Mechanical Properties of Solids Test

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Mechanical Properties of Solids Test - 9

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Weekly Quiz Competition

Question 1
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Two wires are made of the same material and have the same volume. However wire 1 has cross-sectional area $$A$$ and wire 2 has cross- sectional area $$3A$$. If the length of wire 1 increases by $$\Delta \mathrm{x}$$ on applying force $$\mathrm{F}$$, how much force is needed to stretch wire 2 by the same amount ?

A
$$\mathrm{F}$$

B
$$4\mathrm{F}$$

C
$$6\mathrm{F}$$

D
$$9\mathrm{F}$$

Solution

$$\ell_{1}=3\ell_{2}$$ (given)
$$\displaystyle \mathrm{Y}=\frac{\mathrm{F}}{\mathrm{A}}\times\frac{\ell_{1}}{\Delta \mathrm{x}}$$ (i) [symbols have their usual meaning]
$$\displaystyle \mathrm{Y}=\frac{\mathrm{F}^{1}}{3\mathrm{A}}\times\frac{\ell_{1}/3}{\Delta \mathrm{x}}$$ (ii)
$$\displaystyle \frac{\mathrm{F}}{\mathrm{A}}\times\frac{\ell_{1}}{\Delta \mathrm{x}}=\frac{\mathrm{F}'}{3\mathrm{A}}\times\frac{\ell_{1}}{3\Delta \mathrm{x}}$$
$$\mathrm{F'}=9\mathrm{F}$$
Question 2
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A solid sphere of radius r made of a soft material of bulk modulus K is surrounded by a liquid in a cylindrical container. A massless piston of area a floats on the surface of the liquid, covering entire cross section of cylindrical container. When a mass m is placed on the surface of the piston to compress the liquid, the fractional decrement in the radius of the sphere, $$\left(\displaystyle\frac{dr}{r}\right)$$, is?

A
$$\displaystyle\frac{mg}{3 Ka}$$

B
$$\displaystyle\frac{mg}{Ka}$$

C
$$\displaystyle\frac{Ka}{mg}$$

D
$$\displaystyle\frac{Ka}{3 mg}$$

Solution

Volume of sphere $$V = \dfrac{4\pi}{3}r^3$$
Decrease in volume of sphere $$-dV = \dfrac{4\pi}{3} 3r^2 dr$$
Bulk modulus   $$K = \dfrac{P V}{-d V}$$
$$\implies \ $$   $$-dV = \dfrac{PV}{K}$$    ........(1)
Pressure   $$P = \dfrac{Force}{Area} = \dfrac{mg}{a}$$          ......(2)
Using (2) in (1), we get
$$\dfrac{4\pi }{3}3r^2 dr = \dfrac{mg}{a}.\dfrac{4\pi}{3}r^3.\dfrac{1}{K}$$
$$\implies \ \dfrac{dr}{r} = \dfrac{mg}{3Ka}$$
Question 3
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A uniformly tapering conical wire is made from a material of Young's modulus $$Y$$ and has a normal, unextended length $$L$$. The radii, at the upper and lower ends of this conical wire, have values $$R$$ and $$3R$$, respectively. The upper end of the wire is fixed to a rigid support and a mass $$M$$ is suspended from its lower end. The equilibrium extended length, of this wire, would equal to:

A
$$L\left( 1+\dfrac { 1 }{ 3 } \dfrac { Mg }{ \pi \Upsilon { R }^{ 2 } } \right) $$

B
$$L\left( 1+\dfrac { 2 }{ 9 } \dfrac { Mg }{ \pi \Upsilon { R }^{ 2 } } \right) $$

C
$$L\left( 1+\dfrac { 1 }{ 9 } \dfrac { Mg }{ \pi \Upsilon { R }^{ 2 } } \right) $$

D
$$L\left( 1+\dfrac { 2 }{ 3 } \dfrac { Mg }{ \pi \Upsilon { R }^{ 2 } } \right) $$

Solution

Take a cross section at distance $$x$$ from the bottom, hence at this point,

$$r=3R-\dfrac { x }{ L } 2R$$.

$$\dfrac { Mg }{ \pi { r }^{ 2 } } =Y\dfrac { dy }{ dx } =\dfrac { Mg }{ \pi { (3R-\dfrac { 2Rx }{ L } ) }^{ 2 } } $$ where $$dy$$ is elongation of the portion $$dx$$.

