Mechanical Properties of Solids Test

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Mechanical Properties of Solids Test - 67

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Weekly Quiz Competition

Question 1
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The maximum intensity of fringes in Young's experiment is l. If one of the slits is closed , then intensity at that place becomes $${ I }_{ 0 }$$. then relation between I and $${ I }_{ 0 }$$ is

A
$$I={ I }_{ 0 }$$

B
$$I=2{ I }_{ 0 }$$

C
$$I=4{ I }_{ 0 }$$

D
there is no relation between $$I$$ and $$I_0$$

Solution

Suppose slit width's are equal so they produces waves of equal intensity say $$I'$$, Resultant intensity at any point $$I_R =4\ I' \cos^2 \phi$$ where $$\phi$$ is the phase difference between the waves at the point of observation.
For maximum intensity $$\phi =0^o \Rightarrow I_{max} =4\ I' =I......(I)$$
If one of slit is closed. Resultant intensity at the same point will be $$I'$$ only i.e., $$I'=I_0..(ii)$$
Comparing equation (i) and (ii) we get
$$I =4\ I_0$$
Question 2
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A metal wire having Poisson's ratio $$\frac { 1}{ 4 }$$ Young's modulud $$8\times {10 }^{ 10} N/m^2$$ is stretched force, which produces a lateral strain of 0.02 it. The elastic potential energy stored per volume in wire is$$[in J/m^3]$$

A
$$2.56\times 10^4$$

B
$$1.78\times 10^2$$

C
$$3.72\times 10^2$$

D
$$3.18\times 10^5$$

Solution

$$\begin{array}{c}\text { Given poisson ratio }=1 / 4 \text { , young modulus }=8 \times 10^{10} \mathrm{N} /\mathrm{m}^{2}\\\Rightarrow 1 / 4=\dfrac{\text { lateral strain }}{\text { longi} \text {tudinal Strain }\mathrm{}}\end{array}$$

$$\begin{array}{l}\text { given lateral strain }=0.02\%=2\times 10^{-4} \\\Rightarrow \text { longi} \text {tudinal strain }=8 \times 10^{-4}\end{array}$$

$$\begin{aligned}\text {PE density } &=\frac{1}{2} \times Y\times(\operatorname{Strain})^{2} \\&=\frac{1}{2}\times4\times10^{10}\times\left(8\times10^{-4}\right)^{2}\\&=4\times10^{10}\times64\times10^{-8}\\&=256\times10^{2}\\&=2.56 \times 10^{4} \mathrm{~J} / \mathrm{m}^{3}\end{aligned}$$
Question 3
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When the load on a wire is increased from $$3 \mathrm { kg }-wt$$ to $$8\mathrm { kg } -wt$$, the elongation increases from $$0.61 \mathrm { mm }$$to $$1.02 \mathrm { mm } $$ . The required work done during the extension of the wire, is

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

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

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

D
$$31 \times 10 ^ { - 3 } \mathrm { J }$$

Solution

$$ \begin{aligned} &\Delta l_{1}=0.61 \mathrm{~mm} \quad \Delta \mathrm{l}_{2}=1.02 \mathrm{~mm}\\ &\text { (work done) }=\bigcup_{1}-\bigcup_{2} \ldots . .(i)\\ \end{aligned} $$
$$ \begin{aligned} U &=\frac{1}{2} \times \text { stress } x \text { strain } x \text { volume } \\ &=\frac{1}{2} \times \frac{F}{A-1} \times \frac{\Delta L}{-L} \times A \cdot K \\ U &=\frac{1}{2} \times F \times \Delta L \end{aligned} $$
$$ \begin{aligned} U_{1} &=\frac{1}{2} \times 3 \times 10 \times 0.61 \times 10^{-3} \\ &=9.15 \times 10^{-3} \mathrm{~J} \\ &=\text { work done dussing } \\ & \text { case } 1 \end{aligned} $$
$$\begin{aligned} U_{2} &=\frac{1}{2} \times 8 \times 10 \times 1.02 \times 10^{-3} \\ U_{2} &=40.8 \times 10^{-3} \mathrm{~J} \\=& \text { work done during } \\ & \text { case II } \end{aligned}$$
$$ \begin{aligned} (\text { work done) }& U_{2}-U_{1} \text { during case II to case } 1 &=(40.8-9.15) \times 10^{-3} \mathrm{~J} \\ &=31.65 \times 10^{-3} \mathrm{~J} \end{aligned} $$
Question 4
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A copper wire and a steel wire of the same diameter and length 1 m and 2 m respectively are connected end and a force is applied which stretches their combined length by 1 cm. How much each wire is elongated respectively. Y of copper = $$1.2\times { 10 }^{ 10 }$$ $$M{ m }^{ -2 }$$ and Y of steel +$$2.0\times { 10 }^{ 10 }\quad N{ m }^{ -2 }$$

