MCQ Questions for Class 11 Engineering Physics Mechanical Properties Of Solids Quiz 15

Poisson ratio for a material is 0.3 and its bulk modulus is

$1.5\times { 10 }^{ 11 }N/{ m }^{ 2 }$ . Young's modulus of the material is-

0%
$2\times { 10 }^{ 11 }N/{ m }^{ 2 }$
0%
$1.85\times { 10 }^{ 11 }N/{ m }^{ 2 }$
0%
$1.8\times { 10 }^{ 11 }N/{ m }^{ 2 }$
0%
$1.4\times { 10 }^{ 11 }N/{ m }^{ 2 }$

Explanation

$\begin{array}{l} \text { Given ithat } \\ \text { Bulk modulus }(\beta)=1.5 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2} \\ \text { Poisson ratio }(\sigma)=0.3 \\ \text { Young modulus }(y)=? \\ \text { we know velation between } \beta, \sigma, y \end{array}$

$\beta=\frac{y}{3(1-2 \sigma)}$

$y=3 \beta(1-2 \sigma)$

Putting value we get

$\begin{aligned} y &=3 \times 1.5 \times 10^{11}(1-0.6) \\ y &=1.8 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2} \end{aligned}$

The mass and length of a wire are M and L respectively. The density of the material of the wire is d. On applying the force F on the wire, the increase in length is l, then the Young's modulus of the material of the wire will be

0%
$Fdl/Ml$
0%
$FL/Mdl$
0%
$FMl/dl$
0%
$Fd{ L }^{ 2 }/Ml$

Explanation

$\begin{array}{l} \text { Given, } \\ \text { Mass of wire }=M . \\ \text { length of wire }=L \\ \text { Force applied }=F \\ \text { increased length }=l \\ \text { density }=d \\ \text { we know } \\ \text { young modulus }(Y)=\frac{\text { Stress }}{\text { strain }} \end{array}$

$y=\frac{\left(\frac{F}{A}\right)}{\left(\frac{l}{L}\right)}$

$y=\frac{F L}{A}$

$\therefore d=\frac{M}{V}$

$\rightarrow d=\frac{M}{A L} \quad \rightarrow A L=\frac{M}{d}$

$y=\frac{F L x L}{(A \times L) \cdot l}$

$y=\frac{F L^{2} d}{M l}$

Bulk modulus of water is

$2={ 10 }^{ 9 }N/{ m }^{ 2 }$ . The changes is pressure required to increase the density of water by 0.1% is.

0%
$2\times { 10 }^{ 4 }N/{ m }^{ 2 }$
0%
$2\times { 10 }^{ 6 }N/{ m }^{ 2 }$
0%
$2\times { 10 }^{ 8 }N/{ m }^{ 2 }$
0%
$2\times { 10 }^{ 9}N/{ m }^{ 2 }$

A rod AD consisting of three segments AB, BC and CD are joined together is A. The length of the three segment are respectively 0.1 m, 0.2 m and 0.15 m. The cross-section of the rod is uniformly

$10^{-4} m^2$ .A weight of 10 kg is hung from D. Calculate the displacement of point D. If [

$Y_{AB} = 2.5 \times 10^{10} N/{m^2}$ ,

$Y_{BC} = 4 \times 10^{10} N/{m^2}$ and

$Y_{CD} = 1 \times 10^{10} N/{m^2}$ . Neglect the weight of the rod. Take g = 10

$m/s^2$ ]

0%
4 $\times { 10 }^{ -6 }$ m
0%
9 $\times { 10 }^{ -6 }$ m
0%
24 $\times { 10 }^{ -6 }$ m
0%
18 $\times { 10 }^{ -6 }$ m

Explanation

$\begin{array}{l} \text { Eiven, } \\ \text { cross section of wire }(4)=10^{-4} \mathrm{~m}^{2} \\ Y_{A B}(\text { young modulus })=2.5 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2} \\ Y_{B C}=4 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2} \\ Y_{C D}=1 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2} \\ g=10 \mathrm{~m} / \mathrm{s}^{2} \end{array}$

$\begin{aligned} & Y=\frac{F l}{A \Delta l} \\ & \Delta L=\frac{F L}{A Y} \\ \Delta L_{A B} &=\frac{10 \times 10}{10^{-4}} \times \frac{L_{A B}}{Y_{A B}} \\ &=\frac{10^{6} \times 0.1}{2.5 \times 10^{10}} \\ &=4 \times 10^{-6} \mathrm{~m} \end{aligned}$

