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Applications of elasticity - class-XI

Description: applications of elasticity
Number of Questions: 31
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Tags: properties of matter mechanical properties of solids physics properties of bulk matter
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State whether true or false :
The hollow shaft is much stronger than a solid shaft of same mass, same length and same material.

  1. True

  2. False


Correct Option: A
Explanation:

Modulus of elasticity = $\dfrac { stress }{ strain } $     and, stress = $\dfrac { Force }{ Area } $

$\therefore $  Elasticity $\alpha \dfrac { 1 }{ Area } $
As follow shaft, so, elasticity of the follow shaft is more than a solid one. As elasticity measures rigidity. So, hollow shaft is stronger.

In which year Robert Hooke presented his law of elasticity?

  1. 1672

  2. 1674

  3. 1676

  4. 1678


Correct Option: C
Explanation:

In 1676, Robert Hooke presented his law of elasticity, now called Hooke's law.

 A cable that can support a load of 1000 N is cut into equal parts. the maximum load that can be supported by the either part is:-
  1. 1000 N

  2. 2000 N

  3. 500 N

  4. 250 N


Correct Option: A
Explanation:

Maximum load supported by the cable is directly proportional to the breaking stress.

Since,

Breaking stress = F / A

where F is the force and A is the cross-sectional area.

As we see that the breaking stress is independent of the length of the cable. So, if the cable is cut in two equal parts, the maximum load that can be supported by either parts of the cable remain the same as before.

State whether true or false :
The metal used in construction of a bridge should have high Young's modulus.

  1. True

  2. False


Correct Option: A
Explanation:

Young's modulus measures modulus of elasticity of a material. 

Elasticity defines rigidity of the material. 
So, for construction, we need metals which have high elasticity i.e. high young's modulus.

A silver wire of length $10 $ metre and cross-sectional area $10^{-8} m^{2}$ is suspended vertically and a weight of $10 N$ is attached to it. Young's modulus of silver and its resistivity are $7 \times 10^{10} N/m^{2}$ and $1.59 \times 10^{8} N/m^{2}$ \Omega - m$ respectively. The increase in its resistance is equal to:-

  1. $0.0455 \Omega$

  2. $0.455 \Omega$

  3. $0.91 \Omega$

  4. $0.091 \Omega$


Correct Option: C

A steel wire is stretched by 5 kg wt, If the radius of the wire is doubled its Young's modulus

  1. Remains unchanged

  2. Becomes double

  3. Becomes half

  4. Becomes 1/4 times


Correct Option: A

In the system shown in figure pulley is smooth. String is massless and inextensible. The acceleration of the system a, tensions ${T} _{1}\ and {T} _{1}\left (g=10{m/s}^{2}\right)$ are 

  1. $\dfrac { 20 }{ 3 } { m/s }^{ 2 },\dfrac { 50 }{ 3 } N,\dfrac { 60 }{ 3 } N$

  2. $\dfrac { 10 }{ 3 } { m/s }^{ 2 },\dfrac { 100 }{ 3 } N,\dfrac { 60 }{ 3 } N$

  3. $\dfrac { 20 }{ 3 } { m/s }^{ 2 },\dfrac { 100 }{ 3 } N,\dfrac { 60 }{ 3 } N$

  4. $\dfrac { 20 }{ 3 } { m/s }^{ 2 },\dfrac { 100 }{ 3 } N,\dfrac { 50 }{ 3 } N$


Correct Option: C

A stone of mass 'm' s projected from a rubber catapult of length 'l' and cross-sectional area A stretched by an amount 'e'. If Y be the young's modulus of rubber then the velocity of projection of stone?

  1. $Y \sqrt {\dfrac{Ae^2}{lm}}$

  2. $ \sqrt {\dfrac{Ae^2}{lm}}$

  3. $Y \sqrt {\dfrac{YAe^2}{lm}}$

  4. $Y \sqrt {\dfrac{YAe^4}{lm}}$


Correct Option: C

Two wires of equal length and cross section are suspended. their young's modulus are $Y _1$ and $Y _2$ respectively. their equivalent young's modulus of elasticity is

