NCERT Solutions for Class 11 Science Physics Chapter 9 - Mechanical Properties Of Solids

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NCERT Solutions for Class 11 Science Physics Chapter 9 Mechanical Properties Of Solids are provided here with simple step-by-step explanations. These solutions for Mechanical Properties Of Solids are extremely popular among Class 11 Science students for Physics Mechanical Properties Of Solids Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the NCERT Book of Class 11 Science Physics Chapter 9 are provided here for you for free. You will also love the ad-free experience on Meritnation’s NCERT Solutions. All NCERT Solutions for class Class 11 Science Physics are prepared by experts and are 100% accurate.

Page No 65:

Question 9.1:

Modulus of rigidity of ideal liquids is
(a) infinity.
(b) zero.
(c) unity.
(d) some finite small non-zero constant value.

Answer:

The ratio of tangential stress to the shearing strain within limits of proportionality is termed as modulus of rigidity of the material of a given body.
Only solid materials can exhibit shearing as they have a specific shape. Liquids cannot sustain shearing stress and the tangential forces are also zero so they do not develop any tangential stress. Hence, modulus of rigidity is zero.
Hence, the correct answer is option (b).

Page No 65:

Question 9.2:

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.

Answer:

The Breaking force of a wire is given by
Breaking force = Breaking stress ⨯ Area
Breaking stress is a constant for a given material. From the formula we can see that maximum load a wire can withstand is independent of it's length.
Hence, the correct answer is option (d).

Page No 65:

Question 9.3:

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.

Answer:

The Young's modulus decreases with increases temperature. Increase in temperature in the metal material can increase the vibration of atoms in the crystal structure, which will increase the atomic distance and decrease the atomic force.
Hence, the correct answer is option (d).

Page No 66:

Question 9.4:

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.

Answer:

The force acts along the cross sectional area of the spring , instead of being perpendicular, resulting in the development of torsion which leads to the development of shear strain. Also the spring stretches axially, so there is also a longitudinal strain.
Hence, the correct answer is option (c).

Page No 66:

Question 9.5:

A rigid bar of mass M is supported symmetrically by three wires each of length l . 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

(a) Y_copper/Y_iron

(b) $\sqrt{\frac{Y_{iron}}{Y_{copper}}}$

(c) $\frac{Y_{iron}^{2}}{Y_{copper}^{2}}$

(d) $\frac{Y_{iron}}{Y_{copper}} .$

Answer:

As the bar is supported symmetrically by wires the tension (T) and the extension ( $∆$ L) in the wires will be same.
$Y = \frac{S t r e s s}{S t r a i n} = \frac{\frac{F}{A}}{\frac{∆ L}{L}} = \frac{F L}{\frac{π d^{2} ∆ L}{4}} = \frac{4 T L}{π d^{2} ∆ L} since, T and ∆ L are same in all the wires, we can write Y \propto \frac{1}{d^{2}} \Rightarrow d \propto \frac{1}{\sqrt{Y}} on solving we will get, \frac{d_{c o p p e r}}{d_{i r o n}} = \sqrt{\frac{Y_{i r o n}}{Y_{c o p p e r}}}$
Hence, the correct answer is option (b).

Page No 66:

Question 9.6:

A mild steel wire of length 2L and cross-sectional area A is stretched, well within elastic limit, horizontally between two pillars (Fig. 9.1). A mass m is suspended from the mid point of the wire. Strain in the wire is

(a) $\frac{x^{2}}{2 L^{2}}$

(b) $\frac{x}{L}$

(c) $\frac{x^{2}}{L}$

(d) $\frac{x^{2}}{2 L} .$

Answer:

As shown in the diagram, due to the weight at the mid point the wire gets stretched:

$Change in the length of the wire be, = 2 \sqrt{L^{2} + x^{2}} - 2 L = 2 L {[1 + {(\frac{x}{L})}^{2}]}^{\frac{1}{2}} - 2 L = 2 L [1 + \frac{1}{2} {(\frac{x}{L})}^{2}] - 2 L (using binomial expansion and ignoring the higher power terms, as x is very small as compared to the L .) = \frac{x^{2}}{L} The strain in the wire will be given by, Strain = \frac{Change in length}{Original length} = \frac{x^{2}}{2 L^{2}}$
Hence, the correct answer is option (a).

