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NCERT Exemplar Class 11 Physics Chapter 9 Mechanical Properties of Fluids

March 5, 2022

NCERT Exemplar Class 11 Physics Chapter 9 Mechanical Properties of Fluids are part of NCERT Exemplar Class 11 Physics. Here we have given NCERT Exemplar Class 11 Physics Chapter 9 Mechanical Properties of Fluids.

NCERT Exemplar Class 11 Physics Chapter 9 Mechanical Properties of Fluids

Mechanical Properties Of Fluids Numericals NCERT Class 11

Q1. A tall cylinder is filled with viscous oil. A round pebble is dropped from the top with zero initial velocity. From the plot shown in figure, indicate the one that represents the velocity (v) of the pebble as a function of time (t).

Sol: (c) In fluids, when the pebble is dropped from the top of a tall cylinder filled with viscous oil, a variable force called viscous force will act which increases with increase in speed. And at equilibrium this velocity becomes constant, that constant velocity is called terminal velocity.
When the pebble is falling through the viscous oil, the viscous force is F= 6πηrv
where r is the radius of the pebble, v is instantaneous speed, η is coefficient of viscosity. As the force is variable, hence acceleration is also variable so v-t graph will not be a straight line. First velocity increases and then becomes constant known as terminal velocity.

Problems On Mechanical Properties Of Fluids NCERT Class 11

Q2. Which of the following diagrams does not represent a streamline flow?

Sol. (d)
Streamline flow: Streamline flow of a liquid is that flow in which each element of the liquid passing through a point travels along the same path and with the same velocity as the preceding element passes through that point.
A streamline may be defined as the path, straight or curved, the tangent to which at any point gives the direction of the flow of liquid at that point.
The two streamlines cannot cross each other and the greater is the crowding of streamlines at a place, the greater is the velocity of liquid particles at that place. If we consider a cross¬sectional area, then a point on the area cannot have different velocities at the same time
Path ABC is streamline as shown in the figure and v1, v2 and v3 are the velocities of the liquid particles at A, B and C point respectively.

Q3. Along a streamline,
(a) the velocity of a fluid particle remains constant
(b) the velocity of all fluid particles crossing a given position is constant
(c) the velocity of all fluid particles at a given instant is constant
(d) ‘ the speed of a fluid particle remains constant,
Sol:(b) As discussed above for a streamline flow of a liquid velocity of each particle at a particular cross-section is constant, because Av = constant (law of continuity) between two cross-section of a tube of flow. So we can say that along a streamline, the velocity of every fluid particle while crossing a given position is the same.

Mechanical Properties Of Fluids Ncert Exemplar Class 11

Q4. An ideal fluid flows through a pipe of circular cross-section made of two sections with diameters 2.5 cm and 3.75 cm. The ratio of the velocities in the two pipes is
(a) 9:4 (b) 3:2 (c)√3: √2 (d)√2:√3

Sol: (a) The situation is shown in the diagram below in which an ideal fluid is flowing through a pipe of circular cross sections.

Numericals On Viscosity Class 11 NCERT

Q5. The angle of contact at the interface of water-glass is 0°, ethylalcohol-glass is 0°, mercury-glass is 140° and methyliodide-glass is 30°. A glass capillary is put in a trough containing one of these four liquids. It is observed that the meniscus is convex. The liquid in the trough is
(a) water
(b) ethylalcohol
(c) mercury
(d) methyliodide

According to the question, the observed meniscus of liquid in a capillary tube is of convex upward which is only possible when angle of contact is obtuse. It is so when one end of glass capillary tube is immersed in a trough of mercury. Hence, the combination will be of mercury-glass (140°) as shown in the figure.

More Than One Correct Answer Type
Q6. For a surface molecule,
(a) the net force on it is zero
(b) there is a net downward force
(c) the potential energy is less than that of a molecule inside
(d) the potential energy is more than that of a molecule inside
Sol: (b, d)

Key concept: To understand the concept of tension acting on the free
surface of a liquid, let us consider four liquid molecules like A, B, C and D. Their sphere of influence are shown in the figure.
(1) Molecule A is well within the liquid, so it is attracted equally in all directions. Hence the net force on this molecule is zero and it moves freely inside the liquid.
(2) Molecule B is little below the free surface of the liquid and it is atso attracted equally in all directions. Hence the resultant force acts on it is also zero
(3) Molecule C is just below the upper surface of the liquid film and the part of its sphere of influence is outside the free liquid surface. So the number of molecules in the upper half (attracting the molecules upward) is less than the number of molecule in the lower half (attracting the molecule downward). Thus the molecule C experiences a net downward force.
(4) Molecule D is just on the free surface of the liquid. The upper half of the sphere of influence has no liquid molecule. Hence the molecule D experiences a maximum downward force.
Thus all molecules lying on surface film experiences a net downward force. Therefore, free surface of the liquid behaves like a stretched membrane.
From the key concept, it is clear from point (4) that molecules on the surface experiences a net downward force. As shown in figure above, molecule D experiences a net downward force. Because on the above side of this molecule there is no liquid molecule. So, the potential energy is more than that of a molecule inside. Hence option (b) and (d) are correct.
Q7. Pressure is a scalar quantity, because
(a) it is the ratio of force to area and both force and area are vectors.
(b) it is the ratio of the magnitude of the force to area.
(c) it is the ratio of the component of the force normal to the area.
(d) ‘it does not depend on the size of the area chosen.
Sol: (b, c) Pressure is defined as the ratio of magnitude of component of the force normal to the area and the area under consideration.

i.e P = F/A
Pressure is a scalar quantity. Pressure acts normal to a surface and it is always compressive in nature, therefore, only its magnitude is required for its complete description.

Q8. A wooden block with a coin placed on its top, floats in water as shown in figure.
The distance / and h are shown in the figure. After sometime, the coin falls into the water. Then,
(a) l decreases
(b) h decreases
(c) l increases
(d) h increases
Sol: (a, b)
Key concept: When a body of density ρ and volume V is immersed in a liquid of density σ, the forces acting on the body are:
1. Weight of body W = mg = Vpg, acting vertically downwards through centre of gravity of the body

When coin is in water, volume of water displaced by coin is equal to the volume of coin V₁(say).
When the coin falls into the water, weight of the (block + coin) system decreases, which was balanced by the upthrust force earlier. As weight of the system decreases, block moves up. Hence ldecreases.
When coin is at the top of wooden block, it displaces a volume of water V₂, which is more than V₁ Because,
Weight of coin = Weight of volume of water displaced by coin (when coin is at .the top of wooden block)

Hence, upthrust force will also decrease. As volume of water displaced by the block decreases, hence h decreases.

Q9. With increase in temperature, the viscosity of
(a) gases decreases
(b) liquids increases
(c) gases increases
(d) liquids decreases
Sol: (c, d) The viscosity of gases increases with increase of temperature, because on increasing temperature the rate of diffusion increases.
The viscosity of liquid decreases with increase of temperature, because the cohesive force between the liquid molecules decreases with increase of temperature.
Relation between coefficient of viscosity and temperature (Andrade formula)

where T = Absolute temperature of liquid, p = density of liquid, A and C are constants.
Important point: With increase in temperature, the coefficient of viscosity of liquids decreases but that of gases increases. The reason is that as temperature rises, the atoms of the liquid become more mobile, whereas in case of a gas, the collision frequency of atoms increases as their motion becomes more random.

Q10. Streamline flow is more likely for liquids with
(a) high density (b) high viscosity
(c) low density (d) low viscosity
Sol: (b, c) Streamline flow is more likely for liquids having low density. We know that greater the coefficient of viscosity of a liquid more will be the velocity gradient, hence each line of flow can be easily differentiated. Streamline flow is related with critical velocity. The critical velocity is that velocity of liquid flow up to which its flow is streamlined and above which its flow becomes turbulent.
As the critical velocity is related to viscosity ( η) and density (ρ) of the liquid as:

(V_c) α η/ρ

Hence if the density will be low and viscosity will be high, the value of critical velocity will be more. So, option (b) and (c) are correct.

Very Short Answer Type Questions
Q11. Is viscosity a vector?
Sol: Viscosity is not a vector quantity. It is a scalar quantity because viscosity is a property of liquid as it does not have any direction.

Q12. Is surface tension a vector?
Sol: Surface tension of a liquid is measured by the force acting per unit length on either side of an imaginary line drawn on the free surface of liquid, the direction of this force being perpendicular to the line and tangential to the free surface of liquid. So if F is the force acting on one side of imaginary line of length L, then T= (F/L)
It depends only on the nature of liquid and is independent of the area of surface or length of line considered.
It is a scalar quantity as it has a unique direction which is not to be specified

Q14. A vessel filled with water is kept on a weighing pan and the scale adjusted to zero. A block of mass Mand density p is suspended by a massless spring of spring constants k. This block is submerged inside into the water in the vessel. What is the reading of the scale?
Sol: As shown in the diagram, A block of mass M and density p is suspended by a massless spring of spring constants k. This block is submerged into the water in the vessel.
The scale is adjusted to zero, therefore, when the block suspended to a spring is immersed in water, then the reading of the scale will be equal to the upthrust on the block due to water.

Important point: If the scale is not adjusted to zero as said earlier, reading on the scale will be different. Then weight of vessel and weight of water is also added in the reading.

Q15. A cubical block of density ρ is floating on the surface of water. Out of its height L, fraction x is submerged in water. The vessel is in an elevator accelerating upward with acceleration a. What is the fraction immersed?
Sol: Key concept: When a fluid is subjected to constant vertical acceleration, its free surface remains horizontal as the net effective gravity is acting vertically. But the magnitude of pressure at a point in the fluid increases or decreases depending upon the direction of acceleration.

