Tag: properties of material substances

Quizzes (4) Questions (125)

Questions Related to properties of material substances

Multiple choice elastic energy properties of material substances elasticity properties of matter physics

A material capable of absorbing large amount of energy before fracture is known as 

  1. Ductility

  2. Toughness

  3. Resilience

  4. Plasticity
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Toughness is the ability of a material to absorb energy and plastically deform without fracturing. 

It is defined as the amount of energy per unit volume that a material can absorb before rupturing.

The correct option is option(b)

Multiple choice elastic energy properties of material substances elasticity properties of matter physics

A copper wire $1.0$ m and a steel wire of length $0.5$ m having equal cross-sectional areas are joining end to end. The composite wire is stretched by a certain load which stretches the copper wire by $1$ mm. If the Young's modulus  of copper steel are respectively $1.0\times 11^{11}Nm^{-1}$ and $ 2.0 \times 10^{11} Nm^{-2}$, the total extension of the composite wire is

  1. $1.75mm$

  2. $2.0 mm$

  3. $1.50 mm$

  4. $1.25 mm$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Extension deltaL = FL / AY. Since F and A are the same, deltaL is proportional to L/Y. For copper: 1mm = F*1 / (A*1e11). For steel: deltaL_s = F*0.5 / (A*2e11) = 0.25mm. Total extension = 1 + 0.25 = 1.25mm.

Multiple choice elastic energy properties of material substances elasticity properties of matter physics

Two wires of different materials, each $2$m long and of diameter $2\,$mm, are joined in series to form a composite wire. What force will produce a total extension of $0.9$mm. $(Y _1=2\times 10^{11}\ Pa$ & $Y _2=6\times 10^{11}\ Pa)$.

  1. $282.6$ N

  2. $212$ N

  3. $319.8$ N

  4. $382.6$ N
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given.

Length of both wires, $L=2\,m$

Radius of both wires, $d=2mm=2\times {{10}^{-3}}m$

Total extension of joined wire, $\Delta {{L} _{net}}=0.9\,mm=9\times {{10}^{-4}}\,m$

Both wires join in series, so tension in both is equal

$T=\dfrac{{{Y} _{1}}A\Delta {{L} _{1}}}{L}=\dfrac{{{Y} _{2}}A\Delta {{L} _{2}}}{L}$  

Net extension in joined wire.

$\Rightarrow \Delta {{L} _{net}}=\Delta {{L} _{1}}+\Delta {{L} _{2}}=\dfrac{TL}{{{Y} _{1}}A}+\dfrac{TL}{{{Y} _{2}}A}=\dfrac{TL}{A}\left( \dfrac{1}{{{Y} _{1}}}+\dfrac{1}{{{Y} _{2}}} \right)$

$\Rightarrow \Delta {{L} _{net}}=\dfrac{TL}{A}\left( \dfrac{1}{{{Y} _{1}}}+\dfrac{1}{{{Y} _{2}}} \right)$

$\Rightarrow T=\dfrac{A\Delta {{L} _{net}}}{L\left( \dfrac{1}{{{Y} _{1}}}+\dfrac{1}{{{Y} _{2}}} \right)}=\dfrac{\dfrac{\pi }{4}{{\left( 2\times {{10}^{-3}} \right)}^{2}}\times 9\times {{10}^{-4}}}{2\left( \dfrac{1}{2\times {{10}^{11}}}+\dfrac{1}{6\times {{10}^{11}}} \right)}=212.05\ N$

Total force produced in joined wire is $212\ N$ 

Multiple choice elastic energy properties of material substances elasticity properties of matter physics

Four identical hollow cylindrical columns of steel support a big structure of mass $50,000kg$. The inner and outer radii of each column are $30\ cm$ and $60\ cm$ respectively, Assuming the load distribution to be uniform. Calculate the compressional strain of each column,

  1. $7.2\times 10^{-7}$

  2. $3.78\times 10^{-6}$

  3. $2.78\times 10^{-4}$

  4. $3.78\times 10^{-4}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Strain = Stress / Y. Stress = Force / Area. Force = (50000 * 9.8) / 4 columns. Area = pi * (R^2 - r^2). Using Y = 2e11 Pa for steel, the calculation yields approximately 7.2e-7.

Multiple choice elastic energy properties of material substances elasticity properties of matter physics

two wires of different material, each $2m$ long and of diameter $2mm$ are joined in series to form a composite wire.What force will produce a total extension of $0.9mm$ $\left( { Y } _{ 1 }=2\times { 10 }^{ 11 }N/{ m }^{ 2 },{ Y } _{ 2 }=7\times { 10 }^{ 11 }N/{ m }^{ 2 } \right) $

  1. $22 N$

  2. $220 N$

  3. $120 N$

  4. 159 N$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In series, total extension = deltaL1 + deltaL2 = F*L1/(A*Y1) + F*L2/(A*Y2). Given L1=L2=2m, A=pi*(0.001)^2, solve for F. F = 0.9e-3 / (L/A * (1/Y1 + 1/Y2)). Calculation leads to 220 N.

Multiple choice elastic energy properties of material substances elasticity properties of matter physics

A composite wire consists of a steel Wire of length 1 5 and a co uniform cross-sectional area of ${ 2.5\times  }10^{ -5 }{ m }^{ -5 }$.It is loaded with a mass of 200kg. Find the extension produced. Young's modulus of copper is ${ 2.5\times }10^{ 11 }{ Nm }^{ -2 }$ and that of steel ${ 2.0\times  }10^{ 11 }{ Nm }^{ -2 }$

  1. <span>4.156 mm.</span>

  2. <span>2.156 mm.</span>

  3. <span>2.256 mm.</span>

  4. <span>3.156 mm.</span>
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice elastic energy properties of material substances elasticity properties of matter physics

A uniform rod of length L , area of cross-section A , mass m and Young 's modulus Y is pulled on  horizontal surface by a force f , such that the friction acting on it is F/2 . What if the elongation in the rod? 

  1. $\frac { FL }{ 2AY } $

  2. $\frac { FL }{ AY }$

  3. $\frac { 3FL }{ 2AY }$

  4. $\frac { 3FL }{ 4AY }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The tension varies along the rod due to friction. Integrating the strain along the length L with a force F and friction F/2 results in the total elongation being FL / (2AY).