When the tension in a metal wire is $$ T_1$$ its length is $$ l_1 , $$ when the tension is $$ T_2$$ its length is $$ l_$$ The natural length of wire is :

$$ \dfrac {T_2} {T_1} ( l_1 + l_2 ) $$

The relation between Young's modulus $$Y$$, bulk modulus $$K$$ and modulus of elasticity $$\sigma$$ is

$$\cfrac { 3 }{ y } =\cfrac { 1 }{ \eta } +\cfrac { 1 }{ 3k } $$

$$\cfrac { 1 }{ y } =\cfrac { 1 }{ k } +\cfrac { 3 }{ \eta } $$

$$\cfrac { 1 }{ y } =\cfrac { 3 }{ \eta } +\cfrac { 1 }{ 3k } $$

$$\cfrac { 1 }{ \eta } =\cfrac { 3 }{ y } +\cfrac { 1 }{ 3k } $$

One end of a uniform bar of weight $$\displaystyle { w }_{ 1 }$$ is suspended from the roof and a weight $$\displaystyle { w }_{ 2 }$$ is suspended from the other end, the area of cross-section is A. What is the stress at the mid point of the rod?

$$\displaystyle \frac { \left( { { w }_{ 1 } }/{ 2 } \right) +{ w }_{ 2 } }{ A } $$

$$\displaystyle \frac { { w }_{ 1 }+{ w }_{ 2 } }{ A } $$

$$\displaystyle \frac { { w }_{ 1 }-{ w }_{ 2 } }{ A } $$

$$\displaystyle \frac { { { w }_{ 2 } }/{ 2 }+{ w }_{ 1 } }{ A } $$

A uniform slender rod of length $$L$$, cross-sectional area $$A$$ and Young's modulus $$Y$$ is acted upon by the forces shown in the figure. The elongation of the rod is:

$$\displaystyle \dfrac{8FL}{3AY}$$

$$\displaystyle \dfrac{3FL}{5AY}$$

$$\displaystyle \dfrac{2FL}{5AY}$$

$$\displaystyle \dfrac{3FL}{8AY}$$

Identical springs of steel and copper $$\left( { Y }_{ steel }>\quad { Y }_{ copper } \right) $$ are equally stretched

Less work is done on copper spring

Equal work is done on both the springs

Identical springs of steel and copper $$\left( { Y }_{ steel }>{ Y }_{ copper } \right) $$ are equally stretched then:

Less work is done on copper spring

Equal work is done on both the springs

When the load on a wire is increased from $$3kg$$ $$wt$$ to $$5kg$$ $$wt$$ the elongation increases from $$0.61mm$$ to $$1.02mm$$. The required work done during the extension of the wire is:

$$20\times { 10 }^{ -2 }J\quad $$

A wire of length $$L$$ has a linear mass density $$\mu$$ and area of cross-section $$A$$ and Young's modulus $$Y$$ is suspended vertically from a rigid support. The extension produced in the wire due to its own weight is:

$$\cfrac { \mu g{ L }^{ 2 } }{ YA } $$

$$\cfrac { 2\mu g{ L }^{ 2 } }{ YA } $$

$$\cfrac { 2\mu g{ L }^{ 2 } }{ 3YA } $$

The length of a rubber cord is $${l}_{1}$$ when the tension is $$4N$$ and $${l}_{2}m$$ when the tension is $$6N$$. The length when the tension is $$9N$$, is:

$$\left( 6{ l }_{ 2 }-1.5{ l }_{ 1 } \right) m$$

$$\left( 3{ l }_{ 2 }-2{ l }_{ 1 } \right) m$$

$$\left( 3.5{ l }_{ 2 }-2.5{ l }_{ 1 } \right) m$$

MCQ Questions for Class 11 Engineering Physics Mechanical Properties Of Solids Quiz 7

A wire of initial length

L

$L$ and radius

r

$r$ is stretched by a length

l

$l$ . Another wire of same material but with initial length

2 L

$2L$ and radius

2 r

$2r$ is stretched by a length

2 l

$2l$ . The ratio of the stored elastic energy per unit volume in the first and second wire is,

