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

NCERT Exemplar Class 11 Physics Solutions Chapter 9: MCQ I

Question:9.1

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

Answer:

The answer is the option (b) zero.

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 doubled
b) be half
c) be four times
d) remain same

Answer:

The answer is the option (d) remains same.
Stress = force/area
The breaking stress is not dependent on the length, and hence if the cross-sectional area is changed, it will not affect the breaking force.

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 answer is the option (d) will decrease.
$Y\alpha \frac{1}{\mathrm{\Delta }T}$
So as the young’s modulus increases, elasticity decreases.

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 answer is the option (c) longitudinal and shear
The strain produced is shearing and longitudinal in this case as the spring when stretched by a load, the shape and length changes.

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
$(a)\frac{Y_{copper}}{Y_{iron}}$
$(b)\sqrt{\frac{Y_{iron}}{Y_{copper}}}$
$(c)\frac{Y_{iron}^2}{Y_{copper}^2}$
$(d)\frac{Y_{iron}}{Y_{copper}}$

Answer:

The answer is the option (b)
Y = stress/strain
$\frac{FL}{A\mathrm{\Delta }L}=\frac{4FL}{\pi D^2\mathrm{\Delta }L}$$\frac{FL}{A}\mathrm{\Delta }L=\frac{4FL}{\pi D^2\mathrm{\Delta }L}$
$\mathrm{\Delta }L$ copper = $\mathrm{\Delta }L$ iron, F is same in both cases
Hence, $Y\propto \frac{1}{D^2}$
So, $\frac{D_{copper}}{D_{iron}}$= $\sqrt{\frac{Y_{iron}}{Y_{copper}}}$ Hence, option d is correct.

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 Figure mass m is suspended from the mid point of the wire. Strain in the wire is

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

Answer:

$\mathrm{\Delta }L=(AO+BO-AB)$
$\mathrm{\Delta }L=2[AO-L]$
$\mathrm{\Delta }L=2(L^2+x^2{)}^{\frac{1}{2}}-1)$

$\mathrm{\Delta }L=2L[1+\frac{x^2}{2L^2}-1]=\frac{x^2}{L}$
Strain = $\frac{\mathrm{\Delta }L}{2L}=\frac{x^2}{2L^2}$. Hence option a is correct.

Question:9.7

A rectangular frame is to be suspended symmetrically by two strings of equal length on two supports. 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:

The answer is the option (c) least in b)

Let us have a look at the free body diagram for the scenario.
The net forces will amount to zero.
Vertical components
$2Tsin\theta -mg=0$
$T=\frac{mg}{2\sin \theta }$
$T\propto \frac{1}{\sin \theta }$

Hence tension in all the cases is different.
And, option c is the correct choice.

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:

The answer is the option (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.
M is the mass which is being positioned at the mid of the rods made of rubber and steel
Y steel > Y rubber
Hence, $\frac{\mathrm{\Delta }L}{L}$ (rubber) will be larger for the same $\frac{F}{A}$. in the case of steel $\mathrm{\Delta }L$ is insignificant. But, in rubber $\mathrm{\Delta }L$ is significant as the shape changes and hence d is the correct option.

NCERT Exemplar Class 11 Physics Solutions Chapter 9: MCQ II

Question:9.9

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) 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:

The answer is the option (c) and (d)
If we compare the tensile strength, (ii) > (i)
So (ii) is more elastic as compared to (i) and hence option c is correct.
The fracture point for the material (ii) is closer than that of material (i). hence (ii) is more brittle than (i) and hence, option d is correct.

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 $\frac{F}{A}$
b) tensile stress at any cross-section is zero
c) tensile stress at any cross-section A of the wire is$2\frac{F}{A}$
d) tension at any cross-section A of the wire is F

Answer:

The answer is the option (a) and (d)
We know that stress = $\frac{F}{A}$ and hence option a is correct
Force balances the tension, which means T=F and hence option d is correct.

Question:9.11

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. The cross-sectional areas of wires A and B are 1.0 mm² and 2.0 mm² respectively.