Short Answer Type Questions

Q16. The sap in trees, which consists mainly of water in summer, rises in a system of capillaries of radius r = 2.5 x 10^-5 m. The surface-tension of sap is T = 7.28 x 10^-2 Nm^-1 and the angle of contact is 0°. Does surface tension alone account for the supply of water to the top of all trees?
Sol: According to the problem, radius (r) = 2.5 x 10^-5 m Surface tension (S) = 7.28 x 10^-2 N/m
Angle of contact (θ) = 0°

This is the maximum height to which the sap can rise due to surface tension. Many trees have heights much greater than 0.6 m, so only this action is not sufficient for supply of water to the top of such long tree.

Q17. The free surface of oil in a tanker, at rest, is horizontal. If the tanker starts accelerating the free surface will be titled by an angle θ. If the acceleration is a ms ^-2, what will be the slope of the free surface?
Sol: Key concept: The behaviour of a liquid contained in a horizontally accelerated vessel can be understood by understanding the behaviour of a pendulum suspended from the ceiling of a horizontally accelerated trolley.

Every fluid element attains an equilibrium position under the action of gravity and pseudo force. The free surface of the liquid orients itself perpendicular to the direction of net effective gravity.
tan θ = a/g
Suppose tanker accelerates along x-axis with acceleration a, free surface of the tanker will not be horizontal because pseudo force acts as shown in the diagram.

Consider an elementary particle of the oil of mass m.
The acting forces on the particle with respect to the tanker are shown in the figure alongside.
Now, balancing forces (as the particle is in equilibrium) along the inclined direction of surface.
ma = pseudo force
mg = weight of small part of oil.
Along free surface,
Net force = 0

=> ma cos θ = mg sin θ
=> a = g tan θ
=> θ = tan^-1(a/g)

Q18.Two mercury droplets of radii 0.1 cm and 0.2 cm collapse into one single drop. What amount of energy is released? The surface tension of mercury T=5 x 10^-3 Nm^_1Sol: When two drops form a bigger drop, volume remains conserved.
According to the problem, there is two mercury droplets of different radii collapse into one single drop.
Radius of smaller drop = r, = 0.1 cm = 10^-3 m,
Radius of bigger drop = r₂ = 0.2 cm = 2 x 10^-3 m
Surface tension (7) = 435.5 x 10 ^-3 N/m
Let V₁ and V₂ be the volumes of these two mercury droplets and volume of big drop formed by collapsing is V.
Volume of big drop = Volume of small droplets

Therefore, 3.22 x K^-6 J energy will be absorbed. So, the surface area of the water decreases means surface area of bigger drop is less than the sum of surface area of two smaller drops.

Q19. If a drop of liquid breaks into smaller droplets, it results in lowering of temperature of the droplets. Let a drop of radius R, breaks into N small droplets each of radius r. Estimate the drop in temperature.
Sol: The volume remains conserved, when a big drop, breaks into N small droplets.
Volume of liquid drop of radius R = (Volume of liquid droplet of radius r)xN

Due to releasing of this energy, the temperature is lowered.
If c is specific heat of liquid and its temperature is lowered by ΔT, then Energy released, ΔU – me ΔT

Q20. The surface tension and vapour pressure of water at 20°C is 7.28 x 10^-2 Nm^-1and 2.33 x 10³ Pa, respectively. What is the radius of the smallest spherical water droplet which can form without evaporating at 20°C?
Sol:According to the problem, surface tension of water, T = 7.28 x 10^-2 N m^-1Vapour pressure P = 2.33 x 10³ Pa Let r = radius of drop, which formed without evaporating.
The excess pressure (2T/r) should be greater than the vapour pressure. Then, the drop will evaporate.
Vapour pressure = Excess pressure in drop

Long Answer Type Questions

Q22. Surface tension is exhibited by liquids due to force of attraction between molecules of the liquid. The surface tension decreases with increase in temperature and vanishes at boiling point. Given that the latent heat of vaporisation for water L_v = 540 kcal kg^-1, the mechanical equivalent of heat J = 4.2 Jeal^-1, density of water p_w = 10³ kg^-1, Avogadro’s number
N_a = 6.0 xlO²⁶ kmole‘¹ and the molecular weight of water M_A = 10 kg for 1 kmole.

(a) Estimate the energy required for one molecule of water to evaporate

(c) 1 g of water in the vapour state at 1 atm occupies 1601 cm³. Estimate the inter-molecular distance at boiling point, in the vapour state.
(d) During vaporisation a molecule overcomes a force F, assumed constant, to go from an inter-molecular distance d to Estimate the value of F.
(e) Calculate Fid, which is a measure of the surface tension.

Q23. A hot air balloon is a sphere of radius 8 m. The air inside is at a temperature of 60°C. How large a mass can the balloon lift when the outside temperature is20°C? Assume air in an ideal gas, R = 8.314 J mole^-1 K ‘, 1 atm = 1.013 x 10^s Pa, the membrane tension is 5 Nm ^-1.

Sol: Pressure inside the curved surface Will be greater than of outside pressure

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NCERT Exemplar Class 11 Physics Chapter 8 Mechanical Properties of Solids

March 5, 2022

NCERT Exemplar Class 11 Physics Chapter 8 Mechanical Properties of Solids are part of NCERT Exemplar Class 11 Physics. Here we have given NCERT Exemplar Class 11 Physics Chapter 8 Mechanical Properties of Solids. https://www.cbselabs.com/ncert-exemplar-problems-class-11-physics-chapter-8-mechanical-properties-solids/

NCERT Exemplar Class 11 Physics Chapter 8 Mechanical Properties of Solids

Single Correct Answer Type
Mechanical Properties Of Solids Exemplar NCERT Class 11

Q1. Modulus of rigidity of ideal liquids is
(a) infinity
(b) zero
(c) unity
(d) some finite small non-zero constant value
Sol: (b) Key concept: Modulus of Rigidity:
Within limits of proportionality, the ratio of tangential stress to the shearing strain is called modulus of rigidity of the material of the body and is denoted by η

In this case the shape of a body changes but its volume remains unchanged.
Consider a cube of material fixed at its lower face and acted upon by a tangential force F at its upper surface having area A.
Only solids can exhibit a shearing as these have definite shape.

In liquids,η = 0
So, the frictional (viscous) force cannot exist in case of ideal fluid and since they cannot sustain shearing stress or tangential forces are zero, so there is no stress developed.

Mechanical Properties Of Solids Ncert Exemplar Solutions Class 11

Q2. The maximum load a wire can withstand without breaking, when its length is reduced to half of its original length, will
(a) be double (b) be half
(c) be four times (d) remain same
Sol: (d)
When the wire is loaded beyond the elastic limit, then strain increases much more rapidly. The maximum stress corresponding to B (see stress—strain curve) after which the wire begin to flow and breaks, is called breaking stress or tensile strength and the force by application of which the wire breaks is called the breaking force.
(i) Breaking force depends upon the area of cross-section of the wire, i.e. Breaking force ∝ A
Breaking force = P x A
Here P is a constant of proportionality and known as breaking stress.
(ii) Breaking stress is a constant for a given material and it does not depend upon the dimension (length or thickness) of wire.
(iii) If a wire of length L is cut into two or more parts, then again its each part can hold the same weight as breaking force is independent of the length of wire.

Mechanical Properties Of Solids Ncert Exemplar Class 11

Q3. The temperature of a wire is doubled. The Young’s modulus of elasticity
(a) will also double
(b) will become four times
(c) will remain same
(d) will decrease
Sol:(d)
Key concept: Youngs modulus (Y)
It is defined as the ratio of normal stress to longitudinal strain within limit of proportionality.

As Y ∝ 1/ ∆T

When temperature increases ∆T increases, hence Y decreases.

Q4. A spring is stretched by applying a load to its free end. The strain produced in the spring is
(a) volumetric
(b) shear
(c) longitudinal and shear
(d) longitudinal

Sol: (c) According to the diagram where a spring is suspended with a fixed rigid support. Now a load is attached with the lower end of that spring. So, it is stretched by applying a load to its free end. Clearly the length and shape of the spring changes and the weight of the load behaves as a deforming force.
The change in length corresponds to longitudinal strain and change in shape corresponds to shearing strain.

Q5. A rigid bar of mass M is supported symmetrically by three wires each of length /. Those at each end are of copper and the middle one is of iron. The ratio of their diameters, if each is to have the same tension, is equal to

Sol: (b) As the bar is supported symmetrically by the three wires, therefore extension in each wire is same.
Let T be the tension in each wire and diameter of the wire is D, then Young’s modulus is

Q6. A mild steel wire of length 2L and cross-sectional area A is stretched, well within elastic limit, horizontally between two pillars (figure ). A mass m is suspended from the mid-point of the wire. Strain in the wire is

Q7. A rectangular frame is to be suspended symmetrically by two strings of equal length on two supports (figure). It can be done in one of the following three ways:

The tension in the strings will be
(a) the same in all cases
(b) least in (i)
(c) least in (ii)
(d) least in (iii)
Sol:(c) According to the FBD diagram of the rectangular frame. Let M be the mass of rectangular frame and 0be the angle which the tension T in the string makes with the horizontal

Q8. Consider two cylindrical rods of identical dimensions, one of rubber and the other of steel. Both the rods are fixed rigidly at one end to the roof. A mass M is attached to each of the free ends at the centre of the rods.
(a) Both the rods will elongate but there shall be no perceptible change in shape.
(b) The steel rod will elongate and change shape but the rubber rod will only elongate.
(c) The steel rod will elongate without any perceptible change in shape, but the rubber rod will elongate and the shape of the bottom edge will change to an ellipse.
(d) The steel rod will elongate, without any perceptible change in shape, but the rubber rod will elongate with the shape of the bottom edge tapered to a tip at the centre.
Sol: (d)

According to the diagram shown below in which a mass AT is attached at the centre of each rod, then both rods will be elongated. But due to different elastic properties of material the steel rod will elongate without making any perceptible change in shape, but the rubber rod will elongate with the shape of the bottom edge tapered to a tip at the centre.

