Mechanical Properties of Solids Test

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Mechanical Properties of Solids Test - 49

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Weekly Quiz Competition

Question 1
1 / -0

To determine the Young's modulus of a wire, the formula is $$Y = \dfrac {F}{A}, \dfrac {L}{\triangle l}$$; where $$L = length, A =$$ area of cross-section of the wire, $$\triangle L =$$ Change in length of the wire when stretched with a force $$F$$. The conversion factor to change it from $$CGS$$ to $$MKS$$ system is

A
$$1$$

B
$$10$$

C
$$0.1$$

D
$$0.01$$

Solution

CGS unit of Y is dyn/$${ cm }^{ 2 }$$
MKS unit of Y is N/$${ m }^{ 2 }$$
1 dyn/$${ cm }^{ 2 }$$ = $$\dfrac { { 10 }^{ -5 }N }{ { \left( \dfrac { 1 }{ 100 } \right) }^{ 2 }{ m }^{ 2 } } $$ $$\left[ 1N={ 10 }^{ 5 }dyn\\ 1m=100cm \right] $$
= $${ 10 }^{ -5 }\times { 10 }^{ 4 }N/{ m }^{ 2 }$$
= $$0.1N/{ m }^{ 2 }$$
$$\therefore $$ Conversion factor will be 0.1
Question 2
1 / -0

The area of cross section of a steel wire ($$y=2\times { 10 }^{ 11 }N/{ m }^{ 2 }$$) is $$0.1{cm}^{2}$$. The force required to double its length will be:

A
$$2\times { 10 }^{ 12 }N$$

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

C
$$2\times { 10 }^{ 10 }N$$

D
$$2\times { 10 }^{ 6 }N$$

Solution

Given, $$Y=2\times { 10 }^{ 11 }N/{ m }^{ 2 }$$
$$A=0.1{ cm }^{ 2 }=0.1\times { 10 }^{ -4 }{ m }^{ 2 }$$
Let, length of the wire is l. After applying force, length will be 2l.
$$\therefore $$ increase in length = $$\left( 2l-l \right) =l$$
we know, $$Y=\dfrac { \dfrac { F }{ A } }{ \dfrac { \triangle l }{ l } } $$
$$2\times { 10 }^{ 11 }=\dfrac { \dfrac { F }{ 0.1\times { 10 }^{ -4 } } }{ \dfrac { l }{ l } } $$
$$\therefore \quad F=2\times { 10 }^{ 11 }\times 0.1\times { 10 }^{ -4 }$$
$$\therefore \quad F=2\times { 10 }^{ 6 }N$$
Question 3
1 / -0

Identical springs of steel and copper $$\left( { Y }_{ steel }>{ Y }_{ copper } \right) $$ are equally stretched then:

A
Less work is done on copper spring

B
Less work is done on steel spring

C
Equal work is done on both the springs

D
Data is incomplete

Solution

Work done, $$W=\cfrac { 1 }{ 2 } \times F\times \Delta L$$
For a given $$F$$
$$W\propto \Delta L...(i)\quad $$
and $$\Delta L=\cfrac { FL }{ AY } $$
As $$F, A$$ and $$L$$ are constants
$$\therefore \Delta L\propto \cfrac { 1 }{ Y } ...(ii)$$
From (i) and (ii), we get
$$\quad W\propto \cfrac { 1 }{ Y } $$
$$\therefore \cfrac { { W }_{ steel } }{ { W }_{ copper } } =\cfrac { { Y }_{ copper } }{ { Y }_{ steel } } $$

As $${ Y }_{ copper }<{ Y }_{ steel }$$
$$\quad \therefore { W }_{ steel }<{ W }_{ copper }$$
So, less work is done on steel spring
Question 4
1 / -0

Figure shows the strain-stress curve for a given material. The Young's modulus of the material is

