Physics

Solid Mechanics

510 Questions

Solid mechanics questions evaluate the understanding of stress, strain, and material deformation under various loads. Topics include analyzing beams, cantilevers, and structural steel designs using specific industrial standards. These principles are strictly examined in engineering and civil services preliminary tests.

Stress and strainBeam analysisPrestressed concreteMaterial strengthStructural design

Solid Mechanics Questions

Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

A cylindrical tree has a breaking stress of $10^6\ N/m^2$. The maximum possible height of the tree is $5\ m$. the density of material of the tree is (take $g = 10\ m/s^2$)

  1. $10^3\ kg/m^3$
  2. $10^4\ kg/m^3$
  3. $2 \times 10^4\ kg/m^3$
  4. $1\ kg/m^3$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Breaking stress = Force / Area = (Mass * g) / Area = (Volume * density * g) / Area. Since Volume = Area * height, stress = density * g * height. Solving for density: 10^6 = density * 10 * 5, so density = 2 * 10^4 kg/m^3.

Multiple choice chemistry metals physical properties of metals and non-metals some physical properties of metals general characteristics and uses of metals

The property by virtue of which a substance can bear a lot of strain, without breaking is called tensile strength.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Tensile strength is a measurement of the force  required to pull something such as rope, wire or a structural beam to the point where it breaks.

The tensile strength of a material is the maximum amount of tensile stress that it can take before failure, for example breaking.

Multiple choice

Strain =

  1. force / area

  2. length / extension

  3. area / force

  4. extension / length
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Strain is defined as the ratio of the extension of a material to its original length.

Multiple choice

stress=

  1. force / area

  2. extension / length

  3. area / force

  4. length / extention
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Stress is defined as the force applied per unit cross-sectional area.

Multiple choice physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

The pressure on  square plate is measured by measuring the force on the plate and length of sides of plate. If the maximum error in the measurement of force and length are respectively $4$% and $2$%, the maximum error in measurement of pressure is..........

  1. 1%

  2. 2%

  3. 6%

  4. 8%
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Pressure P = Force / Area = F / L^2. dP/P = dF/F + 2*dL/L. dP/P = 4% + 2(2%) = 4% + 4% = 8%.

Multiple choice physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

The pressure on a square plate is measured by measuring the force on the plate and the length of the sides of the plate. If the maximum error in the measurement of force and length are respectively 4% and 2%, the maximum error in the measurement of pressure is:

  1. 1%

  2. 2%

  3. 6%

  4. 8%
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Since $area = l^2$ therefore, error in measurement of area is twice the error in measurement of length. 

Therefore, error in measurement of pressure is error in measurement of force + error in measurement of area
$= 4+2\times 2= 8$%

Multiple choice physics mechanical properties of solids applications of elasticity elastic energy properties of matter

A stone of mass 'm' s projected from a rubber catapult of length 'l' and cross-sectional area A stretched by an amount 'e'. If Y be the young's modulus of rubber then the velocity of projection of stone?

  1. $Y \sqrt {\dfrac{Ae^2}{lm}}$
  2. $ \sqrt {\dfrac{Ae^2}{lm}}$
  3. $Y \sqrt {\dfrac{YAe^2}{lm}}$
  4. $Y \sqrt {\dfrac{YAe^4}{lm}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice physics mechanical properties of solids applications of elasticity elastic energy properties of matter

A breaking stress of a material is ${ 10 }^{ 6 }N/{ m }^{ 2 }$ If density of material is $3\times 10^{ 3 }kg/{ m }^{ 3 }$, what should be the length of the material so that its breaks by it own weight?

  1. 43.3 m

  2. 23.3 m

  3. 13.3 m

  4. 33.3 m
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The wire breaks under its own weight when the stress at the top equals the breaking stress. Stress = (m * g) / A = (rho * V * g) / A = rho * L * g. Therefore, L = breaking_stress / (rho * g) = 10^6 / (3e3 * 10) = 10^6 / 3e4 = 33.33 m.