  1. $Y _1+Y _2$

  2. $Y _1Y _2$

  3. $Y _1-Y _2$

  4. $\dfrac{Y _1+Y _2}2$


Correct Option: C

In the Young's double slit experiment the intestines 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 bright fringe and $P _{2}$ is located at a distance equal to a quarter of fringe width from $P _{1}$, then $I _{1}/I _{2}$ is 

  1. $2$

  2. $1/2$

  3. $4$

  4. $16$


Correct Option: A

Modulus of the wire then the energy density stored in the wire is

  1. $\frac{1}{2}\gamma ^{2}T^{2}Y$

  2. $\frac{1}{3}\gamma ^{2}T^{2}Y^{3}$

  3. $\frac{1}{3}\frac{\gamma ^{2}T^{2}}{Y}$

  4. $\frac{1}{18}\gamma ^{2}T^{2}Y$


Correct Option: D

An Indian rubber cord $L$ metre long and area of cross-section $A$ meter$^2$ is suspended vertically. Density of rubber is $\rho \ kg/$ meter$^3$ and Young's modulus of rubber is $Y$ Newton/metre$^2$. If the cord extends by $l$ metre under its own weight, then extension $l$ is:

  1. $\dfrac{L^2 \rho g}{Y}$

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

  3. $\dfrac{L^2 \rho g}{4Y}$

  4. $\dfrac{Y}{L^2 \rho g}$


Correct Option: B
Explanation:

A small differential element $dx$ at distance $x$ from the bottom of chord

Force acting on this element $ = \dfrac{M}{L} \times x \times g$
If extension  in this element is $dl$
Then
$\begin{array}{l} \dfrac { { dl } }{ { dx } } =\dfrac { f }{ { Ay } } =\dfrac { { Mg } }{ { LAy } } x \ \int  _{ 0 }^{ l }{ dl }=\dfrac { { Mg } }{ { LAy } } \int  _{ 0 }^{ l }{ xdx } \ l=\dfrac { { Mg{ L^{ 2 } } } }{ { 2LAy } } =\dfrac { { Mgl } }{ { 2Ay } }  \ If\, density\, is\, f\, then, \ \rho =\dfrac { m }{ { AL } }  \ So,\, l=\dfrac { { \rho g{ L^{ 2 } } } }{ { 2y } }  \ Hence\, option\, B\, is\, the\, correct\, answer. \end{array}$

A breaking stress of a material is ${ 10 }^{ 6 }N/{ m }^{ 2 }$ If density of material is $3\times 10^{ 3 }kg/{ m }^{ 3 }$, what should be the length of the material so that its breaks by it own weight?

  1. 43.3 m

  2. 23.3 m

  3. 13.3 m

  4. 33.3 m


Correct Option: D

Two wire of same radius and length are subjected to the same load, One wire is of steel and the other is copper. If Young's modulus of steel is twice that of copper, then the ratio of elastic energy stored per unit volume of steel to that of copper wire is

  1. $2:1$

  2. $1:2$

  3. $1:4$

  4. $4:1$


Correct Option: B
Explanation:
${ Y } _{ S }=2{ Y } _{ C }$, then the ratio of elastic energy stored per unit volume.
$E=\dfrac { 1 }{ 2 } \dfrac { { YL }^{ 2 }\alpha  }{ L } $
or,  $\dfrac { E }{ { l }^{ 2 }\alpha  } =\dfrac { 1 }{ 2 } \dfrac { Y }{ L } $
Since same length and same radius.
$\dfrac { { E }^{ 1 } }{ { l }^{ 2 }\alpha  } =\dfrac { 1 }{ 2 } \dfrac { { Y } _{ C } }{ L } $
$\dfrac { { E }^{ 11 } }{ { l }^{ 2 }\alpha  } =\dfrac { 1 }{ 2 } \times \dfrac { 2{ Y } _{ C } }{ L } $
$\dfrac { { E }^{ 1 } }{ { E }^{ 11 } } =\dfrac { 1 }{ 2 } $.