$\begin{aligned} \Delta L_{B C} &=10^{6} \times \frac{L_{B C}}{Y_{B C}} \\ &=10^{6} \times \frac{0.2}{4 \times 10^{10}} \\ &=5 \times 10^{-6 \mathrm{~m}} \\ \Delta L_{C D} &=10^{6} \times \frac{L_{C P}}{Y_{C D}} \\ &=10^{6} \times \frac{0.15}{10^{10}} \\ &=15 \times 10^{-6} \mathrm{~m} \end{aligned}$

$\therefore$ displacement of

$D=\Delta L_{A B} + \Delta L_{B C}+\Delta_{C D}$

$=\left(4 \times 10^{-6}\right)+\left(5 \times 10^{-6}\right)+\left(15 \times 10^{-6}\right)$

$d=24 \times 10^{-6}$

1990423_1322100_ans_259f258eefa44c14aa650afae8940f3e.png

A rubber cord has a cross-sectional area 1

$mm^{ 2 }$ and total unstretched length 10cm. It is stretched to 12cm and then released to project a mass of 80g. The Young's modulus for rubber is 5

$\times 10^{ 8 }Nm^{ -2 }$ . Find the velocity of mass (in m/s)?

0%
5
0%
3
0%
7
0%
2

Explanation

$\begin{array}{l} \text { Civen, } \\ \text { cross sectional Area }(A)=1 m m^{2}=10^{-6} m^{2} \\ \text { Unstreched length }(L)=10 c m=10 \times 10^{-2} \mathrm{~m} \\ \text { mass of missile } \\ \text { young modulus }(y)=5 \times 10^{8} \mathrm{Nm}^{-2} \\ \text {Streched length }(L+\Delta L)=12 \mathrm{~cm} \\ \text { Let us consider rubber cord as spring } \\ \text { then } \\ \qquad \begin{aligned} k &=\frac{y A}{L} \\ &=5 \times 10^{3} \mathrm{~N} / \mathrm{m} \end{aligned} \end{array}$ we know, and this energy is transtered to

$K E$ of missile

$\begin{aligned} \frac{1}{2} k(\Delta L)^{2} &=\frac{1}{2} m v^{2} \\ v &=\sqrt{\frac{K}{m}} \times \Delta L \\ v &=\sqrt{\frac{5 \times 10^{3}}{80 \times 10^{-3}} \times 2 \times 10^{-2}} \\ & v=5 \mathrm{~m} / \mathrm{s} \end{aligned}$

The apparatus used to determine Young's modulus of the material of a given wire is called

0%
Newton's apparatus.
0%
Wheatstone's apparatus
0%
Searle's apparatus
0%
Young's apparatus

Explanation

$\begin{array}{l} \text { The apparatus used to determine Young's modulus of } \\ \text { the material of a given wire is called searle's. } \\ \text { apparatus. } \end{array}$

The Young's experiment is performed with the lights of blue

$\left( \lambda =4360\mathring { A } \right)$ and green colour

$\left( \lambda =5460\mathring { A } \right) ,$ if the distance of the

${ 4 }^{ th }$ fringe from the centre is x, then

0%
$x(Blue)=x(Green)$
0%
$x(Blue)>x(Green)$
0%
$x(green)<X(blue)$
0%
$\dfrac { x(Blue) }{ x(Green) } =\dfrac { 5460 }{ 4360 }$

Explanation

Distance of

$n$ bright fringe

$y_n =\dfrac {n\lambda D}{d}$ i.e,

$y_n \propto \lambda$

$\therefore \dfrac {x_{n_1}}{x_{n_2}}=\dfrac {\lambda_1}{\lambda_2}\Rightarrow \dfrac {x(Blue)}{x(Green)}=\dfrac {4360}{5460}$

$\therefore x(Green) > x(Blue)$

Two metal rods of the same length and area of crosssection are fixed ends to end between rigid supports. The materials of the rods have Young moduli

$\mathrm { Y } _ { 1 } \text { and } \mathrm { Y } _ { 2 }$ , and coefficients of linear expansion