More Than One Correct Answer Type
Q9. The stress-strain graphs for two materials are shown in figure(assume same scale).

(a) Material (ii) is more elastic than material (i) and hence material (ii) is more brittle
(b) Materials (i) and (ii) have the same elasticity and the same brittleness.
(c) Material (ii) is elastic over a larger region of strain as compared to (i).
(d) Material (ii) is more brittle than material (i).
Sol: (c, d)
Key concept: Representation of different points on Stress-Strain graph:

When the strain is small (< 2%) (i.e., in region OP) stress is proportional to strain. This is the region where the so called Hooke’s law is obeyed. The point P is called limit of proportionality and slope of line OP gives the Young’s modulus Y of the material of the wire. If 6is the angle of OP from strain axis, then Y= tan θ.
If the strain is increased a little bit, i.e. in the region PE, the stress is not proportional to strain. However, the wire still regains its original length after the removal of stretching force.
This behaviour is shown up to point E known as elastic limit or yield- point. The region OPE represents the elastic behaviour of the material of wire. ‘
” If the wire is stretched beyond the elastic limit E, i.e. between EA, the strain increases much more rapidly and if the stretching force is removed the wire does not come back to its natural length. Some permanent increase in length takes place.
If the stress is increased further by a very small increase in it a very large increase in strain is produced (region AB) and after reaching point B, the strain increases even if the wire is unloaded and ruptures at C. In the region BC the wire literally flows. The maximum stress corresponding to B after which the wire begins to flow and breaks is called breaking or tensile strength. The region EABC represents the plastic behaviour of the material of wire.
Material having more ultimate tensile strength will be elastic over larger region.
So, from graph (i) linear limit vanishes soon and for small stress there is large strain as compared to graph (ii).
Hence, material (ii) is more elastic over a large region of strain as compared to (i). So, the ultimate tensile strength for material (ii) is greater than (i).
A material is said to be more brittle if its fracture point is more closer to ultimate strength point.
Hence material (ii) is more brittle than material (i).

Q10. A wire is suspended from the ceiling and stretched under the action of a weight F suspended from its other end. The force exerted by the ceiling on it is equal and opposite to the weight.
(a) Tensile stress at any cross-section A of the wire is F/A.
(b) Tensile stress at any cross-section is zero.
(c) Tensile stress at any cross-section A of the wire is 2F/A.
(d) Tension at any cross-section A of the wire is F.
Sol. (a, d)

According to the diagram a wire is stretched under the j
action of a weight F suspended from its other end.
Clearly, forces at each cross-section is F.
Now applying formula,

Tensile stress = Force applied/Area of cross-section = F/A
Tension = Applied force = F

Q11. A rod of length l and negligible mass is suspended at its two ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths (figure).The cross-sectional areas of wires A and B are 1.0 mm² and 2.0 mm², respectively.(Y_Al= 70 x 10⁹ Nm^-2 and T_steel = 200 x 10⁹Nm^-2)

Mass m should be suspended close to wire A to have equal stresses in both the wires.
Mass m should be suspended close to B to have equal stresses in both the wires.
Mass m should be suspended at the middle of the wires to have equal stresses in both the wires.
Mass m should be suspended close to wire A to have equal strain in both wires.

Sol: (b, d) According to the diagram a massless rod is suspended at its two ends by two wires of steel (wire A) and aluminum (wire B) of equal lengths.

Q12. For an ideal liquid,
(a) the bulk modulus is infinite
(b) the bulk modulus is zero
(c) the shear modulus is infinite
(d) the shear modulus is zero

Q13. A copper and a steel wire of the same diameter are connected end to end. A deforming force F is applied to this composite wire which causes a total elongation of 1 cm. The two wires will have
(a) the same stress
(b) different stress
(c) the same strain
(d) different strain

Very Short Answer Type Questions
Q14. The Young’s modulus for steel is much more than that for rubber. For the same longitudinal strain, which one will have greater tensile stress?

Q15. Is stress a vector quantity?
Sol: Stress = Magnitude of internal reaction force / Area of cross-section
Hence, stress is not a scalar quantity not a vector quantity, it is a tensor quantity.

Q16. Identical springs of steel and copper are equally stretched. On which, more work will have to be done?
Sol: Key concept: Work Done in stretching a Wire or Spring:
In stretching a wire work is done against internal restoring forces. This work is stored in the wire as elastic potential energy or strain energy.
If a force F acts along the length L of the wire of cross-section A and stretches it by x, then
Q17. What is the Young’s modulus for a perfect rigid body?
Sol: According to Hooke’s law

Q18. What is the Bulk modulus for a perfect rigid body?
Sol: Bulk modulus is given by

For perfectly rigid body, change in volume AV = 0, therefore volumetric strain is zero.

Hence, bulk modulus for a perfectly rigid body is infinity

Short Answer Type Questions
Q19. A wire of length L and radius r is clamped rigidly at one end. When the other end of the wire is pulled by a force f its length increases by l. Another wire of the same material of length 2L and radius 2r, is pulled by a force 2 f Find the increase in length of this wire.
Sol: We have to apply Hooke’s law to compare the extension in each wire. According to the diagram which shows the situation.
Now, Young’s modulus

Solu:

Q21. To what depth must a rubber ball be taken in deep sea so that its volume is decreased by 0.1%. (The bulk modulus of rubber is 9.8 x 10⁸ N/m²; and the density of sea water is 10³ kg/m³)
Sol: According to the problem, Bulk modulus of rubber (B) = 9.8 x 10⁸ N/m²Density of sea water (p) = 10³ kg/m³Percentage decrease in volume

Q22. A truck is pulling a car out of a ditch by means of a steel cable that is 9.1 m long and has a radius of 5 mm. When the car just begins to move, the tension in the cable is 800 N. How much has the cable stretched? (Young’s modulus for steel is 2 x 10¹¹ N/m²)

Sol. According to the problem,
Length of steel cable l = 9.1 m
Radius r = 5 mm = 5 x 10~³ m
Tension in the cable F = 800 N
Young’s modulus for steel Y= 2 x 10¹¹ N/m²Change in length ∆l = ?

Q23. Two identical solid balls, one of ivory and the other of wet-clay, are dropped from the same height on the floor. Which one will rise to a greater height after striking the floor and why?
Sol: Since, ivory ball is more elastic than wet-clay ball, therefore, it tends to regain its original shape quickly. Due to this reason, more energy and momentum is transferred to the ivory ball in comparison to the wet-clay ball and hence, ivory ball will rise higher after striking the floor even though both are dropped from the same height.

Long Answer Type Questions
Q24. Consider a long steel bar under a tensile stress due to force F acting at the edges along the length of the bar (figure). Consider a plane making an angle 8 with the length. What are the tensile and shearing stresses on this plane?
(a) For what angle is the tensile stress a maximum?
(b) For what angle is the shearing stress a maximum?

Sol: According to the problem force F is applied along horizontal, so we resolve it in two perpendicular components—one is parallel to the inclined plane and other one is perpendicular to the inclined plane as shown in the diagram. Now, we can easily calculate the tensile and shearing stress. Here,

Important point: Here we are not applying direct formula for stress. Because to analyze different types of stresses, we have to divide the force in components. In normal stress, the force is applied normal to the surface. But in shear stress deforming force is applied tangential to one of the faces

Q25. A steel rod of length 21, cross-sectional area A and mass M is set rotating in a horizontal plane about an axis passing through the centre. If Y is the Young’s modulus for steel, find the extension in the length of the rod. (Assume the rod is uniform)

Q26.An equilateral triangle ABC is formed by two Cu rods AB and BC and one At It is heated in such a way that temperature of each rod increases by ∆T. Find change in the angle ABC. [Coefficient of linear expansion for Cu is α₁coefficient of linear expansion for Al is α₂]

Sol: As the temperature of the rods increases length of each side will change, hence the angle corresponding to any vertex also changes as shown in the diagram.

Before heating, AB = BC = CA = l

After heating temperature of each rod is changed by ∆T and sides of ∆ABC is changed.

Let AB = l₁,BC = l₃,CA = l₂Using cosine formula,

Q27. In nature, the failure of structural members usually result from large torque because of twisting or bending rather than due to tensile or compressive strains. This process of structural breakdown is called buckling and in cases of tall cylindrical structures like trees, the torque is caused by its own weight bending the structure. Thus, the vertical through the centre of gravity does not fall within the base. The elastic torque caused because of this bending about the central axis of the tree is given by YπR⁴/ 4R. Y is the Young’s modulus, r is the radius of the trunk and is the radius of curvature of the bent surface along the height of the tree containing the centre of gravity (the neutral surface). Estimate the critical height of a tree for a given radius of the trunk.
Sol: According to the problem, the elastic torque or the bending torque is given and we have to find the torque caused by the weight due to bending.
The diagram of the given situation is as shown.
The bending torque on the trunk of radius r of tree = YπR⁴/ 4R
where R the radius of curvature of the bent surface.