A
0.45 cm, 0.55 cm

B
0.55 cm, 0.45 cm

C
0.045 cm , 0.55 cm

D
0.45 cm, 0.055 cm

Solution

$$ \begin{array}{l} \text { Y of copper }=1.2 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2} \\ \text { Yof steel }=2 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2} \\ \text { L of copper }=\mathrm{tm} \\ \text { L of steel }=2 \mathrm{~m} \\ \text { both have same diameder }=\mathrm{d} \\ \Delta \mathrm{L}_{\text {net }}=1 \mathrm{~cm} \end{array} $$
$$ \text { Let } F \text { force is applied, } $$
$$ \begin{array}{l} \Delta L_{\text {net }}=\Delta L_{1}+\Delta L_{2} \\ \text { we know } \\ Y=\frac{F L}{A \Delta L} \end{array} $$
$$ \Delta L=\frac{F L}{A Y} \quad \Delta L \alpha \frac{L}{Y} \quad \begin{array}{c} (\text { since } F \text { and } A \text { are same } \\ \text { For both }) \end{array} $$
$$ \begin{array}{l} \text { For steel , For copper } \\ \Delta L_{1}=\frac{x L_{1}}{Y_{1}} & \Delta L_{2}=x \frac{L_{2}}{Y_{2}} \end{array} $$
$$ \begin{aligned} \operatorname{1x} 10^{-2} &=\frac{2 \times x}{2 \times 10^{10}}+\frac{x \times 1}{1.2 \times 10^{10}} \\ x &=\frac{6}{11} \times 10^{8} \end{aligned} $$
$$ \begin{aligned} \Delta L_{n e t}=& 1 \mathrm{~cm} \\ &\left\{\begin{array}{c} \{x \text { is proportionality } \\ \text { constant }) \end{array}\right.\\ \therefore \quad \Delta L_{1}=\frac{6}{11} \times \frac{2}{2 \times 10^{10} \times 10^{8}}=0.055 \mathrm{~cm} \\ \Delta L_{2}=\frac{6}{11} \times 10^{8} \times \frac{1}{1.2 \times 10^{10}}=0.045 \mathrm{~cm} \end{aligned} $$
Question 5
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Directions For Questions

In a modified Youngs double slit experimenter, there are three identical parallel slits $$S_{1}, S_{2}$$ and $$ S_{3}$$. A cogerent monochromatic beam of wavelight $$700\ nm$$, having plane wavefronts, falls on the slits. The intensity of the central point $$O$$ on the screen is found to be $$I_{0}W/m^{2}$$. The distance $$S_{1} S_{2}= S_{2} S_{3}=0$$. And screen is $$30\ cm$$ from slits.

...view full instructions

Find the intensity on the screen at $$O$$ if only $$S_{3}$$ is covered

A
$$\dfrac {\sqrt {3}I_{o}}{7}$$

B
$$\dfrac {\sqrt {3}I_{o}}{\sqrt {7}}$$

C
$$\dfrac {3I_{2}}{7}$$

D
$$\dfrac {I_{o}}{2}$$
Question 6
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A 2.0 m long steel cable has cross-sectional area of $$ 0.30 cm^2 $$. A 550 kg load is now hung from the cable. Assume that the cable behaves like a solid rod with an uniform cross-sectional area. Then the stress of the cable is:

A
$$1.8 \times 10^8 Pa$$

B
$$2.8 \times 10^8 Pa$$

C
$$18 \times 10^8 Pa$$

D
$$28 \times 10^8 Pa$$
Question 7
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Two wires of length $$l$$, radius $$r$$ and length $$2l$$, radius $$2r$$ having same Young's modulus $$Y$$ are hung with weight $$mg$$. Net elongation is

A
$$3\ mgl/2\pi r^{2}Y$$

B
$$1\ mgl/2\pi r^{2}Y$$

C
$$2\ mgl/3\pi r^{2}Y$$

D
$$3\ mgl/3\pi r^{2}Y$$

Solution

$$\begin{aligned}\text { Net elongation } &=\Delta L_{1}+\Delta L_{2} \\\Delta L_{1} &=\text { elongation is } 1^{\text {st }} \text { wire } \\\Delta L_{2} &=\text { elongation is } 2^{\text {nd }} \text { wire }\end{aligned}$$
$$\begin{array}{l}\text { For } 1^{\text {st }}\text { wire, }\\\qquad \begin{array}{l}Y=\frac{M g L}{\pi r^{2} \Delta L_{1}}\\\Delta L_{1}=\frac{M g L}{\pi r^{2} Y}\end{array}\end{array}$$
$$\begin{array}{l}\text { For } 2^{\text {nd }}\text { wire } \\Y=\frac{\operatorname{mg} 2 L}{\pi(2 r)^{2} \Delta L_{2}} \\\Delta L_{2}=\frac{2 m g L}{4 \pi r^{2} Y}\end{array}$$
$$\begin{array}{c}\Delta L_{\text {net }}=\frac{m g L}{\pi r^{2} Y}+\frac{m g L}{2 \pi r^{2} Y} \\\Delta L_{\text {net }}=\frac{3}{2}\frac{m g L}{\pi r^{2} Y}\end{array}$$
Hence option A is correct
Question 8
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The time after which the block reaches the position where is laming maximum elongation is

