ T } \) at the point where the curves meet on the y-axis ? If not, what mass of hydrogen yields the same value of \(\frac { PV}{ T } \) (for low pressure-high temperature region of the plot) ?
(Molecular mass of H₂ = 2.02 u, of O₂ = 32.0 u, R = 8.31 mol^-1 K^_1).
Answer:
(a) The dotted plot shows that is a constant quantity \(\frac { PV}{ T } \) . This signifies the ideal gas behaviour.
(b) Here T₁ > T₂. This is because curve at T₁ is close to ideal behaviour of gas which occurs at higher temperature.
(c) At the point where the curve meets the y-axis, we have \(\frac { pv}{ T } \) = μR, where p is the number of moles of oxygen gas.
Here, Mass of oxygen, m = 1.00 x 10^-3 kg
Also, molecular mass, M = 32 X 10^-3 kg

Question 4.
An oxygen cylinder of volume 30 liters has an initial gauge pressure of 15 atm and a temperature of 27°C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drop to 17°C. Estimate the mass of oxygen taken out of the cylinder. (R = 8.31 J mol^-1 K^_1, molecular mass of O₂ = 32 u).
Answer:
Under the initial conditions,
V = 30 liter = 30 x 10^-3 m^-3, P = 15 atm = 15 x 1.01 x 10⁵ Pa
T = 27 °C = 273 + 27 = 300 K.
Also, R = 8.31 J mol^-1 K^-1 and molar mass, M = 32 x 10″³ kg.
∴ Using the relation PV = μRT

Question 5.
An air bubble of volume 1.0 cm³ rises from the bottom of a lake 40 m deep at a temperature of 12°C. To what volume does it grow when it reaches the surface, which is at a temperature of 35 °C ?
Answer:
Volume of the bubble inside, V₁ = 1.0 cm³ = 1 x 10^-6 m³
Pressure on the bubble, P₁ = Pressure of water 4- Atmospheric pressure
= ρgh + 1.01 x 10⁵ = 1000 x 9.8 x 40 + 1.01 x 10⁵
= 3.92 x 10⁵ + 1.01 x 10⁵ = 4.93 x 10⁵ Pa
Temperature, T, = 12°C = 273 + 12 = 285 K
Also, the pressure outside the lake, P2 = 1.01 x 10⁵ N m^-2

Question 6.
Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0 m³ at a temperature of 27°C and 1 atm pressure.
Answer:
Here, P = 1 atm = 1 x 1.01 x 10⁵ Pa,
V = 25.0 m³, T = 27°C = 27 + 273 = 300 K and
R = 8.31 J mol^-1 K^-1

Now, one mole of a gas contains = 6.023 x 10²³ molecules
1.013 x 10³ moles would contain = 6.023 x 10²³ x 1.013 x 10³
= 6.10 x 10²⁶ molecules.
= 6.10 x 10²⁶ molecules.

Question 7.
Estimate the average thermal energy of a helium atom at
(1) room temperature (27°C),
(2) the temperature on the surface of the Sun (6000 K),
(3) the temperature of 10 million kelvin (the typical core temperature in the case of a star).
Answer:

Question 8.
Three vessels of equal capacity have gases at the same temperature and pressure. The first-vessel contains neon (monoatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (polyatomic). Do the vessels contain an equal number of respective molecules? Is the root mean square speed of molecules the same in the three cases? If not, in which case is υ_rms the largest?
Answer:
Equal volumes of all the gases under similar conditions of pressure and temperature contain an equal number of molecules (according to Avogadro’s hypothesis). Therefore, the number of molecules in each case is the same.
The rms velocity of molecules is given by

Since Neon has minimum atomic mass M, its rms velocity is maximum.

Question 9.
At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at – 20°C? (atomic mass of Ar = 39.9 u, of He = 4.0 u).
Answer:
For argon, atomic mass m  = 39.9 u = 39.9 X 1.67 x 10^-27 kg = 6.66 x 10^-26 kg

Question 10.
Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17°C. Take the radius of a nitrogen molecule to be roughly 1.0 A. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of N2 = 28.0 u).
Answer:

Question 11.
A meter-long narrow bore held horizontally (and closed at one end) contains a 76 cm long mercury thread, which traps a 15 cm column of air. What happens if the tube is held vertically with the open end at the bottom?
Answer:

Question 12.
From ascertain apparatus, the diffusion rate of hydrogen has an average value of 28.7 cm³ s^-1. The diffusion of another gas under the same conditions is measured to have an average rate of 7.2 cm³ s^-1. Identify the gas.
Answer:
Here, r₁ = 28.7 cm³ s^_1, r₂ = 7.2 cm³ s^-1, M₁ = 2g and M₂ = ?
∴ Using Graham’s law of diffusion,

