MCQ Questions for Class 11 Engineering Physics Mechanical Properties Of Solids Quiz 8

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

$=9\times { 10 }^{ 8 }N{ m }^{ -2 },g=10m{ s }^{ -2 }$ )

0%
$9m$
0%
$18m$
0%
$90m$
0%
$180m$

Explanation

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

$h=\left( \cfrac { 9\times { 10 }^{ 8 }N\quad { m }^{ -2 } }{ { 10 }^{ 3 }kg\quad { m }^{ -3 }\times { 10 }^{ }m{ s }^{ -2 } } \right) \left( \cfrac { 0.1 }{ 100 } \right) =90m$

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

$\left( { Y }_{ copper }=1.2\times { 10 }^{ 11 }N\quad { m }^{ -2 },{ Y }_{ steel }=2\times { 10 }^{ 11 }N\quad { m }^{ -2 } \right)$

0%
$1.2\times { 10 }^{ 2 }N$
0%
$1.8\times { 10 }^{ 2 }N$
0%
$2.4\times { 10 }^{ 2 }N$
0%
$3.2\times { 10 }^{ 2 }N$

Explanation

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.

1502330_936992_ans_79a8abf4fa2f460a9c63ad2dfd5fbae1.png

The volume change of a solid copper cube

$10cm$ on an edge, when subjected to a pressure of

$7MPa$ is then
(Bulk modulus of copper

$=140GPa$ )

0%
$5\times { 10 }^{ -2 }{ cm }^{ 3 }$ .
0%
$10\times { 10 }^{ -2 }{ cm }^{ 3 }$
0%
$15\times { 10 }^{ -2 }{ cm }^{ 3 }$
0%
$20\times { 10 }^{ -2 }{ cm }^{ 3 }$

Explanation

Here

$L=10cm=10\times { 10 }^{ -2 }m;P=7MPa=7\times { 10 }^{ 6 }Pa;B=140GPa=140\times { 10 }^{ 9 }Pa$
As

$B=\cfrac { P }{ \Delta V/V }$

$\therefore \Delta V=\cfrac { PV }{ B } =\cfrac { P{ L }^{ 3 } }{ B } =\cfrac { \left( 7\times { 10 }^{ 6 }Pa \right) { \left( 10\times { 10 }^{ -2 }m \right) }^{ 3 } }{ 140 } =5\times { 10 }^{ -8 }{ m }^{ 3 }=5\times { 10 }^{ -2 }{ cm }^{ 3 }$

Which of the following apparatus is used to determine the Young's modulus of the material of a given wire?

0%
Searle
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sonometer
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Metre bridge
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Resonance tube

A copper wire of length

$2.4m$ and a steel wire of length

$1.6m$ , both the diameter

$3mm$ , are connected end to end. The ratio fo elongation of steel to the copper wires is then

$\left( { Y }_{ copper }=1.2\times { 10 }^{ 11 }N\quad { m }^{ -2 },{ Y }_{ steel }=2\times { 10 }^{ 11 }N\quad { m }^{ -2 } \right)$

0%
$\cfrac { 5 }{ 2 }$
0%
$\cfrac { 2 }{ 5 }$
0%
$\cfrac { 3 }{ 2 }$
0%
$\cfrac { 2 }{ 3 }$

Explanation

Given:

Length copper wire,

$L_c =2.4m$

Length steel wire,

$L_s =1.6m$

$(Y_{copper}=1.2×1011^{−2}Nm,Y_{steel}=2×1011^{−2}Nm)$

We know that,

Elongation

$\Delta L =\dfrac{FL}{YA}$

So,

$\cfrac { \Delta { L }_{ S } }{ \Delta { L }_{ C } } =\cfrac { { L }_{ S } }{ { L }_{ C } } \times \cfrac { { Y }_{ C } }{ { Y }_{ S } }$

Substitution the given values, we get

$\cfrac { { { \Delta L }_{ S } } }{ { \Delta L }_{ C } } =\cfrac { 1.6m }{ 2.4m } \times \cfrac { 1.2\times { 10 }^{ 11 }N\quad { m }^{ -2 } }{ 2\times { 10 }^{ 11 }N\quad { m }^{ -2 } } =\cfrac { 2 }{ 5 }$

The relation between

$Y, \eta$ and

$B$ where

$Y, \eta ,B$ are Young's Modulus, Shear modulus and bulk modulus respectively.

