Mechanical Properties of Solids Test

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Mechanical Properties of Solids Test - 78

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Question 1
1 / -0

If $$\delta$$ is the depression produced in beam of length $$L$$, breath $$b$$ and thickness $$d$$, when a load is placed at the mid point, then

A
$$\delta \propto L^3$$

B
$$\delta \propto \dfrac{1}{b^3}$$

C
$$\delta \propto \dfrac{1}{d}$$

D
All of these

Solution
Question 2
1 / -0

The normal density of mercury is $$p$$ and its bulk modulus is $$B$$. The increase in density of gold when a pressure $$p$$ is applied uniformly on all sides is:

A
$$\dfrac{p^{P}}{B}$$

B
$$\dfrac{2p^{p}}{B}$$

C
$$\dfrac{p^{B}}{B}$$

D
$$\dfrac{B}{p^{p}}$$

Solution
Question 3
1 / -0

A uniform pressure p is exerted on all sides of a solid cube at temperature $${t^0}C$$. By what amount should the temperature of the cube be raised in order to bring its volume back to the value it had before the pressure was applied? The coefficient of volume expansion of the cube y and the bulk modulus is B.

A
$$\frac{p}{{\sqrt {2yB} }}$$

B
$$\frac{p}{{2yB}}$$

C
$$\frac{{2p}}{{yB}}$$

D
$$\frac{p}{{yB}}$$

Solution

We know that
$$B=\cfrac { -\Delta P }{ \cfrac { \Delta V }{ V } } $$
Let the initial volume be $$V$$. Then
$$\Delta V=\cfrac { -P }{ B } V\quad $$
Expansion to get same volume as initial
$$=-\Delta V=\cfrac { P }{ B } V$$
We know that thermal volume expansion
$$\Delta { V }_{ thermal }=(V)(y)(\Delta T)$$
Here $$\Delta T=\Delta { t }^{ o }C;\quad -\Delta V=\Delta { V }_{ thermal }$$
So, $$\left( \cfrac { P }{ B } \right) V=V(y)(\Delta T)\Rightarrow \cfrac { P }{ By } =\Delta t$$
There is an increment of $$\cfrac { P }{ yB } $$ in temperature to get same volume as initial
Question 4
1 / -0

A thick rope of rubber of density $$1.5 \times {10^3}{\text{kg/}}{{\text{m}}^{\text{3}}}$$ and Young's modulus $$5 \times {10^6}{\text{N/}}{{\text{m}}^2}$$ , $$8 m$$ length is hung from the ceiling of a room , the increases in its length due to its own weight is :
$$\left( {g = 10{\text{m/}}{{\text{s}}^{\text{2}}}} \right)$$

A
$$9.6 \times {10^{{\text{ - 2}}}}{\text{m}}$$

B
$$19.2 \times {10^{{\text{ - 7}}}}{\text{m}}$$

C
$$9.6 \times {10^{{\text{ - 7}}}}{\text{m}}$$

D
$$9.6 m$$

Solution

Given that,

Density of rubber$$\rho=1.5\times10^{3}\ kg/m^3$$

Young's modulus $$Y=5\times10^{6}\ Nm^2$$

Length $$= 8 m$$

We need to calculate the increase length
Mg acts at centre of gravity which is at $$\dfrac{L}{2}$$
so we take length of rope$$=\dfrac{L}{2}$$

we know $$Y=\dfrac{stress}{strain}=\dfrac{F/A}{\Delta l/l}=\dfrac{Fl}{\Delta l.A}$$

$$\Rightarrow \Delta l=\dfrac{Fl}{AY}=\dfrac{mg L}{2AY}=\dfrac{Al\rho g l}{2AY}$$

$$\Delta l=\dfrac{\rho gl^2}{2Y}$$

$$=\dfrac{1.5\times 10^3\times 10\times 8^2}{2\times 5\times 10^6}$$

$$\Rightarrow \Delta l=9.6\times 10^{-2}m$$.
So (A) is correct option.
Question 5
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A force $$F$$ is applied along a rod of transverse sectional area $$A$$. The normal stress to a sectior $$PQ$$ inclined $$\theta$$ to transverse section will be maximum for $$\theta$$ (in degree) is

A
$$0$$

B
$$30$$

C
$$45$$

D
$$90$$
Question 6
1 / -0

Assume that a block of very low shear modulus is fixed on an inclined place as shown. Due to elastic forces it will deform. What will be the shape of the block?

