MCQ Questions for Class 11 Engineering Physics Mechanical Properties Of Solids Quiz 2

Young's modulus of Perfectly elastic body is

0%
Finite value
0%
Zero
0%
One
0%
Infinite value

A wire whose cross-sectional area is

$4\ mm^{2}$ is stretched by

$0.1\ mm$ by a certain load. If a similar wire of double the area of cross-section is under the same load, then the elongation would be

0%
$0.5\ mm$
0%
$0.05\ mm$
0%
$0.005\ mm$
0%
$5\ mm$

Explanation

$0.05mm$

Formula,

$y=\dfrac{Fl}{Ae}$

$\Rightarrow e\propto \dfrac{1}{A}$

$\dfrac{e_2}{0.1}=\dfrac{4}{8}=\dfrac{1}{2}$

$\therefore e_2=0.05mm$

The force required to double the length of the steel wire of area of cross section

$5\times 10^{-5}m^{2}\quad (Y=20\times 10^{10}Pa)$ in

$N$ is:

0%
$10^{17}$
0%
$10^{16}$
0%
$10^{7}$
0%
$10^{15}$

Explanation

$Y=\cfrac{stress}{strain}$

$Y=\cfrac{\cfrac{F}{A}}{\cfrac{\triangle l}{l}}$

$\triangle l=l$

$F=YA$

$=20\times 10^{10}\times 5\times 10^{-5}$

$=100\times 10^{10}\times 10^{-5}$

$=10^7$

Bulk modulus was first defined by

0%
Young
0%
Bulk
0%
Maxwell
0%
None of the above

The only elastic modulus that applies to fluids is

0%
Young's Modulus
0%
Shear Modulus
0%
Modulus of rigidity
0%
Bulk modulus

Minimum and maximum values of Poisson’s ratio for a metal lies between

0%
$-\infty$ to $+\infty$
0%
$0$ to $1$
0%
$-\infty$ to $1$
0%
$0$ to $0.5$

Ratio of transverse to axial strain is

0%
Toricelli ratio
0%
Poisson's ratio
0%
Stoke's ratio
0%
Bernoulli's ratio

Explanation

Hookes law states that stress is proportional to strain up to elastic limit. If p is the stress induced in material and e the corresponding strain, then according to Hooke's law,

$\dfrac{P}{E}$ = E, a constant.

The modulus of elasticity is dimensionally equivalent to

0%
Stress
0%
Surface tension
0%
Strain
0%
Coefficient of viscosity

Explanation

Hookes law establishes the relationship between stress and strain
Stress: The force per unit area
Strain: The elongation or contraction per unit length (dimensionless)
The ratio of stress to strain is known as the elastic modulus of the material
Elastic Modulus

$= \displaystyle \frac{stress}{strain}$ Hence, the modulus of elasticity is dimensionally equivalent to the stress

A wire elongates by

$1 mm$ when a load

$W$ is hung from it. If the wire goes over a pulley and the two weights

$W$ , each are hung at the two ends, then the elongation of the wire will be:

0%
$0.5 mm$
0%
$1 mm$
0%
$2 mm$
0%
$4 mm$

An iron bar of length

$L$ , cross-section

$A$ and Young's modulus

$Y$ is pulled by a force

$F$ from ends so as to produce an elongation

$l$ . Which of the following statements is correct ?

0%
$l \propto \dfrac{1}{L}$
0%
$l \propto A$
0%
$l \propto \dfrac{1}{A}$
0%
$l \propto Y$

Explanation

We know

$\rightarrow l =\dfrac{FL}{AY}$

$l \propto \dfrac{1}{A}$

A wire of length L and radius r fixed at one end and a force F applied to the other end produces and extension

$l$ . The extension produced in another wire of the same material of length 2Land radius 2r by a force 2 F is:

0%
$l$
0%
$2l$
0%
$\dfrac{l}{2}$
0%
$4l$

Explanation

$Y=\dfrac{FL}{Al}$
Y - Young's Modulus
F - Applied Force
A - Cross Section Area
l - Elongation

For 1st Wire

$Y = \dfrac{FL}{(\pi)(r^2)l}$ ---- (M)

For 2nd Wire
z - elongation in 2nd wire

$Y = \dfrac{2F2L}{(\pi)(2^2)(r^2) z}$ ------ (N)

Dividing M by M

$\Rightarrow 1 = \dfrac{z}{l}$
Therefore

$z = l$

Elongation of a wire under its own weight is independent of :

0%
Length
0%
Area of cross section
0%
Density
0%
Young's modulus

Explanation

Mass

$=$ M

Force

$= \rho$ ALg

Let weight of the rod is W.