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$\cfrac { 1 }{ Y } =\cfrac { 1 }{ 3\eta } +\cfrac { 1 }{ 9B }$
0%
$\cfrac { 9 }{ Y } =\cfrac { 1 }{ 3\eta } +\cfrac { 3 }{ B }$
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$\cfrac { 1 }{ \eta } =\cfrac { 1 }{ B } +\cfrac { 1 }{ Y }$
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$\cfrac { 9 }{ Y } =\cfrac { 3 }{ \eta } +\cfrac { 1 }{ B }$

If in the above question, the Young's modulus of the material is Y, the value of extension x is:

0%
$\left ( \frac{El}{YA} \right )^{1/3}$
0%
$\left ( \frac{YA}{Wl} \right )^{1/3}$
0%
$\frac{1}{l}\left [ \frac{WA}{Y} \right ]^{1/3}$
0%
$l\left [ \frac{W}{YA} \right ]^{1/3}$

Explanation

Sln :
We know Y =

$\dfrac{stress}{strain} \,= \,\dfrac{Wl}{2Ax} \, \times \, \dfrac{2l^2}{x^2}$

$Y \, = \, \dfrac{Wl^3}{Ax^3} \, or \, x \, = \, \left(\dfrac{W}{AY} \right )^{1/3} \, \times \, 1$

A metal wire of length

$L_1$ and area of cross section A is attached to a rigid support. Another metal wire of length

$L_2$ and of the same cross-sectional area is attached to the free end of the first wire. A body of mass M is then suspended from the free end of the second wire. If

$Y_1$ and

$Y_2$ are the Young's moduli of the wires respectively, the effective force constant of the system of two wires is

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$\frac{Y_1Y_2A}{2(Y_1L_2+Y_2L_1)}$
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$\frac{Y_1Y_2A}{2(L_1L_2)^{1/2}}$
0%
$\frac{Y_1Y_2A}{(Y_1L_2+Y_2L_1)}$
0%
$\frac{(Y_1Y_2)^{1/2}A}{2(L_1L_2)^{1/2}}$

Explanation

The wires are connected in series

$Y=\dfrac{F l}{A\Delta l}\Rightarrow \dfrac{YA}{l}=\dfrac{F}{\Delta l}\Rightarrow k=\dfrac{YA}{l}$

$K$ is spring constant

In series spring constants are added as

$\dfrac{1}{K_{1}}+\dfrac{1}{K_{2}}=\dfrac{1}{K}\Rightarrow \dfrac{l_{1}}{Y_{1}A}+\dfrac{l_{2}}{Y_2 A}=\dfrac{1}{K}$

$\Rightarrow K=\dfrac{Y_{1}Y_{2}A}{Y_{1}L_{2}+Y_{2}L_{1}}$

The maximum load a wire can withstand without breaking, when its length is reduced to half of its original length, will

0%
be double
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be half
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be four times
0%
remains same

Explanation

Hint:

Stress if given as,

$\sigma = \dfrac{F}{a}$

Correct Option is D.

Explanation for the correct answer:

$\bullet$ If Breaking force for any wire of area of cross section

$A$ is

$F_b$ , then by formula of stress, breaking stress is given as,

$\sigma_b = \dfrac{F_b}{A}$

$F_b = \sigma_b A$

$\bullet$ Thus, breaking force for any wire depends on its breaking stress and area of cross section. It is independent of the length of the wire.

$\bullet$ Thus, maximum load a wore can withstand without breaking when its length is reduced to half of its original length will remain the same.