$\alpha _ { 1 } \text { and } \alpha _ { 2 }$ and When rods are cooled the junction between the rods does not shift if:

0%
$Y _ { 1 } \alpha _ { 1 } = Y _ { 2 } \alpha _ { 2 }$
0%
$Y _ { 1 } \alpha _ { 2 } = Y _ { 2 } \alpha _ { 1 }$
0%
$Y _ { 1 } \alpha _ { 1 } ^ { 2 } = Y _ { 2 } \alpha _ { 2 } ^ { 2 }$
0%
$Y _ { 1 } ^ { 2 } \alpha _ { 1 } = Y _ { 2 } ^ { 2 } \alpha _ { 2 }$

Explanation

$\begin{array}{l} Y_{1} \text { and } Y_{2} \text { are young modulus of rod } \\ \alpha_{1} \text { and } \alpha_{2} \text { are linear expansion coeffeient of rod } \\ \text { Giventhat, } \\ \text { There is no shift in Junction when vod were cooled } \end{array}$

Which means.

Force exerted on both rod due to contraction will be equal

$\begin{aligned} \text { Force } &=\text { Stress } x \text { Area } \\ &=\text { young modulus } x \text { strain } x \text { Area } \end{aligned}$

$=Y_{x} \frac{\Delta L}{L} \times A$

$=Y_{x} \frac{K \alpha \Delta T}{K} \times A \quad \therefore \Delta L=L \alpha \Delta T$ Force

$=Y_{x} \frac{\alpha \Delta T}{1} \times A$

$\begin{array}{l} \text { (Force on rod) } \\ \qquad Y_{1} \alpha_{1} \Delta T A \quad=\quad(\text { Force on rod })_{2} \\ \Delta T \text { and } A \text { are same for both } \\ \qquad Y, \alpha_{1}=Y_{2} \alpha_{2} \end{array}$

1990451_1362845_ans_3c3205a4a2f64ab0b14e82b472c81f23.png

A copper wire and a steel wire of the same diameter and length 1m and 2m respectively are connected end to 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

$\times { 10 }^{ 4 }$

0%
0.45 cm, 0.55 cm
0%
0.55 cm, 0.45 cm
0%
0.45 cm, 0.55 cm
0%
None

Copper of fixed volume V is drawn into a wire of length l. When this wire is subjected to a constant force F, the extension produced in the wire is

$\Delta l$ . If Y represents the Young's modulus, then which of the following graphs is a straight line?

0%
$\Delta l$ verses $V$
0%
$\Delta l$ verses $Y$
0%
$\Delta l$ verses $F$
0%
$\Delta l$ verses $\frac{1}{l}$

The force constant of a wire does not depend on

0%
Nature of the material
0%
Radius of the wire
0%
Length of the wire
0%
none of these

Explanation

$K = \dfrac{YA}{L} =\dfrac{Y \times \pi r^2}{L} \implies K\ \propto\ \dfrac{Yr^2}{L}$

i.e. force constant of a wire depends on Young's Modulus (nature of the material), radius of the wire and length of the wire.

The speed of a transverse wave, going on a wire having a length

$50\ cm$ and mass

$5.0\ g$ , is

$80\ m/s$ . The area of cross-section of the wire is

$1.0\ mm^ {2}$ and its Young's modulus is

$16\times 10^{11}$

$N/m^ {2}$ . Find the extension of the wire over its natural length.

0%
$0.01\ mm$
0%
$0.02\ mm$
0%
$0.002\ mm$
0%
$0.05\ mm$

Explanation

$\begin{array}{l}\text { Given, } \\\text { wave speed on wire }=80 \mathrm{~m} /\mathrm{s} \\\text { Length of wire (L) }=50 \mathrm{~cm} \\\text { mass of wire (M) }=5.0 \mathrm{gram} \\\text { Area of cross section }=1.0 \mathrm{~mm}^{2} \end{array}$

$\text { young's modulus }=1.6\times 10^{11}\mathrm{~N} /\mathrm{m}^{2}$

$\Delta L=?$

$\text { Now,Mass per unit lenght = }\frac{M}{L}$

$\begin{aligned}=&\frac{5\times 10^{-3}\mathrm{Kg}}{50\times 10^{-2}\mathrm{~m}} \\=& 0.01\mathrm{Kg} /\mathrm{m}\end{aligned}$