Consider a small length dx of the rod at distance x from the fixed end. The tension T in element equals the weight of rod below it.

$T=(L-x)\dfrac{W}{L}$

Elongation

$=\dfrac{L\times Stress}{y}=\dfrac{(L-x)Wdx}{LAy}$

Total elongation

$= \int_{O}^{L}\dfrac{(L-x)Wdx}{LAy}$

$=\dfrac{W}{LAy}(Lx-\dfrac{x^{2}}{2})_{0}^{L}=\dfrac{WL}{2Ay}$

Now,

$W=\rho ALg$

Putting it in there:

$\Delta L=\dfrac{\rho AL^2g}{2Ay}$

$=\dfrac{\rho L^2g}{2y}$

Assertion (A): Stress is restoring force per unit area.
Reason (R) : Interatomic forces in solids are responsible for the property of elasticity

0%
Both Assertion and Reason are true and the reason is correct explanation of the assertion
0%
Both Assertion and Reason are true, but reason is not correct explanation of the assertion
0%
Assertion is true, but the Reason is false
0%
Assertion is false, but the reason is true

If stress is numerically equal to young's modulus,the elongation will be

0%
$1/4$ the original length
0%
$1/2$ the original length
0%
equal to the original length
0%
Twice the original length

Explanation

$\textbf{Hint}$ : Young's modulus of elasticity

$\textbf{Step 1}$ :

Young's modulus of elasticity is the ratio of stress to strain.

$Young's \quad modulus= \dfrac{stress}{strain}$

Stress is defined as force per unit area.

$Stress=\dfrac{F}{A}$ where F is force and A is area.

$Strain=\dfrac{\Delta L}{L}$ where 'L' is the length of the wire and

$\Delta L$ is the change in the length of the wire.

$\textbf{Step 2}$ :

When the change in length to the original length is minimum, then modulus of elasticity is dimensionally equivalent to the stress since it is given stress is numerically equivalent to young's modulus ie..,

$Stress=Y$

Thus option C is correct.

A metal string is fixed between rigid supports. It is initially at negligible tension. Its Young's modulus is Y, density is

$\rho$ and coefficient of linear expansion is

$\alpha$ . It is now cooled through a temperature t, transverse waves will move along it with a speed of :

0%
$\sqrt{\dfrac{Y\alpha t}{\rho}}$
0%
$Y\sqrt{\dfrac{\alpha t}{\rho}}$
0%
$\alpha\sqrt{\dfrac{Yt}{\rho}}$
0%
$t\sqrt{\dfrac{\rho}{Y\alpha}}$

Explanation

By the definition of Young's modulus, we have

$y=\left ( \dfrac{F}{A} \right )/\left ( \dfrac{\bigtriangleup L}{l} \right )\ \ \Rightarrow F=\dfrac{YAl(\alpha t)}{l}$

$\bigtriangleup l=l\ \alpha\ t$

$\therefore$ F = YA

$\alpha$ t
The velocity

$V=\sqrt{\dfrac{T}{\mu}}=\sqrt{\dfrac{YA\ \alpha t}{\rho(A)}}$

$\mu = \rho(A)$

$\Rightarrow V = \sqrt{\dfrac{Y\ \alpha t}{\rho}}$

Three wires A, B, C made of different materials elongated by 1.5, 2.5, 3.5 mm, under a load of 5kg. If the diameters of the wires are the same,the most elastic material is that of

0%
A
0%
B
0%
C
0%
All

Consider the following two statements A and B and identify the correct answer.
A) When the length of a wire is doubled, the Young's modulus of the wire is also doubled
B) For elastic bodies Poisson's ratio is + Ve and for inelastic bodies Poissons ratio is -Ve

0%
Both A & B are true
0%
A is true but B is false
0%
A is true but B is true
0%
Both A & B are false

Consider the following two statements A and B and identify the correct answer.
A) The bulk modulus for an incompressible liquid is infinite.
B) Young's modulus increases with raise of temperature.

