Mechanical Properties of Solids

Mechanical Properties of Solids Physics NCERT Exemplar Solutions Class 11th Chapter Nine

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This is a long answer type question as classified in NCERT Exemplar

Consider the diagram according, the bending torque on the trunk of radius r of the tree = $\frac{Y π r^{4}}{4 R}$

When the tree is about to buckle Wd= $\frac{Y π r^{4}}{4 R}$

If R>>h, then the centre of gravity is at a height l $\approx \frac{h}{2} f r o m t h e g r o u n d$

From $? A B C R^{2} \approx (R - d)$ ²+ (h/2)²

If d <2+

So d = h²/8R

If w_o is the weight /volume

$\frac{Y π r^{4}}{4 R} = w_{o} (π r^{2} h) \frac{h^{2}}{8 R}$

h= ( $\frac{2 Y}{w_{o}}$ )^1/3r^2/3

critical height = h= ( $\frac{2 Y}{w_{o}}$ )^1/3r^2/3

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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 ]

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Mechanical Properties of Solids Physics NCERT Exemplar Solutions Class 11th Chapter Nine

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

l₁=AB ,l₂=AC ,l₃=BC

Cos $θ$ = $\frac{l_{3}^{2} + l_{1}^{2} - l_{2}^{2}}{2 l_{3} l_{1}}$

2l₃l₁cos $θ$ = $l_{3}^{2} + l_{1}^{2} - l_{2}^{2}$

Differentiating 2( $l_{3} d l_{1} + l_{1} d l_{3}$ )cos $θ$ -2 $l_{1} l_{3} s i n θ d θ$

= 2 $l_{3} d l_{1} + 2 l_{1} d l_{1} - 2 l_{2} d l_{2}$

= d $l_{1} = l_{1} α_{1} ? t$

=d $l_{2} =$ $l_{2} α_{1} ? t$

=d $l_{3} = l_{3} α_{2} ? t$

$l_{1} = l_{2} = l_{3} = l$

( $l^{2} α_{1} ? t$ + $l^{2} α_{1} ? t$ )cos $θ$ + $l^{2} s i n θ d θ$ = $l^{2} α_{1} ? t$ + $l^{2} α_{1} ? t$ - $l^{2} α_{1} ? t$

sin $θ d θ = 2 α_{1} ? t$ (1-cos $θ$ )- $α_{2} ? t$

$θ = 60$

d $θ * s i n 60 = 2 α_{1} ? t (1 - c o s 60) - α_{2} ? t$

= 2 $α_{1} ? t * \frac{1}{2} - α_{2} ? t = (α_{1} - α_{2}) ? t$

d $θ$ = change in the angle ABC

\frac{(α_{1} - α_{2}) ? t}{s i n 60} = \frac{2 (α_{1} - α_{2}) ? t}{3}

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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.)

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Mechanical Properties of Solids Physics NCERT Exemplar Solutions Class 11th Chapter Nine

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

Consider an element of width dr at r

Let T(r) and T(r+dr) be the tensions at r+dr respectively

So net centrifugal force = w²rdm

= w²r $μ d r$

T(r)-T(r+dr)= $μ$ w²rdr

-dT= $μ$ w²rdr

- $\int_{T = 0}^{T} d T = \int_{r = l}^{r = r} μ w^{2} r d r$

T(r)= $\frac{μ w^{2}}{2} (l^{2} - r^{2})$

Let the increase in length of the element dr be $? r$

So Young's modulus Y= stress/strain= $\frac{\frac{T (r)}{A}}{\frac{? r}{d r}}$

$\frac{? r}{d r} = \frac{T (r)}{A} = \frac{μ w^{2}}{2 Y A}$ (l²-r²)

$? r = \frac{1}{Y A} \frac{μ w^{2}}{2} (l^{2} - r^{2}) d r$

Change in length in right part = $\frac{1}{Y A} \frac{μ w^{2}}{2} \int_{0}^{l} (l^{2} - r^{2}) d r$

= $\frac{1}{Y A} \frac{μ w^{2}}{2} [l^{3} - \frac{l^{3}}{3}] = \frac{1}{3 Y A} μ w^{2} L^{2}$

Total change in length = $\frac{2 μ w^{2} l^{2}}{3 Y A}$

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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*10¹¹ Nm^–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?

