Mechanical Properties of Solids Test

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Mechanical Properties of Solids Test - 50

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Weekly Quiz Competition

Question 1
1 / -0

The following four wires of length $$L$$ and radius $$r$$ are made of the same material. Which of these will have the largest extension, when the same tension is applied?

A
$$L=100cm, r=0.2mm$$

B
$$L=200cm, r=0.4mm$$

C
$$L=300cm, r=0.6mm$$

D
$$L=400cm, r=0.8mm$$

Solution

Young's modulus $$Y=\cfrac { F }{ A } \cfrac { L }{ \Delta L } =\cfrac { T }{ \pi { r }^{ 2 } } \cfrac { L }{ \Delta L } $$
$$\Delta L=\cfrac { T }{ \pi { r }^{ 2 } } \cfrac { L }{ Y } $$\
where the symbols have their ususal meanings
As the four wires are made of the same material, therefore Young's modulus is the same for four wires.
As the $$T$$ and $$Y$$ are the same for the four wires.
$$\therefore \Delta L\propto \cfrac { L }{ { r }^{ 2 } } $$
$$\cfrac { L }{ { r }^{ 2 } } $$
is maximum for wire length $$L=100cm$$ and radius $$r=0.2mm$$
Question 2
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Two wires of the same material and length but diameter in the ratio $$1:2$$ are stretched by the same load. The ratio of elastic potential energy per unit volume for the two wires is:

A
$$1:1$$

B
$$2:1$$

C
$$4:1$$

D
$$16:1$$

Solution

Hint: Elastic energy per unit volume: $$\quad \mathrm{E}=\dfrac{1}{2} \times$$ Stress $$\times$$ Strain

Correct Answer = Option (D)
Step 1: Using Hooke's law: $$\quad$$ Strain $$=\dfrac{\text { Stress }}{\mathrm{Y}}$$

$$\Longrightarrow \quad \mathrm{E}=\dfrac{(\text { Stress })^{2}}{2 \mathrm{Y}}=\dfrac{\mathrm{F}^{2}}{2 \mathrm{YA}^{2}} \quad$$ where $$\mathrm{A}=\dfrac{\pi \mathrm{D}^{2}}{4}$$

$$\Longrightarrow \quad \mathrm{E}=\dfrac{8 \mathrm{~F}^{2}}{\pi^{2} \mathrm{D}^{4} \mathrm{Y}}$$

$$\therefore \dfrac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\dfrac{\mathrm{D}_{2}}{\mathrm{D}_{1}}\right)^{4}$$

$$(\because \mathrm{F}$$ and $$\mathrm{Y}$$ are same for both)

Given : $$\dfrac{\mathrm{D}_{1}}{\mathrm{D}_{2}}=\dfrac{1}{2}$$
on substituing the value of $$\dfrac{\mathrm{D}_{1}}{\mathrm{D}_{2}}=\dfrac{1}{2}$$

$$\dfrac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\dfrac{2}{1}\right)^{4}=\dfrac{16}{1}$$
Question 3
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Among solids, liquids and gases, which posses the greatest bulk modulus?

A
Solids

B
Liquids

C
Gases

D
Both solids and liquids

Solution

Bulk modulus is defining the ability of a material to resist deformation in terms of volume change, under pressure.
Bulk modulus $$=\cfrac { P }{ \Delta V/V } $$
As solids have long-range order by its nature. Because the atoms of a crystal are arranged in an ordered pattern that stretches all the way across its material because of which if the same amount of pressure will be applied on solids, liquid, and gases then minimum fractional change in volume will be for solid.
Thus solids are having the greatest Bulk Modulus.
Option A is correct.
Question 4
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A wire stretches by a certain amount under a load. If the load and radius both are increased to four times. The stretch caused in the wire is then

A
$$l$$

B
$$\cfrac { l }{ 2 } $$

C
$$\cfrac { l }{ 3 } $$

D
$$\cfrac { l }{ 4 } $$

Solution

Young's modulus, $$Y=\cfrac { W }{ A } \times \cfrac { L }{ l } $$
$$\therefore$$ Elongation, $$l=\cfrac { WL }{ AY } =\cfrac { WL }{ \pi { r }^{ 2 }Y } $$
When both load and radius are increased to four times, the elongation becomes
$$l'=\cfrac { 4W\times L }{ \pi { (4r) }^{ 2 }Y } =\cfrac { WL }{ 4\pi { r }^{ 2 }Y } =\cfrac { l }{ 4 } $$
Question 5
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When the load on a wire is increased from $$3kg$$ $$wt$$ to $$5kg$$ $$wt$$ the elongation increases from $$0.61mm$$ to $$1.02mm$$. The required work done during the extension of the wire is:

A
$$16\times { 10 }^{ -3 }J$$

B
$$8\times { 10 }^{ -2 }J$$

C
$$20\times { 10 }^{ -2 }J\quad $$

D
$$11\times { 10 }^{ -3 }J$$

Solution

Given:
Apply Load: $$M_1=3kg \ \ \ \ \Delta L= 0.61mm \\ M_2=5kg \ \ \ \Delta L=1.02mm$$

Work done is stretching the wire through $$0.61mm$$ under the load of $$3kg$$ $$wt$$
$${ W }_{ 1 }=\cfrac { 1 }{ 2 } stretching\quad force\times extension=\cfrac { 1 }{ 2 } \times 3\times 9.8\times 0.61\times { 10 }^{ -3 }=8.967\times { 10 }^{ -3 }J$$
Work done in stretching the wire through $$1.02mm$$ under the load of $$5kg$$ $$wt$$
$${ W }_{ 2 }=\cfrac { 1 }{ 2 } \times 5\times 9.8\times 1.02\times { 10 }^{ -3 }=24.99\times { 10 }^{ -3 }J$$
Hence work done in stretching the wire from $$0.61mm$$ to $$1.02mm$$
$$\Delta W={ W }_{ 2 }-{ W }_{ 1 }=\left( 24.99-8.961 \right) \times { 10 }^{ -3 }\simeq 16\times { 10 }^{ -3 }J$$
Question 6
1 / -0

The length of a rubber cord is $${l}_{1}$$ when the tension is $$4N$$ and $${l}_{2}m$$ when the tension is $$6N$$. The length when the tension is $$9N$$, is:

A
$$\left( 2.5{ l }_{ 2 }-1.5{ l }_{ 1 } \right) m$$

B
$$\left( 6{ l }_{ 2 }-1.5{ l }_{ 1 } \right) m$$

C
$$\left( 3{ l }_{ 2 }-2{ l }_{ 1 } \right) m$$

D
$$\left( 3.5{ l }_{ 2 }-2.5{ l }_{ 1 } \right) m$$

Solution

Let the original unstretched length be $$l$$.
$$\therefore Y=\cfrac { T }{ A } \cfrac { l }{ \Delta l } $$
Now $$Y=\cfrac { 4 }{ A } \cfrac { l }{ \left( { l }_{ 1 }-l \right) } =\cfrac { 6 }{ A } \cfrac { l }{ \left( { l }_{ 2 }-l \right) } =\cfrac { 9 }{ A } \cfrac { l }{ \left( { l }_{ 3 }-l \right) } $$

$$\quad \therefore 4\left( { l }_{ 3 }-l \right) =9\left( { l }_{ 1 }-l \right) \quad or\quad 4{ l }_{ 3 }+5l=9{ l }_{ 1 }...(i)\quad $$

Again $$6\left( { l }_{ 3 }-l \right) =9\left( { l }_{ 2 }-l \right) \quad or\quad 2{ l }_{ 3 }+l=3{ l }_{ 2 }...(ii)$$

To obtain $${l}_{3}$$, solve (i) and (ii) simulatneously
$$\therefore { l }_{ 3 }=\left( 2.5{ l }_{ 2 }-1.5{ l }_{ 1 } \right) m$$
Question 7
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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. When stretched by a load, the net elongation is found to be $$0.7mm$$. The load applied is
$$\left( { Y }_{ copper }=1.2\times { 10 }^{ 11 }N\quad { m }^{ -2 },{ Y }_{ steel }=2\times { 10 }^{ 11 }N\quad { m }^{ -2 } \right) $$