0%
Both A & B are true
0%
A is true but B is false
0%
Both A & B are false
0%
A is false but B is true

Consider the statements A and B, identify the correct answer given below :
(A) : If the volume of a body remains unchanged when subjected to tensile strain, the value of poisson's ratio is 1/2.
(B) : Phosper bronze has low Young's modulus and high rigidity modulus.

0%
A and B are correct
0%
A and B are wrong
0%
A is correct and B is wrong
0%
A is wrong and B is right

Explanation

Experimental value of poisson's ratio is always between

$0$ to

$1/2$ .

As Phosper bronze is solid so, value of young's modulus is also high.

The stress required to double the length of a wire of Young's modulus

$E$ is :

0%
$2E$
0%
$E$
0%
$E/2$
0%
$3E$

Explanation

Young's modulus,

$E$

$= \dfrac{stress}{strain}$

Where,

$strain = \dfrac{change \ in \ length}{original\ length}$
Let the initial length of the wire be

$x$ m.
Thus change in length =

$2x - x = x$
Original length =

$x$
So, strain = 1
So, we get

$E$ = stress/1

The graph shows the behaviour of a steel wire in the region for which the wire obeys Hooke's law.The graph is a part of a parabola. The variables x and y might represent.

0%
x $=$ stress ; y $=$ strain
0%
x $=$ strain ; y $=$ stress
0%
x $=$ strain ; y $=$ elastic energy
0%
x $=$ elastic energy ; y $=$ strain

Explanation

Elastic energy

$= \dfrac{1}{2} \times stress \times strain$
whereas strain deals with change in length / original length
hence, "x' axis depends on depended variable but "y" axis depends on independant variable. So, option "D" is correct.

Assertion (A) : Lead is more elastic than rubber.
Reason (R) : If the same load is attached to lead and rubber wires of the same cross-sectional area, the strain of lead is very much less than that of rubber.

0%
Both assertion and reason are true and the reason is correct explanation of the assertion
0%
Both assertion and reason are true, but reason is not correct explanation of the assertion
0%
Assertion is true, but the reason is false
0%
Assertion is false, but the reason is true

Explanation

More elastic the material higher the Young's Modulus.
Young's Modulus of rubber is 0.01-0.1 Gpa whereas for Lead it is 16 Gpa.

Also,

$Y = (F/A)/l/L$

F - Applied Force
A - Cross Section Area
l - Elongation/Decrease in Length
L - Actual Length
Therefore

$l/L = F/AY$
for F,A same

$l/L$ is inversely proportional to

$Y$
So, higher the Young's Modulus, lesser is the strain.

Three wires A,B, C made of the same material and radius have different lengths. The graphs in the figure shows the elongation-load variation. The longest wire is:

0%
A
0%
B
0%
C
0%
All

Explanation

The slope C > slope B > slope A.

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

Slope

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

$\therefore \dfrac{\Delta L}{F} \propto L$ and , thus,

$C$ is the longest wire.

Consider the following two statements A and B and identify the correct answer .
A) A metal wire held vertically is longer than when it placed on a horizontal table.
B) Due to its own weight, some elongation is produced when it is held vertically.

0%
Both A & B are true
0%
A is false but B is true
0%
A is true but B is false
0%
Both A & B are false

A uniform heavy rod of length

$L$ and area of cross-section area

$A$ is hanging from a fixed support. If Young's modulus of the material of the rod is

$Y$ , then the increase in the length of the rod is (

$\rho$ is a density of the material of the rod) :

0%
$\dfrac{L^{2}Y}{2\rho g}$
0%
$\dfrac{L^{2}\rho g}{2Y}$
0%
$\dfrac{L^{2} g}{2Y\rho}$
0%
$\dfrac{L^{2} g}{3Y\rho}$

Explanation

Consider an element dx at a distance x from the top.