$\begin{array}{l}\text { we know, }\\\text { wave velocity on wive }=\sqrt{\frac{T}{\mu}} ..T=\text { Tension on wive } \\T=v^{2}\mu_{1}=\ldots\text {(i) }\end{array}$

$\begin{array}{l}\text { we also know, } \\\qquad Y=\frac{\text { stress }}{\text { strain }} \\\qquad \begin{aligned}Y &=\frac{T \cdot L}{A\cdot\Delta L}\\\Delta L &=\frac{\text{T L}}{\text {AY}}\\\text { from (i) } \\&=\frac{V^{2}\cdot \mu \cdot L}{A Y}\\&=\frac{80 \times 80 \cdot 0.01\times 50\times 10^{-2}}{1 \times 10^{-6}\times 16\times10^{11}}=0.02\mathrm{~mm}\end{aligned}\end{array}$

The ratio of radius of two wire of same material is 2: 1 Stretched by same force, then the ratio of stress is

0%
2 : 1
0%
1 : 2
0%
1 : 4
0%
4 :1

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

Find the elastic potential energy in a system shown below if the material of wires is same

$(Y=\ Young's modulus)$

0%
$\dfrac {4W^{2}L}{\pi R^{2}}$
0%
$\dfrac {15}{4}\dfrac {W^{2}L}{\pi R^{2}Y}$
0%
$\dfrac {19}{5}\dfrac {W^{2}L}{\pi R^{2}Y}$
0%
$\dfrac {17}{4}\dfrac {W^{2}L}{\pi R^{2}Y}$

A metal wire is first stretched beyond its elastic limit and then released It

0%
loses its elastic property completely and it will not contract.
0%
will contract to its original length
0%
will contract to its length at elastic limit
0%
will contract but final length will be greater than original length

The deformation of a wire under its own weight compared to the deformation of same same wire subjected to the load equal to weight of the wire is

0%
same
0%
one third
0%
half
0%
one fourth

The Bulk modulus for an incompressible liquid is

0%
Zero
0%
Unity
0%
Infinity
0%
Between $0$ to $1$

A block of mass

$'M'$ area of cross-section

$'A'$ and length

$'l'$ is placed on smooth horizontal floor. A force

$'F'$ is applied on the block as shown. If

$'y'$ is young modulus of material, then total extension in the block will be:

0%
$\dfrac{F\ell}{AY}$
0%
$\dfrac{F\ell}{2AY}$
0%
$\dfrac{F\ell}{3AY}$
0%
$cannot\ extend$

Explanation

Acceleration

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

then,

$T=\dfrac{mx}{l}\times \dfrac{F}{m}=\dfrac{Fx}{l}$

Extension in

$dx$ element

$-d\delta=\dfrac{Tdx}{Ay}=\dfrac{Fxdx}{lAy}$

Total extension ,

$\delta=\int_0^l \dfrac{Fxdx}{lAy}$

$\delta= \dfrac{Fl}{2Ay}$

1640153_1391625_ans_4c57ab11c4924a25b205e4974fb9700d.png

Two wires of the same material and length but diameter in the ratio

$1: 2$ are stretched by the same force. The ratio of potential energy per unit volume for the two wires when streched will be:

0%
$1: 1$
0%
$2: 1$
0%
$4: 1$
0%
$16: 1$

What will be the stress at

$-20^{o}C$ , if a steel rod with a cross-sectional area of

$150 mm^2$ is stretched between two fixed points? The tensile load at

$20^{o}C$ is

$5000 N$ : ( Assume

$\alpha = 11.7 \times 10^{-6}/C$ and

$Y = 200 \times 10^{11} N/m^2$ )

0%
$12.7 \times 10^{6} N/m^2$
0%
$1.27 \times 10^{6} N/m^2$
0%
$127 \times 10^{6} N/m^2$
0%
$0.127 \times 10^{6} N/m^2$

A wire is of length 1 m and area

$4\times 10^{ -8 }{ m }^{ 2 }$ having molecules at a distance 2 angstrom apart . It is elongated by 2 mm by applying a force 15 N. The the wire can be supposed to be made of m rows of atoms, the value of m is

0%
$1\times 10^{ 8 }$
0%
$1\times 10^{ 9 }$
0%
$1\times 10^{ 11 }$
0%
$1\times 10^{ 12 }$