Q28. A stone of mass m is tied to an elastic string of negligible mass and spring constant k. The unstretched length of the string is L and has negligible mass. The other end of the string is fixed to a nail at a point P. Initially the stone is at the same level as the point P. The stone is dropped vertically from point P.
(a) Find the distance y from the top when the mass comes to rest for an instant, for the first time.
(b) What is the maximum velocity attained by the stone in this drop?
(c) What shall be the nature of the motion after the stone has reached its lowest point?
Sol:In this problem, the given string is elastic. Consider the diagram the stone is dropped from point P.
(a) When the stone is dropped, then it covers distance Y before coming to rest, for the first instant.

Y=L + (Y-L)
First it covers the distance L equal to length of string distance in free fall and a further distance (Y – L) due to extension in the string. So it covers a total distance Y until it instantaneously comes to rest at Q.
We have to find y, so by applying energy conservation principle,
Loss in potential energy of stone = Gain in elastic potential energy in string

(b) In SHM, the maximum velocity is attained when the body passes, through the “equilibrium, position”, i.e., when the instantaneous acceleration is zero. That is mg -kx = 0, where .r is the extension from L.

(c) When stone is at the lowest point Q, i.e. at instantaneous distance Y from P from where the stone is dropped, then equation of motion of the stone is
mass x acceleration = net force on the stone

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We hope the NCERT Exemplar Class 11 Physics Chapter 8 Mechanical Properties of Solids help you. If you have any query regarding NCERT Exemplar Class 11 Physics Chapter 8 Mechanical Properties of Solids, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Class 11 Physics Chapter 7 Gravitation

March 5, 2022

NCERT Exemplar Class 11 Physics Chapter 7 Gravitation are part of NCERT Exemplar Class 11 Physics. Here we have given NCERT Exemplar Class 11 Physics Chapter 7 Gravitation. https://www.cbselabs.com/ncert-exemplar-problems-class-11-physics-chapter-7-gravitation/

NCERT Exemplar Class 11 Physics Chapter 7 Gravitation

Multiple Choice Questions
Single Correct Answer Type
Ncert Exemplar Class 11 Physics Gravitation

Q1. The earth is an approximate sphere. If the interior contained matter which is not of the same density everywhere, then on the surface of the earth, the acceleration due to gravity
(a) ‘ will be directed towards the centre but not the same every where
(b) will have the same value everywhere but not directed towards the centre
(c) will be same everywhere in magnitude directed towards the centre
(d) cannot be zero at any point
Sol: (d) Acceleration due to gravity g = 0, at the centre if we assume the earth as a sphere of uniform density, then it can be treated as point mass placed at its centre.
But on surface of the earth the acceleration due to gravity cannot be zero at any point. .

Gravitation Class 11 Ncert Exemplar Solutions

Q2. As observed from the earth, the sun appears to move in an approximate circular orbit. For the motion of another planet like mercury as observed from the earth, this would
(a) be similarly true
(b) not be true because the force between the earth and mercury is not inverse square law
(c) not be true because the major gravitational force on mercury is due to the sun
(d) not be true because mercury is influenced by forces other than gravitational forces
Sol: (c) As observed from the earth, the sun appears to move in an approximate circular orbit. The gravitational force of attraction between the earth and the sun always follows inverse square law.
All planets move around the sun due to the huge gravitational force of the sun acting on them. The gravitational force on the mercury due to earth is much smaller as compared to that acting on it due to sun and hence it revolves around the sun and not around the earth.

Gravitation Ncert Exemplar Class 11

Q3. Different points in the earth are at slightly different distances from the sun and hence experience different forces due to gravitation. For a rigid body, we know that if various forces act at various points in it, the resultant motion is as if a net force acts on the CM (centre of mass) causing translation and a net torque at the CM causing rotation around an axis through the CM. For the earth-sun system (approximating the earth as a uniform density sphere)
(a) the torque is zero
(b) the torque causes the earth to spin
(c) the rigid body result is not applicable since the earth is not even approximately a rigid body
(d) the torque causes the earth to move around the sun
Sol: (a) As the earth is revolving around the sun in a circular motion (approximately in actual the path of earth around the sun is elliptical) due to gravitational attraction. When we consider the earth-sun as a single system and we are taking earth as a sphere of uniform density. Then the gravitational force (F) will be of radial nature, i.e. angle between position vector r and force F is zero. So, torque

Q4. Satellites orbiting the earth have finite life and sometimes debris of satellites fall to the earth. This is because
(a) the solar cells and batteries in satellites run out
(b) the laws of gravitation predict a trajectory spiralling inwards
(c) of viscous forces causing the speed of satellite and hence height to gradually decrease
(d) of collisions with other satellites
Sol: (c) We know that the total energy of the earth satellite of radius R bounded system is negative
(-GM\ 2 R)
where Mis mass of the earth. –
Due to the atmospheric friction (viscous force) acting on satellite, energy decreases continuously, radius of the orbit or height decreases gradually and the satellite spirals down with increasing speed till it bums in the denser layers of the atmosphere.

Q5. Both the earth and the moon are subject to the gravitational force of the sun. As observed from the sun, the orbit of the moon
(a) will be elliptical
(b) will not be strictly elliptical because the total gravitational force on it is not central
(c) is not elliptical but will necessarily be a closed curve
(d) deviates considerably from being elliptical due to influence of planets other than the earth
Sol: (b) Moon revolves around the earth in a nearly circular orbit. When it is observed from the sun, two types of forces are acting on the moon one is due to gravitational attraction between the sun and the moon and the other is due to gravitational attraction between the earth and the moon. So moon is moving under the combined gravitational pull acting on it due to the earth and the sun. Hence, total force on the moon is not central.

Q6. In our solar system, the inter-planetary region has chunks of matter (much smaller in size compared to planets) called asteroids. They
(a) will not move around the sun, since they have very small masses compared to the sun
(b) will move in an irregular way because of their small masses and will drift away into outer space
(c) will move around the sun in closed orbits but not obey Kepler’s laws
(d) will move in orbits like planets and obey Kepler’s laws
Sol: (d)
Key concept: The Law of Orbits-. Every planet moves around the sun in an elliptical orbit with sun at one of the foci.
The Law of Area-. The, line joining the sun to the planet sweeps out equal areas in equal intervals of time, i.e. areal velocity is constant. According to this law planet will move slowly when it is farthest from sun and more rapidly when it is nearest to the sun. It is similar to law of conservation of angular momentum.

The Law of Periods: The square of period of revolution (7) of any planet around sun is directly proportional to the cube of the semi-major axis of the orbit.

Asteroids are moving in circular orbits like planets because they are being acted upon by central gravitational forces, they must obey Kepler’s laws. You may consider an asteroid and analyze that it satisfies above laws or not.

Q7. Choose the wrong option.
(a) Inertial mass is a measure of difficulty of accelerating a body by an external force whereas the gravitational mass is relevant in determining the gravitational force on it by an external mass.
(b) That the gravitational mass and inertial mass are equal is an experimental result.
(c) That the acceleration due to gravity on the earth is the same for all bodies is due to the equality of gravitational mass and inertial mass.
(d) Gravitational mass of a particle like proton can depend on the presence of neighbouring heavy objects but the inertial mass cannot.
Sol: (d)
Key concept: Inertial mass: It is the mass of the material body, which measures its inertia.
Gravitational Mass: It is the mass of the material body, which determines the gravitational pull acting upon it.
According to the principle of equivalence, Gravitational mass of proton is equivalent to its inertial mass and is independent of presence of neighboring heavy objects.
Important point: Comparison between inertial and gravitational mass:
• Both are measured in the same units.
• Both are scalars.
• Both do not depend on the shape and state of the body.
• Inertial mass is measured by applying Newton’s second law of motion whereas gravitational mass is measured by applying Newton’s law of gravitation.
• Spring balance measures gravitational mass and inertial balance measure inertial mass.
Q8. Particles of masses 2M, m and M are respectively at points A, B and C with AB = 1/2(BC). m is much-much smaller than M and at time t = 0, they are all at rest as given in figure.
At subsequent times before any collision takes place.

(a) m will remain at rest
(b) m will move towards M
(c) m will move towards 2M
(d) m will have oscillatory motion
Sol:(c) The particle m at B will move towards A with the greater force, due to particle 2M at A.

Force on m at B due to 2M at A is

NCERT Exemplar Class 11 Physics Solutions

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We hope the NCERT Exemplar Class 11 Physics Chapter 7 Gravitation help you. If you have any query regarding NCERT Exemplar Class 11 Physics Chapter 7 Gravitation, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Class 11 Physics Chapter 6 System of Particles and Rotational Motion

March 4, 2022

NCERT Exemplar Class 11 Physics Chapter 6 System of Particles and Rotational Motion are part of NCERT Exemplar Class 11 Physics. Here we have given NCERT Exemplar Class 11 Physics Chapter 6 System of Particles and Rotational Motion. https://www.cbselabs.com/ncert-exemplar-problems-class-11-physics-chapter-6-system-particles-rotational-motion/

NCERT Exemplar Class 11 Physics Chapter 6 System of Particles and Rotational Motion

Multiple Choice Questions
Single Correct Answer Type

Q1. For which of the following does the centre of mass lie outside the body?
(a) A pencil (b) Ashotput
(c) A dice (d) A bangle
Sol: (d) .
Key concept: Center of mass of a system (body) is a point that moves as though all the mass were concentrated there and all external forces were applied there.
Important Points about Center of Mass:
(i) The position of center of mass is independent of the co-ordinate system chosen.
(ii) The position of center of mass depends upon the shape of the body and distribution of mass.
Example: The center of mass of a circular disc is within the material of the body while that of a circular ring is outside the material of the body.
(iii) We can imagine a rigid body also as a system of masses and hence every rigid body has a center of mass. In case of a regularly shaped uniform rigid body, center of mass is simply the geometric centre of the body.
A bangle is in the form of a ring as shown in the diagram below. We know that the position of center of mass depends upon the shape of the body and distribution of mass. So, out of four given bodies, the centre of mass lies at the centre, which is outside the body (boundary) whereas in all other three bodies it lies within the body because they are completely solid.