Let the wt. of rod is W.
The force acting on this because of a part which is below dx is given as

$\dfrac{(L-x)W}{L}$

The elongation in element

$dx$ is

$=(\dfrac{L-x}{L})\dfrac{Wdx}{AY}$

$=\dfrac{(L-x)Wdx}{LAY}$

Total elongation,

$\Delta L=\int_{O}^{L}\dfrac{(L-x)Wdx}{LAY}$

$=\dfrac{WL}{2AY}$

$W=AL\rho g$

$\Delta L=\dfrac{AL\rho gL}{2AY}$

$=\dfrac{L^{2}\rho g}{2Y}$

$20 kg$ load is suspended by a wire of cross section

$0.4 mm$

$^{2}$ . The stress produced in N/m

$^{2}$ is :

0%
4.9 x 10 $^{-6}$
0%
4.9 x 10 $^{8}$
0%
49 x 10 $^{8}$
0%
2.45 x 10 $^{-6}$

Explanation

$stress=\dfrac{Force\, applied }{Area} = \dfrac{20\times 9.81}{0.4\times 10^{-6}}$

$=4.9\times 10^{8} N/m^{2}$

The length of a wire under stress changes by 0.01%. The strain produced is

0%
$10^{-4}$
0%
$0.01$
0%
$1$
0%
$10^4$

Explanation

Strain

$=\dfrac{0.0001\Delta L}{L} = 0.01$ %

$of L=\dfrac{.01\times L}{100}=10^{-4}L$

$=10^{-4}$

A wire of length '

$l$ ' and radius '

$r$ ' is clamped rigidly at one end. When the other end of the wire is pulled by a force '

$F$ ', its length increases by

$'x'$ . Another wire of same material of length '

$2l$ ' and radius

$'2r$ ' is pulled by a force '

$2F'$ , the increase in its length will be :

0%
$x$
0%
$2x$
0%
$x/2$
0%
$4x$

Explanation

In the first case we have

$x=\dfrac{Fl}{\pi r^2Y}$
and in the for the second wire we have the change in length as

$\dfrac{2F\times 2l}{\pi(2r)^2Y}=\dfrac{Fl}{\pi r^2Y}=x$

The length of a wire is

$4m$ . Its length is increased by

$2mm$ when a force acts on it. The strain is:

0%
0.5 x 10 $^{-3}$
0%
5 x 10 $^{-3}$
0%
2 x 10 $^{-3}$
0%
0.05

Explanation

${L}'=4.002 m$

$Strain =\dfrac{0.002}{4}$

$=5\times 10^{-4}$ (using standard result)

Four wires made of same materials are stretched by the same load. Their dimensions are given below. The one which elongates more is ?

0%
Wire of length 1 m and diameter 1 mm
0%
Length 2m, diameter 2 mm
0%
Length 3m, diameter 3 mm
0%
Length 0.5m, diameter 0.5mm

Explanation

$\Delta l=\dfrac{FL}{Ay}$

$A\rightarrow \Delta l=\dfrac{F\times 1}{y\pi (\dfrac{1}{2})^{2}}\times 10^{6} = \dfrac{yF}{\pi y}\times 10^{6}$

$B\rightarrow \Delta l=\dfrac{F\times 2}{y\pi 1^{2}}\times 10^{6} = \dfrac{2F}{\pi y}\times 10^{6}$

$C\rightarrow \Delta l=\dfrac{F\times 3\times 4}{y\pi 1}\times 10^{6} = \dfrac{4}{3}\dfrac{F}{\pi y}\times 10^{6}$

$D\rightarrow \Delta l=\dfrac{F\times \dfrac{1}{2}}{y\pi (\dfrac{1}{4})^{2}}\times 10^{6} = \dfrac{8F}{Y\pi }\times 10^{6}$

$\Delta l$ of wire D is longest.

A steel wire of

$2mm$ in diameter is stretched by applying a force of

$72N$ . Stress in the wire is

0%
$2.29\times 10^7 \ N/m^2$
0%

$1.17\times 10^7 \ N/m^2$
0%

$3.6\times 10^7 \ N/m^2$
0%

$0.8\times 10^7 \ N/m^2$

Explanation

Diameter=

$2\times 10^{-3}$ m

Radius=

$10^{-3}$ m

area of cross section=

$\pi r^{2}$

$A=\dfrac { 22 }{ 7 } \times { 10 }^{ -6 }㎡\\ F=72N\\ stress=\dfrac { F }{ A } \\ =\dfrac { 72 }{ \dfrac { 22 }{ 7 } \times { 10 }^{ -6 } } =\dfrac { 72 }{ 3.14 } \times { 10 }^{ -6 }$