Option D is correct.

A steel bar

$ABCD$

$40cm$ long is made up of three parts

$AB, BC$ and

$CD$ , as shown in the figure The rod is subjected to a pull of

$25kN$ . The total extension of the rod is (Young's modulus for steel

$2\times { 10 }^{ 11 }N{ m }^{ -2 }$ :

0%
$0.0637mm$
0%
$0.0647mm$
0%
$0.0657mm$
0%
$0.0667mm$

Explanation

The axial force

$25kN$ is transmitted to each of the three bars.

Stress in part AB is

$\cfrac { 25000N }{ \cfrac { \pi }{ 4 } { (50) }^{ 2 }{ mm }^{ 2 } } =\cfrac { 40 }{ \pi } =12.73N/{ mm }^{ 2 }$

Stress in oart BC

$=\cfrac { 25000 }{ \cfrac { \pi }{ 4 } { (25) }^{ 2 } } =50.93N/{ mm }^{ 2 }$

Stress in part CD

$\quad =12.73N/{ mm }^{ 2 }$
Therefore, total extension of the rod

$=$ extension in the parts

$AB+BC+CA$

$\Delta L =\Delta L_1 +2\times \Delta L_2$ (extension in AB=CD)

$\Delta L=\dfrac{Stress \times length}{Y}$

$\Delta L =\dfrac{\sigma \times L}{Y}$

$=\left( \cfrac { 12.73N/{ mm }^{ 2 } }{ 2\times { 10 }^{ 5 }N/{ mm }^{ 2 } } \times 100mm \right) \times 2+\cfrac { 50.93N/{ mm }^{ 2 } }{ 2\times { 10 }^{ 5 }N/{ mm }^{ 2 } } \times 200mm=\cfrac { 12732 }{ 2\times { 10 }^{ 5 }N/{ mm }^{ 2 } } mm=0.0637mm$

$\Delta L=0.0637mm$

The mean distance between the atoms of iron is

$3\times10^{-10}m$ and interatomic force constant for iron is

$7 N m^{-1}$ . The Young's modulus of electricity for iron is

0%
$2.33\times 10^5 Nm^{-2}$
0%
$23.3\times 10^{10} Nm^{-2}$
0%
$2.33\times 10^{9} Nm^{-2}$
0%
$2.33\times 10^{10} Nm^{-2}$

Explanation

Given,

Inter atomic distance,

$r=3\times 10^{-10}m$

Interatomic force constant,

$K=7N/m$

$Y=?$

Interatomic force constant,

$K= Y\times r$

$Y=\dfrac{K}{r}$

$Y=\dfrac{7}{3\times 10^{-10}}$

$Y=2.33\times 10^{10}N/m^2$

The young's modulus of elasticity of the iron is

$2.33\times 10^{10}N/m^2$ .

The correct option is D.

Find the stress developed inside a tooth cavity filled with copper when hot tea at temperature of

$57^o C$ is drunk. (Take temperature of tooth to be

$37^o C,\alpha =1.7\times { 10 }^{ -5 }{ }{ /^o C}$ and bulk modulus for copper

$=140\times { 10 }^{ 9 }N{ m }^{ -2 }$ )

0%
$1.43\times { 10 }^{ 8 }N{ m }^{ -2 }$
0%
$4.13\times { 10 }^{ 8 }N{ m }^{ -2 }$
0%
$2.12\times { 10 }^{ 4 }N{ m }^{ -2 }$
0%
$3.12\times { 10 }^{ 4 }N{ m }^{ -2 }$

Explanation

Volumetric strain in tooth cavity

$=\dfrac { \Delta V }{ V }$
Let

$\gamma$ be the coefficient of volume expansion with the change in temperature

$\Delta T$ .
Change in volume is

$\Delta V=\gamma V\Delta T\quad or\quad \dfrac { \Delta V }{ V } =\gamma \Delta T$
Thermal stress in tooth cavity