Q2. Which of the following points is the likely position of Hollow
the centre of mass of the system shown in figure? sphere
(a) A
(b) B
(c) C
(d) D
Sol: (c) The position of centre of mass of the system in this problem is closer to heavier mass or masses or we can say that it depends upon distribution of mass. So it is likely to be at C. In the above diagram, lower part of the sphere containing sand is more heavier than upper part containing air. Hence CM of the system lies below the horizontal diameter.

Q3. A particle of mass m is moving in yz-plane with a uniform velocity v with its trajectory running parallel to +ve y-axis and intersecting z-axis at z = a in figure. The change in its angular momentum about the origin as it bounces elastically from a wall at y = constant is

Q4. When a disc rotates with uniform, angular velocity, which of the following is not true?
(a) The sense of rotation remains same.
(b) The orientation of the axis of rotation remains same.
(c) The speed of rotation is non-zero and remains same.
(d) The angular acceleration is non-zero and remains same.
Sol: (d)
Key concept: The rate of change of angular velocity is defined as angular acceleration. If particle has angular velocity ω₁ at time
t₁, and angular velocity ω₂at time t₂, then
Angular acceleration α = ω₂– ω₁ / t₂– t₁

When the disc is rotated with constant angular velocity, angular acceleration of the disc is zero. Because we know that angular acceleration
α = ∆ ω/∆ t
Here ω is constant, so ∆ ω = 0

Q5. A uniform square plate has a small piece Q of an irregular shape removed and glued to the centre of the plate leaving a hole behind in figure. The moment of inertia about the z-axis is then
(a) increased
(b) decreased
(c) the same
(d) changed in unpredicted manner

Q6. In problem 5, the CM of the plate is now in the following quadrant of x-y plane.
(a) I (b) II (c) III (d) IV
Sol: (c) Let us consider the diagram below, which shows the position of the piece which is removed from the plate. First center of mass is at the centre of the plate (only if its mass is uniformly distributed over the surface) when the piece is removed from quadrant I, therefore the centre of mass is shifted to the

The sum of moments of inertia of a laminar object about two mutually perpendicular axes lying in the plane of lamina is equal to the moment of inertia about an axis normal to the plane of the lamina and passing through the two perpendicular axes. This theorem is applicable only for laminar (thin sheet kind of) object.

I_x and I_y both decreases with the hole. Gluing the removed piece at the centre of the square plate does not affect I_z. The mass comes closer to the z-axis, hence, moment of inertia decreases overall about z-axis.

Q8. A merry-go-round, made of a ring-like platform of radius R and mass M, is revolving with angular speed ω. A person of mass Mis standing on it. At one instant, the person jumps off the round, radially away from the centre of the round (as seen from the round). The speed of the round afterwards is
(a) 2 ω
(b) ω
(c) ω/2
(d) 0
Sol: (b)As no torque is exerted by the person jumping, radially away from the centre of the round (as seen from the round), let the total moment of inertia of the system is 2I (round + Person (because the total mass is 2M) and the round is revolving with angular speed ωSince the angular momentum of the person when it jumps off the round is Iω the actual momentum of round seen from ground is 2 Iω – Iω = Iω
So we conclude that the angular speed remains same, i.e ω

More Than One Correct Answer Type
Q9. Choose the correct alternatives:
(a) For a general rotational motion, angular momentum L and angular velocity ω need not be parallel.
(b) For a rotational motion about a fixed axis, angular momentum L and angular velocity ωare always parallel.
(c) For a general translational motion, momentum p and velocity v are always parallel.
(d) For a general translational motion, acceleration a and velocity v are always parallel.

Sol: (a, c) .
(a) For a general rotational motion where axis of rotation is not symmetric. Angular momentum Z and angular velocity 0) need not be parallel. The wobbly motion of a wheel rotating about an axis inclined at a small angle to the symmetry axis of the wheel represents a situation where angular momentum and angular velocity are not parallel.
(b) Fixed axis should pass through CM of the body, so it is not necessary angular momentum Z and angular velocity ω are always parallel.
(c) As we know in a general translational motion linear momentum is given by, p = mv , hence, direction of p is always along v .
(d) In projectile motion, v and a are not always parallel.

Q10. Figure shows two identical particles 1 and 2, each of mass m, moving in opposite directions with same speed v along parallel lines. At a particular instant r₁ and r₂ are their respective position vectors drawn from point A which is in the plane of the parallel lines. Choose the correct options:

Figure shows two identical particles 1 and 2, each of mass m, moving in opposite directions with same speed v along parallel lines. At a particular instant r₁ and r₂ are their respective position vectors drawn from point A which is in the plane of the parallel lines. Choose the correct options:

Q11. The net external torque on a system of particles about an axis is zero. Which of the following are compatible with it?
(a) The forces may be acting radially from a point on the axis.
(b) The forces may be acting on the axis of rotation.
(c) The forces may be acting parallel to the axis of rotation.
(d) The torque caused by some forces may be equal and opposite to that caused by other forces.
Sol: (a, b, c, d)

Important point: To get the direction where you can use right hand rule: Place the fingers of right hand along r and then curl them into F through the smaller angle between them, n is directed along the (stretched) thumb. For involving two dimensions only, can be replaced by sense of rotation, clockwise or anticlockwise; if the fingers of the right hand curl (while going from clockwise, torque is taken as clockwise and when they curl anticlockwise torque is taken as anticlockwise.)

Sol: (b, d) We can apply the concept of symmetry to calculate the net moment of inertia. Moment of inertia about two symmetrical axes are same.

Very Short Answer Type Questions
Q14. The centre of gravity of a body on the earth coincides with its centre of mass for a small object whereas for an extended object it may not. What is the qualitative meaning of small and extended in this regard? For which of the following two coincides—A building, a pond, a lake, a mountain?
Sol:
Key concept: The center of gravity of a body is that point through which the resultant of the system of parallel forces formed by the weights of all the particles constituting the body passes for all positions of the body. It is denoted as “C.G.” or “G”.
Centre of gravity is centre of a given structure but centre of mass is a point where whole mass of the body can be assumed to be concentrated.
An object is said to be small if its vertical height is very small compared to the radius of the earth, otherwise it is extended.
Building and ponds are small objects so their CG coincides with CM, while a deep lake and a mountain can be considered as extended objects, so the CG does not coincide in their CM.

Q15. Why does a solid sphere have smaller moment of inertia than a hollow cylinder of same mass and radius, about an axis passing through their axes of symmetry?

Sol:Key concept: Moment of inertia of a particle l = mr² where r is the perpendicular distance of particle from rotational axis.
Moment of inertia of a body made up of number of particles (discrete distribution)

I = m₁r₁²+ m₂r₂²+ m₃r₃²
Moment of inertia of a continuous distribution of mass, treating the element of mass dm at position r as particle
dl = dmr²

MI is not constant for a body. It depends on the axis of rotation.
MI depends on the mass of the body. The higher the mass, higher the MI.
MI depends on the distribution of the mass about an axis. The farther the mass is distributed from the axis, higher will be the MI.
Moment of inertia depends on mass, distribution of mass and on the position of axis of rotation.
All the mass in a cylinder lies at distance R from the axis of symmetry but most of the mass of a solid sphere lies at a smaller distance than R. Therefore,
I_{hollowcylinder}> I_sphere

Q16. The variation of angular position , of a point on a rotating rigid body, with time t is shown in figure. Is the body rotating clockwise or anti-clockwise?

Q17. A uniform cube of mass m and side a is placed on a frictionless horizontal surface. A vertical force F is applied to the edge as shown in figure. Match the following (most appropriate choice).

Column I		Column 11
(a)	mg/4 <F< mg/2	(i)	Cube will move up.
(b)	F > mg/2	(ii)	Cube will not exhibit motion.
(c)	F> mg	(iii)	Cube will begin to rotate and slip at A.
(d)	F = mg/4	(iv)	Normal reaction effectively at a/3 from A, no motion.

Sol: Let us first consider the below diagram torque or moment of the force F about point A is given by _l=aF

This is anticlockwise.
Torque of weight mg about A,
₂= q/2
This is clockwise.
N is acting at point A. So, torque due to normal reaction about A will be zero. There is no motion in cube if ₁= ₂

(a) → (ii)
(b) → (iii)
(c) → (i)
(d) → (iv)

Q18. A uniform sphere of mass m and radius R is placed on a rough horizontal surface (figure). The sphere h is struck horizontally at a height h from the floor. ,, Match the following.

Column 1		Column II
(a)	h =R/2	(0	Sphere rolls without slipping with a constant velocity and no loss of energy.
(b)	h = R	(ii)	Sphere spins clockwise, loses energy by friction
(c)	h = 3R/2	(iii)	Sphere spins anti-clockwise, loses energy by friction.
(d)	h = 1R!5	(iv)	Sphere has only a translational motion, looses energy by friction.