0%
1 : 4
0%
1 : 2
0%
2 : 1
0%
1 : 1

Explanation

Elastic energy stored per unit volume

E = \frac{1}{2} \times S t r e s s \times S t r a i n

$E = \dfrac{1}{2}\times Stress \times Strain$

Using Hooke's law,

S t r e s s = Y \times S t r a i n

$Stress = Y\times Strain$

We get

E = \frac{1}{2} Y \times (S t r a i n)^{2}

$E = \dfrac{1}{2}Y\times (Strain)^2$ where Young's modulus

Y

$Y$ is a material property

⟹

$\implies$

E \propto (S t r a i n)^{2}

$E\propto (Strain)^2$

Strain in first wire,

S t r a i n_{1} = \frac{l}{L}

$Strain_1 = \dfrac{l}{L}$

Strain in second wire,

S t r a i n_{2} = \frac{2 l}{2 L} = \frac{l}{L}

$Strain_2 = \dfrac{2l}{2L} = \dfrac{l}{L}$

∴

$\therefore$ Ratio of energy stored per unit volume

\frac{E_{1}}{E_{2}} = \frac{(l / L)^{2}}{(l / L)^{2}} = 1

$\dfrac{E_1}{E_2} = \dfrac{(l/L)^2}{(l/L)^2} =1$

⟹

$\implies$

E_{1} : E_{2} = 1 : 1

$E_1 : E_2 = 1:1$

A copper wire of length 2.2 m and a steel wire of length 1.6 m, both of diameter 3.0 mm are connected end to end. When stretched by a force, the elonation in length 0.50 mm is produced in the copper wire. The stretching force is

(Y_{c u} = 1.1 \times 10^{11} N / m^{2}, Y_{s t e e l} = 2.0 \times 10^{11} N / m^{2})

$(Y_{cu}=1.1\times 10^{11}N/m^2, Y_{steel}=2.0\times 10^{11}N/m^2)$

0%
$5.4\times 10^2N$
0%
$3.6\times 10^2N$
0%
$2.4\times 10^2N$
0%
$1.8\times 10^2N$

Explanation

Given that,

Length of the copper wire

l_{1} = 2.2 m

$l_1=2.2m$

Length of the steel wire

l_{2} = 1.6 m

$l_2=1.6m$

Elongation in length

△ l = 0.5. m m

$\triangle{l}=0.5.mm$

= 0.5 \times 10^{- 3} m

$=0.5\times10^{-3}m$

Radius of the

C u

$Cu$ wire

r_{1} = 1.5 \times 10^{- 3} m

$r_1=1.5\times 10^{-3}m$

Y_{1} = 1.1 \times 10^{11} N / m^{2}

$Y_1=1.1\times 10^{11} N/m^2$

Y_{2} = 2.0 \times 10^{11} N / m^{2}

$Y_2=2.0\times 10^{11}N/m^2$
Let

F

$F$ be the stretching force in both the wires then,
for

C u

$Cu$ wire,

Y_{1} = \frac{F}{π r_{1}^{2}} \times \frac{l_{1}}{△ l_{1}}

$Y_1=\dfrac{F}{\pi r^2_1}\times \dfrac{l_1}{\triangle l_1}$

\Rightarrow F = \frac{Y_{1} π r_{1}^{2} \times △ l_{1}}{l_{1}}

$\Rightarrow F=\dfrac{Y_1\pi r_1^2 \times \triangle l_1}{l_1}$

= \frac{1.1 \times 10^{- 11}}{2.2} \times \frac{22}{7} \times (1.5 \times 10^{- 3})^{2} \times 0.5 \times 10^{3}

$=\dfrac{1.1\times 10^{-11}}{2.2}\times \dfrac{22}{7}\times (1.5\times 10^{-3})^2\times 0.5\times 10^3$

= 1.8 \times 10^{2} N

$=1.8\times 10^2N$

So the streching force in the copper wire be

1.8 \times 10^{2} N

$1.8\times10^2N$

A and B are two steel wires and the radius of A is twice that of B, if they are stretched by the same load, then the stress on B is ...........

0%
Four times that of A.
0%
Two times that of A.
0%
Three times that of A.
0%
Same as that A.

Explanation

\frac{S t r e s s o n B}{S t r e s s o n A} = \frac{r_{A}^{2}}{r_{B}^{2}}

$\dfrac{Stress\, on\, B}{Stress\, on\, A}=\dfrac{r_A^2}{r_B^2}$

\Rightarrow Stress on B / Stress on A = \frac{(2 r_{B})^{2}}{r_{B}^{2}}

$\Rightarrow\text {Stress on B}/\text {Stress on A}=\dfrac{(2r_B)^2}{r_B^2}$

∴

$\therefore$

\frac{S t r e s s o n B}{S t r e s s o n A} = 4

$\dfrac{Stress\, on\, B}{Stress\, on\, A}=4$

Stress on B

= 4 \times

$= 4\times$ Stress of A

A wire of length L and of cross-sectional area A is made of a material of Young's modulus Y. The work done in stretching the wire by an amount

x

$x$ is given by :