Multiple choice physics mechanical properties of solids applications of elasticity elastic energy properties of matter

The depression produced at the end of a $50 cm$ long cantilever on applying a load is $15 mm$. The depression produced at a distance of $30 cm $ from the rigid end will be

  1. 3.24 mm

  2. 1.62 mm

  3. 6.48 mm

  4. 12.96 mm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Length of cantilever, $L= 50cm=0.5m$
Deflection at the end $w(L)= 15mm$
Find: Deflection $w(x)$ at 30cm from rigid end.
Depression at a point in a cantilever beam with load at one end is given by,$ w(x)=\dfrac { P{ x }^{ 2 }(3L-x) }{ 6EI } $
We have,
$\dfrac { w(x) }{ w(L) } =\dfrac { \dfrac { P{ x }^{ 2 }(3L-x) }{ 6EI }  }{ \dfrac { PL^{ 3 } }{ 3EI }  } $

$\dfrac { w(0.3m) }{ w(0.5m) } =\dfrac { \dfrac { P{ (0.3) }^{ 2 }(3(0.5)-0.3) }{ 6EI }  }{ \dfrac { P(0.5)^{ 3 } }{ 3EI }  } $

This gives, $w(0.3m)=6.48mm$

Multiple choice physics mechanical properties of solids applications of elasticity elastic energy properties of matter

A solid cylindrical rod of radius $3 mm$ gets depressed under the influence of a load through $8 mm$. The depression produced in an identical hollow rod with outer and inner radii of $4 mm$ and $2 mm$ respectively, will be

  1. 2.7mm

  2. 1.9mm

  3. 3.2mm

  4. 7.7mm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Depression in Solid cylinder $\delta _{1}= 8mm$
Its radius $r _{1}= 3mm$
Outer radius of Hollow cylinder $R _{2}= 4mm$
Inner radius of Hollow cylinder $r _{2}= 2mm$
Let depression in this cylinder be $\delta _{2}$.
Depression $\delta =\dfrac { W{ l }^{ 3 } }{ 12\pi{r }^{ 4 }Y } $
From the above equation we know that $\delta$ is proportional to $\dfrac{1}{r^{4}}$
Hence, we have
$\dfrac{\delta _{1}}{\delta _{2}}= \dfrac{{R _{2}}^{4}-{r _{2}}^{4}}{r _{1}^{4}}$
Substituting the values in above equation, we get
$\delta _{2}=2.7mm$

Multiple choice physics mechanical properties of solids applications of elasticity elastic energy properties of matter

A beam of cross section area A is made of a material of Young modulus Y. The beam is bent into the arc of a circle of radius R. The bending moment is proportional to

  1. $\displaystyle \frac{Y}{R}$
  2. $\displaystyle \frac{Y}{RA}$
  3. $\displaystyle \frac{R}{Y}$
  4. $YR$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Bending moment $C=\dfrac{Y{I} _{G}}{R}$
Therefore, $C$ is proportional to $\dfrac{Y}{R}$

Multiple choice physics mechanical properties of solids applications of elasticity elastic energy properties of matter

In designing, a beam for its use to support a load. The depression at center is proportional to (where, $Y$ is Young's modulus).

  1. $Y^2$
  2. $Y$
  3. $\dfrac{1}{Y}$
  4. $\dfrac{1}{Y^2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The depression (deflection) of a beam supported at its ends is inversely proportional to its Young's modulus (Y). A stiffer material (higher Y) results in less depression for a given load.

Multiple choice physics mechanical properties of solids applications of elasticity elastic energy properties of matter

For the same cross-section area and for a given load, the ratio of depression for the beam of a square cross-section and circular cross-section is 

  1. $3:\pi$
  2. $\pi :3$
  3. $1:\pi$
  4. $\pi :1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\displaystyle \delta=\displaystyle\frac{Wl^3}{3YI}$, where $W=$load$,\ l=$length of beam$,\ I=$moment of inertia$=\dfrac{b{d}^{3}}{12}$ for rectangular beam, and for square beam$,\ b=d.$ Thus, ${I} _{1}=\dfrac{{b}^{4}}{12}$

Now, for circular cross section, $\displaystyle I _2=\left[\dfrac{\pi r^4}{4}\right]$

$\therefore \delta _1=\dfrac{Wl^3\times 12}{3Yb^4}=\dfrac{4Wl^3}{Yb^4}$

and $\delta _2=\dfrac{Wl^3}{3Y(\pi r^$/4)}=\displaystyle\frac{4Wl^3}{3Y(\pi r^4)}$

Thus, $\dfrac{\delta_1}{\delta_2}=\dfrac{3\pi r^4}{b^4}=\dfrac{3\pi r^4}{(\pi r^2)^2}=\dfrac{3}{\pi}$
$(\because b^2=\pi r^2$ as they have same cross sectional area)