A uniform metal wire ring of mass per unit length

$\lambda$ and radius

$r$ has cross sectional area

$S$ .
If

$Y$ is Young's in modulus of elasticity and it is rotated with and angular velocity

$\omega$ about it own axis then approximate fractional change in radius is

0%
$\left( \dfrac{\lambda r^{2} \omega^{2}}{SY} -1 \right)$
0%
$\dfrac{\lambda r^{2} \omega^{2}}{SY}$
0%
$1-\dfrac{\lambda r^{2} \omega^{2}}{SY}$
0%
$\dfrac{\lambda r \omega^{2}}{SY}$

A steel rod of cross sectional area

$1 m^2$ is acted upon by forces shown in the fig. Determine the total elongation of the bar. (

$Y = 2.0 \times 10^{11} N/m^2$ )

0%
$10 \times 10^{-7}$
0%
$13 \times 10^{-7}$
0%
$23 \times 10^{-7}$
0%
none

In the young's double slit experiments, the intensities at two points

$P_1$ and

$P_2$ on the screen are respectively.

$I_1$ and

$I_2$ . If

$P_1$ is located at the centre of a bright fringe and

$P_2$ is located at a distance equal to a quarter of fringe width from

$P_1$ , then

$\frac {I_1}{I_2}$ is

0%
2
0%
3
0%
4
0%
none of these

Explanation

$\begin{array}{l}\text { Given, } \\\text { } P_{1} \text { and } P_{2} \text { are two points on screen and } \\\text I_{ 1 } { and } I_{2}\text { are their intensities respectively. } \\\text {} P_{2} \text { is located at quarter of fringe width }\end{array}$

$Y_{p_{2}}=\frac{\beta}{4}, \beta=\text { fringe width }$

$y_{p_{2}}=\frac{\lambda D}{4 d}$

$\begin{array}{c}\text { and } p_{1} \text { is located at centre of bright fringe } \\\qquad y_{p_{1}}=0\end{array}$

$\begin{array}{l}\text { now, } \\\qquad\begin{aligned}\frac{Y_{2} d}{D} &=\frac{\lambda}{4} \\\Delta x_{p_{2}} &=\frac{\lambda}{4}\end{aligned}\end{array}$

$\begin{aligned}\text { Now, }\\\text { phase difference }(\phi) &=\frac{2 \pi}{\lambda}\times\text { Path difterence } \\&=\frac{2 \pi}{\lambda} \times \frac{\lambda}{4}\\\phi_{P_{2}} &=\frac{\pi}{2}\end{aligned}$

$\text { So, }\begin{aligned}I_{2} &=4 I_{0} \cos ^{2}\left(\frac{\phi}{2}\right) \\&=4 I_{0} \cos ^{2}(45) \\I_{2} &=\frac{4 I_{0}}{2}\end{aligned}$

$\text { Now for } P_{1} \text { as it lies at centre of fringe }$

$\begin{aligned} & \phi_{p_{1}}=0 \\ I_{1} &=4 I_{0} \cos ^{2}(0) \\ I_{1} &=4 I_{0} \\ \text { Now, } & \frac{I_{1}}{I_{2}}=\frac{4 I_{0}}{2 I_{0}}=2 \end{aligned}$

A tungsten wire, 0.5 mm in diameter , is just stretched between two fixed points at a temperature of

$4{ 0 }^{ 0 }C$ . Determine the tension in the wire when the temperature falls to

$2{ 0 }^{ 0 }C$ . (coefficient of linear expansion of tungsten =

$4.5\times 1{ 0 }^{ -6 }{ / }^{0 }C$ ; Young's modulus of tungsten =

$3.45\times 1{ 0 }^{ 11 }N{ m }^{ -2 }$ )

0%
6.097N
0%
3.097N
0%
5.097N
0%
7.097N

A body of mass 10 kg is attached to a 0.3 m long wire of cross area

${ 10 }^{ 6 }$

${ m }^{ 2 }$ Breaking steers of the wire if

$4.8\times { 10 }^{ 7 }$ What is the maximum angular velocity with which it can be rotated in a horizontal circle?