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Mechanical Properties of Solids Physics NCERT Exemplar Solutions Class 11th Chapter Nine

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

When a small element of length dx is considered at x from the load x=0

(a) letT(x) and T(x+dx) are tensions on the two cross sections a distance dx apart then

T(x+dx)+T(x)=dmg= $μ$ dxg

dT= $μ$ gx+C

at x=0 T(0)=mg

C=mg

T(x)= $μ$ gx+Mg

Let length dx at x increases by dr then

Young's modulus Y= stress/strain

$\frac{\frac{T (x)}{A}}{\frac{d r}{d x}} = Y$

$\frac{d r}{d x} = \frac{T (x)}{Y A}$

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

r= $\frac{1}{2 * 10^{11} * π * 10^{- 6}} [\frac{π * 786 * 10^{- 3} * 10 * 10}{2} + 25 * 10 * 10]$

so r = 4 $* 10^{- 3} m$

(b) tension will be maximum at x=L

T= $μ g L$ +Mg=(m+M)g

The force = (yield strength) area= $250 * π N$

(m+M)g= 250 $π$

Mg $\approx 250 π$

M= 25 $π \approx 75 k g$

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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?

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This is a long answer type question as classified in NCERT Exemplar

Let the cross sectional area A . consider the equilibrium of the plane aa'. A force F must be acting on this plane making an angle of $\frac{π}{2} - θ$ with the normal ON. Resolving F into components along the plane (FP) and normal to the plane.

By resolving into components

We get F_p= Fcos $θ$

And F_N= Fsin $θ$

Let area of the face aa' be A' then

A/A'=sin $θ$ so A=A'sin $θ$

The tensile stress = normal force/area=Fsin $θ / A'$

= $\frac{F s i n θ}{A / s i n θ} = \frac{F}{A}$ sin^{2
$θ$}

Shearing stress = parallel foce/Area

= $\frac{F c o s θ}{A / s i n θ} = \frac{F}{A} s i n θ c o s θ$

$\frac{F}{2 A} (2 s i n θ c o s θ = \frac{F}{2 A} s i n 2 θ$

a) For stress to be maximum , sin^{2
$θ$}=1

So $θ$ = $\frac{π}{2}$

b) Shearing stress to be maximum

sin2 $θ = 1$

So $θ$ = $\frac{π}{4}$

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9.21 The Marina trench is located in the Pacific Ocean, and at one place it is nearly eleven km beneath the surface of water. The water pressure at the bottom of the trench is about 1.1 * 10⁸ Pa. A steel ball of initial volume 0.32 m³ is dropped into the ocean and falls to the bottom of the trench. What is the change in the volume of the ball when it reaches to the bottom?

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Water pressure at the bottom, p = 1.1 $* 10^{8}$ Pa

Initial volume of the steel ball, V = 0.32 $m^{3}$

Bulk modulus of the steel, B = 1.6 $* 10^{11}$ N/ $m^{2}$

The ball falls at the bottom of the Pacific ocean which is 11 km beneath the surface

Let the change in volume of the ball on reaching the bottom of the trench be ΔV

We know, bulk modulus, B = $\frac{p}{\frac{Δ V}{V}}$ or ΔV = $\frac{p V}{B}$

ΔV = $\frac{1.1 * 10^{8} * 0.32}{1.6 * 10^{11}}$ = 2.2 $* 10^{- 4} m^{3}$

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9.20 Two strips of metal are riveted together at their ends by four rivets, each of diameter 6.0 mm. What is the maximum tension that can be exerted by the riveted strip if the shearing stress on the rivet is not to exceed 6.9 * 10⁷ Pa? Assume that each rivet is to carry one quarter of the load.

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Diameter of the metal strip, d = 6.0 mm = 6 $* 10^{- 3}$ m

Radius, r = d/2 = 3 $* 10^{- 3}$ m

Maximum shearing stress = 6.9 $* 10^{7}$ Pa = $\frac{M a x i m u m f o r c e}{A r e a}$

Maximum force = Maximum stress $* A r e a$

= 6.9 $* 10^{7} * π * r^{2}$ = 6.9 $* 10^{7} * π * {(3 * 10^{- 3})}^{2}$ = 1950.93 N

Since each rivet carries 1/4th of load,

Maximum tension on each rivet = 4 $* 1950.93$ N = 7803.72 N

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9.19 A mild steel wire of length 1.0 m and cross-sectional area 0.50 * 10-2 cm2 is stretched, well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the mid-point of the wire. Calculate the depression at the midpoint.

