Tag: forces on solids

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Questions Related to forces on solids

Multiple choice forces on solids elastic and plastic substances forces and matter properties of matter physics

A copper wire of $1mm$ diameter is stretched by applying a force on $10N$. Find the stress in the wire.

  1. $1.273\times 10^7N/m^2$
  2. $1.373\times 10^7N/m^2$
  3. $1.473\times 10^7N/m^2$
  4. $1.573\times 10^7N/m^2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Stress = Force / Area. Area = pi * (d/2)^2 = 3.14 * (0.0005)^2 = 7.85 * 10^-7 m^2. Stress = 10 / 7.85 * 10^-7 = 1.273 * 10^7 N/m^2.

Multiple choice forces on solids elastic and plastic substances forces and matter properties of matter physics

When the inter molecular distance increases due to tensile force, then 

  1. There is no force between the molecules

  2. There is a repulsive force between the molecules

  3. There is an attractive force between the molecules

  4. There is zero resultant force between the molecules
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Intermolecular forces are attractive at larger distances and repulsive at very short distances. When a tensile force increases the distance, the molecules are pulled apart, and the intermolecular force acts to restore the equilibrium, which is an attractive force.

Multiple choice forces on solids elastic and plastic substances forces and matter properties of matter physics

A steel rod of length $5\ m$ is fixed rigidly between two supports, $\alpha$ of steel$=12\times 10^{-6}/^{o}\ C$, $Y=2\times10^{12}Nm^{-2}$. With the increase in its temperature by $40^{o}\ C$, the stress developed in the rod is

  1. $9.6\times10^{5}\ Nm^{-2}$
  2. $9.6\times10^{6}\ Nm^{-2}$
  3. $9.6\times10^{7}\ Nm^{-2}$
  4. $9.6\times10^{8}\ Nm^{-2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Thermal stress developed in a rigidly fixed rod is given by the formula stress = Y * alpha * delta T. Substituting Y = 2 * 10^12 Nm^-2, alpha = 12 * 10^-6 /C, and delta T = 40 C gives 9.6 * 10^8 Nm^-2.

Multiple choice forces on solids elastic and plastic substances forces and matter properties of matter physics

A bar of cross-section A is subjected to equal and opposite tensile forces at its ends. Consider a plane section of the bar whose normal makes an angle $\theta$ with the axis of the bar.
For what value of $\theta$ is the tensile stress maximum?

  1. 0

  2. 1

  3. cant say

  4. 90
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Tensile stress$=\cfrac { force }{ area } =\cfrac { F\cos { \theta  }  }{ A\sec { \theta  }  } $
$=\cfrac { F }{ A } \cos ^{ 2 }{ \theta  } $
Tensile strength will be maximum when $\cos ^{ 2 }{ \theta  } $ is maximum i.e., $\cos { \theta  } =1$ or $\theta =0°$
Multiple choice forces on solids elastic and plastic substances forces and matter properties of matter physics

A composite rod is 1000 mm long, its two ends are 40 $mm^2$ and 30 $mm^2$ in area and length are 300 mm and 200 mm respectively. The middle portion of the rod is 20 $mm^2$ in area and 500 mm long. If the rod is subjected to an axial tensile load of 1000 N, find its total elongation (in mm). (E = 200 GPa).