0%
4 rad/s
0%
8 rad/s
0%
16 rad/s
0%
12 rad/s

A uniform metal wire ring of mass per unit length & radius

$r$ has cross sectional area

$S$ .
If

$Y$ is young's modulus of elasticity and it is rotated with and angular velocity

$\omega$ about it own axis then approx fractional change in radius is :

0%
$\left( \frac { \lambda { r }^{ 2 }{ \omega }^{ 2 } }{ SY } -1 \right)$
0%
$\frac { \lambda { r }^{ 2 }{ \omega }^{ 2 } }{ SY }$
0%
$1-\frac { \lambda { r }^{ 2 }{ \omega }^{ 2 } }{ SY }$
0%
$\frac { \lambda { r }{ \omega }^{ 2 } }{ SY }$

In the question

$22$ , when is the tensile stress maximum ?

0%
$\theta =0^0$
0%
$\theta =30^0$
0%
$\theta =45^0$
0%
$\theta =90^0$

The steel railway track is to be stopped from expanding at the joints due to temperature increases. If the area of cross-section of the track is

$100\ cm^2$ then the elastic temperature increases

$20^oC$ to

$40^oC$ , then the work done per unit length will be: (if a = 1.2

$\times$

$10^{-5}$

$^oC^{-1}$ and

$Y=10^{11}$ N/

$m^2$ )

0%
$57.6\ J$
0%
$14.4\ J$
0%
$28.8\ J$
0%
$7.2\ J$

If uniform rod of length

$5 { m },$ area of cross-section

$50{ cm }^{ { 2 } }$ and Young's modulus

$4 \times 10 ^ { 9 } { Pa }$ is lying at rest under the action of forces as shown, then the total extension in the rod is

0%
$1.5\times 10^{ { -2 } }{ m }$
0%
$10 ^ { - 1 } { m }$
0%
$2 \times 10 ^ { - 2 } { m }$
0%
$5 \times 10 ^ { - 2 } { m }$

What force should be applied to the ends of steel rod of a cross sectional area

$10 { cm } ^ { 2 }$ to prevent it from elongation when heated from

$273 { K }$ to

$303 { K }$ ? (

$\alpha$ of steel

$10^{ { -5 } }\quad ^{ 0 }{ { C }^{ -1 } },Y=2\times 10^{ { 11 } }{ Nm }^{ { -2 } }$ )

0%
$2\times 10^{ { 4 } }{ N }$
0%
$3\times 10^{ { 4 } }{ N }$
0%
$6\times 10^{ { 4 } }{ N }$
0%
$12\times 10^{ { 4 } }{ N }$

A uniform steel rod of length 1m and area of cross section 20

$c{ m }^{2 }$ is hanging from a fixed support. Find the increase in the length of the rod.

$({ Y}_{ steel }=2.0\times 1{0 }^{ 11 }N{m }^{ -2 }, {P }_{ steel }= 7.85\times 1{0 }^{ 3 }kg{m }^{-3 })$

0%
$1.923\times 1{ 0 }^{ -5 }cm$
0%
$2.923\times 1{ 0 }^{ -5 }cm$
0%
$1.123\times 1{ 0 }^{ -5 }cm$
0%
$3.123\times 1{ 0 }^{ -5 }cm$

A uniform wire fixed at its upper end hangs vertically and supports a weight at its lower end. If its radius is r, length is L and the Young's modulus for the material of the wire is E then extension produced in the wire is

0%
directly proportional to E
0%
inversely proportional to r
0%
directly proportional to L
0%
all of these

A steel wire of length 4 m and diameter 5 mm is stretched by 5 kgwt. The increase in its length , if the Young's modulus of steel wire is

$2.4 \times { 10 }^{ 12 } dyne/{ cm }^{ 2 }$ . is

0%
0.003 cm
0%
0.0042 cm
0%
0.00042 cm
0%
0.005 cm

A light rod of length 2 m is suspended from a ceiling horizontally by means of two vertical wires of equal length tied to its ends. One of the wires is made of steel and is of cross-section

$10^{-3} m^2$ and the other is of brass of cross-section

$2 \times 10^{-3} m^2$ . x is the distance from steel wire end, at which a weight may be hung.

$Y_{steel} = 2 \times 10^{11} Pa$ and

$Y_{brass} = 10^{11} Pa$
Which of the following statement is/are correct?