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Length of the mild steel wire, l = 1.0 m

Area of cross-section, A = 0.5 $* 10^{- 2}$ ${c m}^{2}$ = 0.5 $* 10^{- 6} m^{2}$

A mass of 100 gm is suspended at the midpoint.

m = 100 gm= 0.1 kg

Due to the weight, the wire dips, as shown in the figure.

Original length = XZ, depression = l

The final length of the wire after it dips = XO + OZ

Increase in length of the wire, Δl = (XO + OZ) – XZ ……(i)

From Pythagoras theorem

XO = OZ = $\sqrt{{0.5}^{2} + l^{2}}$

From equation (i)

Δl = 2 $* \sqrt{{0.5}^{2} + l^{2}}$ - 1.0 = 2 $* 0.5 * \sqrt{1 + {(\frac{l}{0.5})}^{2}}$ - 1.0 = $\sqrt{1 + {(\frac{l}{0.5})}^{2}}$ - 1.0

Neglecting the smaller terms, we can write, Δl = $\frac{l^{2}}{0.5}$

We know, Strain = $\frac{I n c r e a s e i n l e n g t h}{O r i g i n a l l e n g t h}$

Let T be the tension in the wire, then

mg = 2T $\cos ? θ$

From the figure

$\cos ? θ$ =&

...more

Length of the mild steel wire, l = 1.0 m

Area of cross-section, A = 0.5 $* 10^{- 2}$ ${c m}^{2}$ = 0.5 $* 10^{- 6} m^{2}$

A mass of 100 gm is suspended at the midpoint.

m = 100 gm= 0.1 kg

Due to the weight, the wire dips, as shown in the figure.

Original length = XZ, depression = l

The final length of the wire after it dips = XO + OZ

Increase in length of the wire, Δl = (XO + OZ) – XZ ……(i)

From Pythagoras theorem

XO = OZ = $\sqrt{{0.5}^{2} + l^{2}}$

From equation (i)

Δl = 2 $* \sqrt{{0.5}^{2} + l^{2}}$ - 1.0 = 2 $* 0.5 * \sqrt{1 + {(\frac{l}{0.5})}^{2}}$ - 1.0 = $\sqrt{1 + {(\frac{l}{0.5})}^{2}}$ - 1.0

Neglecting the smaller terms, we can write, Δl = $\frac{l^{2}}{0.5}$

We know, Strain = $\frac{I n c r e a s e i n l e n g t h}{O r i g i n a l l e n g t h}$

Let T be the tension in the wire, then

mg = 2T $\cos ? θ$

From the figure

$\cos ? θ$ = $\frac{O Y}{O X}$ = $\frac{0.5}{\sqrt{{0.5}^{2} + l^{2}}}$ = $\frac{0.5}{0.5 * \sqrt{1 + {(\frac{l}{0.5})}^{2}}}$

Expanding the expression and eliminating the higher terms:

$\cos ? θ$ = $\frac{l}{0.5 * \sqrt{1 + {(\frac{1}{2}) (\frac{l}{0.5})}^{2}}}$

Since ( 1+ $\frac{l^{2}}{0.5}$ ) = 1 for small l, $\cos ? θ$ = $\frac{1}{0.5}$

Then T = $\frac{m g}{2 (\frac{l}{0.5})}$ = $\frac{m g * 0.5}{2 l}$ = $\frac{m g}{4 l}$

Stress = $\frac{T e n s i o n}{A r e a}$ = $\frac{m g}{4 l * A}$

Young's modulus = $\frac{S t r e s s}{S t r a i n}$

Y = $\frac{m g * 0.5}{? l * A * l^{2}}$

l= $\sqrt{\frac{m g * 0.5}{4 Y A}}$

For steel, Y = 2 $* 10^{11}$ Pa, then

l = $\sqrt{\frac{0.1 * 9.8 * 0.5}{4 * 2 * 10^{11} * 0.5 * 10^{- 6}}}$ = 0.0106 m

Hence, the depression at the midpoint is 0.0106 m

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9.18 A rod of length 1.05 m having negligible mass is supported at its ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths as shown in Fig. 9.15. The cross-sectional areas of wires A and B are 1.0 mm2 and 2.0 mm2, respectively. At what point along the rod should a mass m be suspended in order to produce (a) equal stresses and (b) equal strains in both steel and aluminium wires.