The number of independent elastic constant of a solid is=

  1. 1

  2. 2

  3. 3

  4. 4


Correct Option: C

The depression produced at the end of a $50 cm$ long cantilever on applying a load is $15 mm$. The depression produced at a distance of $30 cm $ from the rigid end will be

  1. 3.24 mm

  2. 1.62 mm

  3. 6.48 mm

  4. 12.96 mm


Correct Option: C
Explanation:

Length of cantilever, $L= 50cm=0.5m$
Deflection at the end $w(L)= 15mm$
Find: Deflection $w(x)$ at 30cm from rigid end.
Depression at a point in a cantilever beam with load at one end is given by,$ w(x)=\dfrac { P{ x }^{ 2 }(3L-x) }{ 6EI } $
We have,
$\dfrac { w(x) }{ w(L) } =\dfrac { \dfrac { P{ x }^{ 2 }(3L-x) }{ 6EI }  }{ \dfrac { PL^{ 3 } }{ 3EI }  } $

$\dfrac { w(0.3m) }{ w(0.5m) } =\dfrac { \dfrac { P{ (0.3) }^{ 2 }(3(0.5)-0.3) }{ 6EI }  }{ \dfrac { P(0.5)^{ 3 } }{ 3EI }  } $

This gives, $w(0.3m)=6.48mm$

A solid cylindrical rod of radius $3 mm$ gets depressed under the influence of a load through $8 mm$. The depression produced in an identical hollow rod with outer and inner radii of $4 mm$ and $2 mm$ respectively, will be

  1. 2.7mm

  2. 1.9mm

  3. 3.2mm

  4. 7.7mm


Correct Option: A
Explanation:

Depression in Solid cylinder $\delta _{1}= 8mm$
Its radius $r _{1}= 3mm$
Outer radius of Hollow cylinder $R _{2}= 4mm$
Inner radius of Hollow cylinder $r _{2}= 2mm$
Let depression in this cylinder be $\delta _{2}$.
Depression $\delta =\dfrac { W{ l }^{ 3 } }{ 12\pi{r }^{ 4 }Y } $
From the above equation we know that $\delta$ is proportional to $\dfrac{1}{r^{4}}$
Hence, we have
$\dfrac{\delta _{1}}{\delta _{2}}= \dfrac{{R _{2}}^{4}-{r _{2}}^{4}}{r _{1}^{4}}$
Substituting the values in above equation, we get
$\delta _{2}=2.7mm$

A beam of cross section area A is made of a material of Young modulus Y. The beam is bent into the arc of a circle of radius R. The bending moment is proportional to

  1. $\displaystyle \frac{Y}{R}$

  2. $\displaystyle \frac{Y}{RA}$

  3. $\displaystyle \frac{R}{Y}$

  4. $YR$


Correct Option: A
Explanation:

Bending moment $C=\dfrac{Y{I} _{G}}{R}$
Therefore, $C$ is proportional to $\dfrac{Y}{R}$

A steel wire of length $L$ and area of cross-section A shrinks by $\Delta l$ during night. Find the tension developed at night if Young's modulus is $Y$ and wire is clamped at both ends

  1. $\displaystyle \frac{AYL}{\Delta l}$

  2. $AYL$

  3. $AY\Delta l$

  4. $\displaystyle \frac{AY\Delta l}{L}$


Correct Option: D
Explanation:

$\displaystyle \frac{\Delta l}{L}=\frac{F}{AY}:or:F=\frac{AY\Delta l}{L}$ (using standard equation)

A wire of radius $1 mm$ is bent in the form of a circle of radius $10 cm$. The bending moment will be $(Y = 2\times10^{11}N/m^{2})$

  1. 3.14 N/m

  2. 6.28 N/m

  3. 1.57 N/m

  4. 15.7 N/m


Correct Option: C
Explanation:

Bending Moment $C=\dfrac { Y{ I } _{ G } }{ R } $
where $Y= 2\times { 10 }^{ 11 }N/{m}^{2}$
$R= 10cm= 0.1m$
${ I } _{ G }=\dfrac{\pi{R}^{4}}{4} = \dfrac{\pi{0.1}^{4}}{4} =7.85\times{10}^{-5} m^{4}$
Put these in the expression for $C$,
$C=1.57N/m$

A body of mass 3.14 kg is suspended from one end of a wire of length 10 m. The radius of cross-section of the wire is changing uniformly from $5 \times 10^{-4}$ m at the top (i.e. point of suspension) to $9.8 \times 10^{-4}$ m at the bottom. Young's modulus of elasticity is $2 \times 10^{11} \ N/m^2$. The change in length of the wire is

