Tag: properties of matter

Questions Related to properties of matter

A load of 2 kg produces an extension of 1 mm in a wire of 3 m in length and 1 mm In diameter. The Young's modulus of wire will be 

elastic energy properties of material substances elasticity properties of matter physics

  1. $3.25 \times 10 ^ { 10 } \mathrm { Nm } ^ { - 2 }$

  2. $7.48 \times 10 ^ { 12 } \mathrm { Nm } ^ { 2 }$

  3. $7.48 \times 10 ^ { 10 } \mathrm { Nm } ^ { - 2 }$

  4. $7.48 \times 10 ^ { - 10 } \mathrm { Nm } ^ { - 2 }$

Show answer
Correct Option: C
Explanation:
The applied load $=F = mg = 2 \times 9.8 = 19.6 N$
diameter $= 1 mm =1 \times {10^{ - 3}} m$
=> radius$ = 0.5 \times {10^{ - 7}} m$
Area , $A  = 3.14{r^2} = 3.14 \times 0.25 \times {10^{ - 6}}$
$=0.785 \times {10^{ - 6}}$
We know that$,$
$stress = Force/area$
$= 19.6/0.785 \times {10^{ - 6}}$
$24.96\times { 10^{ -6 } }$
change in length$, ΔL = 1 mm = {10^{ - 3}} m$
Length$, L  = 3 m$
$hence strain = ΔL/L $
$={10^{ - 3}}/3 $
we know that young's modulus is given by the ratio of stress and strain$,$
Hence$,$
$E  = 24.96 \times {10^{ - 6}}/{10^{ - 3}}/3$
$= 74.8 \times {10^9}$
$= 7.48 \times {10^{10}}N{m^{ - 2}}$
Hence, 
option $(C)$ is correct answer.

A wire is suspended by one end. At the other end, a weight equivalent to 20 N force is applied. If the increase in length is I mm, then increase in the f the wire will be

elastic energy properties of material substances elasticity properties of matter physics

  1. 0.01 J

  2. $0.02 \mathrm { J }$

  3. $0.04 J$

  4. $1.00 \mathrm { J }$

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Correct Option: A
Explanation:

$P.E = \frac{1}{2} \times 20 \times 0.001 = 0.01J$

Two wires of same length and same radius one of copper and another of steel are welded to form a long wire. An extension of $3cm$ is produced in it on applying a load at one of its ends. If the Young's modulus of steel is twice that of copper, then the extension in the steel wire will be

elastic energy properties of material substances elasticity properties of matter physics

  1. 1 cm

  2. 2 cm

  3. 1.5 cm

  4. 2.5 cm

Show answer
Correct Option: A
Explanation:

Let the length of copper wire and steel wires be L
Total extension in the joined wire of length 2L= 3cm
Let extension in copper wire be ${ l } _{ 1 }$ and extension in steel wire be ${ l } _{ 2 }$.
Let Young's Modulus of copper wire be Y. So, the Young's Modulus of steel wire is 2Y.(Given)
Since, Stress applied on the long wire is same we can write that,

$\dfrac { { l } _{ 1 } }{ L } \times Y=\dfrac { { l } _{ 2 } }{ L } \times 2Y$
Hence, we get
${ l } _{ 1 }={ 2l } _{ 2 }$                              .....(1)
We also know,
${ l } _{ 1 }{ +l } _{ 2 }=3 cm$                               .....(2)
Solving (1) and (2),
we get
${ l } _{ 2 }=1 cm$

Wire of length $L$ is stretched by length l when a force $F$ is applied at one end. If elastic limit is not exceeded, the amount of energy stored in wire is 

elastic energy properties of material substances elasticity properties of matter physics

  1. $Fl$

  2. $\dfrac{1}{2}Fl$

  3. $\dfrac{Fl^2}{L}$

  4. $\dfrac{1}{2}\dfrac{El^2}{L}$

Show answer
Correct Option: C

A composite rodd consists of a steel rod of length $25cm$ and area $2A$ and a copper rod of length $50cm$ and area $A$. The composite rod is subjected to an axial load $F$. If the Young's modulii of steel and copper are in the ration $2:1$, then

elastic energy properties of material substances elasticity properties of matter physics

  1. The extension produced in copper rod will be more

  2. The extension in copper and steel parts will be in the ratio $1 : 2$

  3. The stress applied to copper rod will be more

  4. No extension will be produced in the steel rod

Show answer
Correct Option: A,B,C
Explanation:
$Stress=\dfrac{Force}{Area}$. Since force applied on both materials is same, but area of steel is more, so stress for copper will be more.
$\Delta l=\dfrac { FL }{ AY } $
Force is same for both.
Thus, $Extension,\Delta l \propto \dfrac{l}{AY}$
Thus$\dfrac{{\Delta l} _{Steel}}{{\Delta l} _{Copper}} \propto \dfrac{{(\dfrac{l}{AY})} _{Steel}}{{(\dfrac{l}{AY})} _{Copper}}$
=$\dfrac{{(\dfrac{25}{2A.2Y})}}{{(\dfrac{50}{AY})}}=1:8$