0%
x = 1.2 m, if the strains of both the wires are to be equal
0%
x = 1.42 m, if the stresses of both the wires are to be equal
0%
x = 1.33 m, if the stresses of both the wires are to be equal
0%
None of the above

A wire of length

$50 cm$ and cross sectional area of

$1 mm^{2}$ is extended by

$1 mm$ . The required work will be

$(Y = 2 \times 10^{10} Nm^{-2})$

0%
$6 \times 10^{-2}\ J$
0%
$4 \times 10^{-2}\ J$
0%
$2 \times 10^{-2}\ J$
0%
$1 \times 10^{-2}\ J$

A student performs an experiment to determine the Young's modulus of wire, exactly

$2 \mathrm { m }$ long , by Searle's method. In a particular reading, the student measures the extension in the length of the wire to be

$0.8 \mathrm { mm }$ with an uncertainty of

$\pm 0.05 \mathrm { mm }$ at a load of exactly

$1.0 \mathrm { kg }$ . The student also measures the diameter of the wire to be

$0.4 \mathrm { mm }$ with a uncertainty of

$\pm 0.01 \mathrm { mm }$ . Take

$g = 9.8 \mathrm { M } / \mathrm { s } ^ { 2 }$ (exact). The Young's modulus obtained from the reading is

0%
$( 2.0 \pm 0.3 ) \times 10 ^ { 11 } \mathrm { N } \mathrm { m } ^ { -2 }$
0%
$( 2.0 \pm 0.2 ) \times 10 ^ { 11 } \mathrm { N } \mathrm { m } ^ { -2 }$
0%
$( 2.0 \pm 0.1 ) \times 10 ^ { 11 } \mathrm { N } \mathrm { m } ^ { -2 }$
0%
$( 2.0 \pm 0.5 ) \times 10 ^ { 11 } \mathrm { N } \mathrm { m } ^ { -2 }$

The bulk modulus of rubber is

$9.1 \times 10 ^ { 8 } \mathrm { N } / \mathrm { m } ^ { 2 }$ .To what depth (approximately) a rubber ball betaken in a lake so that its volume is decreased by0.1

$\%$ ?

0%
25 $\mathrm { m }$
0%
100 $\mathrm { m }$
0%
200 $\mathrm { m }$
0%
500 $\mathrm { m }$

Explanation

Given. Bulk modulues

$(\beta)=9.1 \times 10^{8} \mathrm{~N} / \mathrm{m}^{2}$

$\left|\frac{\Delta V}{V}\right|=0.1 \%$ let it is taken at a be depth of h.

$d P=P g h$

$\begin{array}{l} \text { We know, } \\ \qquad B=-\frac{d P}{\frac{d V}{V}}=\frac{P g h}{10^{-3}}=9.1 \times 10^{8} \end{array}$

$h=\frac{9.1 \times 10^{5}}{10^{3} \times 19.8}=\simeq \text { l0om. }$

glass,rubber,steel,copper in order of increasing the property of elasticity

0%
Copper, glass,rubber,steel
0%
Steel,copper,glass,ruber
0%
Rubber,glass,copper,steel
0%
Glass,rubber,copper,steel

Identical springs of steel and copper are equally stretched. The work done in producing an extension is :

0%
more for copper and less for steel
0%
more for steel and less for copper
0%
work done is equal for both copper as well as steel spring
0%
none of these

A wire is stretched by 0.01 m by a certain force F. Another wire of the same material whose diameter and length are double to the original wire is stretched by the same force. Then its elongation will be:

0%
0.005 m
0%
0.01 m
0%
0.02 m
0%
0.002 m

A thin ring of radius R is made up of a material of density 'p' and Young's modulus Y. If the ring is rotated about its centre in its own place with an angular velocity '

$\omega$ ', then the small increase in radius (

$( \Delta R )$ ) of the ring is

0%
$\frac { \rho \omega ^ { 2 } R ^ { 3 } } { Y }$
0%
$\frac { \rho \omega ^ { 2 } } { \mathrm { YR } ^ { 3 } }$
0%
$\frac { Y } { \rho \omega ^ { 2 } R ^ { 3 } }$
0%
None

Ratio of bulk modulus of elasticity for an ideal gas in isothermal condition to adiabatic condition is given by

0%
$\dfrac {C_V}{C_P}$
0%
${C_V} + {C_P}$
0%
${C_V} - {C_P}$
0%
$\dfrac {C_P}{C_V}$

A block of mass

$m$ produces an extension of 9 cm in an elastic spring of length 60 cm when it is hung by it, and the system is in equilibrium. The spring is cut in two parts of 40 cm and 20 cm lengths. The same block hangs in equilibrium with the help of these two parts connected in parallel. Find the extension (in cm) in this case.