A
$$1.2\times { 10 }^{ 2 }N$$

B
$$1.8\times { 10 }^{ 2 }N$$

C
$$2.4\times { 10 }^{ 2 }N$$

D
$$3.2\times { 10 }^{ 2 }N$$

Solution

Given,
$$l_{copper}=2.4m$$
$$l_{steel}=1.6m$$
$$D=3mm$$
$$r=\dfrac{3}{2}=1.5mm$$
$$Y_{copper}=1.2\times 10^{11}N/m^2$$
$$Y_{steel}=2\times 10^{11}N/m^2$$
$$\Delta l=0.7mm$$
Stress is same, $$\sigma_{copper}=\sigma_{steel}$$
$$Y_{copper}\dfrac{l_{copper}}{\Delta l_{copper}}=Y_{steel}\dfrac{l_{steel}}{\Delta l_{steel}}$$

$$1.2\times 10^{11}\times \dfrac{2.4}{\Delta l_{copper}}=2\times 10^{11}\times \dfrac{1.6}{\Delta l_{steel}}$$

$$\dfrac{\Delta l_{copper}}{\Delta l_{steel}}=2.5$$

$$\Delta l_{copper}=2.5\Delta l_{steel}$$. . . . . . . (1)

$$\Delta l=\Delta l_{copper}+\Delta l_{steel}$$

$$0.7\times 10^{-3}=2.5\Delta l_{steel}+\Delta l_{steel}$$

$$\Delta l_{steel}=\dfrac{0.7\times 10^{-3}}{3.5}=0.2mm$$

$$\Delta l_{copper}=2.5\times 0.2mm=0.5mm$$

$$Y_{copper}=\dfrac{F/A}{\dfrac{\Delta l_{copper}}{l_{copper}}}$$

$$\dfrac{F}{A}=Y_{copper}\dfrac{\Delta l_{copper}}{l_{copper}}$$

$$\dfrac{F}{A}=1.2\times 10^{11}\times \dfrac{0.5\times 10^{-3}}{2.4}$$

$$F=\pi r^2\times 0.25\times 10^{8}$$

$$F=3.14\times 1.5\times 1.5\times 10^{-6}\times 0.25\times 10^8$$

$$F=1.8\times 10^2N$$
The correct option is B.
Question 8
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To what depth must a rubber ball be taken in deep sea so that its volume is decreased by $$0.1$$%
(Take density of sea water $${ 10 }^{ 3 }kg\quad { m }^{ -3 }$$, bulk modulus of rubber $$=9\times { 10 }^{ 8 }N{ m }^{ -2 },g=10m{ s }^{ -2 }$$)

A
$$9m$$

B
$$18m$$

C
$$90m$$

D
$$180m$$

Solution

Let $$h$$ be the depth at which the rubber ball be taken, Then
$$P=h\rho g...(i)$$
By definition of bulk modulus
$$B=-\cfrac { P }{ \Delta V/V } $$
The negative sign shows that with increase in pressure, a decrease in volume occurs.
$$\quad \therefore P=B\cfrac { \Delta V }{ V } $$
Using (i)
$$h\rho g=B\cfrac { \Delta V }{ V } \quad or\quad h=\cfrac { B }{ \rho g } \cfrac { \Delta V }{ V } $$
Substituting the given values, we get
$$h=\left( \cfrac { 9\times { 10 }^{ 8 }N\quad { m }^{ -2 } }{ { 10 }^{ 3 }kg\quad { m }^{ -3 }\times { 10 }^{ }m{ s }^{ -2 } } \right) \left( \cfrac { 0.1 }{ 100 } \right) =90m$$
Question 9
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A $$15kg$$ mass fastened to the end of a steel wire of unstretched length $$1.0m$$ is whirled in a vertical circle with an angular velocity of $$2rev$$ $${s}^{-1}$$ at the bottom of the circle. The cross-section of the wire is $$0.05{cm}^{2}$$. The elongation of the wire when the mass is at the lowest point of its path is
(Take $$g=10m{ s }^{ -2 },{ Y }_{ steel }=2\times { 10 }^{ 11 }N\quad { m }^{ -2 }\quad $$)

A
$$0.52mm$$

B
$$1.52mm$$

C
$$2.52mm$$

D
$$3.52mm$$

Solution

Mass $$(m) = 15\ kg$$
Length of wire $$( L) =1\ m$$
Area of cross section of wire $$(A) = 0.05 cm² = 5 × 10^{-6}\ m^2$$
Angular frequency $$(\nu) = 2\ rev/s$$
So, Angular velocity, $$(\omega) = 2π\nu= 2π(2) = 4π\ rad/s$$