Which of the following are correct?

elastic energy properties of material substances elasticity properties of matter physics

  1. For a small deformation of a material, the ratio (stress/strain)decreases.

  2. For a large deformation of a material, the ratio (stress/strain) decreases

  3. Two wires mad of different materials, having the same diameter and length are connected end to end. A force is applied. This stretches their combined length by $2mm$. Now, the strain is same in both the wire but stress is different.

  4. None of these is correct.

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Correct Option: D
Explanation:

(A)and (B) The ratio(stress/strain) remains constant for a material and is called the modulus of elasticity of the material.
(C)Stress is same on both wires as equal amount of force gets transmitted to each wire. Since, each wire is made up of different material, they will have different strains.

Work done on stretching a rubber will be stored in it as :

elastic energy properties of material substances elasticity properties of matter physics

  1. chemical energy

  2. heat energy

  3. muscular energy

  4. potential energy

Show answer
Correct Option: D
Explanation:

When the rubber is stretched , the work done in stretching the rubber band is converted into elastic strain energy within the rubber.

Elastic strain energy is a form of potential energy.
Hence the correct option is (D).

A brass rod of length 2 m and cross-sectional area 2.0 $\displaystyle cm^{2}$ is attached end to end to a steel rod of length L and cross-sectional area 1.0 $\displaystyle cm^{2}.$ The compound rod is subjected to equal and opposite pulls of magnitude $\displaystyle 5\times 10^{4}N$ at its ends. If the elongations of the two rods are equal the length of the steel rod (L) is
($\displaystyle Y _{Brass}=1.0\times 10^{11}N/m^{2}: : and: : Y _{Steel}=2.0\times 10^{11}N/m^{2}$)

elastic energy properties of material substances elasticity properties of matter physics

  1. 1.5 m

  2. 1.8 m

  3. 1 m

  4. 2 m

Show answer
Correct Option: D
Explanation:

$k=\dfrac{YA}{L}$

$k _S=\dfrac{Y _SA _S}{L},k _B=\dfrac{Y _BA _B}{L _B}$
$F=5\times 10^4 N$
$\Delta l _S=\Delta l _b=\Delta l$
$F=k _B\Delta l$
$\dfrac{Y _SA _S}{L}=\dfrac{Y _BA _B}{L _B}$
$(\dfrac{Y _S}{Y _B}).(\dfrac{L _B}{L})=\dfrac{A _B}{A _S}$
$2\times \dfrac{2}{L}=2$
$L=2m$

If in a wire of Young's modulus $Y$, longitudinal strain $X$ is produced then the potential energy stored in its unit volume will be :

elastic energy properties of material substances elasticity properties of matter physics

  1. $0.5Y{X}^{2}$

  2. $0.5{Y}^{2}X$

  3. $2Y{X}^{2}$

  4. $Y{X}^{2}$

Show answer
Correct Option: A
Explanation:

We know that, the potential energy stored per unit volume is,

$W=\cfrac{1}{2} \times strss \times strain$
And, 
$Y=\cfrac{stress}{Strain}$
$\therefore Stress=Y\times strain$
$\therefore W=\cfrac{1}{2} \times Y \times {(strain)}^{2}$
$\Rightarrow\cfrac{1}{2}Y{X}^{2}=0.5 Y{X}^{2}$ 

A composite wire of a uniform cross-section $5.5\times 10^{-5}m^{2}$ consists of a steel wire of length $1.5\ m$ and a copper wire of length with a mass of $200\ kg$ is [Young's modulus of steel is $2\times 10^{11} N\ m^{-2}$ and that of copper is $1\times 10^{11}Nm^{-2}$. Take $g = 10\ ms^{-2}]$

elastic energy properties of material substances elasticity properties of matter physics

  1. $1\ mm$

  2. $2\ mm$

  3. $3\ mm$

  4. $4\ mm$

Show answer
Correct Option: A
Explanation:
$\Delta Ps=F\cdot \dfrac{P}{A}\cdot Ys$

$=2000\times \dfrac {1.5}{5.5\times10^{-5}}\times 2\times 10^{11}$

$=\dfrac{3\times 10^{-2}}{55\times 2}$

$=0.00027m$

$\Delta Pc=2000\times \dfrac {20}{5.5\times10^{-5}}\times 10\times 10^{11}$

$=0.00072m$

$\Delta P=\Delta Ps+\Delta Pc=1mm$