Page No 67:

Question 9.7:

A rectangular frame is to be suspended symmetrically by two strings of equal length on two supports (Fig. 9.2). 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 (a).
(c) least in (b).
(d) least in (c).

Answer:

Let us take free body diagram for image (c):

$from the free body diagram we can see that the vertical component of tension in string (T) is given by, T \cos θ Since the body is in equillibrium, 2 T \cos θ = m g \Rightarrow T = \frac{m g}{2 \cos θ} A s θ increases (from 0^{ο} to 90^{ο}), \cos θ will decrease, which implies tension (T) increases \Rightarrow T \propto θ \Rightarrow Tension will be minimum in case (b) (θ = 0^{ο}) .$

Hence, the correct answer is option (c).

Page No 67:

Question 9.8:

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.

Answer:

Steel is more elastic than rubber. Steel will elongate without making any perceptible change in shape, but the rubber rod will elongate with the shape of bottom edge tapered to a tip at the centre.
Hence, the correct answer is option (d).

Page No 68:

Question 9.9:

The stress-strain graphs for two materials are shown in Fig.9.3 (assume same scale).

(a) Material (ii) is more elastic than material (i) and hence material (ii) is more brittle.
(b) Material (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).

Answer:

From graph (i) linear limit vanishes soon and for small stress there is large strain as compared to graph (ii).
Thus, material (ii) is more elastic over a large region of strain as compared to (i).

A material is said to be more brittle if it's fracture point is more closer to ultimate strength point.
Thus, material (ii) is more brittle than material (i).
Hence, the correct options are (c) and (d).

Page No 68:

Question 9.10:

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.

Answer:

According to the diagram a wire is stretched under the action of weight F suspended from its other end. Clearly, forces at each cross-section is F.
$Tensile stress = \frac{Force}{Area of cross - section}$

Hence, the correct options are (a) and (d).

Page No 68:

Question 9.11:

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 (Fig. 9.4). The cross-sectional areas of wires A and B are 1.0 mm²and 2.0 mm², respectively.
(Y_Al = 70 × 10⁹ Nm^–2 and Y_steel = 200 × 10⁹ Nm^–2)

(a) Mass m should be suspended close to wire A to have equal stresses in both the wires.
(b) Mass m should be suspended close to B to have equal stresses in both the wires.
(c) Mass m should be suspended at the middle of the wires to have equal stresses in both the wires.
(d) Mass m should be suspended close to wire A to have equal strain in both wires.

Answer:

Tension in the wire will be more if the mass m is suspended closer to it.
$by formula for stress, stress = \frac{Tension}{Area} let T_{A} be tension in wire A, and T_{B} be tension in wire B for equal stress, \frac{T_{A}}{A_{A}} = \frac{T_{B}}{A_{B}} \frac{T_{A}}{T_{B}} = \frac{A_{A}}{A_{B}} = \frac{1}{2} Thus T_{A} < T_{B} \Rightarrow for equal stress mass m should be suspended closer to wire B by formula for strain, strain = \frac{stress}{Y} = \frac{T}{YA} for equal strain, \frac{T_{A}}{Y_{A} A_{A}} = \frac{T_{B}}{Y_{B} A_{B}} \frac{T_{A}}{T_{B}} = \frac{Y_{A} A_{A}}{Y_{B} A_{B}} = \frac{200 \times 10^{9} \times 1}{70 \times 10^{9} \times 2} = \frac{10}{7} Thus T_{A} > T_{B} \Rightarrow for equal strain mass m should be suspended closer to wire A$
Hence, the correct options are (b) and (d).

Page No 69:

Question 9.12:

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.