A
$$5\times { 10 }^{ 9 }N\quad { m }^{ -2 }$$

B
$$5\times { 10 }^{ 10 }N\quad { m }^{ -2 }$$

C
$$7.5\times { 10 }^{ 9 }N\quad { m }^{ -2 }$$

D
$$7.5\times { 10 }^{ 10 }N\quad { m }^{ -2 }$$

Solution

From the given graph for a stress of $$150\times {10}^{6}N$$ $${m}^{-2}$$ when strain is $$0.002$$
$$\therefore$$ Young's modulus $$Y=\cfrac { stress }{ strain } $$
$$Y=\cfrac { 150\times { 10 }^{ 6 } }{ 0.002 } N\quad { m }^{ -2 }=7.5\times { 10 }^{ 10 }N\quad { m }^{ -2 }$$
Question 5
1 / -0

Let $${Y}_{S}$$ and $${Y}_{A}$$ represent Young's modulus for steel and aluminium respectively It is said that steel is more elastic than aluminium. Therefore, it follows that

A
$${ Y }_{ S }={ Y }_{ A }$$

B
$${ Y }_{ S }< { Y }_{ A }$$

C
$${ Y }_{ S }> { Y }_{ A }$$

D
$$\cfrac { { Y }_{ S } }{ { Y }_{ A } } =0$$

Solution

Steel is more elastic than aluminium
It means $${ Y }_{ S }> { Y }_{ A }$$
Question 6
1 / -0

If the work done in stretching a wire by $$1mm$$ is $$2J$$, the work necessary for stretching another wire of same material but with double radius of cross-section and half the length by $$1mm$$ is:

A
$$16J$$

B
$$8J$$

C
$$4J$$

D
$$\cfrac { 1 }{ 4 } J$$

Solution

Stretching force, $$F=\cfrac { Y\pi { r }^{ 2 }\Delta L }{ L } $$
where the symbols have their usual meanings.
Both the wires are of same material, so $$Y$$ will be equal, extension in both the wires is same, so $$\Delta L$$ will be equal
$$\therefore F\propto \cfrac { { r }^{ 2 } }{ L } $$
$$\quad \therefore \cfrac { F' }{ F } =\cfrac { { (2r) }^{ 2 } }{ (L/2) } \times \cfrac { L }{ { r }^{ 2 } } =8$$
or $$F'=8F...(i)\quad $$
Work done is stretching a wire,
$$W=\cfrac { 1 }{ 2 } \times F\times \Delta L\quad $$
For sam extension
$$W\propto F$$
$$\therefore \cfrac { W' }{ W } =\cfrac { F' }{ F } =8$$ [Using (i)]
$$W'=8W=8\times 2J=16J$$
Question 7
1 / -0

Stress is a ______ quantity.

A
scalar

B
vector

C
tensor

D
dimensionless

Solution

According to terms in Physics,

Scalar is a quantity having only magnitude.

A vector is a quantity that has magnitude and direction both.

Tensor is a quantity having magnitude, direction, and plane too.

Stress is a tensor as it obeys the coordinate transformation law. Stresses are not vectors because they do not combine according to the parallelogram law of addition. Instead, stresses are much more complex quantities than the vectors and are called tensors. Stress is a second-order tensor. Also, it has a dimension too.

Thus option C is correct.
Question 8
1 / -0

A steel rod of length $$1m$$ and radius $$10mm$$ is stretched by a force $$100kN$$ along its length. The percentage strain in the rod is then
$$\left( { Y }_{ steel }=2\times { 10 }^{ 11 }N\quad { m }^{ -2 } \right) $$

A
$$0.04$$%

B
$$0.08$$%

C
$$0.16$$%

D
$$0.24$$%

Solution

Given:
force, $$F =100KN$$
radius, $$ L= 10mm$$
length, $$L= 1m$$
Area $$A=\pi r^2$$
Stress $$\sigma =\dfrac{F}{A}=\dfrac{10^5}{\pi (10^{-2})^2}$$
Elongation
$$\quad \Delta L=\cfrac { (F/A)L }{ Y } =\cfrac { \left( \dfrac{10^5}{\pi (10^{-2})^2}\right) \left( 1m \right) }{ 2\times { 10 }^{ 11 }N\quad { m }^{ -2 } } =1.59\times { 10 }^{ -3 }m=1.59mm\quad $$