  1. $4 \times 10^{-3}$ m

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

  3. $ 10^{-3}$ m

  4. $2 \times 10^{-3}$ m


Correct Option: C

A wire of cross section $A$ is stretched horizontally between two clamps located $2lm$ apart. A weight $Wkg$ is suspended from the mid-point of the wire.If the Young's modulus of the material is $Y$, the value of extension $x$ is

  1. $ { \left( \cfrac { Wl }{ YA } \right) }^{ 1/3 }$

  2. $ { \left( \cfrac { YA }{ WI } \right) }^{ 1/3 }$

  3. $\cfrac { 1 }{ l } { \left( \cfrac { Wl }{ YA } \right) }^{ 2/3 }$

  4. $ l{ \left( \cfrac { W }{ YA } \right) }^{ 2/3 }$


Correct Option: D
Explanation:
Let $M$, $L$ and $T$ represent the dimensions of mass, length and time respectively. Then:
${ [(\dfrac { Wl }{ YA } ) }^{ \dfrac { 1 }{ 3 }  }]=\dfrac { [{ M }^{ \dfrac { 1 }{ 3 }  }{ L }^{ \dfrac { 2 }{ 3 }  }{ T }^{ -\dfrac { 2 }{ 3 }  }] }{ [{ M }^{ \dfrac { 1 }{ 3 }  }{ L }^{ \dfrac { 2 }{ 3 }  }{ T }^{ -\dfrac { 2 }{ 3 }  }] } =[{ M }^{ 0 }{ L }^{ 0 }{ T }^{ 0 }]$
Since extension has dimensions of length, the only option which fits the requirement is the last one where an additional length term is multiplied.

Relation among elastic contents $Y, G, B, \sigma $

  1. $\dfrac{9}{Y} = \dfrac{1}{B} + \dfrac{3}{G}$

  2. $Y = 2G (1 + \sigma)$

  3. $Y = 3B (1 - 2\sigma)$

  4. $\sigma = \dfrac{3B - 2G}{2(G + 3B)}$


Correct Option: A

You are given three wires $  \mathrm{A}, \mathrm{B}  $ and $ \mathrm{C}  $ of the same length and cross section. They are each stretched by applying the same force to the ends. The wire A is stretched least comes back to its original length when the stretching force is removed. The wire $  B  $ is stretched more than $  A  $ and also comes back to its original length when the stretching force is removed. The wire C is stretched most and remains stretched even when the stretching force is removed. The greatest Young's modulus of elasticity is possessed by the material of a wire

  1. A

  2. B

  3. C

  4. All have the same elasticity


Correct Option: A

In designing, a beam for its use to support a load. The depression at center is proportional to (where, $Y$ is Young's modulus).

  1. $Y^2$

  2. $Y$

  3. $\dfrac{1}{Y}$

  4. $\dfrac{1}{Y^2}$


Correct Option: C

A light rod of length $2\ m$ is suspended from the 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 $0.1\ cm^{2}$. A weight is suspended from a certain point of the rod such that equal stress are produced in both the wires. Which of the following are correct?

  1. The ratio of tension in the steel and brass wires is $0.5$

  2. The load is suspended at a distance of $400/3cm$ from the steel wire

  3. Both (a) and (b) are correct

  4. Neither (a) nor (b) are correct


Correct Option: A,B,C
Explanation:
As $Stress=Force(Tension\ here)/Area$
As, $stress _{steel}=stress _{brass}$
$\Rightarrow Tension _{steel}/Tension _{brass}=Area _{steel}/Area _{brass}=0.1/0.2=0.5$
option A is correct.

As tension is inversely proportional to the distance of suspension of load. So distance of suspension for brass=0.5 x distance of suspension for steel.
As distance of suspension for steel+distance of suspension for brass$=2m=200cm$
So distance of suspension for steel+0.5xdistance of suspension for steel$=2m=200cm$
distance of suspension for steel$=\dfrac{2}{1.5}m=\dfrac{400}{3}cm$
option B is correct.

Both option A and B is correct.

