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Forces on solids - class-XI

Description: forces on solids
Number of Questions: 20
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Tags: properties of matter forces and matter physics
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A cable that can support a load of 800 N is cut into two equal parts. The maximum load that can be supported by either part is 

  1. 100 N

  2. 400 N

  3. 800 N

  4. 1600 N


Correct Option: C
Explanation:

Breaking stress $= \dfrac{800 N}{A} \Rightarrow BA$

  $F = 800 N$
Breaking stress doesn't depend upon the length of the cable.

A uniform steel bar of cross-sectional area A and length L. is suspended so that it hangs vertically. The stress at the middle point of the bar is ( $\rho $ is the density of steel)

  1. $\frac{L}{2A} \rho g$

  2. $\frac{L\rho g}{2} $

  3. $\frac{LA}{\rho g}$

  4. $L\rho g$


Correct Option: C

On suspending a weight $Mg$ the length $l$ of elastic wire and area of cross section $A$ its length becomes double the initial length. The instantaneous stress action on the wire is:

  1. $\dfrac{Mg}{A}$

  2. $\dfrac{Mg}{2A}$

  3. $\dfrac{2Mg}{A}$

  4. $\dfrac{4Mg}{A}$


Correct Option: C

The velocity of the transverse waves in a wire of density $8000kg/m^3$ is $300 m/s$. The tensile stress in the wire is then

  1. $7.2\times10^8 N/m^2$

  2. $6.8\times10^8 N/m^2$

  3. $5.2\times10^8 N/m^2$

  4. $8.4\times10^8 N/m^2$


Correct Option: A
Explanation:

Given density = 8000 $kg/{ m }^{ 3 }$ and velocity =300 m/s

Also we know velocity of transverse wave  $\text{V}=\sqrt { \dfrac { T }{ \mu  }  } \text{where T= tension and}\quad \mu =\text{mass per unit lenght}\quad =\dfrac { m }{ L } \quad $
also $\text{Density}\quad \rho =\dfrac { mass }{ Area\times Lenght } \ \rho \times A=\dfrac { m }{ L } $
$\text{Tension T}=\rho A{ V }^{ 2 }$
We know stress $\sigma =\dfrac { Force }{ Area } =\dfrac { Tension }{ Area } =\dfrac { { V }^{ 2 }\rho A }{ A } ={ V }^{ 2 }\rho $
${ 300 }^{ 2 }\times 8000=7.2\times { 10 }^{ 8 }\dfrac { N }{ { m }^{ 2 } } $ 

An external force of $10\ N$ acts normally on a square area of each side $50\ cm$. The stress produced in equilibrium state is

  1. $10\ N/m^{2}$

  2. $20\ N/m^{2}$

  3. $40\ N/m^{2}$

  4. $50\ N/m^{2}$


Correct Option: C

The length of wire is increased by $0.06\%$ by a load of $40N$ whose tensile modulus is $20\times10^{10}N/M^2$.The subjected stress is 

  1. $12\times10^{10}N/m^2$

  2. $1.2\times10^{8}N/m^2$

  3. $120N/m^2$

  4. $1.25\times10^6N/m^2$


Correct Option: B
Explanation:

$\cfrac{\triangle l}{l}\times 100=0.06$

$\implies \cfrac{\triangle l}{l}=\cfrac{0.06}{100}$
Now stress $=\cfrac{20\times 10^{10}\times 0.06}{1000}\=1.2\times 10^8\ N /m^2$

According to $C.E$ van der Waal, the interatomic potential varies with the average interatomic distance $(R)$ as 

  1. $R^{-1}$

  2. $R^{-2}$

  3. $R^{-4}$

  4. $R^{-6}$


Correct Option: D
Explanation:

According to the relation

$ V(r)=\dfrac{-3}{4}\dfrac{{{\alpha }^{2}} _{0}l}{{{(4\pi {{\varepsilon } _{0}})}^{2}}{{R}^{6}}} $

$ V(r)\propto \dfrac{1}{{{R}^{6}}} $

$ V(r)\propto {{R}^{-6}} $


Overall changes in volume and radii of a uniform cylindrical steel wire are $0.2\% $ and $0.002\%$ respectively when subjected to some suitable force. Longitudinal tensile stress acting on the wire is $\left( {Y = 2.0 \times {{10}^{11}}N{m^{ - 2}}} \right)$

  1. $3.2 \times {10^9}N{m^{ - 2}}$

  2. $3.2 \times {10^7}N{m^{ - 2}}$

  3. $3.6 \times {10^9}N{m^{ - 2}}$

  4. $3.6 \times {10^7}N{m^{ - 2}}$


Correct Option: C

A steel wire has an ultimate strength of above $2.0 \times 10 ^ { 7 } \mathrm { kg } - \mathrm { w } \mathrm { J } / \mathrm { m } ^ { 2 }$ . How large a load can a
0.7$\mathrm { cm }$ in diameter steel wire hold before breaking?

  1. $700 \mathrm { kg } - \mathrm { wt }$

  2. $770 \mathrm { kg } - \mathrm { wt }$

  3. $300 \mathrm { kg } - \mathrm { wt }$

  4. None


Correct Option: A

If equal and opposite forces applied to a body tend to elongate it, the stress so produced is called

  1. Tensile stress

  2. Compressive stress

  3. Tangential stress

  4. Working stress


Correct Option: A

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
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