Strain produced in the rod is
$$Strain=\cfrac { \Delta L }{ L } =\cfrac { 1.59\times { 10 }^{ -3 }m }{ 1m } =1.59\times { 10 }^{ -3 }=0.16\quad $$

Question 9

1 / -0

Match the Column I with Column II

Column I	Column II
(A) A body which regains its original shape after the removal of external forces	(p) Elasticity
(B) A body which does not regain its original shape after the removal of external forces.	(q) Elastic body
(C) A body which does not show any deformation on applying external forces	(r) Plastic body
(D) The property of the body to regain its original configuration when the deforming forces are removed.	(s) Rigid body

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

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

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

A-s, B-p, C-q, D-r.

Solution

A body which regains its original shape after the removal of external force is called elastic body.
A-q
A body which does not regain its original shape after the removal of external force is called plastic body
B-r
A body which does not show any deformation on applying external forces is called rigid body.
C-s
The property of the body to regain its original configuration when the deforming forces are removed is called elasticity

Question 10
1 / -0

Which of the following statements is correct regarding Poisson's ratio?

A
It is the ratio of the longitudinal strain to the lateral strain

B
Its value is independent of the nature of the material

C
It is unitless and dimensionless quantity

D
The practical value of Poisson's ratio lies between $$0$$ and $$1$$

Solution

The ratio of the lateral strain to longitudinal strain is called Poisson's ratio.
Hence, option (a) is an incorrect statement.
Its value depends only on the nature of the material.
Hence, option (b) is an incorrect statement.
It is the ratio of two like physical quantities.
Therefore, it is unitless and dimensionless quantity.
Hence option (c) is a correct statement
The practical value of Poisson's ratio lies between $$0$$ and $$0.5$$
hence option (d) is an incorrect statement.

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Mechanical Properties of Solids Test - 49

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Mechanical Properties of Solids Test - 49

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Question 1

$$1$$

$$10$$

$$0.1$$

$$0.01$$

Solution

Question 2

$$2\times { 10 }^{ 12 }N$$

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

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

$$2\times { 10 }^{ 6 }N$$

Solution

Question 3

Less work is done on copper spring

Less work is done on steel spring

Equal work is done on both the springs

Data is incomplete

Solution

Question 4

$$5\times { 10 }^{ 9 }N\quad { m }^{ -2 }$$

$$5\times { 10 }^{ 10 }N\quad { m }^{ -2 }$$

$$7.5\times { 10 }^{ 9 }N\quad { m }^{ -2 }$$

$$7.5\times { 10 }^{ 10 }N\quad { m }^{ -2 }$$

Solution

Question 5

$${ Y }_{ S }={ Y }_{ A }$$

$${ Y }_{ S }< { Y }_{ A }$$

$${ Y }_{ S }> { Y }_{ A }$$

$$\cfrac { { Y }_{ S } }{ { Y }_{ A } } =0$$

Solution

Question 6

$$16J$$

$$8J$$

$$4J$$

$$\cfrac { 1 }{ 4 } J$$

Solution

Question 7

scalar

vector

tensor

dimensionless

Solution

Question 8

$$0.04$$%

$$0.08$$%

$$0.16$$%

$$0.24$$%

Solution

Question 9

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

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

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

A-s, B-p, C-q, D-r.

Solution

Question 10

It is the ratio of the longitudinal strain to the lateral strain

Its value is independent of the nature of the material

It is unitless and dimensionless quantity

The practical value of Poisson's ratio lies between $$0$$ and $$1$$

Solution

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