$=\beta \times volumetric \ strain=\beta \gamma \Delta T$

$=\beta \times 3\alpha \Delta T$

$\left( \because \gamma =3\alpha \right)$

$=140\times { 10 }^{ 9 }\times 3\times 1.7\times { 10 }^{ -5 }\times \left( 57-37 \right)$

$=1.43\times { 10 }^{ 8 }N{ m }^{ -2 }$

The adjacent graph shows the extension (

$\Delta l$ ) of a wire of length

$1m$ suspended from the top of a roof at one end and with a load

$W$ connected to the other end. If the cross-sectional area of the wire is

${ 10 }^{ -6 }{ m }^{ 2 }$ , the Young's modulus of the material of the wire is

0%
$2\times { 10 }^{ 11 }N{ m }^{ 2 }$
0%
$2\times { 10 }^{ -11 }N{ m }^{ 2 }$
0%
$3\times { 10 }^{ -12 }N{ m }^{ 2 }$
0%
$2\times { 10 }^{ -13 }N{ m }^{ 2 }$

Explanation

Given:

Length,

$L=1m$

Area of crossection,

$A= 10^{−6}m$

From above graph at load

$20N$ extention

$\Delta L=1\times 10^{-4}m^2$

$Y=\dfrac{F \times l}{\Delta A \times \Delta l}$

putting values in above equation

$Y= \dfrac{20}{10^{-6} \times 10^{-4}}$

$Y= 2 \times 10^{11}$

Y=F/AΔl/l=FAlΔl

The ratio of diameters of two wires of same material is n:The length of each wire is 4 m. On applying the same load, the increases in the length of the thin wire will be (n > l)

0%
$n^2$ times
0%
n times
0%
2n times
0%
(2n + 1) times

Explanation

Consider the expression for Young’s modulus.

$Y=\dfrac{FL}{A\Delta L}$

$\Delta L=\dfrac{FL}{AY}$

Where A is the area, F is the applied force, $\Delta L$ is the increase in length.

So,

$\Delta L=\dfrac{FL}{\pi {{r}^{2}}Y}$

$\Delta L\propto \dfrac{1}{{{r}^{2}}}$

$\dfrac{{{l}_{2}}}{{{l}_{1}}}={{\left( \dfrac{{{r}_{1}}}{{{r}_{2}}} \right)}^{2}}$

$={{\left( \dfrac{{{d}_{1}}/2}{{{d}_{2}}/2} \right)}^{2}}$

$={{(n)}^{2}}$

${{l}_{2}}=n^2\,{{l}_{1}}$

the length is increased by a factor of ${{n}^{2}}$ .

The dimensions of four wires of the same material are given below. In which wire the increase in the length will be maximum?

0%
Length 100 cm, diameter 1 mm
0%
Length 200 cm, diameter 2 mm
0%
Length 300 cm. diameter 3 mm
0%
Length 50 cm, diameter 0.5 mm

Explanation

Sln :

$Y \, = \, \dfrac{Fla}{\Delta lll} \, = \, \dfrac{Fl}{a \Delta l} \, or \, \Delta l \, \, or \, \dfrac{1}{D^2}$

$\dfrac{100}{l^2} \, = \, 100$

$\dfrac{200}{4} \, = \, 50$

$\dfrac{300}{9} \, = \, 33.33$

$\dfrac{50}{(1/2)^2} \, = \, 200$

A rubber ball of bulk modulus B is taken to a depth h of a liquid of density p. Find the fractional change in the radius of the ball.

