Mechanical Properties of Solids Test

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Mechanical Properties of Solids Test - 51

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Weekly Quiz Competition

Question 1
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Which of the following apparatus is used to determine the Young's modulus of the material of a given wire?

A
Searle

B
sonometer

C
Metre bridge

D
Resonance tube

Solution

Normally, we use Searle's method to measure Young's modulus of some material. Young's modulus is independent of the shape of the material, so we can utilize any shape for its calculation. Searle’s apparatus is used for the measurement of Young’s modulus experimentally. This apparatus consists of two equal length wires that are attached to a rigid support.
Question 2
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A copper wire of length $$2.4m$$ and a steel wire of length $$1.6m$$, both the diameter $$3mm$$, are connected end to end. The ratio fo elongation of steel to the copper wires is then
$$\left( { Y }_{ copper }=1.2\times { 10 }^{ 11 }N\quad { m }^{ -2 },{ Y }_{ steel }=2\times { 10 }^{ 11 }N\quad { m }^{ -2 } \right) $$

A
$$\cfrac { 5 }{ 2 } $$

B
$$\cfrac { 2 }{ 5 } $$

C
$$\cfrac { 3 }{ 2 } $$

D
$$\cfrac { 2 }{ 3 } $$

Solution

Given:
Length copper wire, $$L_c =2.4m$$
Length steel wire, $$L_s =1.6m$$
$$(Y_{copper}=1.2×1011^{−2}Nm,Y_{steel}=2×1011^{−2}Nm)$$

We know that,
Elongation $$\Delta L =\dfrac{FL}{YA}$$
So,
$$\cfrac { \Delta { L }_{ S } }{ \Delta { L }_{ C } } =\cfrac { { L }_{ S } }{ { L }_{ C } } \times \cfrac { { Y }_{ C } }{ { Y }_{ S } } $$

Substitution the given values, we get
$$\cfrac { { { \Delta L }_{ S } } }{ { \Delta L }_{ C } } =\cfrac { 1.6m }{ 2.4m } \times \cfrac { 1.2\times { 10 }^{ 11 }N\quad { m }^{ -2 } }{ 2\times { 10 }^{ 11 }N\quad { m }^{ -2 } } =\cfrac { 2 }{ 5 } $$

Question 3

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Match the column I with column II

Column I	Column II
(A) The of shape rubber heel changes under stress	(p) Young's modulus of elasticity is involved
(B) In a suspended bridge, there is a strain in the ropes by the load of the bridge	(B) Bulk modulus of elasticity is involved
(C) In an automobile tyre, when air is compressed, the shape of tyre changes	(r) Modulus of rigidity is involved
(D) A solid body is subjected to a deforming force	(s) All the moduli of elasticity are involved

A-q; B-r; C-s; D-p

A-p; B-q; C-r; D-s

A-r; B-q; C-p; D-s

A-r; B-p; C-q; D-s

Solution

The shape of rubber heel changes under stress involves modulus of rigidity as shape changes without change in volume.
A-r
The strain in the ropes of a suspended bridge involves Young's modulus of elasticity as the extension of ropes takes place due to the load of the bridge.
B-p
In an automobile tyre, bulk modulus of elasticity is involved, as the volume of air in the tyre changes
C-q
A solid body when subjected to a deforming force, all the moduli of elasticity are involved.
D-s

Question 4
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If in the above question, the Young's modulus of the material is Y, the value of extension x is:

A
$$\left ( \frac{El}{YA} \right )^{1/3}$$

B
$$\left ( \frac{YA}{Wl} \right )^{1/3}$$

C
$$\frac{1}{l}\left [ \frac{WA}{Y} \right ]^{1/3}$$

D
$$l\left [ \frac{W}{YA} \right ]^{1/3}$$

Solution

Sln :
We know Y = $$\dfrac{stress}{strain} \,= \,\dfrac{Wl}{2Ax} \, \times \, \dfrac{2l^2}{x^2}$$

$$Y \, = \, \dfrac{Wl^3}{Ax^3} \, or \, x \, = \, \left(\dfrac{W}{AY} \right )^{1/3} \, \times \, 1$$
Question 5
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A metal wire of length $$L_1$$ and area of cross section A is attached to a rigid support. Another metal wire of length $$L_2$$ and of the same cross-sectional area is attached to the free end of the first wire. A body of mass M is then suspended from the free end of the second wire. If $$Y_1$$ and $$Y_2$$ are the Young's moduli of the wires respectively, the effective force constant of the system of two wires is