Answer:

An ideal fluid is not compressible. Hence change in volume,
$∆ V = 0 . Bulk modulus (b) is given by, B = \frac{Volume stress}{Volume strain} = \frac{\frac{F}{A}}{\frac{∆ V}{v}} = \frac{F V}{A ∆ V} = \infty A liquid cannot sustain tangential force . It may contain tangential viscous, so η = \frac{shear stress}{shear strain} = 0$

Hence, the correct options are (a) and (d).

Page No 69:

Question 9.13:

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 1cm. The two wires will have
(a) the same stress.
(b) different stress.
(c) the same strain.
(d) different strain.

Answer:

$Since the wires are in series the deforming force (F) will be same, hence the stress (\frac{F}{A}) will also be the same in both the wires . From the formula of Young' s modulus, Y = \frac{F L}{A ∆ L}, we get ∆ L = \frac{F L}{Y A} for steel ∆ L_{s t e e l} = \frac{F L}{Y_{s t e e l} A} and for copper ∆ L_{c o p p e r} = \frac{F L}{Y_{c o p p e r} A} since, Y_{s t e e l} \neq Y_{c o p p e r} \Rightarrow ∆ L_{s t e e l} \neq ∆ L_{c o p p e r} \Rightarrow The wires will have different strain (\frac{∆ L}{L})$
Hence, the correct options are (a) and (d).

Page No 69:

Question 9.14:

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?

Answer:

$Young' s modulus (Y) = \frac{Tensile stress}{Longitudnal strain} Tensile stress = Young' s modulus \times Longitudnal strain It is given that, {strain}_{steel} = {strain}_{rubber} Since, Y_{steel} ≫ Y_{rubber} \Rightarrow {stress}_{steel} ≫ {stress}_{rubber}$
Hence, for same longitudinal strain, steel will have greater tensile stress than that of the rubber.

Page No 69:

Question 9.15:

Is stress a vector quantity?

Answer:

$stress = \frac{magnitude of deforming force}{magnitude of area of cross - section}$
Stress is not a vector quantitty

Page No 69:

Question 9.16:

Identical springs of steel and copper are equally stretched. On which, more work will have to be done?

Answer:

$Total work done in increasing the length by ∆ L is given by, W = \frac{Y A {(∆ L)}^{2}}{2 L} This work done is stored in form of potential energy in the wire . If the the wires are identical then L and A will be same for both the wires It is also given that ∆ L is same for both the wires . Since, W \propto Y we know that, Y_{s t e e l} > Y_{c o p p e r} \Rightarrow W_{s t e e l} > W_{c o p p e r}$
Hence, more work will be done in case of spring made of steel.

Page No 69:

Question 9.17:

What is the Young’s modulus for a perfect rigid body?

Answer:

$Young' s modulus (Y) = \frac{stress}{strain} for a perfectly rigid body, strain = 0, no matter what the stress is, \Rightarrow Y = \infty$
Hence for perfectly rigid body, Young modulus will be infinity.

Page No 69:

Question 9.18:

What is the Bulk modulus for a perfect rigid body?

Answer:

$Bulk modulus (Y) = \frac{v o l u m e stress}{v o l u m e strain} for a perfectly rigid body, v o l u m e strain = 0, no matter what the stress is, \Rightarrow B = \infty$

Hence for a perfectly rigid body, Bulk modulus will be infinity (∞).
Hence, for a perfectly rigid body, Bulk modulus will be infinity (∞).

Page No 69:

Question 9.19:

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 2f. Find the increase in length of this wire.

Answer:

$Y = \frac{stress}{strain} = \frac{\frac{F}{A}}{\frac{∆ L}{L}} = \frac{F L}{A ∆ L} for the first wire, Y = \frac{f L}{π r^{2} l} . . . (1) for the second wire, Y = \frac{(2 f) (2 L)}{π {(2 r)}^{2} l'} . . . (2) equating (1) to (2) \frac{f L}{π r^{2} l} = \frac{(2 f) (2 L)}{π {(2 r)}^{2} l'} \Rightarrow l' = l$
Hence, the required increase in length of the wire be l.