0%
$\, \dfrac{\rho gh}{4B}$
0%
$\, \dfrac{\rho gh}{3B}$
0%
$\, \dfrac{\rho gh}{2B}$
0%
$\dfrac{\rho gh}{5B}$

Explanation

$\dfrac{\delta V}{V} \, = \, -\dfrac{P}{B}$
Where

$P \, = \, \rho gh$
Then

$-\dfrac{\delta V}{V}\, = \, \dfrac{\rho gh}{B}$
Since, the volume of the sphere is V =

$\dfrac{4}{3}\pi r^{3}$ , we have

$\dfrac{\delta V}{V}\, = \, \dfrac{3\delta r}{r}$
Using Eqs, (i) and (ii), we have

$\dfrac{\delta r}{r}\, = \, \dfrac{\rho gh}{3B}$

A wire is stretched 1 mm by a force of 1 kN. How far would a wire of the same material and length but of four times that diameter he stretched by the same force?

0%
$\dfrac{1}{2}$ mm
0%
$\dfrac{1}{4}$ mm
0%
$\dfrac{1}{8}$ mm
0%
$\dfrac{1}{16}$ mm

Explanation

$Y \, = \, \dfrac{Fl}{a\Delta l}$
Y, l and F are constants.

$\therefore \, \Delta l \, \, or \, \dfrac{1}{D^2}$

$\dfrac{\Delta l_2}{\Delta l_1} \, = \, \dfrac{d_1^2}{D_2^2} \, = \, \dfrac{1}{16}$

$\therefore \, \Delta l_2 \, = \, \dfrac{1}{16} \, mm$

A steel wire is stretched by 1 kg wt. If the radius of the wire is doubled, its Young's modulus will:

0%
remain unchanged
0%
become half
0%
become double
0%
become four times

Plastic deformation results from the following

0%
Slip
0%
Twinning
0%
Both slip and twinning
0%
creep

A gas undergoes a process in which its pressure

$p$ and value

$v$ are related as

$Vp^2 =$ constant. The bulk modulus for the gas in this process is:

0%
$\dfrac{p}{2}$
0%
$2p$
0%
$2pv$
0%
$3p$

Explanation

$Vp^2 = C$

$\beta = -V \left(\dfrac{dp}{dv}\right)$

$p^2 + 2vp \dfrac{dp}{dv} = 0$

$\dfrac{dp}{dv} = \dfrac{-p}{2v}$

$\beta = -v \times \left(\dfrac{-p}{2v}\right) = \dfrac{P}{2}$

A wire has a tensile strength of 70MPa, and breaks under 100N of force. What is the cross-sectional area of the wire just before breaking?

0%
$14.2 \times 10^{-6} m^2$
0%
$1.42 \times 10^{-6} m^2$
0%
$1.42 \times 10^{-3} m^2$
0%
$1.42 \times 10^{-2} m^2$

Explanation

The tensile strength = stress = force/area

Thus,

$area = Force/ Stress =100/ (70 \times 10^6) = 1.42 \times 10^{-6} m^2$

The correct option is option (b)

A thin metal sheet is being bent by or pounded in to a new shape. The process of being elastic to plastic behaviour is known as

0%
Yield
0%
Creep
0%
Welding
0%
Tinkering

An elastic spring is given a force of 1000 N over an area of

$0.2 m^2.$

0%
$3000 N m^{-2}$
0%
$5000 N m^{-2}$
0%
$500 N m^{-2}$
0%
$2500 N m^{-2}$

Explanation

Stress is defined by Stress =

$1000/0.2=5000 Nm^{-2}$

The correct option is option (b)

The plasticity behaviour of a material determines the

0%
elastic behavior of the material
0%
resistance of the material to electric fields
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viscous behavior of the material and is irrecoverable.
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resistance of the material to magnetic fields

Substances that elongate considerably and undergo plastic deformation before they break are known as

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brittle substances
0%
breakable substances
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ductile substances
0%
all of these

Plastic deformation in a material begins at

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Q point
0%
Yield point
0%
Proportionality limit
0%
Elastic limit

At yield point, Hooke's law doesn't hold good

0%
True
0%
False

Elasticity is defined as the ability of a body to

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Resist linear motion in a hard surface
0%
Resist rolling motion in a hard surface
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Resist a distorting influence and to return to its original size and shape when that influence or force is removed.
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Resist electric current in a magnetic field

Which of the following is the dimension of Bulk Modulus?

