Tag: elastic energy

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Questions Related to elastic energy

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

A steel wire of length $L$ and area of cross-section A shrinks by $\Delta l$ during night. Find the tension developed at night if Young's modulus is $Y$ and wire is clamped at both ends

  1. $\displaystyle \frac{AYL}{\Delta l}$

  2. $AYL$

  3. $AY\Delta l$

  4. $\displaystyle \frac{AY\Delta l}{L}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\displaystyle \frac{\Delta l}{L}=\frac{F}{AY}:or:F=\frac{AY\Delta l}{L}$ (using standard equation)

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

A wire of radius $1 mm$ is bent in the form of a circle of radius $10 cm$. The bending moment will be $(Y = 2\times10^{11}N/m^{2})$

  1. 3.14 N/m

  2. 6.28 N/m

  3. 1.57 N/m

  4. 15.7 N/m
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Bending Moment $C=\dfrac { Y{ I } _{ G } }{ R } $
where $Y= 2\times { 10 }^{ 11 }N/{m}^{2}$
$R= 10cm= 0.1m$
${ I } _{ G }=\dfrac{\pi{R}^{4}}{4} = \dfrac{\pi{0.1}^{4}}{4} =7.85\times{10}^{-5} m^{4}$
Put these in the expression for $C$,
$C=1.57N/m$

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

A body of mass 3.14 kg is suspended from one end of a wire of length 10 m. The radius of cross-section of the wire is changing uniformly from $5 \times 10^{-4}$ m at the top (i.e. point of suspension) to $9.8 \times 10^{-4}$ m at the bottom. Young's modulus of elasticity is $2 \times 10^{11} \ N/m^2$. The change in length of the wire is

  1. $4 \times 10^{-3}$ m

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

  3. $ 10^{-3}$ m

  4. $2 \times 10^{-3}$ m
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The change in length is calculated by integrating the strain over the length of the wire, where the radius r(x) varies linearly from r1 at the top to r2 at the bottom. Using the formula delta L = integral of (F dx) / (pi * r(x)^2 * Y), the result for the given values yields 10^-3 m.

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

A wire of cross section $A$ is stretched horizontally between two clamps located $2lm$ apart. A weight $Wkg$ is suspended from the mid-point of the wire.If the Young's modulus of the material is $Y$, the value of extension $x$ is

  1. $ { \left( \cfrac { Wl }{ YA } \right) }^{ 1/3 }$

  2. $ { \left( \cfrac { YA }{ WI } \right) }^{ 1/3 }$

  3. $\cfrac { 1 }{ l } { \left( \cfrac { Wl }{ YA } \right) }^{ 2/3 }$

  4. $ l{ \left( \cfrac { W }{ YA } \right) }^{ 2/3 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Let $M$, $L$ and $T$ represent the dimensions of mass, length and time respectively. Then:
${ [(\dfrac { Wl }{ YA } ) }^{ \dfrac { 1 }{ 3 }  }]=\dfrac { [{ M }^{ \dfrac { 1 }{ 3 }  }{ L }^{ \dfrac { 2 }{ 3 }  }{ T }^{ -\dfrac { 2 }{ 3 }  }] }{ [{ M }^{ \dfrac { 1 }{ 3 }  }{ L }^{ \dfrac { 2 }{ 3 }  }{ T }^{ -\dfrac { 2 }{ 3 }  }] } =[{ M }^{ 0 }{ L }^{ 0 }{ T }^{ 0 }]$
Since extension has dimensions of length, the only option which fits the requirement is the last one where an additional length term is multiplied.
Multiple choice physics mechanical properties of solids applications of elasticity elastic energy properties of matter

Relation among elastic contents $Y, G, B, \sigma $

  1. $\dfrac{9}{Y} = \dfrac{1}{B} + \dfrac{3}{G}$

  2. $Y = 2G (1 + \sigma)$

  3. $Y = 3B (1 - 2\sigma)$

  4. $\sigma = \dfrac{3B - 2G}{2(G + 3B)}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The relationship between Young's modulus (Y), Bulk modulus (B), and Shear modulus (G) is given by the standard elastic constant formula 9/Y = 1/B + 3/G. This is a fundamental identity in the theory of elasticity.

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

You are given three wires $  \mathrm{A}, \mathrm{B}  $ and $ \mathrm{C}  $ of the same length and cross section. They are each stretched by applying the same force to the ends. The wire A is stretched least comes back to its original length when the stretching force is removed. The wire $  B  $ is stretched more than $  A  $ and also comes back to its original length when the stretching force is removed. The wire C is stretched most and remains stretched even when the stretching force is removed. The greatest Young's modulus of elasticity is possessed by the material of a wire

  1. A

  2. B

  3. C

  4. All have the same elasticity
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Young's modulus is defined as stress divided by strain. For a given force and cross-section, the wire that stretches the least (A) has the highest Young's modulus, as Y = (F * L) / (A * delta L).

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.