Mechanical Properties of Solids Test

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Mechanical Properties of Solids Test - 58

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Question 1
1 / -0

The length of a rubber cord is $$l_1$$ metres when the tension in it is $$4N$$ and $$l_2$$ metres when the tension is $$5N$$. then the length in meters when the tension is $$9N$$ is

A
$$3l_2 +4l_1$$

B
$$3l_2 +2l_1$$

C
$$5l_2 - 4l_1$$

D
$$3l_2 - 2l_1$$

Solution

Let, $${ x }_{ 1, }\quad { x }_{ 2 },\quad and\quad { x }_{ 3 }\quad be\quad the\quad extension\quad with\quad 4N,\quad 5N\quad and\quad 9N\quad tension\quad respectively.\\ 4=\quad k{ x }_{ 1, },\quad 5=k{ x }_{ 2 }\quad and\quad 9=k{ x }_{ 3 }\quad \\ $$
Let l be the natural length of the rubber chord.
Therefore, we can say,
$${ x }_{ 1 }+l=\quad { l }_{ 1 }-------(a)\\ { x }_{ 2 }+l=\quad { l }_{ 2 }--------(b)\\ { x }_{ 3 }+l=\quad { l }_{ 3 }--------(c)\\ \dfrac { 4 }{ 5 } =\dfrac { { x }_{ 1 } }{ { x }_{ 2 } } =\dfrac { { l }_{ 1 }-l }{ { l }_{ 2 }-l } \\ \quad \quad \quad =>l=5{ l }_{ 1 }-{ 4 }l_{ 2 }-------(1)\\ \\ \dfrac { 9 }{ 5 } =\dfrac { { x }_{ 3 } }{ { x }_{ 2 } } =\dfrac { { l }_{ 3 }-l }{ { l }_{ 2 }-l } \\ \qquad =>9{ l }_{ 2 }-9l=\quad 5{ l }_{ 3 }-5l\\ \qquad =>4l=9{ l }_{ 2 }-5{ l }_{ 3 }\\ From\quad (1)\quad =>\quad 4(5{ l }_{ 1 }-4{ l }_{ 2 })=9{ l }_{ 2 }-5{ l }_{ 3 }\\ \qquad \qquad \qquad =>20{ l }_{ 1 }=25{ l }_{ 2 }-5{ l }_{ 3 }------(2)\\ $$
$$From\quad euation\quad (2,)\quad { l }_{ 3 }=5{ l }_{ 2 }-4{ l }_{ 1 }$$
Therefore at 9N tension, extension is ($$5{ l }_{ 2 }-4{ l }_{ 1 }$$)m
Question 2
1 / -0

A sphere contracts in volume by 0.01% when taken to the bottom of sea 1 km deep. Find bulk modulus of the material of sphere

A
$$ 9.8 \times 10^6 M/M^2 $$

B
$$ 1.2 \times 10^{10} N/M^2 $$

C
$$ 9.8 \times 10^{10} N/M^2 $$

D
$$ 9.8 \times 10^{11} N/M^2 $$

Solution

Given that,

Sphere contracts in volume $$0.01$$%

Change in Depth = $$1$$ km

Acceleration due to gravity = $$9.8$$ m/s²

Density of water = $$1000$$ kg/m³

Now, the volume of sphere is $$V$$

So, $$dV= 0.01$$% of $$V$$

Now,

$$ dV=\dfrac{0.01V}{100} $$

$$ \dfrac{dV}{V}=\dfrac{0.01}{100} $$

$$ \dfrac{dV}{V}={{10}^{-4}} $$

Now, the change in pressure

$$ dP=\rho g\left( dh \right) $$

$$ dP={{10}^{3}}\times 9.8\times {{10}^{3}} $$

$$ dP=9.8\times {{10}^{6}}\,N/{{m}^{2}} $$

Now, the bulk modulus is

$$ K=\dfrac{dP}{\dfrac{dV}{V}} $$

$$ K=\dfrac{9.8\times {{10}^{6}}}{{{10}^{-4}}} $$

$$ K=9.8\times {{10}^{10}}\,N/{{m}^{2}} $$

Hence, the bulk modulus of the material of sphere is $$9.8\times10^{10}N/m^{2}$$
Question 3
1 / -0

The extension of wire by application of load is $$0.3cm$$ The extension in a w wire of same material but of double the length and half the radius of cross section by the same load will be in (cm)