0%
2 cm
0%
3 cm
0%
4 cm`
0%
9cm

A wire of length

$L_0$ is supplied heat to raise its temperature by

$T$ , if

$\gamma$ is the coefficient of volume expansion of the wire then the energy density stored in the wire is

0%
$\dfrac{1}{2}\gamma^2 T^2 Y$
0%
$\dfrac{1}{3}\gamma^2 T^2 Y^3$
0%
$\dfrac{1}{18}\gamma^2 T^2 Y$
0%
$\dfrac{1}{2}\dfrac{\gamma^2 T^2}{Y}$

A bar is subjected to equal and opposite forces as shown in the figure.

$PQRS$ is a plane making angle

$\theta$ with the cross-section of the bar. If the area of cross-section be

$A$ , then what is the tensile stress on

$PQRS$ .

0%
$F/A$
0%
$F\cos { \theta } /A$
0%
$F\cos { ^{ 2 }\theta } /A$
0%
$F/A\cos { \theta }$

The length of wire is 2.01 m when 5 kg is hanging from it and 2.02 m when 10 kg hanging, then natural length of wire is

0%
2m
0%
1.95m
0%
1.9m
0%
2.005m

A solid sphere of radius 'R' made of a material of bulk modulus B is surrounded by a liquid in a cylindrical container.A massless pistion of area 'A' floats on the surface of the liquid. Find the fractional change in the radius of the sphere

$(\dfrac{dR}{R})$ , when a mass M is placed on the piston to compress the liquid.

0%
$\dfrac{Mg}{2 AB}$
0%
$\dfrac{Mg}{3 AB}$
0%
$\dfrac{2Mg} { AB}$
0%
$\dfrac{3 Mg}{AB}$

Explanation

Radius of sphere,

$R$ Mass placed on massless piston =

$M$ Area of piston =

$A$

* Volume of sphere,

$V=\frac{4}{3} \pi R^{3}$ diff wrt

$R$ both side

$\begin{aligned} \therefore \quad-d V &=4 \pi R^{2} d R \\ \text { Bulk Moduls, } B &=\frac{d P}{-\frac{d V}{V}} \end{aligned}$

$\begin{aligned}B &=\frac{M g}{A} \\B &=\frac{M g}{-\frac{4 \pi R^{2} d R}{\frac{4}{3} \pi R^{3}}} \\&-\frac{d R}{R}=\frac{M g}{3 A B}\end{aligned}$

$\begin{array}{l}\text { Fractional decrease in radius of sphere } \\\text { is } \frac{M g}{3 R B}\end{array}$

A uniform steel wire of length

$3$ m and area of cross section

$2\ mm^2$ is extended through

$3$ mm. Calculate the energy stored in the wire, if the elastic limit is not exceeded.

$(Y_{\rm steel}=20\times 10^{10}\ N/m^2)$

0%
0.6 J
0%
1.6 J
0%
0.8 J
0%

1.2 J

Explanation

$\begin{array}{l} L_{0}=3 m, \quad A=2 \times 10^{-6} \mathrm{~m}^{2} \\ \Delta l=3 \times 10^{-3} \mathrm{~m}, \quad y=20 \times 10^{10} \mathrm{Nm}^{-2} \end{array}$

$\begin{aligned} \text { Energy stored in wire } &=\frac{1}{2} \times \text { Stress } \times \text { Strain } \times \text { (Volume of wire) } \\ & \text { or, } \frac{1}{2} \times(\text { Strain })^{2} \times \text { Young's modulus } \times \text { Volume } \\ &=\frac{1}{2} \times\left(\frac{3 \times 10^{-3}}{3}\right)^{2} \times\left(20 \times 10^{10}\right) \times\left(3 \times 2 \times 10^{-6}\right) \mathrm{J} \\ &=0.6 \mathrm{~J} \end{aligned}$

CBSE Questions for Class 11 Engineering Physics Mechanical Properties Of Solids Quiz 15 - 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 15 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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