0%
$\dfrac{YAx^2}{L}$
0%
$\dfrac{YAx^2}{2L}$
0%
$\dfrac{YAL^2}{x}$
0%
$\dfrac{YAL^2}{2x}$

Explanation

Work done is equal to

\frac{1}{2} \times

$\frac{1}{2} \times$ stress

\times

$\times$ strain

\times

$\times$ volume

Let the work done be

W

$W$ , strain be

S

$S$ and volume be

V

$V$

We have

W = \frac{1}{2} \times Y \times S^{2} \times V

$W= \frac{1}{2} \times Y \times S^{2} \times V$

\Rightarrow W = \frac{1}{2} \times Y \times (\frac{x}{L})^{2} \times A L

$\Rightarrow W=\cfrac{1}{2} \times Y \times (\cfrac{x}{L})^{2} \times AL$

\Rightarrow W = \frac{Y A x^{2}}{2 L}

$\Rightarrow W=\cfrac{YAx^{2}}{2L}$

Therefore option

B

$B$ is correct

If longitudinal strain for a wire is

0.03

$0.03$ and its poisson ratio is

0.5

$0.5$ , then its lateral strain is

0%
$0.003$
0%
$0.0075$
0%
$0.015$
0%
$0.4$

Explanation

Poisson's ratio is defined as the ratio of lateral strain to the longitudinal strain.

ν = \frac{Lateral Strain}{Longitudinal Strain}

$\nu = \cfrac{\text{Lateral Strain}}{\text{Longitudinal Strain}}$

Lateral Strain = 0.015

$\text{Lateral Strain} = 0.015$

When the tension in a metal wire is

T_{1}

$T_1$ its length is

l_{1},

$l_1 ,$ when the tension is

T_{2}

$T_2$ its length is

l_

$l_$ The natural length of wire is :

0%
$T_1l_1 + T_2 l_2$
0%
$\dfrac {l_1 T_2 - l_2 T_1}{T_2 - T_1}$
0%
$\dfrac {l_1 T_2 - l_2 T_1}{T_2 + T_1}$
0%
$\dfrac {T_2} {T_1} ( l_1 + l_2 )$

Explanation

y = \frac{F l}{A △ l}

$y = \dfrac {Fl}{ A \triangle l }$
y is constant

∴ △ l \propto F

$\therefore \triangle l \, \propto \, F$

l_{t} - l \propto T_{1}

$l_t - l \,\propto \,T_1$ and

l_{2} - l \propto T_{2}

$l_2 - l\, \propto \,T_2$

∴ \frac{l_{1} - l}{l_{2} - l} = \frac{T_{1}}{T_{2}}

$\therefore \dfrac {l_1 - l}{l_2 - l} = \dfrac {T_1}{T_2}$

l_{1} T_{2} - l T_{2} = l_{2} T_{1} - l T_{1}

$l_1 T_2 - l T_2 = l_2 T_1 - lT_1$

l (T_{1} - T_{2}) = l_{2} T_{1} - l_{1} T_{2}

$l (T_1 - T_2) = l_2 T_1 - l_1 T_2$

l = \frac{l_{2} T_{1} - L_{1} T_{2}}{T_{1} - T_{2}} \Rightarrow= \frac{l_{1} T_{2} - l_{2} T_{1}}{T_{2} - t_{1}}

$l = \dfrac {l_2 T_1 - L_1 T_2 }{T_1 - T_2} \Rightarrow = \dfrac { l_1 T_2 - l_2 T_1 }{T_2 - t_1}$

Stretching of a rubber band results in ___________.

0%
No change in potential energy
0%
Zero value of potential energy
0%
Increase in potential energy
0%
Decrease in potential energy

A liquid of bulk modulus

k

$k$ is compressed by applying an external pressure such that its density increases by

0.01

$0.01$ %. The pressure applied on the liquid is:

0%
$\cfrac { k }{ 10000 }$
0%
$\cfrac { k }{ 1000 }$
0%
$1000k$
0%
$0.01k$

Explanation

Bulk modulus is given by:

k = - \frac{P}{\frac{Δ V}{V}} \Rightarrow \frac{P}{k} = - \frac{Δ V}{V} = \frac{Δ ρ}{ρ}

$k=-\cfrac { P }{ \cfrac { \Delta V }{ V } } \Rightarrow \cfrac { P }{ k } =-\cfrac { \Delta V }{ V } =\cfrac { \Delta \rho }{ \rho }$

P = k \frac{Δ ρ}{ρ} = \frac{k}{10000}

$P=k\cfrac { \Delta \rho }{ \rho } =\cfrac { k }{ 10000 }$

One end of a slack wire (Youngs modulus

Y

$Y$ , length

L

$L$ and cross -sectional area A) is clamped to a rigid wall and the other end to a block (mass m) which rests on a smooth horizontal plane. The block is set in motion with a speed v. What is the maximum distance the block will travel after the wire becomes taut?