A
$$0.3$$

B
$$0.6$$

C
$$0.2$$

D
$$2.4$$

Solution

given, extension of wire by application of a load is $0.3 \mathrm{~cm}$$
Now, on doutling length and making the radics half of its origenal, we get
$$\begin{aligned}A_{f} &=\frac{A_{i}}{4} \quad L_{f}=2 L \\F &=Y A_{i}\left(\frac{0 \cdot 3}{L}\right) \\F &=Y \frac{A_{i}}{4}\left(\frac{\Delta L}{2 L}\right) \Rightarrow \frac{F}{0.3} \times \frac{\Delta L}{8} \\& \therefore \quad \Delta L=2 \cdot 4 \mathrm{~cm}\end{aligned}$$
Question 4
1 / -0

An aluminium rod has a breaking strain $$0.2\%$$. The minimum cross-sectional area of the rod in $$m^{2}$$ in order to support a load of $$10^{4}N$$ is if (Young's modulus is $$7\times 10^{9}Nm^{-2}$$)

A
$$1.7\times 10^{-4}$$

B
$$1.7\times 10^{-3}$$

C
$$7.1\times 10^{-4}$$

D
$$1.4\times 10^{-4}$$

Solution

We know
Young's modulus $$(\gamma)=\cfrac{stress}{strain}$$
For breaking (maximum) strain $$0.2$$% and load of $${10}^{4}N$$, let the minimum supportive cross-sectional area be $$A$$ then,
$$\gamma =\cfrac{\cfrac{L}{A}}{breaking\quad strain}$$
$$\Rightarrow$$ $$A=\cfrac{L}{r\times B.S}=\cfrac{{10}^{4}}{7\times{10}^{9}\times (\cfrac{2}{1000})}$$
$$A=7.1\times {10}^{-4}{m}^{2}$$
Question 5
1 / -0

Two wires A and B have young's modulii in the ratio $$1:2$$ and ratio of the lengths is $$1:1$$. under the application of same stress the ratio of elongation is

A
$$1:1$$

B
$$1:2$$

C
$$2:1$$

D
$$1:4$$

Solution

Given,
$$Y_A:Y_B=1:2$$
$$l_A:l_B=1:1$$
Stress, $$s=\dfrac{F}{A}$$
Young's modulus,
$$Y=\dfrac{F/A}{\Delta l/l}$$
$$Y\propto \dfrac{l}{\Delta l}$$
$$\Delta l \propto \dfrac{l}{Y}$$
$$\dfrac{\Delta l_A}{\Delta l_B}=\dfrac{l_A}{l_B}.\dfrac{Y_B}{Y_A}$$
$$\dfrac{\Delta l_A}{\Delta _B}=\dfrac{1}{1}.\dfrac{2}{1}=\dfrac{2}{1}$$
$$\Delta l_A:\Delta l_B=2:1$$
The correct option is B.
Question 6
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A steel ring of radius $$r$$ and cross sectional area $$A$$ is fitted on to a wooden disc of radius $$R(R> r)$$. If Young;s modulus be $$Y$$, then the force with which the steel ring is expanded, is

A
$$AY\cfrac{R}{r}$$

B
$$AY(\cfrac{R-r}{r})$$

C
$$\cfrac{Y}{A}\cfrac{(R-r)}{r}$$

D
$$\cfrac{Yr}{AR}$$

Solution

Young modulus $$\begin{array}{l} \left( Y \right) =\frac { { FL } }{ { Ax } } \\ F=\dfrac { { AxY } }{ L } \end{array}$$
But length $$=2\pi r$$ (circumference of ring)
Change in length $$\left( x \right) =2\pi R-2\pi r=2\pi r=2\pi \left( { R-r } \right) $$
Hence force of expansions is
$$F=\dfrac { { YA\times 2\pi \left( { R-r } \right) } }{ { 2\pi r } } =\dfrac { { YA\left( { R-r } \right) } }{ r } $$
Question 7
1 / -0

Volume of a liquid when compressed by additional pressure of $$10^5 N/m^2$$ is $$196$$cc and when compressed by a pressure of $$1.5 \times 10 ^5 N/m^2$$, the volume is $$194cc$$. The bulk modulus of the liquid is:

A
$$10^5N/m^2$$

B
$$1.5 \times 10^6$$

C
$$5 \times10^5N/m^2$$

D
$$5 \times 10^6N/m^2$$

Solution

Bulk modulus is the ratio of the infinitesimal increase of pressure to the relative decrease in volume.
When calculating the pressure changes and volume changes we get the ratio as 1.5 × $$10^6$$ .
So the correct answer is ' 1.5 × $$10^6$$ '.
Question 8
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Consider the bar in Fig. Cross-sectional area $$A_{e} = 1.2 in.^{2}$$, and Young's modulus $$E = 30\times 10^{6} psi$$. If $$q_{1} = 0.02\ in.$$ and $$q_{2} = 0.025\ in.$$, determine the following (by hand calculation).