So, stress=

$22.92\times 10^{6}M/m^{2}$

or stress=

$2.29\times10^{7}N/m^{2}$

The lengths of two wires of the same material and diameter are

$100 cm$ and

$125 cm$ . If same force is applied on them, the elongation in the first wire is

$4 mm$ . The elongation in the second wire (in mm) is

0%
$4$
0%
$5$
0%
$0.8$
0%
$1.25$

Explanation

$L_{1}=100 cm ,L_{2}=125 cm$

$\Delta L_{1}=4 cm , \Delta L_{2}=?$

$A_{1}=A_{2} Y_{1}=Y_{2} , F_{1}=F_{2}$

$\dfrac{\Delta L_{1}}{L_{1}}=\dfrac{\Delta L_{2}}{L_{2}}$

$\dfrac{1250\times 4}{1000}=\Delta L_{2}$

$=5 mm$

A force of

$30N$ acts on a rod of area of cross section 5 x 10

$^{-6}$

$m^{2}$ . The stress produced in dyne/cm

$^{2}$ is :

0%
$6 \times 10$ $^{6}$
0%
$6 \times 10$ $^{7}$
0%
$6 \times 10$ $^{5}$
0%
$0.16 \times 10$ $^{-6}$

Explanation

$1N=10^{5} dyne$

$stress=\dfrac{30\times 10^{5} dyne}{5\times 10^{-6}\times 10^{4} cm^{2}}$

$=6\times 10^{7} dyne/cm^{2}$

The force required to double the length of a steel wire of area of cross-section

$5 \times 10^{-5} m^{2}$ (in N) is :
(

$Y=20 \times 10^{10} Pa$ )

0%
$10^{7}$
0%
$10^{6}$
0%
$10^{-7}$
0%
$10^{5}$

Explanation

$L=\Delta L$

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

$=5\times 10^{-5}\times 20\times 10^{10}$

$=10^{7} N$

The length of two wires are in the ratio

$3 : 4$ .Ratio of the diameters is

$1:2$ ; young's modulus of the wires are in the ratio

$3:2$ ; If they are subjected to same tensile force, the ratio of the elongation produced is

0%
1 :1
0%
1 :2
0%
2 : 3
0%
2 : 1

Explanation

$\displaystyle \frac{L_{1}}{L_{2}}=\frac{3}{4}, \quad \frac{D_{1}}{D_{2}}=\frac{1}{2} , \quad \frac{A_{1}}{A_{2}}=\frac{1}{4}$

$\dfrac{y_{1}}{y_{2}}=\dfrac{3}{2}$ same force

$\dfrac{\Delta L_{1}}{\Delta L_{2}}=\dfrac{F.L_{1}}{A_{1}y_{1}}\times \dfrac{A_{2}y_{2}}{FL_{2}}$

$=\dfrac{3}{4}\times \dfrac{4}{1}\times \dfrac{02}{3}$

$\dfrac{\Delta L_{1}}{\Delta L_{2}}=\dfrac{2}{1}$

A solid sphere hung at the lower end of a wire is suspended from a fixed point so as to give an elongation of

$0.4mm$ . When the first solid sphere is replaced by another one made of same material but twice the radius, the new elongation is

0%
$0.8mm$
0%
$1.6mm$
0%
$3.2mm$
0%
$1.2mm$

Explanation

As radius is doubled volume becomes 8 times and, hence, the wt. of sphere.
The force becomes 8 times.

$F \propto \Delta L$

$\therefore \Delta L$ also becomes 8 times

$\Delta {L}'=8\times 0.4 mm$

$=3.2 mm$

The ratio of lengths of two wires made of same material is

$2 :3$ . The ratio of their respective longitudinal stress to produce same elongation is

0%
$4 : 9$
0%
$9 : 4$
0%
$2 : 3$
0%
$3: 2$

Explanation

$\dfrac{L_{1}}{L_{2}}=\dfrac{2}{3}$

$\Delta L_{1}=\Delta L_{2}$ (given)

$y_{1}=y_{2}$ (given)