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Cross-sectional area of wire A, $a_{1}$ = 1 ${m m}^{2}$ = 1 $* 10^{- 6} m^{2}$

Cross-sectional area of wire B, $a_{2}$ = 2 ${m m}^{2}$ = 2 $* 10^{- 6} m^{2}$

Young's modulus for steel, $Y_{1}$ = 2 $* 10^{11}$ N/ $m^{2}$

Young's modulus for aluminium, $Y_{2}$ = 7 $* 10^{10}$ N/ $m^{2}$

Stress in the wire = $\frac{F o r c e}{a r e a}$ = $\frac{F}{a}$

If the two wires have equal stresses, then

$\frac{F_{1}}{a_{1}}$ = $\frac{F_{2}}{a_{2}}$ or $\frac{F_{1}}{F_{2}}$ = $\frac{a_{1}}{a_{2}}$ = $\frac{1}{2}$ ………(i)

Where $F_{1}$ is the force exerted on steel wire and $F_{2}$ is the force exerted on aluminium wire

Taking a moment around the point of suspension, we get

$F_{1} y$ = $F_{2} (1.05 - y)$

$\frac{F_{1}}{F_{2}}$ = $\frac{(1.05 - y)}{y}$ ……(ii)

Using equation (i) and (ii), we can

...more

Cross-sectional area of wire A, $a_{1}$ = 1 ${m m}^{2}$ = 1 $* 10^{- 6} m^{2}$

Cross-sectional area of wire B, $a_{2}$ = 2 ${m m}^{2}$ = 2 $* 10^{- 6} m^{2}$

Young's modulus for steel, $Y_{1}$ = 2 $* 10^{11}$ N/ $m^{2}$

Young's modulus for aluminium, $Y_{2}$ = 7 $* 10^{10}$ N/ $m^{2}$

Stress in the wire = $\frac{F o r c e}{a r e a}$ = $\frac{F}{a}$

If the two wires have equal stresses, then

$\frac{F_{1}}{a_{1}}$ = $\frac{F_{2}}{a_{2}}$ or $\frac{F_{1}}{F_{2}}$ = $\frac{a_{1}}{a_{2}}$ = $\frac{1}{2}$ ………(i)

Where $F_{1}$ is the force exerted on steel wire and $F_{2}$ is the force exerted on aluminium wire

Taking a moment around the point of suspension, we get

$F_{1} y$ = $F_{2} (1.05 - y)$

$\frac{F_{1}}{F_{2}}$ = $\frac{(1.05 - y)}{y}$ ……(ii)

Using equation (i) and (ii), we can write

$\frac{1}{2} = \frac{(1.05 - y)}{y}$ or y = 2.1 – 2y or y = 2.1/3 = 0.7 m

We know Young's modulus Y = Stress / Strain

Y = (F/A)/ Strain or Strain = (F/A)/Y

In case of Steel wire, ${S t r a i n}_{s t e e l}$ = $\frac{F_{1}}{a_{1} * Y_{1}}$

In case of Aluminium wire, ${S t r a i n}_{a l u}$ = $\frac{F_{2}}{a_{2} * Y_{2}}$

Since both the strains are equal, we get

$\frac{F_{1}}{F_{2}}$ = $\frac{a_{1} * Y_{1}}{a_{2} * Y_{2}}$ = $\frac{1 * 10^{- 6} * 2 * 10^{11}}{2 * 10^{- 6} * 7 * 10^{10}}$ = 1.429

From (ii) we get

$\frac{(1.05 - y)}{y}$ = 1.429

y = 0.432m

In order to produce an equal strain in the two wires. The mass should be suspended at a distance of 0.432m from the end A

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9.17 Anvils made of single crystals of diamond, with the shape as shown in Fig. 9.14, are used to investigate behavior of materials under very high pressures. Flat faces at the narrow end of the anvil have a diameter of 0.50 mm, and the wide ends are subjected to a compressional force of 50,000 N. What is the pressure at the tip of the anvil?

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