Page No 70:

Question 9.20:

A steel rod (Y = 2.0 × 10¹¹Nm^–2; and α = 10^–5C^–1) of length 1 m and area of cross-section 1 cm²is heated from 0°C to 200°C, without being allowed to extend or bend. What is the tension produced in the rod?

Answer:

$According to the question as temperature increases the length of the rod should had increased, but due to rigid support it does not, which creates strain in the rod . ∆ L = α L ∆ T strain = \frac{∆ L}{L} = α ∆ T = 10^{- 5} \times 200 = 2 \times 10^{- 3} we know that, Y = \frac{\frac{T}{A}}{\frac{∆ L}{L}} T = Y \times A \times \frac{∆ L}{L} = 2 \times 10^{11} \times 10^{- 4} \times 2 \times 10^{- 3} = 4 \times 10^{4} N$

Page No 70:

Question 9.21:

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 × 10⁸ N m^–2; and the density of sea water is 10³ kg m^–3.)

Answer:

Let h be the required depth,
$By defination, B = - \frac{∆ P}{\frac{∆ v}{v}}, change in pressure (∆ P) = ρ g h volume strain (\frac{∆ V}{V}) = - 0.1 % = - \frac{1}{1000} ∆ P = - B \times \frac{∆ V}{V} ρ g h = - B \times \frac{∆ V}{V} 10^{3} \times 9.8 \times h = - (9.8 \times 10^{8}) \times (- \frac{1}{1000}) \Rightarrow h = 10^{2} m$

Page No 70:

Question 9.22:

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 × 10¹¹Nm^–2.)

Answer:

$According to the question, length of steel cable, L = 9.1 m radius, r = 5 mm = 5 \times 10^{- 3} m Tension, F = 800 N Young' s modulus, F = 2 \times 10^{11} N / m^{2} Y is given by the formula, Y = \frac{\frac{F}{A}}{\frac{∆ L}{L}} we get, ∆ L = \frac{F L}{Y (π r^{2})} = \frac{800 \times 9.1}{(2 \times 10^{11}) \times 3.14 \times {(5 \times 10^{- 3})}^{2}} on solving we get, ∆ L = 5 \times 10^{- 4} m = 0.5 mm$

Page No 70:

Question 9.23:

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?

Answer:

Ivory ball is more elastic than a wet-clay ball, it has a higher tendency to regain its original shape. 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 same height.

Page No 70:

Question 9.24:

Consider a long steel bar under a tensile stress due to forces F acting at the edges along the length of the bar (Fig. 9.5). Consider a plane making an angle θ 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?

Answer:

$In the given problem force (F) acts along the horizontal, so we resolve it in two perpendicular components, one is parallel to the plane (α α') which will be given by, F_{∥} = F \cos θ the other will be perpendicular to the plane (α α'), F_{⊥} = F \sin θ let the cross sectional area of bar be A, then area of plane α α' will be given by, \frac{A}{\sin θ} Tensile stress = \frac{perpendicular force}{Area} = \frac{F_{⊥}}{\frac{A}{\sin θ}} = \frac{F \sin^{2} θ}{A} Shear stress = \frac{parallel force}{Area} = \frac{F_{⊥}}{A} = \frac{F \cos θ}{\frac{A}{\sin θ}} = \frac{F \sin 2 θ}{2 A} (a) maximum tensile stress will be when \sin^{2} θ = 1, \Rightarrow θ = 90^{o} (b) maximum shear stress will be when \sin 2 θ = 1, \Rightarrow 2 θ = 90^{o} \Rightarrow θ = 45^{o}$

Page No 70:

Question 9.25:

(a) A steel wire of mass µ per unit length with a circular cross section has a radius of 0.1 cm. The wire is of length 10 m when measured lying horizontal, and hangs from a hook on the wall. A mass of 25 kg is hung from the free end of the wire. Assuming the wire to be uniform and lateral strains << longitudinal strains, find the extension in the length of the wire. The density of steel is 7860 kg m^–3(Young’s modules Y = 2 × 10¹¹Nm^–2).
(b) If the yield strength of steel is 2.5 × 10⁸ Nm^–2, what is the maximum weight that can be hung at the lower end of the wire?