$\dfrac{\sigma _{1}}{\epsilon _{1}}=\dfrac{\sigma _{2}}{\epsilon _{2}}$ (as inferred from question)

$\dfrac{\sigma _{1}}{\sigma _{2}}=\dfrac{\epsilon _{1}}{\epsilon _{2}}$

$\dfrac{\sigma _{1}}{\sigma _{2}}=\dfrac{\Delta L_{1}}{L_{1}}\times \dfrac{L_{2}}{\Delta L_{2}}$

$=\dfrac{3}{2}\times 1$

Ratio of lengths of two brass wires is 3 : 4; their areas of cross section are in the ratio 2:When same force is applied on them, the elongations produced will be in the ratio:

0%
9 : 8
0%
8 : 9
0%
2 $\sqrt{2}$ :3
0%
1 : 1

Explanation

$\dfrac{L_{1}}{L_{2}}=\dfrac{3}{4} , \dfrac{A_{1}}{A_{2}}=\dfrac{2}{3}$

$\dfrac{\Delta L_{1}}{\Delta L_{2}}=\dfrac{3}{4}\times \dfrac{3}{2}$

$=\dfrac{9}{8}$

The length of a wire of cross-sectional area

$1 \times$ 10

$^{-6} m^{2}$ is 10m. The young's modulus of the material of the wire is 25 G.pa. When the wire is subjected to a tensile force of

$100N$ , the elongation produced in

$mm$ is:

0%
$0.04$
0%
$0.4$
0%
$4$
0%
$40$

Explanation

$\Delta L=\dfrac{100\times 10}{10^{-6}\times 25\times 10^{9}}$

$=0.04 m$

$=40 mm$ (using standard result

$Y=\dfrac{(\dfrac{F}{A})}{(\dfrac{\text{change in length}}{\text{actual length}})}$ )

A steel wire of length 1 m has cross sectional area

$1cm$

$^{2}$ . If young's modulus of steel is

$10^{11}N / m^{2}$ ,then force required to increase the length of wire by 1 mm will be :

0%
$10^{11}N$
0%
$10^{7}N$
0%
$10^{4}N$
0%
$10^{2}N$

Explanation

Given,

$L=1 m, A=10^{-4} m^{2}$ ,

$\Delta l = 1mm$ ,

$y= 10 ^{11} N/m^2$

We know,

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

$.001=\dfrac{F\times 1}{10^{-4}\times 10^{11}}$

$F=10^{4}N$

Two steel wires have equal volumes. Their diameters are in the ratio 2 :When same force is applied on them, the elongation produced will be in the ratio of:

0%
$1:8$
0%
$8:1$
0%
$1:16$
0%
$16:1$

Explanation

Volume is equal.

i.e.

$\pi r_{1}^{2}L_{1} = \pi r_{2}^{2}L_{2}$

given

$\dfrac{r_{1}}{r_{2}}=\dfrac{2}{1}$

i.e.

$\dfrac{L_{1}}{L_{2}}=\dfrac{1}{4}$

applying same force.

$\dfrac{\Delta L_{1}}{\Delta L_{2}}=\dfrac{L_{1}}{L_{2}}\times \dfrac{A_{2}}{A_{1}}$

$=\dfrac{1}{4}\times \dfrac{1^{2}}{2^{2}}$

$=\dfrac{1}{16}$

The elongation produced in a copper wire of length 2m and diameter 3mm, when a force of 30N is applied is [Y

$=$ 1x10

$^{11}$ N.m

$^{-2}$ ]

0%
$8.5mm$
0%
$0.85mm$
0%
$0.085mm$
0%
$85mm$

Explanation

$L=2m,$

$d=3mm, A=\dfrac{9\pi }{4}\times 10^{-6} m^{2}$

$\Delta L=\dfrac{30\times 2}{\dfrac{9\pi }{4}\times 10^{-6}\times 10^{11}}$

$=8.48\times 10^{-5} m$

$=0.085 mm$

The diameters of two steel wires are in the ratio 2:Their lengths are equal. When same force is applied on them, the ratio of the elongation produced is

