Mechanical Properties of Solids Test

Result

Mechanical Properties of Solids Test - 46

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

A and B are two steel wires and the radius of A is twice that of B, if they are stretched by the same load, then the stress on B is ...........

A
Four times that of A.

B
Two times that of A.

C
Three times that of A.

D
Same as that A.

Solution

$$\dfrac{Stress\, on\, B}{Stress\, on\, A}=\dfrac{r_A^2}{r_B^2}$$
$$\Rightarrow\text {Stress on B}/\text {Stress on A}=\dfrac{(2r_B)^2}{r_B^2}$$
$$\therefore$$ $$\dfrac{Stress\, on\, B}{Stress\, on\, A}=4$$
Stress on B$$= 4\times $$ Stress of A
Question 2
1 / -0

If longitudinal strain for a wire is $$0.03$$ and its poisson ratio is $$0.5$$, then its lateral strain is

A
$$0.003$$

B
$$0.0075$$

C
$$0.015$$

D
$$0.4$$

Solution

Poisson's ratio is defined as the ratio of lateral strain to the longitudinal strain.
$$\nu = \cfrac{\text{Lateral Strain}}{\text{Longitudinal Strain}}$$
$$\text{Lateral Strain} = 0.015$$
Question 3
1 / -0

When the tension in a metal wire is $$ T_1$$ its length is $$ l_1 , $$ when the tension is $$ T_2$$ its length is $$ l_2. $$ The natural length of wire is :

A
$$ T_1l_1 + T_2 l_2 $$

B
$$ \dfrac {l_1 T_2 - l_2 T_1}{T_2 - T_1}$$

C
$$ \dfrac {l_1 T_2 - l_2 T_1}{T_2 + T_1}$$

D
$$ \dfrac {T_2} {T_1} ( l_1 + l_2 ) $$

Solution

$$ y = \dfrac {Fl}{ A \triangle l } $$
y is constant
$$ \therefore \triangle l \, \propto \, F $$
$$ l_t - l \,\propto \,T_1 $$ and $$ l_2 - l\, \propto \,T_2 $$
$$ \therefore \dfrac {l_1 - l}{l_2 - l} = \dfrac {T_1}{T_2} $$

$$ l_1 T_2 - l T_2 = l_2 T_1 - lT_1 $$

$$ l (T_1 - T_2) = l_2 T_1 - l_1 T_2 $$

$$ l = \dfrac {l_2 T_1 - L_1 T_2 }{T_1 - T_2} \Rightarrow = \dfrac { l_1 T_2 - l_2 T_1 }{T_2 - t_1} $$
Question 4
1 / -0

The bulk modulus of an ideal gas at constant temperature is :

A
Equal to its pressure

B
Equal to its volume

C
Equal to $${ p }/{ 2 }$$

D
Cannot be determined
Question 5
1 / -0

The area of cross-section of a steel wire $$\left( Y=2.0\times { 10 }^{ 11 }{ N }/{ { m }^{ 2 } } \right)$$ 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

Young's modulus of elasticity
$$Y=\dfrac { FL }{ Al }$$
To double the length $$l = L$$
$$\therefore Y=\dfrac { F }{ A } \Rightarrow F=YA$$
$$=2\times { 10 }^{ 11 }\times 0.1\times { 10 }^{ -4 }=2\times { 10 }^{ 6 }N$$
Question 6
1 / -0

Two wires of same material and same diameter have lengths in the ratio 2 : 5. They are stretched by the same force. The ratio of work done in stretching them is :

A
5 : 2

B
2 : 5

C
1 : 3

D
3 : 1

Solution

Hint: Use the formula of work done $$W= \dfrac {1}{2} Kx^2$$
Explanation:
Step 1: Concept used:
Work done is given by-

$$Work\ Done \ = \frac { 1 }{ 2 } \ \times \ { Force }\ \times \ { Elongation } $$ ........... (1)

By definition of Young's modulus:

$$Y= \dfrac {Stress}{Strain}= \dfrac {F/A}{ \Delta l/L}= \dfrac {FL}{A \Delta l}$$
So, elongation $$ \Delta l= \dfrac {Fl}{AY}$$ ......... (2)
Putting the value from eq.(2) in eq.(1) and we get-
$$Work\ Done\ =\ \dfrac { 1 }{ 2 } \ \times \ { F }\ \times \ \dfrac { Fl }{ AY } \ =\ \dfrac { { F }^{ 2 }L }{ 2AY }$$

Since Force $$(F)$$, material $$(Y)$$ and area $$(A)$$ are same

$$\\ \therefore \ Work\ Done\ \propto \ length\ $$

Step 2: Finding the ratio:

$$\\ \dfrac { { W }_{ 1 } }{ { W }_{ 2 } } \ =\ \dfrac { { l }_{ 1 } }{ { l }_{ 2 } } \ =\ \dfrac { 2 }{ 5 } $$

Hence, the answer is option B
Question 7
1 / -0

A liquid of bulk modulus $$k$$ is compressed by applying an external pressure such that its density increases by $$0.01$$%. The pressure applied on the liquid is:

A
$$\cfrac { k }{ 10000 } $$

B
$$\cfrac { k }{ 1000 } $$

C
$$1000k$$

D
$$0.01k$$

Solution

Bulk modulus is given by:
$$k=-\cfrac { P }{ \cfrac { \Delta V }{ V } } \Rightarrow \cfrac { P }{ k } =-\cfrac { \Delta V }{ V } =\cfrac { \Delta \rho }{ \rho } $$
$$P=k\cfrac { \Delta \rho }{ \rho } =\cfrac { k }{ 10000 } $$
Question 8
1 / -0

One end of a slack wire (Youngs modulus $$Y$$, length $$L$$ and cross -sectional area A) is clamped to a rigid wall and the other end to a block (mass m) which rests on a smooth horizontal plane. The block is set in motion with a speed v. What is the maximum distance the block will travel after the wire becomes taut?

