Tag: forces and matter

Questions Related to forces and matter

A rubber cord 10 m is suspended vertically . How much does is stretch under its own weight (density of rubber is $1500kg{ m }^{ -3 },Y=5\times { 10 }^{ 8 }{ Nm }^{ -2 },g={ ms }^{ -2 }$)

  1. $15\times { 10 }^{ -4 }m\quad $

  2. $7.5\times { 10 }^{ -4 }m\quad $

  3. $12\times { 10 }^{ -4}m $

  4. $25\times { 10 }^{ -4 }m\quad $


Correct Option: B

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. $1.573\times 10^7N/m^2$


Correct Option: A

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


Correct Option: C

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. $9.6\times10^{8}\ Nm^{-2}$


Correct Option: C

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


Correct Option: A
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 $\theta =0°$

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


Correct Option: C
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.
$\dfrac{1}{k _{eq}}=\dfrac{1}{k _1}+\dfrac{1}{k _2}+\dfrac{1}{k _3}$
$=\dfrac{3\times 10^4}{4E}$+$\dfrac{2\times 10^4}{3E}$+$\dfrac{5\times 10^4}{2E}$
$=\dfrac{10^4}{E}(\dfrac{3}{4}+\dfrac{2}{3}+\dfrac{5}{2})$
$=\dfrac{10^4}{E} \times \dfrac{47}{12}$
$k _{eq}=\dfrac{12E}{47} \times 10^{-4}$
$F=1000N$
$1000=\dfrac{12E}{47}\times 10^{-4} \Delta l$
$\Delta l=\dfrac{1000 \times 47}{12E}\times 10^4$
$\Delta l=\dfrac{1000\times 47 \times 10^4}{12\times 200 \times 10^9}$
$=19.6\times 10^{-5}$
$=0.196 mm$

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


Correct Option: C

A metal wire of length L, area of cross-section A and Young modulus Y behaves as a spring of spring constant

  1. $K= \frac{YA}{L}$

  2. $K= \frac{2YA}{L}$

  3. $K= \frac{YA}{2L}$

  4. $K= \frac{YL}{A}$


Correct Option: A

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 $


Correct Option: C
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 $

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. $W _{1}+(W)/S$


Correct Option: C