Young's modulus for steel $$(Y) = 2 × 10^{11}\ N/m^2$$
At the lowest point of the vertical circle,
$$T - mg = m\omega^2\ L$$
$$T = mg + m\omega^2 L$$
$$\ \ \ = 15[{ 9.8 + (4π)²×1 }]$$
$$\ \ \ = 15( { 9.8 + 16π²})$$
$$\ \ \ = 15 × 167.72\ N$$
$$\ \ \ = 2515.8\ N$$

Now,
Youngs modulus $$=\dfrac{\text{stress}}{\text{strain}}$$

$$Y= \dfrac{TL}{A∆L}$$

$$∆L =\dfrac{FL}{AY}$$

$$\ \ \ =\dfrac{ 2515.8 × 1}{(5×10^{-6})×(2×10¹¹)}$$
$$\ \ \ = 2.52 × 10^-{3}\ m$$
$$\Delta L= 2.52\ mm$$

Hence, option $$(C)$$ is correct.
Question 10
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A wire of length $$L$$ has a linear mass density $$\mu$$ and area of cross-section $$A$$ and Young's modulus $$Y$$ is suspended vertically from a rigid support. The extension produced in the wire due to its own weight is:

A
$$\cfrac { \mu g{ L }^{ 2 } }{ YA } $$

B
$$\cfrac { \mu g{ L }^{ 2 } }{ 2YA } $$

C
$$\cfrac { 2\mu g{ L }^{ 2 } }{ YA } $$

D
$$\cfrac { 2\mu g{ L }^{ 2 } }{ 3YA } $$

Solution

Consider a small element of length $$dx$$ at a distance $$x$$ from the free end of wire as shown in the figure. Tension in the wire at distance $$x$$ from the lower end is
$$T(x)=\mu gx....(i)$$
Let $$dl$$ be increase in length of the element. Then
$$T=\cfrac { T(x)/A }{ dl/dx } $$

$$dl=\cfrac { T(x)dx }{ YA } =\cfrac { \mu gxdx }{ YA } $$ [Using (i)]

Total extension produced in the wire is
$$\quad l=\displaystyle \int _{ 0 }^{ L }{ \cfrac { \mu gx }{ YA } } =\cfrac { \mu g }{ YA } { \left[ \cfrac { { x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ L }=\cfrac { \mu g{ L }^{ 2 } }{ 2YA } $$

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Re-Attempt Weekly Quiz Competition

Mechanical Properties of Solids Test - 50

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Mechanical Properties of Solids Test - 50

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SHARING IS CARING

Question 1

$$L=100cm, r=0.2mm$$

$$L=200cm, r=0.4mm$$

$$L=300cm, r=0.6mm$$

$$L=400cm, r=0.8mm$$

Solution

Question 2

$$1:1$$

$$2:1$$

$$4:1$$

$$16:1$$

Solution

Question 3

Solids

Liquids

Gases

Both solids and liquids

Solution

Question 4

$$l$$

$$\cfrac { l }{ 2 } $$

$$\cfrac { l }{ 3 } $$

$$\cfrac { l }{ 4 } $$

Solution

Question 5

$$16\times { 10 }^{ -3 }J$$

$$8\times { 10 }^{ -2 }J$$

$$20\times { 10 }^{ -2 }J\quad $$

$$11\times { 10 }^{ -3 }J$$

Solution

Question 6

$$\left( 2.5{ l }_{ 2 }-1.5{ l }_{ 1 } \right) m$$

$$\left( 6{ l }_{ 2 }-1.5{ l }_{ 1 } \right) m$$

$$\left( 3{ l }_{ 2 }-2{ l }_{ 1 } \right) m$$

$$\left( 3.5{ l }_{ 2 }-2.5{ l }_{ 1 } \right) m$$

Solution

Question 7

$$1.2\times { 10 }^{ 2 }N$$

$$1.8\times { 10 }^{ 2 }N$$

$$2.4\times { 10 }^{ 2 }N$$

$$3.2\times { 10 }^{ 2 }N$$

Solution

Question 8

$$9m$$

$$18m$$

$$90m$$

$$180m$$

Solution

Question 9

$$0.52mm$$

$$1.52mm$$

$$2.52mm$$

$$3.52mm$$

Solution

Question 10

$$\cfrac { \mu g{ L }^{ 2 } }{ YA } $$

$$\cfrac { \mu g{ L }^{ 2 } }{ 2YA } $$

$$\cfrac { 2\mu g{ L }^{ 2 } }{ YA } $$

$$\cfrac { 2\mu g{ L }^{ 2 } }{ 3YA } $$

Solution

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