A
$$v\sqrt {\dfrac {mL}{AY}}$$

B
$$v\sqrt {\dfrac {2mL}{AY}}$$

C
$$v\sqrt {\dfrac {mL}{2AY}}$$

D
$$L\sqrt {\dfrac {mv}{AY}}$$

Solution

The strain energy of a elastic wire is given by $$\dfrac{1}{2}VY\varepsilon^2$$
where V is the volume of wire and $$\varepsilon$$ is the strain.
Let the maximum elongation be x. At this point block will come to rest and all the kE energy will be converted to strain energy .
$$\varepsilon=\dfrac{x}{L}$$
$$V=AL$$
Now we know
$$\dfrac{1}{2}ALY{(\dfrac{x}{L})}^2=\dfrac{1}{2}mv^2$$
$$x=v\sqrt{\dfrac{mL}{AY}}$$
Question 9
1 / -0

The relation between Young's modulus $$Y$$, bulk modulus $$K$$ and modulus of elasticity $$\sigma$$ is

A
$$\cfrac { 1 }{ y } =\cfrac { 1 }{ k } +\cfrac { 3 }{ \eta } $$

B
$$\cfrac { 3 }{ y } =\cfrac { 1 }{ \eta } +\cfrac { 1 }{ 3k } $$

C
$$\cfrac { 1 }{ y } =\cfrac { 3 }{ \eta } +\cfrac { 1 }{ 3k } $$

D
$$\cfrac { 1 }{ \eta } =\cfrac { 3 }{ y } +\cfrac { 1 }{ 3k } $$

Solution

We know that,
$$Y=3K(1-2\sigma)$$
or, $$\sigma=\dfrac{1}{2}(1-\dfrac{Y}{2K})$$ ----------(i)
Also, $$Y=2\eta(1+\sigma)$$
or, $$\sigma=\dfrac{Y}{2\eta}-1$$ ----------(ii)
Now, From (i) and (ii),
$$\dfrac{1}{2}(1-\dfrac{Y}{3K})=\dfrac{Y}{2\eta}-1$$
or, $$1-\dfrac{Y}{3K}=\dfrac{Y}{\eta}-2$$
or, $$\dfrac{3}{Y}=\dfrac{1}{\eta}+\dfrac{1}{3K}$$
Question 10
1 / -0

A metallic rod of length $$l$$ and cross sectional area $$A$$ is made of a material of Young's modulus $$Y$$. If the rod is elongated by an amount $$y$$, then the work done is proportional to

A
$$y$$

B
$$1/y$$

C
$${ y }^{ 2 }$$

D
$$1/{ y }^{ 2 }$$

Solution

strain=ElongationOriginallength=yl
Young's modulus Y=StressStrain
work done, W=12×stress×strain×volume
W=12×Y×(Strain)2×Al=12(YAl)y2
W∝y2

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Mechanical Properties of Solids Test - 46

Result

Mechanical Properties of Solids Test - 46

Score

-

Rank

-

SHARING IS CARING

Question 1

Four times that of A.

Two times that of A.

Three times that of A.

Same as that A.

Solution

Question 2

$$0.003$$

$$0.0075$$

$$0.015$$

$$0.4$$

Solution

Question 3

$$ T_1l_1 + T_2 l_2 $$

$$ \dfrac {l_1 T_2 - l_2 T_1}{T_2 - T_1}$$

$$ \dfrac {l_1 T_2 - l_2 T_1}{T_2 + T_1}$$

$$ \dfrac {T_2} {T_1} ( l_1 + l_2 ) $$

Solution

Question 4

Equal to its pressure

Equal to its volume

Equal to $${ p }/{ 2 }$$

Cannot be determined

Question 5

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

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

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

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

Solution

Question 6

5 : 2

2 : 5

1 : 3

3 : 1

Solution

Question 7

$$\cfrac { k }{ 10000 } $$

$$\cfrac { k }{ 1000 } $$

$$1000k$$

$$0.01k$$

Solution

Question 8

$$v\sqrt {\dfrac {mL}{AY}}$$

$$v\sqrt {\dfrac {2mL}{AY}}$$

$$v\sqrt {\dfrac {mL}{2AY}}$$

$$L\sqrt {\dfrac {mv}{AY}}$$

Solution

Question 9

$$\cfrac { 1 }{ y } =\cfrac { 1 }{ k } +\cfrac { 3 }{ \eta } $$

$$\cfrac { 3 }{ y } =\cfrac { 1 }{ \eta } +\cfrac { 1 }{ 3k } $$

$$\cfrac { 1 }{ y } =\cfrac { 3 }{ \eta } +\cfrac { 1 }{ 3k } $$

$$\cfrac { 1 }{ \eta } =\cfrac { 3 }{ y } +\cfrac { 1 }{ 3k } $$

Solution

Question 10

$$y$$

$$1/y$$

$${ y }^{ 2 }$$

$$1/{ y }^{ 2 }$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.