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:

The answer is the option (b) and (d)

$T_{b}x=T_a(l-x)$
$\frac{T_b}{T_a}=(\frac{l}{x}-1)$ ------------------(1)
For wire A, stress
$=\frac{T_a}{T_b}$
For wire B, stress
$\frac{T_b}{A_b}=\frac{T_b}{2A_a}$
Since stress on steel = stress on Al
$\frac{T_a}{A_a}=\frac{T_b}{2A_a}\to T_a=\frac{T_b}{2}$
$\frac{T_b}{T_a}=2$
$\frac{l}{x}-1=\frac{2}{1}$
$x=\frac{l}{3}$ from point b
so, distance from A = $l-x=l-\frac{l}{3}=\frac{2l}{3}$
hence m is closer to B than A. so, option b is correct.
the rod remains in a horizontal and balanced. so, the strain will also be equal.
Strain (A) = Strain (B)
$\frac{S_a}{Y_a}=\frac{S_b}{Y_b}$
$Ysteel/YAl=$$(\frac{T_a}{T_b})(\frac{A_b}{A_a})=(\frac{x}{l}-x)(\frac{2Aa}{Aa})$
$\frac{200\times {10}^9}{70\times {10}^9}=\frac{2x}{l-x}$
$14x=20l-20x$
$34x=20l$
$x=\frac{20l}{34}=\frac{10l}{17}$
$(l-x)=\frac{7}{17}$
So, for an equal amount of strain, mass m will be closer to A
Hence option d is correct.

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:

The answer is the option (a) and (d).
A property of an ideal liquid is that it is not compressible. (K)Bulk modulus = $\frac{-p(V)}{\mathrm{\Delta }V}$
$\mathrm{\Delta }V=0$ and K = infinite for an ideal liquid.
No tangential forces act on this liquid, the shearing strain Δθ =0 and F =0
$n=\frac{\frac{F}{A}}{\mathrm{\Delta }\theta }=\frac{0}{0}$ = indeterminate
hence a and d are true.

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

Answer:

The answer is the option (a) and (d).
As we know stress =
$\frac{F}{A}$
The areas of cross-sections of both wires are the same, and the force with which they get stretched is also the same. Hence the total stress for both is the same. Hence option a.
Strain = Stress/Y
As stresses are same on both,
Strain $\propto \frac{1}{Y}$ (steel) and Strain $\propto \frac{1}{Y}$ (Al)
Strain (steel)/ Strain (Al) = Y (Al) / Y(steel)
Now, Y(Al) < Y (Steel). Hence, strain(steel) < Strain (Al)
Hence, option d is correct.

NCERT Exemplar Class 11 Physics Solutions Chapter 9: Very Short Answer

Question:9.14

The Young’s modulus for steel is more than that for rubber. For the same longitudinal strain, which one will have greater tensile stress?

Answer:

Y = stress/strain and Y $\propto$ stress
Y (steel)/ Y(rubber) = Stress(steel)/Stress(rubber)
We know that Y steel > Y rubber
Y steel/Y rubber > 1
Hence, stress (steel)/stress (rubber) > 1
So, stress (steel) > stress (rubber)

Question:9.15

Is stress a vector quantity?

Answer:

Stress = force/area
The directions of restoring and deforming forces are equal and opposite and hence there is no net direction of force. So, stress is not a vector quantity.

Question:9.16

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

Answer:

$\frac{\mathrm{\Delta }{L}_1}{\mathrm{\Delta }{L}_2}=\frac{Ycu}{Ys}$
Work done $(WD)=F.\mathrm{\Delta }L$
WD steel/ WD copper = Y cu/ Ys
Ycu/Ys < 1, so WD steel < WD copper

Question:9.17

What is the Young&rsquo;s modulus for a perfect rigid body?

Answer:

Young’s modulus
$Y=\frac{FL}{A\mathrm{\Delta }L}$
Since the body is rigid, it cannot be deformed or reshaped.
Hence, $\mathrm{\Delta }L=0$
Y (rigid body) = $\frac{FL}{0}$ = infinite

Question:9.18

What is the Bulk modulus for a perfect rigid body?