Integrating this equation for $$x$$ in 0 to L, we get,

$$y=L(1+\dfrac { 1 }{ 3 } \dfrac { Mg }{ \pi Y{ R }^{ 2 } } )$$
Question 4
1 / -0

The Bulk moduli of Ethanol, Mercury and water are given as $$0.9, 25$$ and $$2.2$$ respectively in units of $$10^9/ Nm^{-2}$$. For a given value of pressure, the fractional compression in volume is $$\displaystyle \frac{\Delta V}{V}$$. Which of the following statements about $$\displaystyle \frac{\Delta V}{V}$$ for these three liquids is correct?

A
$$Mercury > Ethanol > Water$$

B
$$Ethanol > Mercury > Water$$

C
$$Water > Ethanol > Mercury$$

D
$$Ethanol > Water > Mercury$$

Solution

The bulk modulus of elasticity for a fluid is given by $$K=\displaystyle \frac{\Delta P}{\Delta V/V} $$
Thus higher is the value of $$K$$, lower is the value of $$\dfrac{\Delta V}{V}$$
Hence, the order will depend on their bulk modulus of elasticity.
Hence, it follows the order,
$$Mercury < Water < Ethanol$$
Question 5
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If speed(V), acceleration(A) and force(F) are considered as fundamental units, the dimension of Young's modulus will be?

A
$$V^{-2}A^2F^2$$

B
$$V^{-4}A^2F$$

C
$$V^{-4}A^{-2}F$$

D
$$V^{-2}A^2F^{-2}$$

Solution

$$\dfrac{F}{A}=y\cdot \dfrac{\Delta l}{l}$$
$$[Y]=\dfrac{F}{A}$$
Now from dimension
$$F=\dfrac{ML}{T^2}$$
$$L=\dfrac{F}{M}\cdot T^2$$
$$L^2=\dfrac{F^2}{M^2}\left(\dfrac{V}{A}\right)^4$$
$$\because T=\dfrac{V}{A}$$
$$L^2=\dfrac{F^2}{M^2A^2}\dfrac{v^4}{A^2}$$ $$F=MA$$
$$L^2=\dfrac{V^4}{A^2}$$
$$[Y]=\dfrac{[F]}{[A]}=F^1V^{-4}A^2$$.
Question 6
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A thin 1 m long rod has a radius of 5 mm.1 A force of 50 $$\pi kN$$ is applied at one end to determine its Young's modulus. Assume that the force is exactly known. If the least count in the measurement of all lengths is 0.01 mm, which of the following statements is false ?

A
The maximum value of Y that can be determined is $$10^{14} N/m^2$$.

B
$$\frac{\Delta Y}{Y}$$ gets minimum contribution . from the uncertainty in the length.

C
$$\frac{\Delta Y}{Y}$$ gets its maximum contribution from the uncertainty in strain

D
The figure of merit is the largest for the length of the rod.

Solution

$$Y=\dfrac{Fl}{\pi r^2\Delta l}$$

$$= \dfrac{50\times \pi \times 10^3\times 1}{\pi \times (0.005)^2\times 0.01\times 10^{-3}}$$

$$=2\times 10^{14}N/m^2$$
Question 7
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A bottle has an opening of radius $$a$$ and length $$b$$. A cork of length $$b$$ and radius $$(a+ \Delta a)$$ where $$(\Delta a << a)$$ is compressed to fit into the opening completely (See figure). If the bulk modulus of cork is B and the frictional coefficient between the bottle and cork is $$\mu$$ then the force needed to push the cork into the bottle is :

A
$$(\pi \mu B b)a$$

B
$$(2\pi \mu B b)\Delta a$$

C
$$(\pi \mu B b)\Delta a$$

D
$$(4\pi \mu B b)\Delta a$$

Solution

$$P=\dfrac{N}{A}=\dfrac{N}{(2\pi a)b}$$

$$\Rightarrow$$ Stress = $$B$$ $$\times$$ strain

$$\dfrac{N}{(2\pi a)b}=B\dfrac{2\pi a\Delta a\times b}{\pi a^2b}$$

$$f=\mu N=\mu 4\pi b\,\Delta aB$$
Question 8
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A wire fixed at the upper end stretches by length $$l$$ by applying a force $$F$$. The work done in stretching is:

A
$$\dfrac{F}{2l}$$

B
$$Fl$$

C
$$\dfrac{2F}{l}$$

D
$$\dfrac{Fl}{2}$$

Solution

Let the area be $$A$$ and length $$L$$ of the string then the work done by the force is :
$$W = \dfrac{1}{2} \times stress \times strain \times volume = \dfrac{1}{2} \times \dfrac{F}{A} \times \dfrac{l}{L} \times A \times L = \dfrac{Fl}{2}$$
Question 9
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A wire suspended vertically from one of its ends stretched by attaching a weight of $$200 N$$ to the lower end. The weight stretches the wire by $$1 mm$$. Then the elastic energy stored in the wire is:

