Tag: properties of matter

Questions Related to properties of matter

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

physics mechanical properties of solids applications of elasticity elastic energy properties of matter

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

  2. $AYL$

  3. $AY\Delta l$

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

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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 $(Y = 2\times10^{11}N/m^{2})$

physics mechanical properties of solids applications of elasticity elastic energy properties of matter

  1. 3.14 N/m

  2. 6.28 N/m

  3. 1.57 N/m

  4. 15.7 N/m

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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

physics mechanical properties of solids applications of elasticity elastic energy properties of matter

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

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

  3. $ 10^{-3}$ m

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

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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

physics mechanical properties of solids applications of elasticity elastic energy properties of matter

  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 }$

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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 $

physics mechanical properties of solids applications of elasticity elastic energy properties of matter

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

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

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

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

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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

physics mechanical properties of solids applications of elasticity elastic energy properties of matter

  1. A

  2. B

  3. C

  4. All have the same elasticity

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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).

physics mechanical properties of solids applications of elasticity elastic energy properties of matter

  1. $Y^2$

  2. $Y$

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

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

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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?

physics mechanical properties of solids applications of elasticity elastic energy properties of matter

  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

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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 

physics mechanical properties of solids applications of elasticity elastic energy properties of matter

  1. $3:\pi$

  2. $\pi :3$

  3. $1:\pi$

  4. $\pi :1$

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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 

physics mechanical properties of solids applications of elasticity elastic energy properties of matter

  1. $Y^2$

  2. $Y$

  3. $1/Y$

  4. $1/Y^2$

Show answer
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 } $