Answer:

Consider an element of length dx at a distance x from the load (x = 0)

$(a) If tension T (x) and T (x + d x) are tensions on the cross sections a distance d x apart, then T (x + d x) - T (x) = μ g d x \frac{d T}{d x} \times d x = μ g d x T (x) = μ g x + C at x = 0, T (0) = M g \Rightarrow C = M g T (x) = μ g x + M g Let the length d x at x extend by d r, then Y = \frac{\frac{T (x)}{A}}{\frac{d r}{d x}} d r = \frac{T (x)}{Y A} r = \frac{1}{Y A} \int_{0}^{L} (μ g x + M g) d x = \frac{1}{Y A} {[\frac{μ g x^{2}}{2} + M g x]}_{0}^{L} = \frac{1}{Y A} [\frac{μ g L^{2}}{2} + M g L] = \frac{1}{Y A} [\frac{m g L^{2}}{2} + M g L] (m is the mass of the wire) A = π \times {(10^{- 3})}^{2} m^{2}, Y = 200 \times 10^{9} {Nm}^{- 2} m = π \times {(10^{- 3})}^{2} \times 10 \times 7860 kg r = \frac{1}{2 \times 10^{11} \times π \times 10^{- 6}} [\frac{π \times 786 \times 10^{- 7} \times 10 \times 10}{2} + 25 \times 10 \times 10] = [196.5 \times 10^{- 6} + 3.98 \times 10^{- 3}] = 4 \times 10^{- 3} m$

$(b) The maximum tension would be at x = L T = μ g L + M g = (m + M) g the yield force = 250 \times 10^{6} \times π \times {(10^{- 3})}^{2} = 250 π N At yield (m + M) g = 250 π m = π \times {(10^{- 3})}^{2} \times 10 \times 7860 = 0.025 kg thus, m < < M M g = 250 π M = 25 π = 75 kg$

Page No 71:

Question 9.26:

A steel rod of length 2l, 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.)

Answer:

Consider an element of length dr at a distance r from the centre of the rod,

$Mass of the element, d m = \frac{M}{2 l} d r Centripetal force on this element, d F = d m . r ω^{2} \Rightarrow d F = \frac{M}{2 l} r ω^{2} d r Let F be the tension in the rod at distance r from centre then, F = \int_{r}^{l} \frac{M}{2 l} r ω^{2} d r = \frac{M}{4 l} ω^{2} (l^{2} - r^{2}) If ∆ r is the extension in the element of length d r at position r, then ∆ r = \frac{F d r}{Y A} Hence extension in half of the rod is given by : ∆ r = \int_{0}^{l} \frac{F d r}{Y A} = \int_{0}^{l} \frac{M}{4 l Y A} ω^{2} (l^{2} - r^{2}) d r = \frac{M ω^{2}}{4 l Y A} (l^{3} - \frac{l^{3}}{3}) = \frac{M ω^{2} l^{2}}{6 Y A} total extension in entire rod of length will be 2 ∆ r Hence, the total change in length = \frac{M ω^{2} l^{2}}{3 Y A}$

Page No 71:

Question 9.27:

An equilateral triangle ABC is formed by two Cu rods AB and BC and one Al rod. It is heated in such a way that temperature of each rod increases by ∆T. Find change in the angle ABC. [Coeff. of linear expansion for Cu is α₁, Coeff. of linear expansion for Al is α₂]

Answer:

Due to the increase in length of the rod, each side of the triangle will change, and hence the angle corresponding to any vertex will also change. Initial length of each side be, AB = BC = CA = l
Now after heating, the temperature of each rod is changed by $∆ T$ , and hence the sides of ABC are changed accordingly.
Let the new length be, AB = l₁, CA = l₂ and BC = l₃
Let the angle B of the triangle be $θ$ , then
$\cos (θ) = \frac{{(l_{1})}^{2} + {(l_{3})}^{2} - {(l_{2})}^{2}}{2 (l_{1} \times l_{3})} \Rightarrow 2 (l_{1} \times l_{3}) \cos (θ) = {(l_{1})}^{2} + {(l_{3})}^{2} - {(l_{2})}^{2} Differentiating both side, and using expansion in length as (d l = l α ∆ T), we get s i n (θ) d θ = 2 α_{1} ∆ T (1 - c o s (θ)) - α_{2} ∆ T For equilateral triangle, θ = 60^{o} \Rightarrow d θ = \frac{2 (α_{1} - α_{2}) ∆ T}{\sqrt{3}} = change in angle B of the triangle ABC .$

Page No 71:

Question 9.28:

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 $\frac{Y π r^{4}}{4 R} .$ Y is the Young’s modulus, r is the radius of the trunk and R 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.

Answer:

Let W be the weight of the tree, which acts in downward direction
When the tree bent , then,
Torque on the trunk of the tree due to its weight,
$W d = \frac{Y π r^{4}}{4 R}$
$If R > > h, then its centre of gravity is at at a height \frac{h}{2} from the ground . From, triangle ABC, R^{2} = {(R - d)}^{2} + {(\frac{h}{2})}^{2} If d < < R, then d = \frac{h^{2}}{8 R} \Rightarrow \frac{Y π r^{4}}{4 R} = ω_{o} (π r^{2} h) \frac{h^{2}}{8 R} (W h e r e, ω_{o} = \frac{w e i g h t}{v o l u m e}) \Rightarrow h = {(\frac{2 Y}{ω_{o}})}^{1 / 3} \times r^{2 / 3} = Critical height$

Page No 71:

Question 9.29:

A stone of mass m is tied to an elastic string of negligble 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?

Answer:

Till the stone drops through a length L it will be in free fall. After that it will execute simple harmonic motion as the restoring force starts acting on it.
(a) Let stone covers a distance y before coming to rest when the stone is dropped,
So, loss in potential energy of the stone = gain in potential energy in string
$\Rightarrow m g y = \frac{1}{2} k {(y - L)}^{2} Solving we get, y = \frac{(k L + m g) + \sqrt{2 m g k L + {(m g)}^{2}}}{k}$
(b) The maximum velocity is attained when the body passes through the “equilibrium, position”.
$At equilibrium position, \Rightarrow m g - k x = 0 \Rightarrow x = \frac{m g}{k} . . . . . (1) Also, its mechanical energy must be conserved, \Rightarrow \frac{1}{2} m v^{2} + \frac{1}{2} k x^{2} = m g (L + x) . . . . . (2) (where, v = velocity at that position) Solving eq (1) and eq (2), we get \Rightarrow v = \sqrt{2 g L + \frac{m g^{2}}{k}}$
(c) The stone is at the lowest point Q, then
$Angular frequency of body in SHM, ω = \sqrt{\frac{k}{m}} At mean position (y = y_{o}), \frac{k}{m} (y_{o} - L) - g = 0 y_{o} = L + \frac{m g}{k} So, the stone will execute simple harmonic motion with angular frequency ω about a point y_{o} = L + \frac{m g}{k} . \Rightarrow \frac{d^{2} y}{d t^{2}} + \frac{k}{m} (y - L) - g = 0 Let, \frac{k}{m} (y - L) - g = z then, \frac{d^{2} y}{d t^{2}} = \frac{m}{k} \frac{d^{2} z}{d t^{2}} Hence, the equation becomes, \frac{m}{k} \frac{d^{2} z}{d t^{2}} + z = 0 \Rightarrow \frac{d^{2} z}{d t^{2}} = - \frac{k}{m} z = - ω^{2} z \Rightarrow ω = \sqrt{\frac{k}{m}} For mean position, z = 0 \Rightarrow \frac{k}{m} (y_{o} - L) - g = 0 \Rightarrow y_{o} = L + \frac{m g}{k}$
Hence, the stone will execute SHM, about point $y_{o} = L + \frac{m g}{k}$

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