Answer:

Bulk modulus = $\frac{-p(V)}{\mathrm{\Delta }V}$
Since the body is rigid, it cannot be deformed or reshaped and cannot be stretched.
Hence, $\mathrm{\Delta }V=0$
B (rigid body) = $\frac{pV}{0}$= infinite.

NCERT Exemplar Class 11 Physics Solutions Chapter 9: Short Answer

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:

$\frac{l}{\mathrm{\Delta }{L}_2}=\left(\frac{\frac{F_2{L}_2}{A_2{Y}_2}}{\frac{F_1{L}_1}{A_1{Y}_1}}\right)=\frac{2f2L}{fL}\times \pi r^2\times \frac{Y}{4\pi r^2\times Y}$
$\frac{\mathrm{\Delta }{L}_2}{l}=\frac{4}{4}=1$
Hence, $\mathrm{\Delta }{L}_2=l$, which means the change of length in the second wire is the same.

Question:9.20

A steel rod of length 1 m and area of cross-section 1 cm² is heated from $0^{\circ }C$ to ${200}^{\circ }C$ without being allowed to extend or bend. What is the tension produced in the rod?

Answer:

$L_t=L(1+\alpha \mathrm{\Delta }t)$
$\mathrm{\Delta }L=1\times {10}^{-5}\times 200$
$Y=\frac{FL}{A\mathrm{\Delta }L}(L=1m,A=0.0001m^2)$
$Y=2\times {10}^{11}N/m^2$
$F=2\times {10}^{11}\times {10}^{-5}\times 200\times \frac{0.0001}{1}=4\times {10}^{4}N$

Question:9.21

To what depth must a rubber ball be taken in deep sea so that its volume is decreased by 0.1%.

Answer:

B = Bulk modulus = $9.8\times 108N/m^2$
p = density of water = 1000 kg/m³
percentage change in volume = $\frac{\mathrm{\Delta }V}{V}\times 100=0.1$
let h be the depth to which the rubber ball is being taken = hpg
$p=B\times \frac{\mathrm{\Delta }V}{V}$
$h\times 9.8\times {10}^3=9.8\times {10}^8\times \frac{1}{1000}$
hence, h=100 m

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?

Answer:

The cable length is 9.1 m, r = 5 mm,
Tension in cable = 800 N
$Y=2\times {10}^{11}N/m^2$
$\mathrm{\Delta }L=\frac{FL}{AY}=800\times \frac{9.10}{3.14\times {10}^{-6}\times 25\times 2\times {10}^{11}}$
$\mathrm{\Delta }L=4.64\times {10}^{-5}m$

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 flood and why?

Answer:

Both balls are dropped from the same height. This means the velocity with which they approach and strike the ground will be same. But the ivory ball is more elastic in nature than the wet clay ball, so it tries to get back to its original form as quickly as possible as compared to the wet clay ball. Hence, the amount energy transferred to the ivory ball is higher than that of clay ball and as a result it rises higher.

NCERT Exemplar Class 11 Physics Solutions Chapter 9: Long Answer

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. Consider a plane making an angle $\theta$ with the length. What are the tensile and shearing stresses on this planet?

a) for what angle is the tensile stress a maximum?
b) for what angle is the shearing stress a maximum?

Answer:

a) Tensile stress = normal forces to the surface of plane F/ area
A = area of cross-section which is perpendicular to the bar
A’ = the area of plane cut along aa’ of the cross section
$\sin \theta =\frac{A}{A^{{}^{\prime }}}$

The force perpendicular component along A or $A{A}^{{}^{\prime }}=\sin \theta$
Tensile stress = $\left(\frac{F\sin \theta }{A}\right)\times \sin \theta$
For max value, sin θ should be equal to 1. Hence, $\theta =\frac{\pi }{2}$
b) Shearing stress = forces along the plane F/ area = $F\frac{\cos \theta }{A^{{}^{\prime }}}=\left(\frac{F}{2A}\right)\times 2\sin \theta \cos \theta =\left(\frac{F}{2A}\right)\sin 2\theta$
for shearing stress to be maximum, $\sin 2\theta =1$
which gives us $\theta =\frac{\pi }{4}$

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/m³.
b) If the yield strength of steel is 2.5 × 10⁸ N/m², what is the maximum weight that can be hung at the lower end of the wire?