This is the molecular mass of oxygen. Therefore, the other gas is oxygen

Question 13.
A gas in equilibrium has uniform density and pressure throughout its volume. This is strictly true only if there are no external influences. A gas column under gravity, for example, does not have a uniform density (and pressure). As you might expect, its density decreases with height. The precise dependence is given by the so-called law of atmospheres

where n₂, n₁ refer to number density at heights h₂ and h_x
respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column

where ρ is the density of the suspended particle, and ρ’ that of the surrounding medium. [N_A is Avogadro’s number, and R the universal gas constant].
Answer:

Consider the molecules of a gas uniformly distributed in a given region of space. Let p be die density of the gas and it is in thermodynamical equilibrium.

Consider an imaginary cylinder of this gas having a unit area of cross-section and placed vertically. Let Y direction be the vertical direction so that the lower cap I of the cylinder is parallel to XZ plane at a height y + dy above the XZ plane as shown in Fig. Let P and P + dP be the pressures at the caps I and cap II respectively

Then the force acting on the cap I due to gravity is the weight of cylinder
W = Mass of cylinder X g = Vρg = 1 X dy pg

Question 14.
Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms :

Answer:

We hope the NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory, help you. If you have any query regarding . NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 11 Physics Chapter 6 Work, Energy and Power

These Solutions are part of NCERT Solutions for Class 11 Physics. Here we have given NCERT Solutions for Class 11 Physics Chapter 6 Work, Energy and Power

Question 1.
The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative:

1. work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket,
2. work done by the gravitational force in the above case,
3. work done by friction on a body sliding down an inclined plane,
4. work done by an applied force on a body moving on a rough horizontal plane with uniform velocity,
5. work done by the resistive force of air on a vibrating pendulum in bringing it to rest.

Answer:
1. We know that: W = \(\overrightarrow{\mathrm{F}}, \overrightarrow{\mathrm{S}}\) = FS cos θ
‘Positive’
Reason: Force is acting in the direction of displacement (θ = 0°)
2. ’Negative’
Reason: Force is acting in the opposite direction to displacement (θ =180°)
3. ’Negative’
Reason: Force of friction is opposite to the displacement (θ = 180°)
4. ‘Positive’
Reason: The body mover in the direction of force applied (θ = 0°)
5. ‘Negative’
Reason: The resistive force opposes the motion (θ = 0°)

Question 2.
A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1. Compute the
(a) work done by the applied force in 10 s
(b) work done by friction in 10 s
(c) work done by the net force on the body in 10 s
(d) change in kinetic energy of the body in 10 s and interpret your results.
Answer:
Here M = 2 kg; u = 0; μ= 0.1; applied force, F = 7N, t = 10s
Force of friction, f= μMg = 0.1 x 2 x 9.8 = 1.96 N
.’. net force under which body moves, F’ = F- f = 7 – 1.96 = 5.04 N

Question 3.
Given in Fig. are examples of some potential energy functions in one dimension. The total energy of the particle is indicated by a cross on the ordinate axis. In each case, specify the regions, if any, in which the particle cannot be found for the given energy. Also, indicate the minimum total energy the particle must have in each case. Think of simple physical contexts for which these potential energy shapes are relevant.


Answer:
(a) We know that Total energy E = KE + PE, kinetic energy can never be negative. In the region between x = 0 & x = a.
Potential energy is ‘0’. So, kinetic energy y is positive. In region x > a the potential energy has a value greater than ‘E’. So kinetic energy will be negative in this region. Hence the particle cannot be present in the region x > a.

(b) Here PE > E, the total energy of the object and as such the kinetic energy of the object would be negative. Thus object cannot be present in any region on the graph.

(c) Here x = 0 to x = a & x > b, the P E is more then E so, K E is negative. The particle cannot be present in these portions.

(d) The object cannot exist in the region between
x = \(\frac{-b}{2}\) to x = \(\frac{-a}{2}\) & x = \(\frac{-a}{2}\) to x = \(\frac{-b}{2}\)
Because in this region P E > E.

Question 4.
The potential energy function for a particle executing linear simple harmonic motion is given by V (x) = Kx² /2, where A is the force constant of the oscillator. For A = 0.5 N nr¹, the graph of V (x) versus x is shown in Fig. Show that a particle of total energy 1 J moving under this potential must ‘turn back’ when it reaches x = ± 2 m.