A
$$0.2 \ J$$

B
$$10 \ J$$

C
$$20 \ J$$

D
$$0.1\ J$$

Solution

Given : $$F = 200 N$$ $$\Delta L = 1mm = 0.001$$ m
Elastic energy $$E = \dfrac{1}{2}\times Stress \times Strain \times Volume$$

$$\therefore$$ $$E =\dfrac{1}{2} \times \dfrac{F}{A} \times \dfrac{\Delta L}{L} \times AL = \dfrac{F\Delta L}{2}$$

OR $$E = \dfrac{200 \times 0.001}{2} = 0.1$$ J
Question 10
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A wire elongates by $$1 mm$$ when a load W is hanged from it. lf the wire goes over a pulley and two weights $$\mathrm{W}$$ each are hung at the two ends, the elongation of the wire will be (in mm):

A
$$1/2$$

B
$$1 $$

C
$$2 $$

D
zero

Solution

Initially the length of wire is $$L$$
$$\therefore$$ $$Y = \dfrac{W L}{A \delta L}$$ where $$\delta L = 1 mm$$

$$\implies$$ $$\dfrac{WL}{AY} =1 mm$$ ..................(1)

Now the wire goes over the pulley.
Thus the length of wire on each side of the pulley $$L' = \dfrac{L}{2}$$
Let the extensions in the wires on the two sides of the pulley be $$\delta L_1$$ and $$\delta L_2$$ respectively.

For one side     $$Y = \dfrac{W L/2}{A \delta L_1}$$
$$\implies $$    $$ \delta L_1=\dfrac{WL}{2AY} =0.5 mm$$ (using 1)
For another side    $$Y = \dfrac{W L/2}{A \delta L_2}$$
$$\implies $$    $$ \delta L_2=\dfrac{WL}{2AY} =0.5 mm$$ (using 1)
  Thus total elongation $$\delta L_T = 0.5 + 0.5 =1 mm$$

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Re-Attempt Weekly Quiz Competition

Mechanical Properties of Solids Test - 9

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Mechanical Properties of Solids Test - 9

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Question 1

$$\mathrm{F}$$

$$4\mathrm{F}$$

$$6\mathrm{F}$$

$$9\mathrm{F}$$

Solution

Question 2

$$\displaystyle\frac{mg}{3 Ka}$$

$$\displaystyle\frac{mg}{Ka}$$

$$\displaystyle\frac{Ka}{mg}$$

$$\displaystyle\frac{Ka}{3 mg}$$

Solution

Question 3

$$L\left( 1+\dfrac { 1 }{ 3 } \dfrac { Mg }{ \pi \Upsilon { R }^{ 2 } } \right) $$

$$L\left( 1+\dfrac { 2 }{ 9 } \dfrac { Mg }{ \pi \Upsilon { R }^{ 2 } } \right) $$

$$L\left( 1+\dfrac { 1 }{ 9 } \dfrac { Mg }{ \pi \Upsilon { R }^{ 2 } } \right) $$

$$L\left( 1+\dfrac { 2 }{ 3 } \dfrac { Mg }{ \pi \Upsilon { R }^{ 2 } } \right) $$

Solution

Question 4

$$Mercury > Ethanol > Water$$

$$Ethanol > Mercury > Water$$

$$Water > Ethanol > Mercury$$

$$Ethanol > Water > Mercury$$

Solution

Question 5

$$V^{-2}A^2F^2$$

$$V^{-4}A^2F$$

$$V^{-4}A^{-2}F$$

$$V^{-2}A^2F^{-2}$$

Solution

Question 6

The maximum value of Y that can be determined is $$10^{14} N/m^2$$.

$$\frac{\Delta Y}{Y}$$ gets minimum contribution . from the uncertainty in the length.

$$\frac{\Delta Y}{Y}$$ gets its maximum contribution from the uncertainty in strain

The figure of merit is the largest for the length of the rod.

Solution

Question 7

$$(\pi \mu B b)a$$

$$(2\pi \mu B b)\Delta a$$

$$(\pi \mu B b)\Delta a$$

$$(4\pi \mu B b)\Delta a$$

Solution

Question 8

$$\dfrac{F}{2l}$$

$$Fl$$

$$\dfrac{2F}{l}$$

$$\dfrac{Fl}{2}$$

Solution

Question 9

$$0.2 \ J$$

$$10 \ J$$

$$20 \ J$$

$$0.1\ J$$

Solution

Question 10

$$1/2$$

$$1 $$

$$2 $$

zero

Solution

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