0%
$4 : 9$
0%
$9 : 4$
0%
$3 : 2$
0%
$2 : 3$

Explanation

$\dfrac{r_{1}}{r_{2}}=\dfrac{2}{3}$

$\dfrac{A_{1}}{A_{2}}=\dfrac{4}{9}$

$\dfrac{\Delta L_{1}}{\Delta L_{2}}=\dfrac{A_{2}}{A_{1}}$

$=9/4$

A copper wire and a steel wire of radii in the ratio

$1:2$ lengths in the ratio

$2:1$ are stretched by the same forces. If young's modulus of copper

$=1.1\times 10^{11}N / m^{2}$ , young's modulus of steel

$= 2\times 10^{11}N /m^{2}$ . Ratio of their extensions is :

0%
$160 : 11$
0%
$16 : 110$
0%
$1 : 61$
0%
$6 : 11$

Explanation

Given,

$\dfrac{r_{c}}{r_{s}}=\dfrac{1}{2} , \quad \dfrac{L_{c}}{L_{s}}=\dfrac{2}{1} , \quad \dfrac{Area_{c}}{Area_{s}}=\dfrac{1^{2}}{2^{2}}=\dfrac{1}{4}$

(

$r_c$ ,

$L_c$ and

$Area_c$ denote the corresponding radius,length and area of specimen used)

$y_{c}=1.1\times 10^{11} N/m^{2} , \quad y_{s}=2\times 10^{11} N/m^{2}$

$\Delta L=\dfrac{FL}{Ay}$ (F is same)

$\therefore \dfrac{\Delta L_{C}}{\Delta L_{S}}=\dfrac{FL_{C}}{A_{C}y_{C}}\times \dfrac{A_{S}y_{S}}{FL_{S}}$

$=\dfrac{2}{1}\times \dfrac{4}{1}\times \dfrac{2\times 10^{11}}{1.1\times 10^{11}}$

$=\dfrac{16}{1.1}$

$=\dfrac{160}{11}$

The force that must be applied to a steel wire

$6m$ long and diameter

$1.6mm$ to produce an extension of 1mm [

$y=2.0 \times 10^{11}N.m^{-2}$ ] is approximate.

0%
$100N$
0%
$50N$
0%
$67N$
0%
$33.5N$

Explanation

$L = 6m , d = 1.6\times 10^{-3} m$
(

$d$ ,and

$L$ denote the corresponding radius,and length of specimen used)

$\Delta L=1\times 10^{-3} m , y=2\times 10^{11} N/m^{2}$

$F=\dfrac{\Delta L Ay}{L}$

$=\dfrac{1\times 10^{-3}\times \pi \times (1.6\times 10^{-3})^2\times 2\times 10^{11} }{6\times 4}$

$=67 N$

A volume of

$10^{-3}m^{3}$ is subjected to a pressure of 10 atmospheres. The change in volume is

$10^{-6}m^{3}$ . Bulk modulus of water is (Atmosphere pressure = 1x10

$^{5} N / m^{2}$ ) :

0%
1x10 $^{9}$ N / m $^{2}$
0%
1x10 $^{10}$ N / m $^{2}$
0%
1x10 $^{12}$ N / m $^{2}$
0%
1x10 $^{7}$ N / m $^{2}$

Explanation

Given,

$\Delta V=-10^{-6}m^3$

$V=10^{-3}m^3$

$P=10\times 10^5 N/m^2$

Bulk modulus,

$B=-\dfrac{F/A}{\Delta V/V}$

$B=-\dfrac{P}{\Delta V/V}$

$B=-\dfrac{10\times 10^5}{-10^{6}/10^{-3}}=1\times 1-^9$

$B=1\times 10^9 N/m^2$

The correct option is A.

Two exactly similar wires of steel (y

$=$ 20 x 10

$^{11}$ dyne/cm

$^{2}$ ) and copper (y

$=$ 12 x 10

$^{11}$ dyne/cm

$^{2}$ )are stretched by equal forces. If the total elongation is 1cm, elongation of copper wire is

0%
$3/5 cm$
0%
$5/3 cm$
0%
$3/8 cm$
0%
$5/8 cm$

Explanation

$\triangle l = \dfrac{F l}{A Y}$

$\triangle l_{s} = \dfrac{F l}{A Y_{s}} - 1$ (different cases have different subscripts)