0%
$v\sqrt {\dfrac {mL}{AY}}$
0%
$v\sqrt {\dfrac {2mL}{AY}}$
0%
$v\sqrt {\dfrac {mL}{2AY}}$
0%
$L\sqrt {\dfrac {mv}{AY}}$

Explanation

The strain energy of a elastic wire is given by

\frac{1}{2} V Y ε^{2}

$\dfrac{1}{2}VY\varepsilon^2$

where V is the volume of wire and

ε

$\varepsilon$ is the strain.

Let the maximum elongation be x. At this point block will come to rest and all the kE energy will be converted to strain energy .

ε = \frac{x}{L}

$\varepsilon=\dfrac{x}{L}$

V = A L

$V=AL$

Now we know

\frac{1}{2} A L Y {(\frac{x}{L})}^{2} = \frac{1}{2} m v^{2}

$\dfrac{1}{2}ALY{(\dfrac{x}{L})}^2=\dfrac{1}{2}mv^2$

x = v \sqrt{\frac{m L}{A Y}}

$x=v\sqrt{\dfrac{mL}{AY}}$

The relation between Young's modulus

Y

$Y$ , bulk modulus

K

$K$ and modulus of elasticity

σ

$\sigma$ is

0%
$\cfrac { 1 }{ y } =\cfrac { 1 }{ k } +\cfrac { 3 }{ \eta }$
0%
$\cfrac { 3 }{ y } =\cfrac { 1 }{ \eta } +\cfrac { 1 }{ 3k }$
0%
$\cfrac { 1 }{ y } =\cfrac { 3 }{ \eta } +\cfrac { 1 }{ 3k }$
0%
$\cfrac { 1 }{ \eta } =\cfrac { 3 }{ y } +\cfrac { 1 }{ 3k }$

Explanation

We know that,

Y = 3 K (1 - 2 σ)

$Y=3K(1-2\sigma)$

or,

σ = \frac{1}{2} (1 - \frac{Y}{2 K})

$\sigma=\dfrac{1}{2}(1-\dfrac{Y}{2K})$ ----------(i)

Also,

Y = 2 η (1 + σ)

$Y=2\eta(1+\sigma)$

or,

σ = \frac{Y}{2 η} - 1

$\sigma=\dfrac{Y}{2\eta}-1$ ----------(ii)

Now, From (i) and (ii),

\frac{1}{2} (1 - \frac{Y}{3 K}) = \frac{Y}{2 η} - 1

$\dfrac{1}{2}(1-\dfrac{Y}{3K})=\dfrac{Y}{2\eta}-1$

or,

1 - \frac{Y}{3 K} = \frac{Y}{η} - 2

$1-\dfrac{Y}{3K}=\dfrac{Y}{\eta}-2$

or,

\frac{3}{Y} = \frac{1}{η} + \frac{1}{3 K}

$\dfrac{3}{Y}=\dfrac{1}{\eta}+\dfrac{1}{3K}$

The area of cross-section of a steel wire

(Y = 2.0 \times 10^{11} N / m^{2})

$\left( Y=2.0\times { 10 }^{ 11 }{ N }/{ { m }^{ 2 } } \right)$ is

0.1 {c m}^{2}

$0.1 { cm }^{ 2 }$ . The force required to double its length will be :

0%
$2\times { 10 }^{ 12 }N$
0%
$2\times { 10 }^{ 11 }N$
0%
$2\times { 10 }^{ 10 }N$
0%
$2\times { 10 }^{ 6 }N$

Explanation

Young's modulus of elasticity

Y = \frac{F L}{A l}

$Y=\dfrac { FL }{ Al }$

To double the length

l = L

$l = L$

∴ Y = \frac{F}{A} \Rightarrow F = Y A

$\therefore Y=\dfrac { F }{ A } \Rightarrow F=YA$

= 2 \times 10^{11} \times 0.1 \times 10^{- 4} = 2 \times 10^{6} N

$=2\times { 10 }^{ 11 }\times 0.1\times { 10 }^{ -4 }=2\times { 10 }^{ 6 }N$

Two wires of same material and same diameter have lengths in the ratio 2 :They are stretched by the same force. The ratio of work done in stretching them is :