Answer:

Question 5.
Answer the following :
(a) The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy required for burning obtained? The rocket or the atmosphere ?
(b) Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not normal to the comet’s velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why?
(c) An artificial satellite orbiting the earth in very thin atmosphere loses its energy gradually due to dissipation against atmospheric resistance, however small. Why then does its speed increase progressively as it comes closer and closer to the earth ?
(d) In Fig.

1. The man walks 2 m carrying a mass of 15 kg on his hands. In Fig.
2. He walks the same distance pulling the rope behind him. The rope goes over a, pulley, and a mass of 15 kg hangs at its other end. In which case is the work done greater ?
   

Answer:
(a) Heat energy required for the burning of the casing of a rocket is obtained by a rocket. Since the work is done against the friction, the kinetic energy of the rocket decreases continuously and this work against friction reappears as heat energy.

(b) This is because of the conservative nature of the gravitational force. Work done by the gravitational force in a closed path is zero.

(c) As an artificial satellite gradually loses its energy due to dissipation against atmospheric resistance, its potential decreases rapidly. As a result, the kinetic energy of the satellite slightly increases i.e. its speed increases progressively.

(d) In fig (i) the force applied by the man is perpendicular to the direction of movement of mass, i.e. θ = 90°
W = Fs cos θ = Fs cos 90° = 0
In figure (ii) the force applied is in direction of the movement of mass i.e. θ = o°
∴ W = Fs cos θ
= Mgs cos θ
= 15 × 9.8 × 2 × 1
= 294 J.

Question 6.
Underline the correct alternative:
(a) When a conservative force does positive work on a body, the potential energy of the body increases/decreases/ remains unaltered.
(b) Work done by a body against friction always results in a loss of its kinetic/potential energy.
The rate of change of total momentum of many-particle system is proportional to the external force/sum of the internal forces on the system.
(c) In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy total linear momentum!total energy of the system of two bodies.
Answer:
(a) Work done by conservative force is equal to the negative of potential energy. When work done is positive, potential energy decreases.
(b)  Kinetic energy, because friction does work against motion of the body.
(c) External force, because in many-particle systems, the internal forces in the system cancel each other and hence cannot change the net momentum of the system.
(d) In inelastic collision, total energy and linear momentum are conserved. However, kinetic energy decreases.

Question 7.
State if each of the following statements is true or false. Give reasons for your answer.

1. In an elastic collision of two bodies, the momentum and energy of each body are conserved.
2. The total energy of a system is always conserved, no matter what internal and external forces on the body are present.
3. Work done in the motion of a body over a closed loop is zero for every force in nature.
4. In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system.

Answer:

1. False, the momentum and energy of each body are conserved.
2. False, the external force on the system may increase or decrease the total energy of the system.
3. False, for the nonconservative forces (friction) the work done in closed-loop is not zero.
4. True, usually in an inelastic collision the final kinetic energy is always less than the initial kinetic energy of the system.

Question 8.
Answer carefully, with reasons :
(a) In an elastic collision of two billiard balls, is the total kinetic energy conserved during the short time of collision of the balls (i.e., when they are in contact) ?
(b) Is the total linear momentum conserved during the short time of an elastic collision of two balls ?
(c) What are the answers to (a) and (b) for an inelastic collision ?
(d) If the potential energy of two billiard balls depends only on the separation distance between their centers, is the collision elastic or inelastic ? (Note, we are talking here of potential energy corresponding to the force during collision, not gravitational potential energy).
Answer:
(a) In this case total kinetic energy is not conserved because when the bodies are in contact during elastic collision even, the kinetic energy is converted into internal energy.
(b) Yes, because total momentum conserves as per law of conservation of momentum.
(c) In inelastic collision, K.E. is not conserved but linear momentum is conserved.
(d) It is a case of elastic collision because in this case the forces are of conservative nature

Question 9.
A body is initially at rest. It undergoes one-dimensional motion with constant acceleration. The power delivered to it at time t is proportional to
(1) t^m
(2) t
(3) t^3/2
(4) t²
Answer:

Question 10.
A body is moving uni-directionally under the influence of a source of constant power. Its displacement in time t is proportional to
(1) t^m
(2) t
(3) t^3/2
(4) f²
Answer:

Question 11.
A body constrained to move along the z-axis of a coordinate system is subject to a constant force F given by \( \vec { F }= \) \( \hat {i}+{ 2j }+{ 3K }\)
where \( \hat {i}+{ j }+{ K }\)are unit vectors along the x-, y- and z-axis of the system respectively. What is the work done by this force in moving the body a distance of 4 m along the z-axis ?
Answer:

Question 12.
An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy 10 keV, and the second with 100 keV. Which is faster, the electron or the proton? Obtain the ratio of their speeds, (electron mass = 9.11 x 10^-31 kg, proton mass = l.67 x 10^-27 kg, 1 eV = 1.60 x 10^-19 J).
Answer:
Let _υe = Speed of electron
_υp = speed of the proton
mg = mass of electron
and mp = mass of proton

Question 13.
A rain drop of radius 2 mm falls from a height of 500 m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height, it attains its maximum (terminal) speed, and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first and second half of its journey ? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is 10 ms^-1?
Answer:

Question 14.
A molecule in a gas container hits a horizontal wall with speed 200 ms^-1 and angle 30° with the normal, and rebounds with the same speed. Is momentum conserved in the collision ? Is the collision elastic or inelastic ?
Answer:
In all types of collisions, momentum is conserved. Let us check the conservation of kinetic energy.As the wall is too heavy, the recoiling molecule produces no velocity in the wall. l%m is mass of the gas molecule and M is mass of wall, then total K.E. after collision,

Question 15.
A pump on the ground floor of a building can pump up water to fill a tank of volume 30 m³ in 15 min. If the tank is 40 m above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump ?
Answer:

Question 16.
Two identical ball bearings in contact with each other and resting on a frictionless. table are hit head-on by another ball bearing of the same mass moving initially with a speed V. If the collision is elastic, which of the following (Fig.) is a possible result after collision ?

Answer:
Collision is elastic, so K.E. of the system is conserved.

Question 17.
The bob A of a pendulum released from 30° to the vertical hits another bob B of the same mass at rest on a table as shown in Fig. How high does the bob A rise after the collision? Neglect the size of the bobs and assume the collision to be elastic.
Answer:

Since the collision is elastic, therefore A would come to rest and B would begin to move with the velocity of A to conserve the linear momentum.

Question 18.
The bob of a pendulum is released from a horizontal position A as shown in Fig. If the length of the pendulum is 1.5 m, what is the speed with which the bob arrives at the lowermost point B, given that it dissipated 5% of its initial energy against air resistance ?
Answer:

P.E. of the bob at position A = mgh = m x 9.8 x 1.5
Since 5% of energy is lost when reach at B, so K.E. at the lowermost point

Question 19.
A trolley of mass 300 kg carrying a sandbag of 25 kg is moving uniformly with a speed of 27 km/h on a frictionless track. After a while, sand starts leaking out of a hole on the trolley’s floor at the rate of 0.05 kgs^_1. What is the speed of the trolley after the entire sandbag is empty ?
Answer:
The system of trolley and sandbag is moving at a uniform speed. Clearly, the system is not being acted upon by external force. If the sand leaks out, even when no external force acts. So there shall no change in the speed of the trolley.

Question 20.
A particle of mass 0.5 kg travels in a straight line with velocity υ = ax^3/2, where a = 5 m^-1/2 s^-1. What is the work done by the net force during its displacement from x = 0 to x = 2m?
Answer:
Here m = 0.5 kg

Question 21.
The blades of a windmill sweep out a circle of area A.
(a) If the wind flows at a velocity v perpendicular to the circle, what is the mass of the air passing through it in time t ?
(b) What is the kinetic energy of the air ?
(c) Assume that the windmill converts 25% of the wind’s energy into electrical energy, and that A = 30 m², υ= 36 km/h and the density of air is 1.2 kg m^-3. What is the electrical power produced ?
Answer:
(a) Volume of wind flowing per sec = Aυ
Mass of wind flowing per sec = Aυp
Mass of air passing in time t = Aυpt

Question 22.
A person trying to lose weight (dieter) lifts a 10 kg mass 0.5 m, 1000 times. Assume that the potential energy lost each time she lowers the mass is dissipated,
(a) How much work does she do against the gravitational force ?
(b) Fat supplies 3.8 x 10⁷ J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up ?
Answer:
Here m = 10 kg, h = 0.5 m, n = 1000
(a) Work done against gravitational force = W = n mgh
= 1000 x (10 x 9.8 x 0.5)
= 49000 J

Question 23.
A large family uses 8 kW of power,
(a) Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square meter. If 20% of this energy can be converted to useful electrical energy, how large an area is needed to supply 8 kW?
(b) Compare this area to that of the roof of a house constructed on a plot of size 20 m x 15 m with a permission to cover upto 70%.
Answer:
(a) Energy incident per square meter = 200 W
Let A be the area needed to supply 8 kW
.’. Energy incident on area A = (200 A) W
Energy converted into useful electrical energy