A
$$\frac{Y_1Y_2A}{2(Y_1L_2+Y_2L_1)}$$

B
$$\frac{Y_1Y_2A}{2(L_1L_2)^{1/2}}$$

C
$$\frac{Y_1Y_2A}{(Y_1L_2+Y_2L_1)}$$

D
$$\frac{(Y_1Y_2)^{1/2}A}{2(L_1L_2)^{1/2}}$$

Solution

The wires are connected in series
$$Y=\dfrac{F l}{A\Delta l}\Rightarrow \dfrac{YA}{l}=\dfrac{F}{\Delta l}\Rightarrow k=\dfrac{YA}{l}$$
$$K$$ is spring constant
In series spring constants are added as
$$\dfrac{1}{K_{1}}+\dfrac{1}{K_{2}}=\dfrac{1}{K}\Rightarrow \dfrac{l_{1}}{Y_{1}A}+\dfrac{l_{2}}{Y_2 A}=\dfrac{1}{K}$$
$$\Rightarrow K=\dfrac{Y_{1}Y_{2}A}{Y_{1}L_{2}+Y_{2}L_{1}}$$
Question 6
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A steel bar $$ABCD$$ $$40cm$$ long is made up of three parts $$AB, BC$$ and $$CD$$, as shown in the figure The rod is subjected to a pull of $$25kN$$. The total extension of the rod is (Young's modulus for steel $$2\times { 10 }^{ 11 }N{ m }^{ -2 }$$:

A
$$0.0637mm$$

B
$$0.0647mm$$

C
$$0.0657mm$$

D
$$0.0667mm$$

Solution

The axial force $$25kN$$ is transmitted to each of the three bars.

Stress in part AB is $$\cfrac { 25000N }{ \cfrac { \pi }{ 4 } { (50) }^{ 2 }{ mm }^{ 2 } } =\cfrac { 40 }{ \pi } =12.73N/{ mm }^{ 2 }$$

Stress in oart BC$$=\cfrac { 25000 }{ \cfrac { \pi }{ 4 } { (25) }^{ 2 } } =50.93N/{ mm }^{ 2 }$$

Stress in part CD$$\quad =12.73N/{ mm }^{ 2 }$$
Therefore, total extension of the rod $$=$$ extension in the parts $$AB+BC+CA$$
$$\Delta L =\Delta L_1 +2\times \Delta L_2$$ (extension in AB=CD)

$$\Delta L=\dfrac{Stress \times length}{Y}$$

$$\Delta L =\dfrac{\sigma \times L}{Y}$$

$$=\left( \cfrac { 12.73N/{ mm }^{ 2 } }{ 2\times { 10 }^{ 5 }N/{ mm }^{ 2 } } \times 100mm \right) \times 2+\cfrac { 50.93N/{ mm }^{ 2 } }{ 2\times { 10 }^{ 5 }N/{ mm }^{ 2 } } \times 200mm=\cfrac { 12732 }{ 2\times { 10 }^{ 5 }N/{ mm }^{ 2 } } mm=0.0637mm$$

$$\Delta L=0.0637mm$$
Question 7
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The mean distance between the atoms of iron is $$3\times10^{-10}m$$ and interatomic force constant for iron is $$7 N m^{-1}$$. The Young's modulus of electricity for iron is

A
$$2.33\times 10^5 Nm^{-2}$$

B
$$23.3\times 10^{10} Nm^{-2}$$

C
$$2.33\times 10^{9} Nm^{-2}$$

D
$$2.33\times 10^{10} Nm^{-2}$$

Solution

Given,
Inter atomic distance, $$r=3\times 10^{-10}m$$
Interatomic force constant, $$K=7N/m$$
$$Y=?$$
Interatomic force constant, $$K= Y\times r$$
$$Y=\dfrac{K}{r}$$
$$Y=\dfrac{7}{3\times 10^{-10}}$$
$$Y=2.33\times 10^{10}N/m^2$$
The young's modulus of elasticity of the iron is $$2.33\times 10^{10}N/m^2$$.
The correct option is D.
Question 8
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Find the stress developed inside a tooth cavity filled with copper when hot tea at temperature of $$ 57^o C $$ is drunk. (Take temperature of tooth to be $$ 37^o C,\alpha =1.7\times { 10 }^{ -5 }{ }{ /^o C} $$ and bulk modulus for copper $$ =140\times { 10 }^{ 9 }N{ m }^{ -2 } $$ )

A
$$ 1.43\times { 10 }^{ 8 }N{ m }^{ -2 } $$

B
$$ 4.13\times { 10 }^{ 8 }N{ m }^{ -2 } $$

C
$$ 2.12\times { 10 }^{ 4 }N{ m }^{ -2 } $$

D
$$ 3.12\times { 10 }^{ 4 }N{ m }^{ -2 } $$

Solution

Volumetric strain in tooth cavity $$ =\dfrac { \Delta V }{ V } $$
Let $$ \gamma $$ be the coefficient of volume expansion with the change in temperature $$ \Delta T $$ .
Change in volume is
$$ \Delta V=\gamma V\Delta T\quad or\quad \dfrac { \Delta V }{ V } =\gamma \Delta T $$
Thermal stress in tooth cavity
$$ =\beta \times volumetric \ strain=\beta \gamma \Delta T $$
$$ =\beta \times 3\alpha \Delta T $$ $$ \left( \because \gamma =3\alpha \right) $$
$$ =140\times { 10 }^{ 9 }\times 3\times 1.7\times { 10 }^{ -5 }\times \left( 57-37 \right) $$
$$ =1.43\times { 10 }^{ 8 }N{ m }^{ -2 } $$
Question 9
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The adjacent graph shows the extension ($$\Delta l$$) of a wire of length $$1m$$ suspended from the top of a roof at one end and with a load $$W$$ connected to the other end. If the cross-sectional area of the wire is $${ 10 }^{ -6 }{ m }^{ 2 }$$, the Young's modulus of the material of the wire is