0%
5 : 2
0%
2 : 5
0%
1 : 3
0%
3 : 1

Explanation

Hint: Use the formula of work done $W= \dfrac {1}{2} Kx^2$
Explanation:
Step 1: Concept used:
Work done is given by-

W o r k D o n e = \frac{1}{2} \times F o r c e \times E l o n g a t i o n

$Work\ Done \ = \frac { 1 }{ 2 } \ \times \ { Force }\ \times \ { Elongation }$ ........... (1)

By definition of Young's modulus:

Y = \frac{S t r e s s}{S t r a i n} = \frac{F / A}{Δ l / L} = \frac{F L}{A Δ l}

$Y= \dfrac {Stress}{Strain}= \dfrac {F/A}{ \Delta l/L}= \dfrac {FL}{A \Delta l}$
So, elongation

Δ l = \frac{F l}{A Y}

$\Delta l= \dfrac {Fl}{AY}$ ......... (2)
Putting the value from eq.(2) in eq.(1) and we get-

W o r k D o n e = \frac{1}{2} \times F \times \frac{F l}{A Y} = \frac{F^{2} L}{2 A Y}

$Work\ Done\ =\ \dfrac { 1 }{ 2 } \ \times \ { F }\ \times \ \dfrac { Fl }{ AY } \ =\ \dfrac { { F }^{ 2 }L }{ 2AY }$

Since Force

(F)

$(F)$ , material

(Y)

$(Y)$ and area

(A)

$(A)$ are same

∴ W o r k D o n e \propto l e n g t h

$\\ \therefore \ Work\ Done\ \propto \ length\$

Step 2: Finding the ratio:

\frac{W_{1}}{W_{2}} = \frac{l_{1}}{l_{2}} = \frac{2}{5}

$\\ \dfrac { { W }_{ 1 } }{ { W }_{ 2 } } \ =\ \dfrac { { l }_{ 1 } }{ { l }_{ 2 } } \ =\ \dfrac { 2 }{ 5 }$

Hence, the answer is option B

A metallic rod of length

l

$l$ and cross sectional area

A

$A$ is made of a material of Young's modulus

Y

$Y$ . If the rod is elongated by an amount

y

$y$ , then the work done is proportional to

0%
$y$
0%
$1/y$
0%
${ y }^{ 2 }$
0%
$1/{ y }^{ 2 }$

Young's modulus of rubber is

10^{4}

$10^4$ N

/ m^{2}

$/m^2$ and area of cross-section is

2

$2$

c m^{2}

$cm^2$ . If force of

2 \times 10^{5}

$2\times 10^5$ dyne is applied along its length, then its final length becomes.

0%
$3L$
0%
$4L$
0%
$2L$
0%
None of these

Explanation

Given,

Y = 10^{4} N / m^{2}, A = 2 c m^{2} = 2 \times 10^{- 4} m^{2}

$Y=10^4N/m^2, A=2cm^2=2\times 10^{-4}m^2$

F = 2 \times 10^{5}

$F=2\times 10^5$ dyne

= 2

$=2$ N

∴

$\therefore$ Initial length

I = \frac{F L}{A Y}

$I=\displaystyle\frac{FL}{AY}$

= \frac{2 \times L}{2 \times 10^{- 4} \times 10^{4}} = L

$=\displaystyle\frac{2\times L}{2\times 10^{-4}\times 10^4}=L$

∴

$\therefore$ Final length

=

$=$ Initial length

+

$+$ Increment

= 2 L

$=2L$ .

A rubber cord catapult has cross-sectional area 25 mm

^{2}

$^2$ and initial length of rubber cord is 10 cm. It is stretched to 5 cm and then released to project a missile of mass 5 g. Taking

Y_{r u b b e r} = 5 \times 10^{8} N m^{- 2}

$Y_{rubber} = 5 \times 10^8 Nm^{-2}$ , velocity of projected missile is:

0%
20 ms $^{-1}$
0%
100 ms^{-1}$$
0%
250 ms $^{-1}$
0%
200 ms $^{-1}$

Explanation

Potential energy stored in the rubber cord catault will be converted into kinetic energy of mass