Answer:

An element dx is of mass dm taken from a wire of length L from a distance x. mass per unit length here is u.
$dm=u.dx,r=0.1cm=0.001m$
$A=\pi r^2=\pi (10{)}^{-6}$
M = 25 kg, L =10m

a) The force that acts in the downward direction on dx = wt
$T(x)=(x.u)g+Mg$
$Y=T\frac{\frac{x}{A}}{\frac{dr}{dx}}$
dr is the increase in the length of wire
$Y=\frac{T(x).dx}{A.dr}$
$dr=\frac{1}{AY}(xug+Mg)dx$
integrating both sides we get,
r (change in length of wire) = 1/AY [ug x2/g + Mgx] L0
extension in wire of length L =
$\frac{gL}{2\pi r^{2}Y}[ul+2M]$
extension=
$10\times 10[0.25+2\times 0.25]/(2\times 3.14\times {10}^{-6}\times 2\times {10}^{11})$
extension = $400\times {10}^{-5}$ m mass m = volume x density =
$(A.L)\times P=\pi r^2\times 10\times 7860=0.25kg$
b) at x = L, the wire experiences maximum tension
T(x) = ugx + Mg
T(L) = ugL + Mg
T = (m + M) g {as uL = m}
Yield force = Yield strength x A
$=250\times 3.14$
Max tension = yield force
$(m+M)g=250\times 3.14$
$M=\frac{785}{10}-0.25=78.35Kg$

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.

Answer:

Dr element of the rod is at a distance of r from the centre. We assume Tr and T(r+dr) to be the tensions at points A and B.
Dr has a centrifugal force acting on it equal to T(r+dr) - T(r)
Centrifugal force = -dT in an outward direction.
Centripetal force as a result of the rotation, $dr=dmr{w}^2$
Hence, $-dT=dmr{w}^2$
If u is the mass per unit length, we have
$-dT=u{w}^{2}r.dr$
On integrating both sides we get,
$-{\int }_0^T=dT=u{w}^2{\int }_r^{1}r.dr$
tension in rod,
$-T(r)=\frac{u{w}^2}{2}(l^2-r^2)$ --------------------------(1)
Y = stress/strain =
$\frac{\frac{T(r)}{A}}{\frac{\delta r}{dr}}$
$\frac{\delta r}{dr}=\frac{T(r)}{AY}=\frac{u{w}^2}{2AY}(l^2-r^2)dr$
${\int }_0^{\delta }\delta r={\int }_0^l\frac{(u{w}^2)}{2AY}(l^2-r^2)dr$
$\delta =\left(\frac{u{w}^2}{2AY}\right)\left(\frac{2}{3}\right)l^3$
$\delta =\frac{u{w}^2}{3AY}(l^3)$Since the extension is on both sides, total extension =
$2\delta =2\frac{u{w}^2}{3AY}(l^3)$

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 the temperature of each rod increases by $\mathrm{\Delta }$T. Find the change in the angle ABC.

Answer:

$\cos \theta =\frac{l_1^2+l_3^2-l_2^2}{2l_1{l}_3}$

$2l_1{l}_3\cos \theta =l_1^2+l_3^2-l_2^2$

Differentiating on both sides of the equation,
$2[d(l_1{l}_3)\cos \theta +l_1{l}_{3}d(\cos \theta )]=2l_{1}d{l}_1+2l_{3}d{l}_3-2l_{2}d{l}_2$
$(l_{1}d{l}_3+l_{3}d{l}_1)\cos \theta -l_1{l}_3\cos \theta d\theta =l_{1}d{l}_1+l_{3}d{l}_3-l_{2}d{l}_2$ --------------(1)
$d{l}_1=l_1{\alpha }_1\mathrm{\Delta }t$
$d{l}_2=l_2{\alpha }_2\mathrm{\Delta }t$
$d{l}_3=l_3{\alpha }_3\mathrm{\Delta }t$
L1 = L2 = L3
$d{l}_1=d{l}_3=l{\alpha }_1\mathrm{\Delta }t$ also $d{l}_2=l{\alpha }_2\mathrm{\Delta }t$
substituting in 1, we get
$\cos \theta (l^2{\alpha }_1\mathrm{\Delta }t+l^2{\alpha }_1\mathrm{\Delta }t)-l^2\sin \theta d\theta =l^2{\alpha }_1\mathrm{\Delta }t+l^2{\alpha }_1\mathrm{\Delta }t-l^2{\alpha }_2\mathrm{\Delta }t$
$2{\alpha }_1\mathrm{\Delta }t\times \frac{1}{2}-2{\alpha }_1\mathrm{\Delta }t+{\alpha }_2\mathrm{\Delta }t=\frac{\sqrt{3}}{2}d\theta$
$d\theta =2(\alpha 2-\alpha 1)\frac{\mathrm{\Delta }t}{\sqrt{3}}$

