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 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. $L\rho g$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The stress at any point in a hanging bar is caused by the weight of the portion of the bar hanging below that point. At the middle point, the length of the bar suspended below is L/2, so the weight of that section is (L/2) * A * rho * g. Dividing this force by the cross-sectional area A yields the stress as (L * rho * g) / 2.

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

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. $\dfrac{4Mg}{A}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Stress is defined as the internal restoring force per unit area. When the wire is stretched to double its length, the instantaneous cross-sectional area decreases due to the Poisson effect, but in standard textbook problems of this type, it is assumed the force is applied to the original area or the question implies the stress at the point of doubling. However, based on standard stress-strain definitions for this specific problem type, 2Mg/A is the accepted answer.

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

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. $8.4\times10^8 N/m^2$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 } } $ 

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

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. $50\ N/m^{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Stress = Force / Area. The force is 10 N. The side of the square is 50 cm = 0.5 m, so the area is 0.5 * 0.5 = 0.25 m^2. Stress = 10 / 0.25 = 40 N/m^2.

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

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. $1.25\times10^6N/m^2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

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

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

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. $R^{-6}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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}} $


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

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. $3.6 \times {10^7}N{m^{ - 2}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Volumetric strain = deltaV/V = 0.002. Longitudinal strain = deltaL/L. Using the relation deltaV/V = (1 - 2*nu) * longitudinal_strain and the radial strain relation, one can find the longitudinal strain. Given the values, the calculation leads to 3.6 * 10^9 N/m^2.

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

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
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Stress = Force / Area. Area = pi * r^2 = 3.14 * (0.0035 m)^2. Force = Stress * Area. Using the given ultimate strength, the load is approximately 700 kg-wt.

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

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
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Tensile stress occurs when equal and opposite forces act to pull or elongate a body. Compressive stress occurs when forces act to shorten a body.