\frac{1}{2} m v^{2} = \frac{1}{2} \frac{Y A l^{2}}{L}

$\dfrac{1}{2} mv^2 = \dfrac{1}{2} \dfrac{YAl^2}{L}$

where

l = 5 \times 10^{- 2}

$l = 5\times 10^{-2} \$ m,

A = 25 \times 10^{- 6} m^{2},

$A = 25\times 10^{-6} \ m^2,$

L = 10 \times 10^{- 2}

$L = 10\times 10^{-2} \$ m,

m = 5 \times 10^{- 3} k g

$m = 5\times 10^{-3} \ kg$

\Rightarrow v = \sqrt{\frac{Y A l^{2}}{m L}}

$\Rightarrow v = \sqrt{\dfrac{YAl^2}{mL}}$

v = \sqrt{\frac{5 \times 10^{8} \times 25 \times 10^{- 6} \times (5 \times 10^{- 2})^{2}}{5 \times 10^{- 3} \times 10 \times 10^{- 2}}}

$v= \sqrt{\dfrac{5 \times 10^8 \times 25 \times 10^{-6} \times (5 \times 10^{-2})^2}{5 \times 10^{-3} \times 10 \times 10^{-2}}}$

v = 250 m s^{- 1}

$v= 250 ms^{-1}$

The bulk modulus of an ideal gas at constant temperature is :

0%
Equal to its pressure
0%
Equal to its volume
0%
Equal to ${ p }/{ 2 }$
0%
Cannot be determined

A ball falling-in a lake of depth 400 m has a decrease of

0.2 %

$0.2\%$ in its volume at the bottom. The bulk modulus of the material of the ball is (in

N m^{-} 2

$Nm^-{2}$ )

0%
$9.8 \times 10^9$
0%
$9.8 \times 10^{10}$
0%
$1.96 \times 10^{10}$
0%
$9.8 \times 10^{11}$
0%
$1.96 \times 10^9$

Explanation

Given,

h = 400 m, \frac{△ V}{V} = \frac{0.2}{100}

$h=400m, \dfrac{\triangle V}{V}=\dfrac{0.2}{100}$
We know that the bulk modulus of the material

B = \frac{h p g}{(\begin{matrix} \frac{△ V}{V} \end{matrix})}

$B=\dfrac{hpg}{\begin{pmatrix}\dfrac{\triangle V}{V}\end{pmatrix}}$

B = \frac{400 \times 10^{3} \times 9.8}{\frac{0.2}{100}}

$B=\dfrac{400\times 10^3\times 9.8}{\dfrac{0.2}{100}}$

B = \frac{400 \times 10^{3} \times 9.8 \times 100}{0.2}

$B=\dfrac{400\times 10^3\times 9.8\times 100}{0.2}$

B = 2000 \times 10^{3} \times 9.8 \times 100

$B=2000\times 10^3\times 9.8\times 100$

B = 1.96 \times 10^{9} N m^{- 2}

$B=1.96\times 10^9 Nm^{-2}$

One end of a uniform bar of weight

w_{1}

$\displaystyle { w }_{ 1 }$ is suspended from the roof and a weight

w_{2}

$\displaystyle { w }_{ 2 }$ is suspended from the other end, the area of cross-section is A. What is the stress at the mid point of the rod?

0%
$\displaystyle \frac { { w }_{ 1 }+{ w }_{ 2 } }{ A }$
0%
$\displaystyle \frac { { w }_{ 1 }-{ w }_{ 2 } }{ A }$
0%
$\displaystyle \frac { \left( { { w }_{ 1 } }/{ 2 } \right) +{ w }_{ 2 } }{ A }$
0%
$\displaystyle \frac { { { w }_{ 2 } }/{ 2 }+{ w }_{ 1 } }{ A }$

Explanation

The downward tension at mid-point of the bar is due to weight of lower half of the bar and the hanging weight.

The downward tension at mid-point is,

(w_{1}) / 2 + w_{2}

$(w_1)/2 + w_2$

As,

s t r e s s = f o r c e / a r e a = \frac{(w_{1}) / 2 + w_{2}}{A}

$stress = force/area = \dfrac{(w_1)/2 + w_2}{A}$

Option C is correct.

The Poisson's ratio of a material is

0.8

$0.8$ . If a force is applied to a wire of this material decreases its cross-sectional area by

4 %

$4\%$ , then the percentage increase in its length will be.