Sol: Mass of the sphere = m
Radius = R
h = height from the floor
The sphere will roll without slipping when

ω = V/R
where, v is linear velocity and to is angular velocity of the sphere.
Now, angular momentum of sphere, about centre of mass [We are applying conservation of angular momentum just before and after struck.]

Short Answer Type Questions

Q19. The vector sum of a system of non-collinear forces acting on a rigid body is given to be non-zero. If the vector sum of all the torques due to the system of forces about a certain point is found to be zero, does this mean that it is necessarily zero about any arbitrary point?

Q20. A wheel in uniform motion about an axis passing through its centre and perpendicular to its plane is considered to be in mechanical (translational plus rotational) equilibrium because no net external force or torque is required to sustain its motion. However, the particles that constitute the wheel do experience a centripetal acceleration directed towards the centre. How do you reconcile this fact with the wheel being in equilibrium?How would you set a half-wheel into uniform motion about an axis passing through the centre of mass of the wheel and perpendicular to its plane? Will you require external forces to sustain the motion?
Sol: Internal elastic forces gives rise to the centripetal acceleration of its particles in a wheel. These forces are in pairs and cancel each other because they are part of a symmetrical system.
In a half wheel, the distribution of mass about its centre of mass (through which axis of rotation passes) is not symmetrical. Therefore, the direction of angular momentum of the wheel does not coincide with the direction of its angular velocity. Hence, an external torque is required to maintain the motion of the half wheel.

Q21. A door is hinged at one end and is free to rotate about a vertical axis (figure). Does its weight cause any torque about this axis? Give reason for your answer.
Sol: According to the diagram, where weight of the door acts along negative y-axis.
Torque is not produced by weight about y-axis.
Because the direction of weight is parallel to y-axis (axis of rotation).
A force can produce torque only along direction normal to itself because f = r x F. So, when the door is in the xy-plane, the torque produced by gravity can only be along ±z-direction never about an axis passing through y-direction.
Hence, the weight will not produce any torque about y-axis.

Q22. (n – 1) equal point masses each of mass m are placed at the vertices of a regular n-polygon. The vacant vertex has a position vector a with respect to the centre of the polygon. Find the position vector of centre of mass.

Hence the center of mass of n particles is a weighted average of the position vectors of n particles making up the system.
The centre of mass of a regular n-polygon lies at its geometrical centre.
Let position vector of each centre of mass or regular n polygon is r .
(n – 1) equal point masses each of mass m are placed at (n – 1) vertices of the regular n-polygon, therefore, for its centre of mass

Long Answer Type Questions
Q23. Find the centre of mass of a uniform (a) half-disc, (b) quarter-disc.

If Mis mass of the half-disc of radius R, then mass per unit area of the half-disc

Q24. Two discs of moments of inertia I₁, and I₂ about their respective axes (normal to the disc and passing through the centre), and rotating with angular speed ω₁ and ω₂ and are brought into contact face to face with their axes of rotation coincident.

(a) Does the law of conservation of angular momentum apply to the situation? Why?
(b) Find the angular speed of the two discs system.
(c) Calculate the loss in kinetic energy of the system in the process.
(d)Account for this loss.

Q25. A disc of radius R is rotating with an angular co0 about a horizontal axis. It is placed on a horizontal table. The coefficient of kinetic friction is jJ.K.
(a) What was the velocity of its centre of mass before being brought in contact with the table?
(b) What happens to the linear velocity of a point on its rim when placed in contact with the table?
(c) What happens to the linear speed of the centre of mass when disc is placed in contact with the table?
(d) Which force is responsible for the effects in (b) and (c)?
(e) What condition should be satisfied for rolling to begin?
(f) Calculate the time taken for the rolling to begin.
Sol:
Key concept:When the axis of rotation is not fixed (Stationary in space), the motion of a rigid body is considered as combination of motion of centre of mass plus a rotation about centre of mass. The two components of motion are described by

How the translational motion and rotational motion about the centre of mass are superimposed to get the motion of a rigid body (say a disc of radius R) are illustrated in the following figures.

To get the instantaneous velocity of any point on the rigid body, we calculate the instantaneous velocity of that point in pure translation and in pure rotation and add them vectorially.

(a) Disc is rotating only about its horizontal axis before being brought in contact with the table. Hence its CM is at rest; v_CM = 0
(b) When the disc is placed in contact with the table due to friction velocity of a point on the rim
(c) Linear speed of the CM of disc increases when disc is placed in contact with the table, because its acceleration becomes
a_CM = _kg
(d) Friction is responsible for the effects in (b) and (c) because friction is disturbing the velocity of the point which is in contact with table, hence velocity at all the points of disc is disturbed

Important point: In pure rolling motion, frictional force will support. Or we can say that it just opposes the relative motion of point of contact at any instant.

Q26. Two cylindrical hollow drums of radii R and 2R, and of a common height h, are rotating with angular velocities (anti-clockwise) and (clockwise), respectively. Their axes, fixed are parallel and in a horizontal plane separated by (3R + ). They are now brought in contact ( → 0).
(a) Show the frictional forces just after contact.
(b) Identify forces and torques external to the system just after contact.
(c) What would be the ratio of final angular velocities when friction ceases?
Sol: (a) The frictional forces acting between two cylindrical hollow drums are as shown in the diagram below.

Force F upward shows the friction force on left drum.
Force F downward shows the friction force on right drum.

(b) F¹ = F =F” where F¹ and F” are external forces through support.
=> F_net = 0 (one each cylinder)
Net external torque to the system about any axis=Fx3R, anticlockwise

(c) Let _l and ₂ be final angular velocities of smaller and bigger drum respectively (anticlockwise and clockwise respectively).

Finally, there will be no friction. When friction ceases at the point of contact, then both drums has equal linear velocity at that point.

^VA = ^VB
Hence, R _l = 2R ₂⇒_l/ ₂= 2

Important point: Friction force just opposes the relative motion of point of contacts at any instant. So, we should be very careful while indicating direction of frictional forces.

Q27. A uniform square plate S (side c) and a uniform rectangular plate R (sides b, a) have identical areas and masses.

Q28. A uniform disc of radius R, is resting on a table on its rim. The coefficient of friction between disc and table is μ , (Figure). Now, the disc is pulled with a force F as shown in the figure. What is the maximum value of F for which the disc rolls without slipping?

Sol:In this problem friction force on the disc will act in opposite direction of F at the point which in contact with surface at any instant of time and supports the rotation of the disc in clockwise direction.

NCERT Exemplar Class 11 Physics Solutions

NCERT Exemplar Problems Maths Physics Chemistry Biology

We hope the NCERT Exemplar Class 11 Physics Chapter 6 System of Particles and Rotational Motion help you. If you have any query regarding NCERT Exemplar Class 11 Physics Chapter 6 System of Particles and Rotational Motion, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Class 11 Physics Chapter 5 Work, Energy and Power

March 4, 2022

NCERT Exemplar Class 11 Physics Chapter 5 Work, Energy and Power are part of NCERT Exemplar Class 11 Physics. Here we have given NCERT Exemplar Class 11 Physics Chapter 5 Work, Energy and Power. https://www.cbselabs.com/ncert-exemplar-problems-class-11-physics-chapter-5-work-energy-power/

NCERT Exemplar Class 11 Physics Chapter 5 Work, Energy and Power

Multiple Choice Questions
Single Correct Answer Type

Q1. An electron and a proton are moving under the influence of mutual forces. In calculating the change in the kinetic energy of the system during motion, one ignores the magnetic force of one on another. This is, because
(a) the two magnetic forces are equal and opposite, so they produce no net ‘ effect
(b) the magnetic forces do not work on each particle
(c) the magnetic forces do equal and opposite (but non-zero) work on each particle
(d) the magnetic forces are necessarily negligible
Sol: (b)
Key concept: To calculate the change in kinetic energy of the system during motion we have to apply work-energy, theorem. According to this theorem, Net work done = Final kinetic energy – Initial kinetic energy of the object The above statement shows the connection between work and kinetic energy as: “The work done by the net force acting on an object is equal to the change in the kinetic energy of that object”.
Net work done (IF) on a particle equals change in kinetic energy of the particle.

ΣW=K₂-K₁

According to the problem as the electron and proton are moving under the influence of mutual forces, the magnetic forces will be perpendicular to their motion, hence, it acts as a centripetal force for the particle. In this way the particle performs the uniform circular morion, this implies speed will remain constant. So, there is no change in kinetic energy of the particle. Hence no work is done by these forces.

(magnetic force) will be perpendicular to both B and v, where B is the external magnetic field and v is the velocity of particle. That is why one ignores the magnetic force of one particle on another.

Q2. A proton is kept at rest. A positively charged particle is released from rest at a distance d in its field. Consider two experiments; one in which the charged particle is also a proton and in another, a positron. In the same time t, the work done on the two moving charged particles is
(a) same as the same force law is involved in the two experiments
(b) less for the case of a positron, as the positron moves away more rapidly and the force on it weakens
(c) more for the case of a positron, as the positron moves away a larger distance
(d) same as the work done by charged particle on the stationary proton
Sol: (c) Force between two protons is equal to the force between proton and a
positron because their charges are same. As the mass of positron is much lesser than proton, (1/1840 times) it moves away through much larger distance compared to proton.
Change in their momentum will be same. So, velocity of lighter particle will be greater than that of a heavier particle. So, positron is moved through a larger distance.
As work done = force x distance. As forces are same in case of proton and positron but distance moved by positron is larger, hence, work done will be more.

