Tag: applications of elasticity

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Questions Related to applications of elasticity

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.

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

A light rod of length $2\ m$ is suspended from the ceiling horizontally by means of two vertical wires of equal length tied to its ends. One of the wires is made of steel and is of cross section $0.1\ cm^{2}$. A weight is suspended from a certain point of the rod such that equal stress are produced in both the wires. Which of the following are correct?

  1. The ratio of tension in the steel and brass wires is $0.5$

  2. The load is suspended at a distance of $400/3cm$ from the steel wire

  3. Both (a) and (b) are correct

  4. Neither (a) nor (b) are correct
Reveal answer Fill a bubble to check yourself
A,B,C Correct answer
Explanation
As $Stress=Force(Tension\ here)/Area$
As, $stress _{steel}=stress _{brass}$
$\Rightarrow Tension _{steel}/Tension _{brass}=Area _{steel}/Area _{brass}=0.1/0.2=0.5$
option A is correct.

As tension is inversely proportional to the distance of suspension of load. So distance of suspension for brass=0.5 x distance of suspension for steel.
As distance of suspension for steel+distance of suspension for brass$=2m=200cm$
So distance of suspension for steel+0.5xdistance of suspension for steel$=2m=200cm$
distance of suspension for steel$=\dfrac{2}{1.5}m=\dfrac{400}{3}cm$
option B is correct.

Both option A and B is correct.
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)

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

The buckling of a beam is found to be more if __________.

  1. The breadth of the beam is large

  2. The beam material has large value of Young's modulus

  3. The length of the beam is small

  4. The depth of the beam is small
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Critical buckling stress of a column formula is given by 

$\sigma=\dfrac{F}{A}=\dfrac{{\pi}^2 r^2 E}{L^2}$
where $\sigma$ = critical stress
$L$= unsupported length of the column
$r=$ least radius 
So if the depth of the beam i small, buckling of a beam will be more.

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

Assertion: When a wire is stretched to three times its length, its resistance becomes 9 times

Reason: $R = {{\rho l} \over a}$

  1. both, Assertion and Reason are true and the reason is correct explanation of the Assertion

  2. both, Assertion and Reason are true and the reason is not correct explanation of the Assertion

  3. Assertion is true, but the reason is false.

  4. Both, Assertion and reason and false
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

When a wire is stretched to 3 times its length, its volume remains constant, so the cross-sectional area becomes 1/3 of the original. Since R = rho * (L/A), the new resistance becomes rho * (3L) / (A/3) = 9 * (rho * L/A) = 9R.