0%
$1\%$
0%
$2\%$
0%
$2.5\%$
0%
$4\%$

Explanation

The Poisson's ratio

σ

$\sigma$ is given by

σ = \frac{l a t e r a l s t r a i n}{l o n g i t u d i n a l s t r a i n} = \frac{Δ d / d}{Δ l / l}

$\sigma =\displaystyle\frac{lateral strain}{longitudinal strain}=\frac{\Delta d/d}{\Delta l/l}$ ........(i)
where, l

=

$=$ length of wire
d

=

$=$ diameter of wire
The cross-sectional area of wire is given by

A = π r^{2} = \frac{π d^{2}}{4}

$A=\displaystyle \pi r^2=\frac{\pi d^2}{4}$

∴ \frac{Δ A}{A} = \frac{2 Δ d}{d}

$\therefore \displaystyle\frac{\Delta A}{A}=\frac{2\Delta d}{d}$

\Rightarrow \frac{Δ d}{d} = \frac{Δ A}{2 A}

$\Rightarrow \displaystyle\frac{\Delta d}{d}=\frac{\Delta A}{2A}$ ........(ii)
From Eqs. (i) and (ii), we get

σ = \frac{\frac{Δ A}{2 A}}{Δ l / l}

$\displaystyle \sigma =\frac{\displaystyle\frac{\Delta A}{2A}}{\Delta l/l}$

\Rightarrow \frac{Δ l}{l} = \frac{Δ A}{2 σ \cdot A} = \frac{1}{2 \times σ} (\frac{Δ A}{A})

$\Rightarrow \displaystyle\frac{\Delta l}{l}=\frac{\Delta A}{2\sigma \cdot A}=\dfrac{1}{2\times \sigma }\left(\displaystyle\frac{\Delta A}{A}\right)$

= \frac{1}{2 \times 0.8} \times 4 %

$=\displaystyle\frac{1}{2\times 0.8}\times 4\%$

\frac{Δ l}{l} = 2.5 %

$\displaystyle\frac{\Delta l}{l}=2.5\%$ .

An aluminium rod (Young's modulus

= 7 \times 10^{9} N / m^{2})

$= 7\times 10^{9} N/m^{2})$ has a breaking strain of

0.2

$0.2$ %. The minimum cross-sectional area of the rod in order to support a load of

10^{4}

$10^{4}$ Newton's is:

0%
$1\times 10^{-2}m^{2}$
0%
$1.4\times 10^{-3}m^{2}$
0%
$3.5\times 10^{-3}m^{2}$
0%
$7.1\times 10^{-4}m^{2}$

Explanation

No correct option.

Given,

Young's modulus

= 7 \times 10^{9} N / m^{2}

$=7\times10^9N/m^2$ Breaking strain

= 0.2 % f = 10^{4} N

$=0.2\% f=10^4N$

So from young's modulus formula.

y = \frac{f I^{'}}{A I}

$y=\dfrac{fI'}{AI}$ Where

I^{'} = \frac{2}{1000}

$I'=\dfrac{2}{1000}$

By putting the value then we get,

7 \times 10^{9} = 10^{4} \times \frac{2}{1000} \times A \Rightarrow A = 3.5 \times 10^{8} m^{2}

$7\times10^9=10^4\times\dfrac{2}{1000}\times A\Rightarrow A=3.5\times10^8m^2$

The compressibility of water

4 \times 10^{- 5}

$4\times 10^{-5}$ per unit atmospheric pressure. The decrease in volume of

100 c u b i c

$100\ cubic$ centimeter of water under a pressure of

100

$100$ atmosphere will be:

0%
$0.4\ cc$
0%
$4\times 10^{-5} cc$
0%
$0.025\ cc$
0%
$0.004\ cc$

Explanation

Compressibility

(β) = \frac{\frac{△ V}{V}}{P}

$(\beta)=\cfrac{\cfrac{\triangle V}{V}}{P}$

Here,

β = 4 \times 10^{- 5} / a t m

$\beta=4 \times {10}^{-5}/atm$

V = 100 c c

$V=100\quad cc$

P = 100 a t m

$P=100\quad atm$

∴ △ V = β P V

$\therefore \triangle V=\beta PV$

= 4 \times 10^{- 5} \times 100 \times 100

$=4 \times {10}^{-5} \times 100 \times 100$

∴ △ V = 0.4 c c

$\therefore \triangle V=0.4cc$

One end of a long metallic wire of length

L

$L$ is tied to the ceiling. The other end is tied to massless spring of spring constant

k

$k$ . A mass

m

$m$ hangs freely from the free end of the spring. The area of cross-section and Young's modulus of the wire are

A

$A$ and

Y

$Y$ respectively. If the mass is slightly pulled down and released, it will oscillate with a time period

T

$T$ equal to

0%
$2\pi \sqrt {m/k}$
0%
$2\pi \sqrt {\dfrac {m(YA + kL)}{YAk}}$
0%
$2\pi \sqrt {\dfrac {mYA}{kL}}$
0%
$2\pi \sqrt {\dfrac {mL}{YA}}$