0%
$[M^1L^{-1}T^{-1}]$
0%
$[M^1L^{-1}T^{-2}]$
0%
$[M^1L^{1}T^{2}]$
0%
$[M^1L^{1}T^{-1}]$

Explanation

Bulk modulus is pressure/volumetric strain. The strain dosen't have any units and hence bulk's modulus takes the units of pressure. Thus, dimensions of bulk's modulus is

$[B]=[MLT^{-2}]/[L^2] =[ML^{-1}T^{-2}]$

The correct option is (b)

Longitudinal strain is calculated using the formula

0%
Change in length/ original length
0%
Original length/Change in length
0%
Original length $\times$ Change in length
0%
Original length - Change in length

Poisson' ratio is defined as the ratio of

0%
longitudinal stress and longitudinal strain
0%
longitudinal stress and lateral stress
0%
lateral stress and longitudinal stress
0%
lateral stress and lateral strain

The difference between pressure and stress is

0%
pressure and stress have different units
0%
pressure and stress have different dimensions
0%
Force cannot be determined using stress, but in pressure it can be done
0%
Pressured is applied to a body, while stress is induced

Which one of the following is true about Bulk Modulus of elasticity?

0%
It is the ratio of compressive stress to volumetric strain
0%
It is the ratio of compressive stress to linear strain
0%
It is the ratio of tensile stress to volumetric strain
0%
It is the ratio of tensile stress to linear strain

A force of 10 N is applied to an object, whose area is

$5 cm^2$ at an angle of 30 degrees with the vertical. What kind of stress can be found from this data

0%
Normal and areal stress can be found
0%
only normal stress can be found
0%
only areal stress can be found
0%
Stress cannot be found from this data, since applied force is neither along the horizontal or vertical

A steel wire is suspended from a fixed end, while the other end is loaded with a weight W. This produced an extension x. As the weight is increased, the extension was also increased. A plot of extension vs load within elastic limits will give rise to

0%
a curve
0%
an ellipse
0%
a straight line
0%
a hyperbola

The radius of a copper wire is 4 mm. What force is required to stretch the wire by 20% of its length, assuming that the elastic limit is not exceeded (Y=

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

0%
$7.23 \times 10^5 N$
0%
$7.23 \times 10^7 N$
0%
$7.23 \times 10^6 N$
0%
$7.23 \times 10^8 N$

Explanation

We know that

$Y=\dfrac{F/A}{\delta L/L}=\dfrac{FL}{A \delta L}$

Extension = L+0.2L=1.2 L. Substituting, we get,

$F= 1.2YA=1.2 \times 12 \times 10^{10} \times (\pi 16 \times 10^{-6})=723.16 \times 10^4 N$

The correct option is (c)

A wire of length L can support a load W. If the wire is broken in to two equal parts , then how much load can be suspended by one of those cut wires?

0%
Half
0%
Same
0%
Double
0%
One fourth

A rubber cord 10 m long is suspended vertically. How much does it stretch under its own weight. ( [Density of rubber is $1500 (kg / m^3), Y = 5 \times 10^8 N/m^2)$

0%
0.3 mm
0%
0.15 mm
0%
0.015 mm
0%
0.03 mm

Explanation

$\begin{array}{l}\text { According to the question, } \\\text { length of cord }=10 \mathrm{~m} \\\text { Density of rubber }=1500\mathrm{~kg}/ \mathrm{m}^{3} \\\text { young's Modulus }=5 \times 10^{8} \mathrm{~N} / \mathrm{m}^{2}\end{array}$

$Stress = Weight/area = mg/A = (\rho A g)/A=\rho g$ and

$strain=\delta L/L \implies$ change in length =

$\rho g L/Y$
Also the length to be considered here is from the centre of gravity to the point of suspension, since the entire mass is assumed to be concentrated at the centre