Question 24.
A bullet of mass 0.012 kg and horizontal speed 70 ms^-1 strikes a block of wood of mass 0.4 kg and instantly comes to rest with respect to the block. The block is suspended from the ceding by means of thin wires. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block.
Answer:
By using law of conservation of momentum,
m₁u₁+ m₂u₂ ⁼ (m₁ +m₂)υ

Question 25.
Two inclined frictionless tracks, one gradual and the other steep meet at A from where two stones are allowed to slide down from rest, one on each track (Fig.). Will the stones reach the bottom at the same time ? Will they reach there with the same speed ? Explain. Given θ₁ = 30°,θ₂ = 60°, and h = 10 m, what are the speeds and times taken by the two stones?

Answer:
Here, K.E. at bottom = P.E. at top

Since vertical height of both planes is same, so they will reach the bottom with same speed. Acceleration of a body sliding down an inclined plane, a = g sin θ
Let t be the time taken by stone 1 to travel AB distance.

Question 26.
A 1 kg block situated on a rough incline is connected to a spring of spring constant 100 N m^-1 as shown in Fig. The block is released from rest with the spring in the unstretched position. The block moves 10 cm down the incline before coming to rest. Find the coefficient of friction between the block and the incline. Assume that the spring has negligible mass and the pulley is frictionless.

Answer:
From fig, R = mg cos θ
Force of friction, F = μR = μ mg cosθ
Net force on the block down the incline
= mg sinθ – F = mg sinθ – (a mg cosθ = mg (sinθ – μcos θ)
Distance moved, x = 10 cm = 0.1 m
In equilibrium, Work done = P.E. of stretched spring

Question 27.
A bolt of mass 0.3 kg falls from the ceiling of an elevator moving down with a uniform speed of 7 ms^-1. It hits the floor of the elevator (length of the elevator = 3 m) and does not rebound. What is the heat produced by the impact ? Would your answer be different if the elevator were stationary ?
Answer:
Potential energy of bolt = mgh = 0.3 × 9.8 × 3 = 8.82 J.
Since the bolt does not rebound, the while energy is converted into heat. Since the value of acceleration due to gravity is the same in all inertial systems, the answer will not change even if the elevator is stationary.

Question 28.
A trolley of mass 200 kg moves with a uniform speed of 36 km/h on a frictionless track. A child of mass 20 kg bins on the trolley from one end to the other (10 m away) with a speed of 4 ms^-1 relative to the trolley in a direction opposite to the trolley’s motion and jumps out of the trolley. What is the final speed of the trolley ? How much has the trolley moved from the time the child begins to run ?
Answer:
Initial total momentum = p_i = (m₁ + m₂) u₁

Question 29.
Which of the following potential energy curves in Fig. cannot possibly describe the elastic collision of two billiard balls ? Here r is the distance between centers of the balls.

Answer:
During short time of collision, the kinetic energy converts into potential energy. Since potential energy of a system of two masses varies inversely as the distance between them i.e., as 1/r, all the potential energy curves except the one shown in fig (v) cannot describe an elastic collision.

Question 30.
Consider the decay of a free neutron at rest: n → p + e^–. Show that the two-body decay of this type must necessarily give an electron of fixed energy and, therefore, cannot account for the observed continuous energy distribution in the (β-decay of a neutron or a nucleus (Fig.) [Note: The simple result of this exercise was one among the several arguments advanced by W. Pauli to predict the existence of a third particle in the decay products of (3-decay. This particle is known as neutrino. We now know that it is a particle of intrinsic spin 1/2 (like e^–, p or n), but is neutral, and either massless or having an extremely small mass (compared to electron’s mass) and which interacts very weakly with matter. The correct decay process of neutron is : n —> p + e^– + v]

Answer:
If the decay of a neutron (inside the nucleus) into proton and electron is according to the given scheme, then the available energy in the decay must be carried by the electron coming out of the nucleus and therefore the emitted electrons should always possess a fixed value of kinetic energy. However the graph shows that the emitted electron can have any value of energy between zero and the maximum value. Therefore, the given decay mode cannot account for the observed continuous energy spectrum in the β decay.

We hope the NCERT Solutions for Class 11 Physics Chapter 6 Work, Energy and Power, help you. If you have any query regarding NCERT Solutions for Class 11 Physics Chapter 6 Work, Energy and Power, drop a comment below and we will get back to you at the earliest.