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 $\frac{\pi r^4}{4R}$. 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 us consider triangle ABC which is a right-angled triangle. We use Pythagoras theorem here.
$R^2=(R-d{)}^2+{\left(\frac{h}{2}\right)}^2$
$2Rd=\frac{h^2}{4}$
$d=\frac{h^2}{8R}$ ------------(1)
we assume W to be the weight of trunk per unit volume.
Weight of trunk = total vol. x W = $\pi r^{2}h\times W$
Torque exerted by bending the trunk, T = $\pi r^{2}h\times W\times d$
Given value of torque = $\frac{\pi r^{4}Y}{4R}$
Equating both values we get,
$\pi r^{2}h\times W\times \frac{h^2}{8R}=\frac{\pi r^{4}Y}{4R}$
$h^3=\pi r^{4}Y\times \frac{8R}{4R\pi r^{2}W}$
$h^3=[\frac{2Y}{W}{]}^{\frac{1}{3}}{r}^{2/3}$

Question:9.29

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?

Answer:

The string is of length L, and stone is tied at P. The nail O is the point where the string is fixed. L is the height through which the stone is lifted. The stone tries to follow a path PP’ when it falls under the effect of gravity. But, due to elastic property of the string, it will follow a circular path from P to Q which gives rise to centrifugal force. The potential energy of the stone is converted to mechanical energy in the string with a spring constant K. PE of stone = Mechanical energy of the string
a) $mgy=\frac{1}{2}K(y-L{)}^2$
$mgy=\frac{1}{2}K(y^2+L^2-2yL)$
$2mgy=K{y}^2-2KyL+K{L}^2$
$K{y}^2-2(KL+mg)y+K{L}^2=0$

$Y=\frac{[KL+mg]}{K}\pm \frac{\sqrt{mg(mg+2KL)}}{K}$

b) At the lowest point, the acceleration is zero.
So, F = 0
Here the force of the string is balances by the force of gravity which makes, mg = Kx
If we assume v to be equal to maximum velocity then by the law of conservation of energy,
Kinetic energy of stone + Potential energy gained by the string = potential energy lost from P to Q’
$\frac{1}{2}mv2+\frac{1}{2}K{x}^2=mg(L+x)$
$m{v}^2+K{x}^2=2mg(L+x)$
$x=\frac{mg}{K}$
$m{v}^2=2mgL+\frac{2m^2{g}^2}{K}-\frac{m^2{g}^2}{K}$
$v=[2gL+\frac{m{g}^2}{K}{]}^{\frac{1}{2}}$

c) From part a, at the lowest point
$m\frac{d^{2}z}{d{t}^2}=mg-K(y-L)$
$\frac{d^{2}z}{d{t}^2}-g+\frac{K}{m(y-L)}=0$
$z=[(y-L)-\frac{mg}{K}]$ {through transformation of variables}
$\frac{d^{2}z}{d{t}^2}+\frac{k}{m}(z)=0$
This is a second order differential equation of the kind of simple harmonic motion.
w is the angular frequency, $w=\sqrt{\frac{k}{m}}$
The solution of such differential equation is of type $z=A\cos (wt+\theta )$
$z=(L+\frac{K}{m}g)+A^{{}^{\prime }}\cos (wt+\theta )$
Hence, the stone undergoes simple harmonic motion about point y=0,
Hence, $z_0=(L+\frac{mg}{K})$