Substituting the values, we get, extension =

$(1500 \times 10 \times 5)/(5 \times 10^8) =0.15 mm$

The correct option is (b)

When the temperature of a gas is

$20^0C$ and pressure is changed from

$P_1=1.01\times 10^{5}\, Pa$ to

$P_2=1.165\times 10^5\,Pa$ , then the volume changes by

$10$ %. The bulk modulus is

0%
$1.55\times10^5 \,Pa$
0%
$1.05\times10^5 \,Pa$
0%
$1.4\times10^5 \,Pa$
0%
$0.115\times10^5 \,Pa$

Explanation

Bulk Modulus,

$B=\dfrac{\triangle P}{\triangle V/ V}$

$\dfrac{\triangle V}{V}\times 100$ %

$=10$ %

$\dfrac{\triangle V}{V}=\dfrac{10}{100}=\dfrac{1}{10}$

$\triangle P=(1.165-1.01)\times10^5\,Pa$

$=0.155\times10^5 \,Pa$

$B=\dfrac{0.155\times 10^5}{1/10}\,Pa$

$=1.55\times 10^5\,Pa$

A student measures the poisson's ratio to be greater than 1 in an experiment. The meaning of this statement would be

0%
An increase in length would also result in decrease in area of cross section of the wire
0%
An increase in length would also result in increase in area of cross section of the wire
0%
An decrease in length would also result in decrease in area of cross section of the wire
0%
An increase in length will not change the area of cross section of the wire

The theoretical limits of poisson's ratio lies between -1 to 0.5 because

0%
Shear modulus and bulk's modulus should be positive
0%
Bulk's modulus is negative during compression
0%
Shear modulus is negative during compression
0%
Young's modulus should be always positive

Explanation

Let Y, K, n and

$\sigma$ be the Young's Modulus, Bulk modulus, Modulus of Rigidity and Poisson's Ratio, respectively.
Y = 3K (1 - 2

$\sigma$ ) [Standard formula]
Y = 2n (1 +

$\sigma$ ) [Standard formula]
Hence, 3K (1 - 2

$\sigma$ ) = 2n (1 +

$\sigma$ )
Now K and n are always positive, so
i) If

$\sigma$ be +ve, then RHS is always +ve. So LHS must also be +ve. Therefore, 2

$\sigma$ < 1 or

$\sigma$ <1/2
ii) If

$\sigma$ be -ve, then LHS will always be +ve. Therefore, 1+

$\sigma$ > 0 or

$\sigma$ > -1
Thus the limiting values of Poisson's ratio are -1 <

$\sigma$ < 1/2

The correct option is (a)

$1$ m long metal wire of cross sectional area

$10^{-6}m^2$ is fixed at one end from a rigid support and a weight W is hanging at its other end. The graph shows the observed extension of length

$\Delta l$ of the wire as a function of W. Young's modulus of material of the wire in SI units is?

0%
$5\times 10^4$
0%
$2\times 10^5$
0%
$2\times 10^{11}$
0%
$5\times 10^{11}$

A uniform wire (Young's modulus

$2\times 10^{11}Nm^{-2}$ ) is subjected to longitudinal tensile stress of

$5\times 10^7\ Nm^{-2}$ . If the overall volume change in the wire is

$0.02\%$ , the frictional decrease in the radius of the wire is close to

0%
$1.0 \times 10^{-4}$
0%
$1.5 \times 10^{-4}$
0%
$0.25 \times 10^{-4}$
0%
$5 \times 10^{-4}$

Explanation

$\dfrac{\triangle V}{V}=\dfrac{\triangle l}{l}+ \dfrac{2\triangle r}{r}, \dfrac{\triangle l}{l}= \dfrac{stress}{Y} = 2.5 \times 10^{-4}$

The ratio of the coefficient of volume expansion of glass container to that of a viscous liquid kept inside the container is

$1:4$ . What fraction of the inner volume of the container should the liquid occupy so that the volume of the remaining vacant space will be same at all the temperature?