A
$$2\pi \sqrt { \dfrac { m }{ k } } $$

B
$$\pi \sqrt { \dfrac { m }{ k } } $$

C
$$\frac { \pi }{ 2 } \sqrt { \dfrac { m }{ k } } $$

D
$$\pi \sqrt { \dfrac { m }{ 2k } } $$
Question 9
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Directions For Questions

In a modified Youngs double slit experimenter, there are three identical parallel slits $$S_{1}, S_{2}$$ and $$ S_{3}$$. A cogerent monochromatic beam of wavelight $$700\ nm$$, having plane wavefronts, falls on the slits. The intensity of the central point $$O$$ on the screen is found to be $$I_{0}W/m^{2}$$. The distance $$S_{1} S_{2}= S_{2} S_{3}=0$$. And screen is $$30\ cm$$ from slits.

...view full instructions

All three sliits are now uncovered and a transparent plate of thickness $$1.4\ \mu$$ and refreactive index $$1.25$$ is placecd in front of $$S_{2}$$. Rresultant intensity at point $$O$$ is

A
$$3I_{0}/7$$

B
$$4I_{0}/7$$

C
$$5I_{0}/7$$

D
$$6I_{0}/7$$
Question 10
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Directions For Questions

In a modified Youngs double slit experimenter, there are three identical parallel slits $$S_{1}, S_{2}$$ and $$ S_{3}$$. A cogerent monochromatic beam of wavelight $$700\ nm$$, having plane wavefronts, falls on the slits. The intensity of the central point $$O$$ on the screen is found to be $$I_{0}W/m^{2}$$. The distance $$S_{1} S_{2}= S_{2} S_{3}=0. And screen is $$30\ cm$$ from slits.

...view full instructions

Find the intensity on the screen at $$O$$ if $$S_{1}$$ and $$S_{3}$$ are covered.

A
$$\dfrac {I_{0}}{\sqrt {7}}$$

B
$$\dfrac {I_{0}}{{7}}$$

C
$$\dfrac {I_{0}}{\sqrt {6}}$$

D
$$\dfrac {I_{0}}{{6}}$$

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Re-Attempt Weekly Quiz Competition

Mechanical Properties of Solids Test - 67

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Mechanical Properties of Solids Test - 67

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SHARING IS CARING

Question 1

$$I={ I }_{ 0 }$$

$$I=2{ I }_{ 0 }$$

$$I=4{ I }_{ 0 }$$

there is no relation between $$I$$ and $$I_0$$

Solution

Question 2

$$2.56\times 10^4$$

$$1.78\times 10^2$$

$$3.72\times 10^2$$

$$3.18\times 10^5$$

Solution

Question 3

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

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

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

$$31 \times 10 ^ { - 3 } \mathrm { J }$$

Solution

Question 4

0.45 cm, 0.55 cm

0.55 cm, 0.45 cm

0.045 cm , 0.55 cm

0.45 cm, 0.055 cm

Solution

Question 5

$$\dfrac {\sqrt {3}I_{o}}{7}$$

$$\dfrac {\sqrt {3}I_{o}}{\sqrt {7}}$$

$$\dfrac {3I_{2}}{7}$$

$$\dfrac {I_{o}}{2}$$

Question 6

$$1.8 \times 10^8 Pa$$

$$2.8 \times 10^8 Pa$$

$$18 \times 10^8 Pa$$

$$28 \times 10^8 Pa$$

Question 7

$$3\ mgl/2\pi r^{2}Y$$

$$1\ mgl/2\pi r^{2}Y$$

$$2\ mgl/3\pi r^{2}Y$$

$$3\ mgl/3\pi r^{2}Y$$

Solution

Question 8

$$2\pi \sqrt { \dfrac { m }{ k } } $$

$$\pi \sqrt { \dfrac { m }{ k } } $$

$$\frac { \pi }{ 2 } \sqrt { \dfrac { m }{ k } } $$

$$\pi \sqrt { \dfrac { m }{ 2k } } $$

Question 9

$$3I_{0}/7$$

$$4I_{0}/7$$

$$5I_{0}/7$$

$$6I_{0}/7$$

Question 10

$$\dfrac {I_{0}}{\sqrt {7}}$$

$$\dfrac {I_{0}}{{7}}$$

$$\dfrac {I_{0}}{\sqrt {6}}$$

$$\dfrac {I_{0}}{{6}}$$

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