For the same cross-section area and for a given load, the ratio of depression for the beam of a square cross-section and circular cross-section is 

  1. $3:\pi$

  2. $\pi :3$

  3. $1:\pi$

  4. $\pi :1$


Correct Option: A
Explanation:

$\displaystyle \delta=\displaystyle\frac{Wl^3}{3YI}$, where $W=$load$,\ l=$length of beam$,\ I=$moment of inertia$=\dfrac{b{d}^{3}}{12}$ for rectangular beam, and for square beam$,\ b=d.$ Thus, ${I} _{1}=\dfrac{{b}^{4}}{12}$

Now, for circular cross section, $\displaystyle I _2=\left[\dfrac{\pi r^4}{4}\right]$

$\therefore \delta _1=\dfrac{Wl^3\times 12}{3Yb^4}=\dfrac{4Wl^3}{Yb^4}$

and $\delta _2=\dfrac{Wl^3}{3Y(\pi r^$/4)}=\displaystyle\frac{4Wl^3}{3Y(\pi r^4)}$

Thus, $\dfrac{\delta_1}{\delta_2}=\dfrac{3\pi r^4}{b^4}=\dfrac{3\pi r^4}{(\pi r^2)^2}=\dfrac{3}{\pi}$
$(\because b^2=\pi r^2$ as they have same cross sectional area)

A beam of metal supported at the two edges is loaded at the centre. The depression at the centre is proportional to 

  1. $Y^2$

  2. $Y$

  3. $1/Y$

  4. $1/Y^2$


Correct Option: C
Explanation:

Young's modulus: $ Y=\dfrac { \dfrac { F }{ a }  }{ \dfrac { \Delta l }{ l }  } =\dfrac { Fl }{ a(\Delta l) } $

$ \Delta l\rightarrow \dfrac { 1 }{ Y } $

The buckling of a beam is found to be more if __________.

  1. The breadth of the beam is large

  2. The beam material has large value of Young's modulus

  3. The length of the beam is small

  4. The depth of the beam is small


Correct Option: D
Explanation:

Critical buckling stress of a column formula is given by 

$\sigma=\dfrac{F}{A}=\dfrac{{\pi}^2 r^2 E}{L^2}$
where $\sigma$ = critical stress
$L$= unsupported length of the column
$r=$ least radius 
So if the depth of the beam i small, buckling of a beam will be more.

Assertion: When a wire is stretched to three times its length, its resistance becomes 9 times

Reason: $R = {{\rho l} \over a}$

  1. both, Assertion and Reason are true and the reason is correct explanation of the Assertion

  2. both, Assertion and Reason are true and the reason is not correct explanation of the Assertion

  3. Assertion is true, but the reason is false.

  4. Both, Assertion and reason and false


Correct Option: A

A light rod of length $2.00 m$ is suspended from the 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 $2\times10^{-3}m^{2}$ . Find out the position along the rod at which a weight may be hung to produce.(Youngs modulus for steel is 2x10$^{11}$N /m$^{2}$ and for brass is 10$^{11}$N / m$^{2}$ )
a) equal stress in both wires
b) equal strains on both wires

  1. $1.33 m, 1m$

  2. $1m, 1.33 m$

  3. $1.5 m, 1.33 m$

  4. $1.33m, 1.5 m$


Correct Option: A
Explanation:
For  equal  stress
$ \dfrac{F _{1}}{A _{1}} = \dfrac{F _{2}}{A _{2}}$
$ \dfrac{F _{1}}{F _{2}} = \dfrac{A _{1}}{A _{2}} = \dfrac{10^{-3} m^{2}}{2 \times 10^{-3}m^{2}} = \dfrac{1}{2}$
$ 2F _{1} = F _{2}$
For  balance  of  rod
$ W = F _{1} + F _{2}$
$W = \dfrac{3 F _{2}}{2}$
$ F _{2} = \dfrac{2}{3} W$
Now equating torque
$Wx = F _{2} \times 2$
$x = \dfrac{2}{3} \times 2 = \dfrac{4}{3} = 1.33m$
For equal strain
$ \dfrac{\triangle l _{1}}{l} = \dfrac{\triangle l _{2}}{l}$
or
$\dfrac{\sigma _1}{Y _1}=\dfrac{\sigma _2}{Y _2}$
or
$\dfrac{F _1}{10^{-3}\times 2\times 10^{11}}=\dfrac{F _2}{2\times 10^{-3}\times 10^{11}}$
Thus we get $F _1=F _2$.
So, weight  will  be  hanging  mid - way 1m
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