0%
$2/5$
0%
$1/4$
0%
$1/64$
0%
$1/8$

Explanation

We know that,

When there is no change in liquid level in vessel then,

$\gamma _{real}^{'}=\gamma _{vessel}^{'}$

Change in volume in liquid relative to vessel

$\Delta {{V}_{app}}=V\gamma _{app}^{'}\Delta \theta$

$\Delta {{V}_{app}}=V\left( \gamma _{real}^{'}-\gamma _{vessel}^{'} \right)$

Hence, the fraction is $1:4$

A metal rod of Young's modulus

$Y$ and coefficient of thermal expansion

$\alpha$ is held at its two ends such that its length remains invariant. If its temperature is raised by

${t}^{o}C$ , the linear stress developed in it is:

0%
$1/\left( Y\alpha t \right)$
0%
$\alpha t/Y$
0%
$Y/\alpha t$
0%
$Y\alpha T$

For which material the poisson's ratio is greater than 1

0%
Steel
0%
Copper
0%
Aluminium
0%
None of the above

The maximum strain energy that can be stored in a body is known as:

0%
impact energy
0%
toughness
0%
proof resilience
0%
none of the above

A rubber ball is taken to depth

$1$ km inside water so that its volume reduces by

$0.05\%$ .What is the bulk modulus of the rubber:

0%
$2 \times 10^{10} N/m^2$
0%
$2 \times {10^{9\,}}\,N/{m^2}$
0%
$2 \times 10^{7}N/m^2$
0%
$2 \times {10^{11\,}}\,N/{m^2}$

Explanation

Pressure at

$1\ km$ depth inside water,

$P=1000\times 10\times 1000=10^7\ Nm^{_2}$

say, V=volume of ball

$\dfrac{0.05}{100}\times V$ = Reduction in volume (i.e)

$\Delta V$

Bulk Modulus,

$\beta=\dfrac{-PV}{\Delta V}=\dfrac{-10^7\times V}{\dfrac{-0.05V}{100}}$

$=2\times 10^{10}\ Nm^{-2}$

Hence Option

$\textbf A$ is correct.

The proportional limit of steel is

$8\times { 10 }^{ 8 }N/{ m }^{ 2 }$ and its Young's modulus is

$2\times { 10 }^{ 11 }N/{ m }^{ 2 }$ . The maximum elongation, a one metre long steel wire can be given without exceeding the proportional limit is

0%
2 mm
0%
4 mm
0%
1 mm
0%
8 mm

Explanation

The correct option is B

Given,

Proportional limits of steel are

$8\times10^8N/m^2$

Young's modulus is

$2\times10^{11}N/m^2$

So, Stress up to which stress and strain are proportional.

$\dfrac{stress}{strain}=\dfrac{8\times10^8}{strain}$

Since

$strain = \dfrac{stress}{modulus}$

$\Rightarrow strain=\dfrac{4\times10^8}{2\times10^{11}}$

$=4\times10^{-3}$

Thus,

$\dfrac{\Delta L}{L}=4\times10^{-3}$ Where

$\Delta L$ is stress

But,

$L=1m$

So,

$\Delta L=4\times10^{-3}m$

$=4mm$

Overall changes in volume and radii of a uniform cylindrical steel sire are

$0.2\%$ and

$0.002\%$ respectively when subjected to some suitable force. Longitudinal tensile stress acting on the wire is :-

$(Y=2.0\times 10^{11}\ NM^{-2})$

0%
$3.2\times 10^{9}\ Nm^{-2}$
0%
$3.2\times 10^{7}\ Nm^{-2}$
0%
$3.6\times 10^{9}\ Nm^{-2}$
0%
$3.9\times 10^{8}\ Nm^{-2}$

CBSE Questions for Class 11 Engineering Physics Mechanical Properties Of Solids Quiz 8 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Mechanical Properties Of Solids Quiz 8 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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