$\triangle l_{c} = \dfrac{F l}{A Y_{c}} - 2$

Dividing 1 and 2

$\dfrac{\triangle l_{s}}{\triangle l_{c}} = \dfrac{Y_{c}}{Y_{s}} = \dfrac{12 \times 10^{11}}{20 \times 10^{11}} = \dfrac{3}{5}$

$1 cm = \triangle l_{s} + \triangle l_{c}$

$1 cm = \dfrac{3}{5} \triangle l_{c} + \triangle l_{c}$

$1 cm = \dfrac{8}{5} \triangle l_{c}$

$So, \triangle l_{c} = \dfrac{5}{8}cm$

The radii and Young's modulus of two uniform wires

$A$ &

$B$ are in the ratio

$2:\ 1$ and

$1:\ 2$ respectively. Both the wires are subjected to the same longitudinal force. If increase in the length of wire

$A$ is

$1\%$ . Then the percentage increase in length of wire

$B$ is :

0%
$1$
0%
$1.5$
0%
$2$
0%
$3$

Explanation

$Y = \dfrac{F/A}{\triangle l/l}$

$\triangle l = \dfrac{F l}{\pi r^{2}Y}$

$\triangle l_{A} = \dfrac{F l}{\pi r_{A}^{2}Y_{A}} .............(1)$

$\triangle l_{B} = \dfrac{F l}{\pi r_{B}^{2}Y_{B}} .............(2)$

Dividing 1 by 2

$\dfrac{\triangle l_{A}}{\triangle l_{B}} =\dfrac{r_{B}^{2}Y_{B}}{r_{A}^{2}Y_{A}} = (\dfrac{r_{B}}{r_{A}})^{2} (\dfrac{Y_{B}}{Y_{A}}) = (\dfrac{1}{2})^{2} (\dfrac{2}{1}) = \dfrac{1}{2}$

$\triangle l_{B} = 2\triangle l_{A} = 2 \times 1$ %

$= 2$ %

A load of

$4.0\ kg$ is suspended from a ceiling through a steel wire of length

$20\ m$ and radius

$2.0\ mm$ . It is found that the length of the wire increases by

$0.031\ mm$ as equilibrium is achieved. If

$g=3.1\ \pi\ ms^{-2}$ , the value of young's modulus in

$Nm^{-2}$ is

0%
$2.0 \times 10^{12}$
0%
$4.0 \times 10^{11}$
0%
$2.0 \times 10^{11}$
0%
$0.02 \times 10^{9}$

Explanation

For equilibrium
Weight = Tension

$mg = T$

$\therefore T = 4 \times 3.1 \pi$

$= 12.4\pi N$ (as can be inferred from the question)

$Y = \dfrac{T/A}{\triangle l/l}$

$= \dfrac{12.4 \pi / \pi (\dfrac{2}{1000})^{2}}{\dfrac{0.031}{1000}/20}$

$= \dfrac{12.4 \times 20 \times 1000 \times (1000)^{2}}{4 \times 0.031}$

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

Two wires of equal cross section, but one made up of steel and the other copper, are joined end to end. When the combination is kept under tension, the elongations in the two wires are found to be equal. If

$Y_{steel} =$ 2.0 x

$10^{11} Nm^{-2}$ and

$Y_{copper} =$ 1.1 x 10

$^{11} Nm^{-2}$ , the ratio of the lengths of the two wires is :

0%
$20 : 11$
0%
$11:20$
0%
$5 : 4$
0%
$4 : 5$

Explanation

$Y = \dfrac{F/A}{\triangle l/l}$ (standard result)

$\triangle l = \dfrac{F l}{A Y}$

$\triangle l_{S} = \dfrac{F L_{s}} {A Y_{s}}$ (for different cases different subscripts are used)

$\triangle l_{c} = \dfrac{F l_{c}}{A Y_{c}}$

$By\ problem \triangle l_{S} = \triangle l_{c}$

$\dfrac{F l_{s}}{A Y_{s}} = \dfrac{F l_{c}}{A Y_{c}}$

$\dfrac {l_{S}} {l_{c}} = \dfrac {Y_{s}} {Y_{c}} = \dfrac {2 \times 10^{11}} {1.1 \times 10^{11}} = \dfrac{20}{11}$

CBSE Questions for Class 11 Engineering Physics Mechanical Properties Of Solids Quiz 2 - 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 2 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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