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

To break a wire of 1 m length, minimum 40 kg weight is required. Then the wire of the same material of double radius and 6 m length will require breaking weight 

  1. 80 kg weight

  2. 240 kg weight

  3. 200 kg weight

  4. 160 kg weight
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Breaking weight is proportional to the cross-sectional area of the wire, meaning it scales with the square of the radius (r^2), and is independent of length. Doubling the radius increases the breaking weight by a factor of 2^2 = 4. Thus, 40 kg * 4 = 160 kg.

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

Which of the following shows greater increment in length when subjected to same load to wires made of same material:

  1. $L = 1 m$ and $r = 1 mm$
  2. $L = 1 m$ and $r = 2 mm$
  3. $L = 2 m$ and $r = 1 mm$
  4. $L = 2 m$ and $r = 2 mm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

From Young's modulus relation, delta L = (F * L) / (Y * A) = (F * L) / (Y * pi * r^2). The extension is directly proportional to length L and inversely proportional to the square of the radius r^2. Maximizing L/r^2 gives the greatest increment; for option C, L = 2 and r = 1, giving a ratio of 2/1 = 2, which is the highest among the choices.

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. 4.156 mm.

  2. 2.156 mm.

  3. 2.256 mm.

  4. 3.156 mm.
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).