A

B

C
none

D
all of these
Question 7
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A steel wire of length $$1.5\ m$$ and area of cross section $$1.5\ mm^{2}$$ is stretched by $$1.5\ cm$$. then the work done per unit volume. $$(Y =2\times 10^{11}Nm^{2})$$.

A
$$1\times 10^{7}J/m^{3}$$

B
$$2\times 10^{7}J/m^{3}$$

C
$$3\times 10^{7}J/m^{3}$$

D
$$4\times 10^{7}J/m^{3}$$

Solution
Question 8
1 / -0

A horizontal rod is supported at both ends and loaded at the middle. It L and Y are length and Young's modulus respectively, then depression at the middle is directly proportional to

A
L

B
$${ L }^{ 2 }$$

C
Y

D
$$\dfrac { 1 }{ Y } $$

Solution

$$\begin{array}{l}\text { Depression } \delta=\frac{W L^{3}}{4 b d^{3} y} \\\therefore\ \delta \alpha {L}{}\end{array}$$
Hence option A is correct
Question 9
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A-3 Two wire of equal length and cross-sectional area suspended as shown in figure. Their Young's modulus are $$ Y_1 and Y_2 $$ respectively. The equivelent Young's modulus will be

A
$$ Y_1 + Y_2 $$

B
$$ \frac { Y_ 1+Y_ 2 }{ 2 } $$

C
$$ \frac { Y_ 1Y_ 2 }{ Y_1 + Y_2 } $$

D
$$ \sqrt { Y_ 1Y_ 2 } $$

Solution

Consider whole combination a single rod
$$\begin{array}{l} Y=\frac { { FL } }{ { 2A\ell } } =\frac { { \left( { { F_{ 1 } }+{ F_{ 2 } } } \right) L } }{ { 2A\ell } } =\frac { { { F_{ 1 } }L } }{ { 1A\ell } } \\ =\frac { { { Y_{ 1 } } } }{ 2 } +\frac { { { Y_{ 2 } } } }{ 2 } =\frac { { { Y_{ 1 } }+{ Y_{ 2 } } } }{ 2 } \end{array}$$
Question 10
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If a rubber ball is taken down to a 100 m deep lake, its volume decreases by 0.1%. If $$g=10\quad m/{ s }^{ 2 }$$ then the bulk modulus of elasticity for rubber, in N/$${ m }^{ 2 }$$, is

A
$${ 10 }^{ 8 }$$

B
$${ 10 }^{ 9 }$$

C
$${ 10 }^{ 11 }$$

D
$${ 10 }^{ 10 }$$

Solution

A rubber ball is taken 1 km depth inside
water
Pressure on rubber ball ,P = $$ P_0+\rho{gh}$$
Here, Po is atmospheric pressure, p is
a density of water, g is acceleration due
to gravity and his depth inside water.
Given, h 1 km-1000 m
g 10 m/s2, $$\rho$$= 1000 kg/m3 density of
water

Now , pressure act on rubber ball , P = 1.03$$\times{10^5} + 1000 \times 10 \times{1000}$$

$$1.03\times10^5 + 100\times10^5 = 101.03\times10^5 N/m^2$$

Given $$\dfrac{\Delta{V}}{V}=\dfrac{0.05}{100}$$

and$$\Delta{Ρ}$$ =$$P - P_o$$ = $$10^7$$$$ N/m^2$$

$$\frac{\Delta{V}}{V} = 5 \times 10^{-4}$$

Now, bulk modulus = $$\Delta{Ρ}/\Delta{V}/V$$

= $$\dfrac{10^7}{5\times 10^{-4}}$$

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