Q3. A man squatting on the ground gets straight up and stand. The force of reaction of ground on the man during the process is
(a) constant and equal to mg in magnitude
(b) constant and greater than mg in magnitude
(c) ‘ variable but always greater than mg
(d) at first greater than mg and later becomes equal to mg
Sol: (d) In the process of squatting on the ground he gets straight up and stand. Then he is tilted somewhat, the man exerts a variable force on the ground to balance his weight, hence he also has to balance frictional force besides his weight in this case.
N = Normal reaction force = friction + mg => N > mg
Once the man gets straight up that variable force = 0 =>
Normal reaction force = mg

Q4. A bicyclist comes to a skidding stop in 10 m. During this process, the force on the bicycle due to the road is 200 N and is directly opposed to the motion. The work done by the cycle on the road is
(a) +2000 J
(b) -200 J
(c) zero
(d) -20,000 J
Sol: (c) As the friction is present in fhis problem, so mechanical energy is not conserved. So energy will be lost due to dissipation by friction.^Here, work is done by the frictional force on the cycle and is equal to __
200 x 10 = -2000 J
As the road does not move at all, therefore, work done by the cycle on the road is zero.
Important point: We should be aware that here the energy of bicyclist is lost during the motion, but it is lost due to friction in the form of heat.

Q5. A body is falling freely under ‘the action of gravity alone in vacuum. Which of the following quantities remain constant during the fall?
(a) Kinetic energy
(b) Potential energy
(c) Total mechanical energy
(d) Total linear momentum
Sol: (c) As the body is falling freely under gravity, the potential energy decreases continuously and kinetic energy increases continuously as all the conservative forces are doing work. So, total mechanical energy (PE + KE) of the body will be constant.
Let us discuss this in detail:
In the given diagram an object is dropped from-a height H from ground.
At point A total mechanical energy will be EA = K.E + P.E

So, total mechanical energy will remain same (if we neglect the air friction).

Q6. During inelastic collision between two bodies, which of the following quantities always remain conserved?
(a) Total kinetic energy
(b) Total mechanical energy
(c) Total linear momentum
(d) Speed of each body
Sol: (c) If in a collision kinetic energy after collision is not equal to kinetic energy before collision, the collision is said to inelastic.
Coefficient of restitution 0 < e < 1
When we are considering the two bodies as system the total external force on the system will be zero.
Hence, total linear momentum of the system remain conserved.
Here kinetic energy appears in other forms, i.e. energy may be lost in the form of heat and sound etc. In some cases

(KE)_final < (KE)_initialsuch as when initial KE is converted into intertial energy of the product (as heat, elastic or excitation) while in other cases (KE)_final > (KE)_initialsuch as when internal energy stored in the colliding particles is released.
Examples’. (1) Collision between two billiard balls.
(2) Collision between two automobiles on a road.
In fact all majority of collisions belong to this category.

Q7. Two inclined ffictionless 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 as shown in figure.
Which of the following statement is correct?
(a) Both the stones reach the bottom at the same time but not with the same speed.
(b) Both the stones reach the bottom with the same speed and stone I reaches the bottom earlier than stone II.
(c) Both the stones reach the bottom with the same speed and stone II reaches the bottom earlier than stone I.
(d) Both the stones reach the bottom at different times and with different speeds.

Sol: (c) As shown in diagram AB and AC are two smooth planes inclined to the angle θ₁ and θ₂ respectively. As friction is absent here, hence, mechanical energy will be conserved. As both the tracks having common height h,
From conservation of mechanical energy,

(a) V=0,K = E
(b) V=E,K=0
(c) V< E, K= O
(d) V= 0,K<E

Sol:(b)

Q9. Two identical ball bearings in contact with each other frictionless table are hit head-on by another ball bearing of the same mass moving initially with a speed v as shown in figure.
If the collision is elastic, which of the following (figure) is a possible result after collision?

Sol: (b)
Key concept: In a collision if the motion of colliding particles before and after the collision is along the same line, the collision is said to be head on or one dimensional.
When two bodies of equal masses collides elastically, their velocities are interchanged.
Kinetic energy and linear momentum remains conserved Total kinetic energy of the system before collision

Q10. A body of mass 0.5 kg travels in a straight line with velocity v = ax^3/2 where a = 5 m^-1/2 s^_1. The work done by the net force during its displacement from x = 0 to; x = 2m is

(a) 5 J
(b) 50 J
(c) 10 J
(d) 100 J
Sol:

Q11. A body is moving unidirectionally under the influence of a source of constant power supplying energy. Which of the diagrams shown in figure correctly shown the displacement-time curve for its motion?

Q12. Which of the diagrams shown in figure most closely shows the variation in kinetic energy of the earth as it moves once around the sun in its elliptical orbit?

Sol: (d) As the earth moves once around the sun in its elliptical orbit, when the earth is closest to the sun, speed of the earth is maximum, hence KE is maximum. When the earth is farthest from the sun speed is minimum, hence KE is minimum but never zero and negative.
This variation of KE vs t is correctly represented by option (d).

Q13. Which of the diagrams in figure represents varation of total mechanical energy of a pendulum oscillating in air as functon of time?

Sol: (c) When a pendulum oscillates in air, its total mechanical energy decreases continuously in overcoming resistance due to air. Therefore, total mechanical energy of the pendulum decreases exponentially with time.
The variation of E vs t is correctly represented by curve (c) in which the relation between energy and time is shown.

Q14. A mass of 5 kg is moving along a circular path of radius 1 m. If the mass moves with 300 rev/min, its kinetic energy would be

Q15. A raindrop falling from a height h above ground, attains a near terminal velocity when it has fallen through a height (3/4)h. Which of the diagrams shown in figure correctly shows the change in kinetic and potential energy of the drop during its fall up to the ground?

Sol:(b) At height h from ground raindrop have maximum potential energy. And kinetic velocity will be zero (at the instant when it dropped its velocity will be zero), then as the rain drop falls its PE starts decreasing and kinetic energy start increasing.
The total mechanical energy will remain conserved if we neglect the air resistance. If there is some air resistance, there is some force called upthrust (in fluids) which opposes its motion. It depends upon velocity of object as the velocity increases, upthrust also increases. Hence during fall of raindrop first its velocity increases and then become constant after some time.
This constant velocity is called terminal velocity, hence KE also become constant. PE decreases continuously as the drop is falling continuously.
The variation in PE and KE is best represented by (b).

Q16. In a shotput event an athlete throws the shotput of mass 10 kg with an initial speed of 1 m s^-1 at 45° from a height 1.5 m above ground. Assuming air resistance to be negligible and acceleration due to gravity to be 10 m s ^-2, the kinetic energy of the shotput when it just reaches the ground will be
(a) 2.5 J
(b) 5.0 J
(c) 52.5 J
(d) 155.0 J

Sol: (d) If air resistance is negligible, total mechanical energy of the system will remain constant. And let us take ground as a reference where potential energy will be zero.
According to the problem, h = 1.5 m, v = 1 m/s, m = 10 kg, g = 10 ms^– ^2

23

Q17. Which of the diagrams in figure correctly shows the change in kinetic energy of an iron sphere falling freely in a lake having sufficient depth to impart it a terminal velocity?

Sol: (b) When an iron sphere is falling freely in the lake, it will accelerate by the force due to gravity then its velocity increases continuously and resistance due to water cause a force called viscous force or upthrust (in fluids) which opposes its motion. It depends upon velocity of object as the velocity increases upthrust increases. Hence during fall of sphere first its velocity increases and then become constant after some depth.
This constant velocity is called terminal velocity, hence KE become constant beyond this depth, which is best represented by (b).

Q18. A cricket ball of mass 150 g moving with a speed of 126 km/h hits at the middle of the bat, held firmly at its position by the batsman. The ball moves straight back to the bowler after hitting the bat. Assuming that collision between ball and bat is completely elastic and the two remain in contact for0.001 s, the force that the batsman had to apply to hold the bat firmly at its place would be

More Than One Correct Answer Type
Q19. A man of mass m, standing at the bottom of the staircase, of height L climbs it and stands at its top.
(a) Work done by all forces on man is equal to the rise in potential energy mgL.
(b) Work done by all forces on man is zero.
(c) Work done by the gravitational force on man is mgL.
(d) The reaction force from a step does not do work because the point of application of the force does not move while the force exists.
Sol: (b, d) When a man of mass m climbs up the staircase of height L, work done by the gravitational force on the man = -mgL
Work done by internal muscular forces = -Work done against gravitational force = mgL
Work done by all the forces = mgL – mgL = 0
As the point of application of the contact forces does not move, hence work done by reaction forces will also be zero.
And work done by friction will also be zero as there a no dissipation or rubbing is involved.

Q20. A bullet of mass m fired at 30° to the horizontal leaves the barrel of the gun with a velocity v. The bullet hits a soft target at a height h above the ground while it is moving downward and emerge out with half the kinetic energy it had before hitting the target.
Which of the following statements are correct in respect of bullet after it emerges out of the target?
(a) The velocity of the bullet will be reduced to half its initial value.
(b) The velocity of the bullet will be more than half of its earlier velocity
(c) The bullet will continue to move along the same parabolic path.
(d) The bullet will move in a different parabolic path.
(e) The bullet will fall vertically downward after hitting the target.
(f) The internal energy of the particles of the target will increase.
Sol:(b, d, f) The given situation is shown in the diagram. Let speed after emerging from the target is v”, v’ is speed of the bullet just before hitting the target, v is the initial speed of bullet

(a) The velocity of the bullet after it emerges out of the target will be reduced to half its initial value.