  1. 0.165

  2. 0.111

  3. 0.196

  4. none of the above
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

$k _1=\dfrac{EA _1}{L _1}$,$k _2=\dfrac{EA _2}{L _2}$,$k _3=\dfrac{EA _3}{L _3}$
$k _1=E\times \dfrac{40}{300} \times 10^{-3}$
$k _2=E\times \dfrac{30}{200} \times 10^{-3}$
$k _3=E\times \dfrac{20}{500} \times 10^{-3}$
$k _1=\dfrac{4E}{3} \times 10^{-4}$
$k _1=\dfrac{3E}{3} \times 10^{-4}$
$k _1=\dfrac{2E}{3} \times 10^{-4}$
$k _1,k _2,k _3$ are in series.
$\dfrac{1}{k _{eq}}=\dfrac{1}{k _1}+\dfrac{1}{k _2}+\dfrac{1}{k _3}$
$=\dfrac{3\times 10^4}{4E}$+$\dfrac{2\times 10^4}{3E}$+$\dfrac{5\times 10^4}{2E}$
$=\dfrac{10^4}{E}(\dfrac{3}{4}+\dfrac{2}{3}+\dfrac{5}{2})$
$=\dfrac{10^4}{E} \times \dfrac{47}{12}$
$k _{eq}=\dfrac{12E}{47} \times 10^{-4}$
$F=1000N$
$1000=\dfrac{12E}{47}\times 10^{-4} \Delta l$
$\Delta l=\dfrac{1000 \times 47}{12E}\times 10^4$
$\Delta l=\dfrac{1000\times 47 \times 10^4}{12\times 200 \times 10^9}$
$=19.6\times 10^{-5}$
$=0.196 mm$

Multiple choice forces on solids elastic and plastic substances forces and matter properties of matter physics

A steel wire AB of length 100 cm is fixed rigidly at points A and B in an aluminium frame as shown in the figure If the temperature of the system increases through 100C, then the excess stress produced in the steel wire relative to the aluminium? ${ \alpha } _{ \mu }=22\times { 10 }^{ -6 }{ / }^{ 0 }Cand{ \alpha } _{ stret }=11\times { 10 }^{ -6 }{ / }^{ 0 }C$ young 's modulus of steel is $2\times { 10 }^{ 31 }{ Nm }^{ -2 }$

  1. $2.2\times { 10 }^{ 5 }$Pa
  2. $22\times { 10 }^{ 2 }$Pa
  3. $2.2\times { 10 }^{ 2 }$Pa
  4. $220\times { 10 }^{ 2 }$Pa
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice forces on solids elastic and plastic substances forces and matter properties of matter physics

A steel rod of length $1m$ and radius $10mm$ is stretched by a force $100kN$ along its length. The stress produced in the rod is then
 $\left( { Y } _{ steel }=2\times { 10 }^{ 11 }N\quad { m }^{ -2 } \right) $

  1. $3.18\times { 10 }^{ 6 }N\quad { m }^{ -2 }$
  2. $3.18\times { 10 }^{ 7 }N\quad { m }^{ -2 }\quad $
  3. $3.18\times { 10 }^{ 8 }N\quad { m }^{ -2 }$
  4. $3.18\times { 10 }^{ 9 }N\quad { m }^{ -2 }\quad $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Here $r=10mm=10\times { 10 }^{ -3 }m={ 10 }^{ -2 }m$
$L=1m,F=100kN=100\times { 10 }^{ 3 }N={ 10 }^{ 5 }N$

Stress produced in the rod is:
$Stress=\cfrac { F }{ A } =\cfrac { F }{ \pi { r }^{ 2 } } =\cfrac { 100\times { 10 }^{ 3 }N }{ 3.14\times { \left( { 10 }^{ -2 }m \right)  }^{ 2 } } =3.18\times { 10 }^{ 8 }N\quad { m }^{ -2 }\quad $

Multiple choice forces on solids elastic and plastic substances forces and matter properties of matter physics

One end of a uniform wire of length L and of weight W is attached rigidly to a point in the roof and
a weight $W _{1}$ is suspended from its lower end. If S is the are of cross-section of the wire, the stress in
the wire at a height (3 L /4) from its lower end is :

  1. $W _{1}/S$
  2. $[W _{1}+(W/4)]/S$
  3. $[W _{1}+(3W/4)]/S$
  4. $W _{1}+(W)/S$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The stress at a point is the total force below that point divided by the area. The weight of the wire below the point at 3L/4 from the bottom is (3/4)W. The total force is W1 + (3/4)W. Stress = (W1 + 3W/4) / S.