A
$$2\times { 10 }^{ 11 }N{ m }^{ 2 }$$

B
$$2\times { 10 }^{ -11 }N{ m }^{ 2 }$$

C
$$3\times { 10 }^{ -12 }N{ m }^{ 2 }$$

D
$$2\times { 10 }^{ -13 }N{ m }^{ 2 }$$

Solution

Given:
Length, $$L=1m $$
Area of crossection, $$A= 10^{−6}m$$
From above graph at load $$20N$$ extention $$\Delta L=1\times 10^{-4}m^2$$
$$Y=\dfrac{F \times l}{\Delta A \times \Delta l}$$
putting values in above equation
$$Y= \dfrac{20}{10^{-6} \times 10^{-4}}$$

$$Y= 2 \times 10^{11}$$
Question 10
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The ratio of diameters of two wires of same material is n: 1. The length of each wire is 4 m. On applying the same load, the increases in the length of the thin wire will be (n > l)

A
$$n^2$$ times

B
n times

C
2n times

D
(2n + 1) times

Solution

Consider the expression for Young’s modulus.

$$ Y=\dfrac{FL}{A\Delta L} $$

$$ \Delta L=\dfrac{FL}{AY} $$

Where A is the area, F is the applied force, $$\Delta L$$ is the increase in length.

So,
$$ \Delta L=\dfrac{FL}{\pi {{r}^{2}}Y} $$

$$ \Delta L\propto \dfrac{1}{{{r}^{2}}} $$

$$ \dfrac{{{l}_{2}}}{{{l}_{1}}}={{\left( \dfrac{{{r}_{1}}}{{{r}_{2}}} \right)}^{2}} $$

$$ ={{\left( \dfrac{{{d}_{1}}/2}{{{d}_{2}}/2} \right)}^{2}} $$

$$ ={{(n)}^{2}} $$

$$ {{l}_{2}}=n^2\,{{l}_{1}} $$

Hence, on applying the same load, the length is increased by a factor of $${{n}^{2}}$$.

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Re-Attempt Weekly Quiz Competition

Mechanical Properties of Solids Test - 51

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Mechanical Properties of Solids Test - 51

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SHARING IS CARING

Question 1

Searle

sonometer

Metre bridge

Resonance tube

Solution

Question 2

$$\cfrac { 5 }{ 2 } $$

$$\cfrac { 2 }{ 5 } $$

$$\cfrac { 3 }{ 2 } $$

$$\cfrac { 2 }{ 3 } $$

Solution

Question 3

A-q; B-r; C-s; D-p

A-p; B-q; C-r; D-s

A-r; B-q; C-p; D-s

A-r; B-p; C-q; D-s

Solution

Question 4

$$\left ( \frac{El}{YA} \right )^{1/3}$$

$$\left ( \frac{YA}{Wl} \right )^{1/3}$$

$$\frac{1}{l}\left [ \frac{WA}{Y} \right ]^{1/3}$$

$$l\left [ \frac{W}{YA} \right ]^{1/3}$$

Solution

Question 5

$$\frac{Y_1Y_2A}{2(Y_1L_2+Y_2L_1)}$$

$$\frac{Y_1Y_2A}{2(L_1L_2)^{1/2}}$$

$$\frac{Y_1Y_2A}{(Y_1L_2+Y_2L_1)}$$

$$\frac{(Y_1Y_2)^{1/2}A}{2(L_1L_2)^{1/2}}$$

Solution

Question 6

$$0.0637mm$$

$$0.0647mm$$

$$0.0657mm$$

$$0.0667mm$$

Solution

Question 7

$$2.33\times 10^5 Nm^{-2}$$

$$23.3\times 10^{10} Nm^{-2}$$

$$2.33\times 10^{9} Nm^{-2}$$

$$2.33\times 10^{10} Nm^{-2}$$

Solution

Question 8

$$ 1.43\times { 10 }^{ 8 }N{ m }^{ -2 } $$

$$ 4.13\times { 10 }^{ 8 }N{ m }^{ -2 } $$

$$ 2.12\times { 10 }^{ 4 }N{ m }^{ -2 } $$

$$ 3.12\times { 10 }^{ 4 }N{ m }^{ -2 } $$

Solution

Question 9

$$2\times { 10 }^{ 11 }N{ m }^{ 2 }$$

$$2\times { 10 }^{ -11 }N{ m }^{ 2 }$$

$$3\times { 10 }^{ -12 }N{ m }^{ 2 }$$

$$2\times { 10 }^{ -13 }N{ m }^{ 2 }$$

Solution

Question 10

$$n^2$$ times

n times

2n times

(2n + 1) times

Solution

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