A
The displacement at point $$P$$

B
The strain $$\epsilon$$ and stress $$\sigma$$

C
The element stiffness matrix, and

D
The strain energy in the element
Question 9
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A uniform rod of length $$L$$ has a mass per unit length $$\lambda$$ and area of cross section $$A$$. If the Young's modulus of the rod is $$Y$$. The elongation in the rod due to its own weight is

A
$$\dfrac{2\lambda gL^2}{AY}$$

B
$$\dfrac{\lambda gL^2}{2AY}$$

C
$$\dfrac{\lambda gL^2}{4 AY}$$

D
$$\dfrac{\lambda gL^2}{6AY}$$

Solution

$$F=\frac { { Mg } }{ 2 } $$ (mass of rod will at $$\frac{L}{2}$$ only
$$\begin{array}{l} \Delta l=\frac { { FL } }{ { AY } } \Rightarrow \Delta l=\frac { { MgL } }{ { 2AY } } =\frac { { Mg{ L^{ 2 } } } }{ { 2AYL } } \\ \Rightarrow \frac { { \lambda g{ L^{ 2 } } } }{ { 2Ay } } \left( { \lambda =\frac { M }{ G } } \right) \end{array}$$
Question 10
1 / -0

A uniform cubical block is subjected to volumetric compression, which decreases its each side of $$2$$%. The Bulk strain produced in it is:

A
$$0.03$$

B
$$0.02$$

C
$$0.06$$

D
$$0.12$$

Solution

$$\begin{array}{l}\text { Given, a cutical block whose sides } \\\text { are reduced by } 2 \% \\\text { New side }=l-\frac{2 l}{100}=\frac{981}{100} \\\text { New volume}=\left(\frac{98 l}{100}\right)^{3} \\\Rightarrow \quad V_{f}=\left(\frac{98}{100}\right)^{3} \mathrm{~V}_{i} \\\end{array}$$
$$\Rightarrow \quad \frac{\Delta V}{V}=\begin{array}{l}\text { BUlR } \\\text { Strain }\end{array}\begin{aligned}=& 1-\left(\frac{98}{100}\right)^{3} =& 0.06 \end{aligned}$$

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Re-Attempt Weekly Quiz Competition

Mechanical Properties of Solids Test - 58

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Mechanical Properties of Solids Test - 58

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SHARING IS CARING

Question 1

$$3l_2 +4l_1$$

$$3l_2 +2l_1$$

$$5l_2 - 4l_1$$

$$3l_2 - 2l_1$$

Solution

Question 2

$$ 9.8 \times 10^6 M/M^2 $$

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

$$ 9.8 \times 10^{10} N/M^2 $$

$$ 9.8 \times 10^{11} N/M^2 $$

Solution

Question 3

$$0.3$$

$$0.6$$

$$0.2$$

$$2.4$$

Solution

Question 4

$$1.7\times 10^{-4}$$

$$1.7\times 10^{-3}$$

$$7.1\times 10^{-4}$$

$$1.4\times 10^{-4}$$

Solution

Question 5

$$1:1$$

$$1:2$$

$$2:1$$

$$1:4$$

Solution

Question 6

$$AY\cfrac{R}{r}$$

$$AY(\cfrac{R-r}{r})$$

$$\cfrac{Y}{A}\cfrac{(R-r)}{r}$$

$$\cfrac{Yr}{AR}$$

Solution

Question 7

$$10^5N/m^2$$

$$1.5 \times 10^6$$

$$5 \times10^5N/m^2$$

$$5 \times 10^6N/m^2$$

Solution

Question 8

The displacement at point $$P$$

The strain $$\epsilon$$ and stress $$\sigma$$

The element stiffness matrix, and

The strain energy in the element

Question 9

$$\dfrac{2\lambda gL^2}{AY}$$

$$\dfrac{\lambda gL^2}{2AY}$$

$$\dfrac{\lambda gL^2}{4 AY}$$

$$\dfrac{\lambda gL^2}{6AY}$$

Solution

Question 10

$$0.03$$

$$0.02$$

$$0.06$$

$$0.12$$

Solution

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