Compare the kinetic energies before and after hitting the target.Just before hitting the target the mechanical energy will be conserved. After hitting its energy will be loose. So by conservation of mechanical energy between “O” and “A” (A is the point where the target is placed),

Hence, after emerging from the target velocity of the bullet (v”) is more than half of its earlier velocity v’ (velocity before emerging into the target). So option (b) is correct.

(c) Since velocity of the bullet is changed after hitting the target, so it follows a different parabolic path.
(d) As the velocity of the bullet changes to v’which is less than v’ hence path followed will change and the bullet reaches at point B instead of A’, as shown in the figure.
(e) After emerging from the target, bullet follows a parabollic path, the bullet will not fall vertically downward, so it is an incorrect option.
(f) As the bullet is passing through the target, the loss in energy of the • bullet is transferred to particles of the target. Therefore, their internal energy increases.

Q21. Two blocks M₁ and M₂ having equal mass are free to move on a horizontal frictionless surface. M₂ is attached to a massless spring as shown in figure. Initially M₂ is at rest and A/, is moving toward M₂ with speed v and collides head-on with M₂.
(a) While spring is fully compressed all the KE of Myis stored as PE of spring.
(b) While spring is fully compressed the system momentum is not conserved, though final momentum is equal to initial momentum.
(c) If spring is massless, the final state of the M₁ is state of rest.
(d) If the surface on which blocks are moving has friction, then collision cannot be elastic

Sol: (c, d) If there is not specified we always consider the collision elastic.
When two bodies of equal masses collides elastically, their velocities are interchanged in these types of collision.
Kinetic energy and linear momentum remain conserved.
According to the above diagram when m₁ comes in contact with the spring, m₁ is retarded by the spring force and m₂ is accelerated by the spring force.

(a) The spring will continue *to compress until the two blocks acquire common velocity. So some of kinetic energy of block M_x store into P.E and some part of it stores into K.E of block M₂. So option (a) is incorrect.
(b) As surfaces are frictionalless momentum of the system will be conserved. So option (b) is also incorrect.
(c) The two bodies of equal mass exchange their velocities in a head on elastic collision between them. So, if spring is massless, the final state of the M₁ is state of rest.
(d) Since there is a loss of K.E when the blocks collides on the rough surface. Hence, the collision is inelastic.

Very Short Answer Type Questions
Q22. A rough inclined plane is placed on a cart moving with a constant velocity u on horizontal ground. A block of mass M rests on the incline. Is any work done by force of friction between the block and incline? Is there then a dissipation of energy?
Sol: Let us consider the diagram according to the situation. Just imagine the situation that wedge is moving with velocity u. As the block Mis at rest with respect to the inclined plane. There is no pseudo force acting on the block because the wedge is moving with constant velocity. So,
f = frictional force = Mg sinθ (upward as shown in the figure)

As the block rests on the incline, the force of friction acting between the block and the incline opposes the tendency of sliding of the block. Since, block is not in motion with respect to incline, therefore, work done by the force of friction between the block and the inclined plane is zero. Also due to this reason there is no dissipation of energy.
Q23. Why is electrical power required at all when the elevator is descending? Why should there be a limit on the number of passengers in this case?
Sol: When the elevator is descending, then electric motor (mainly induction motor are used) provides some force to overcome the weight of elevator to prevent it from falling freely under gravity, then electric power is required to run electric motor which is holding the elevator via cable to prevent it from falling freely. This cable is able to sustain some limiting value of tension developed in it. Due to this reason, it is needed to limit the number of passengers in the elevator.

Q24. A body is being raised to a height h from the surface of earth. What is the sign of work done by
(a) applied force and (b) gravitational force?
Sol:(a) External force is applied on the body to lift it in upward direction against its weight, therefore, angle between the applied force and displacement is = 0°

Work done by the applied force
W= F. S =Fs cos = Fs cos 0° = Fs ( cos 0° = 1)

i.e., the sign of work done by applied force is positive.

(b) As shown in figure the gravitational force acts in downward direction and displacement in upward direction, therefore, angle between them is = 180°.
Work done by the gravitational force
W = Fs cos 180° = -Fs (cos 180° = -1)

Q25. Calculate the work done by a car against gravity in moving along a straight horizontal road. The mass of the car is 400 kg and the distance moved is 2 m.
Sol:Weight of the car (mg) vertically downward and car is moving along horizontal road, so displacement of the car is in the along horizontal, i.e. angle between them is 90°.
Work done by weight of the car
W = Fs cos 90° = 0 (cos 90° = 0)
Hence, the work done by car against gravity will also be zero.

Q26. A body falls towards earth in air. Will its total mechanical energy be conserved during the fall? Justify.
Sol: If we neglect the air resistance then the total mechanical energy of the body is conserved but if there is some air resistance then a small part of its energy is utilized against resistive force of air, which is non-conservative force. So, total mechanical energy of the body falling freely under gravity in this case is not conserved. In this condition, gain in KE < loss in PE.
Let E be the total mechanical energy.
Initial mechanical energy of the body

Q27. A body is moved along a closed loop. Is the work done in moving the body necessarily zero? If not, state the condition under which work done over a closed path is always zero.
Sol:
Key concept: If the work done by a force on a body depends upon the initial and final positions only of that body, then the force is conservative e.g., gravitational, electrostatic, magnetic forces.
If the work done by a force on a body which has moved in closed path and has come back to its initial position is zero, the force is conservative.
Work done in moving along a closed loop is not always zero. Work done in moving a closed path is zero when forces acting on the body are conservative but for non-conservative forces like friction force, viscous force etc. work done in a closed path is not zero.
Q28. In an elastic collision of two billiard balls, which of the following quantities remain conserved during the short time of collision of the balls (i.e., when they are in contact)?
(a) Kinetic energy
(b) Total linear momentum
Give reason for your answer in each case.
Sol: During collision no external force is acting on the balls, therefore total linear momentum of the system of two balls is always conserved.
During collision when the balls are in contact, there may be deformation, hence some kinetic energy of system will be transformed to potential energy of system and consequently kinetic energy will not be conserved.
Important point: Though kinetic energy of system will not be conserved but total energy of the system will be conserved.

Q29. Calculate the power of a crane, in watts, which lifts a mass of 100 kg to a height of 10 m in 20 s.
Sol: According to the problem, mass = m= 100 kg height = h- 10 m, time interval, t = 20 s Power is the rate of doing work with respect to time.

Q30. The average work done by a human heart while it beats once is 0.5 J. Calculate the power used by heart if it beats 72 times in a minute.
Sol: According to the problem, average work done by a human heart per beat = 0.5 J
Total work done during 72 beats in 1 minute
= 72 x 0.5 J =36 J
Power = Work done = 36 J / 60s = 0.6 w

Q31. Give example of a situation in which an applied force does not result in a change in kinetic energy.
Sol: Assume a ball tied to a string and is moving in a vertical circle. Work done by tension force will be zero and hence tension force will not cause any change in KE of ball. Because at any instant of time the displacement is tangential and the force is central in nature, i.e., tension in the string and the small displacement at any instant are perpendicular to each other.

Q32. Two bodies of unequal mass are moving in the same direction with equal kinetic energy. The two bodies are brought to rest by applying retarding force of same magnitude. How would the distance moved by them before coming to rest compare?
Sol: According to work-energy theorem,
Change in KE is equal to work done by all the forces acting on the body. Let us assume that only one force (retarding force) is acting on the body, therefore,
KE of the body = Work done by retarding force KE of the body = Retarding force x Displacement
As KE of the bodies and retarding forces applied on them are same, therefore, both bodies will travel equal distances before coming to rest.

Q33. A bob of mass m suspended by a light string of length L is whirled into a vertical circle as shown in figure. What will be the trajectory of the particle, if the string is cut at

Point B?
Point C?
Point A?

Sol: Key concept: According to the situation shown above that a bob of mass m is whirled into a vertical circle, the required centripetal force is obtained from the net force towards center at any point of time in the string. Tension in the string is variable and it is always towards center. But the gravitational force on the bob is always towards center. The speed of the body will be different at different points. So the equations of dynamical equilibrium (F_c = ma_c, F_t = ma_t) must be satisfied at all the points. Let when the string makes an angle 9 with vertical, the speed of mass is v.
Apply Newton’s law perpendicular to the string:
Mg sin = ma, => a,= g sin
The above equation gives tangential acceleration as a function of angle . At lowest point = 0° and at highest point = 180°. So at both points sin 9=0. Hence a, = 0 at both points L and H.
At point M, = 90°, then a₁ = g. It is the maximum value of a_tApply Newton’s law along the string: T – mg cos = ma_cor T=mgcos + m v² /r…(i)

As the body goes up, its velocity will go on decreasing and angle θ will go on increasing. Maximum speed of the body will be at lowest point L and minimum at highest point H. Then from above relation we can find that tension will be maximum at lowest point and minimum at highest point.

When string is cut, tension in string becomes zero and centripetal force is not provided. Hence, bob tends to move in along the direction of its velocity.
(a) If the string is cut at any point, then velocity of body of mass m is along the tangent to the circle. Tangent at point B is vertically downward so the trajectory of the particle is the straight line.
(b) Tangent at point C is horizontally towards right.
So the trajectory of the particle is the parabola.
(c) Tangent at point X makes some angle with the horizontal. Again bob will follow a parabolic path with vertex higher than C.

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