Explanation

young's modulus

Y = \frac{F \times L}{A \times △ L}

$Y = \dfrac {F\times L}{A\times \triangle L}$ or

\frac{F}{△ L} = \frac{Y A}{L}

$\dfrac {F}{\triangle L} = \dfrac {YA}{L}$
Force constant of wire

k_{1} = \frac{F}{△ L} = \frac{Y A}{L}

$k_{1} = \dfrac {F}{\triangle L} = \dfrac {YA}{L}$
Force constant of wire

k_{1}

$k_{1}$ and spring of force constant

k

$k$ are in series

\frac{1}{k_{2}} = \frac{1}{k} + \frac{1}{k_{1}}

$\dfrac {1}{k_{2}} = \dfrac {1}{k} + \dfrac {1}{k_{1}}$
or

k_{2} = \frac{k k_{1}}{k + k_{1}} = \frac{k (\frac{Y A}{L})}{k + \frac{Y A}{L}} = \frac{k Y A}{k L + Y A}

$k_{2} = \dfrac {kk_{1}}{k + k_{1}} = \dfrac {k\left (\dfrac {YA}{L}\right )}{k + \dfrac {YA}{L}} = \dfrac {kYA}{kL + YA}$
The time period of the combination is

T^{'} = 2 π \sqrt{\frac{m}{k_{2}}} = 2 π \sqrt{\frac{m (k L + Y A)}{k Y A}}

$T' = 2\pi \sqrt {\dfrac {m}{k_{2}}} = 2\pi \sqrt {\dfrac {m(kL + YA)}{kYA}}$ .

The load versus elongation graph for four wires of the same material and same length is shown in the figure. The thinnest wire is represented by the line.

0%
$OA$
0%
$OB$
0%
$OC$
0%
$OD$

Explanation

As slop of the load-elongation graph given, young's modulus. So, OA has least value of young's modulus.

And, Young's modulus =

\frac{s t r e s s}{s t r a i n}

$\dfrac { stress }{ strain }$

As, Young's modulus of OA is the least, so, strain is maximum. As strain is maximum, so, it is the thinnest wire.

A cubical ball is taken to a depth of

200 m

$200 m$ in a sea. The decrease in volume observed to be

0.1

$0.1$ %. The bulk modulus of the ball is -

(g = 10 / m / s^{- 2})

$(g = 10/ m/s^{-2})$

0%
$2\times 10^{7} Pa$
0%
$2\times 10^{6} Pa$
0%
$2\times 10^{9} Pa$
0%
$1.2\times 10^{9} Pa$

Explanation

Let the bulk modulus be

B

$B$ . The initial volume be

v

$v$ and final volume is

(v) (0.1 / 100) = 0.001 v

$(v)(0.1/100)=0.001\ v$

Change in volume

\to d v = 0.001 v - v = - 0.99

$\rightarrow dv=0.001\ v-v=-0.99$

Pressure on the body,

P = ρ g h

$P=\rho gh$

= 1000 \times 10 \times 200

$=1000\times 10\times 200$

= 2000000

$=2000000$

= 2 \times 10^{6} P a

$=2\times 10^6\ Pa$

A uniform rod of length

^{'} L^{'}

$'L'$ and density

^{'} ρ^{'}

$'\rho'$ is being pulled along a smooth floor with horizontal acceleration

α

$\alpha$ as shown in the figure. The magnitude of the stress at the transverse cross-section through the mid-point of the rod is _________.

0%
$\dfrac {\rho l \alpha}{4}$
0%
$4\rho l\alpha$
0%
$2\rho l\alpha$
0%
$\dfrac {\rho l\alpha}{2}$

Explanation

Given:

The length of the rod is

L

$L$ .

The density of the rod is

ρ

$\rho$ .

The angular acceleration of the rod is

α

$\alpha$ .

The mid-point will experience the stress due to the half length of the rod.

The mass of the half length of the rod is given as:

M a s s = ρ \times V

$Mass=\rho\times V$

The volume of the half of the rod is given as:

V = a r e a \times \frac{L}{2}

$V=area\times \dfrac L2$

So,

M a s s = ρ \times A \times \frac{L}{2}

$Mass=\rho\times A\times \dfrac L2$

The force experienced by the rod is given as:

F = m \times a

$F=m\times a$

F = ρ \times A \times \frac{L}{2} \times α

$F=\rho\times A\times \dfrac L2\times\alpha$

= \frac{ρ A L α}{2}

$=\dfrac{\rho AL\alpha}{2}$

Now, the stress experienced by the mid-point of the rod is :

S t r e s s = \frac{F o r c e}{a r e a}

$Stress=\dfrac{Force}{area}$

= \frac{ρ A L α}{2 A}

$=\dfrac{\rho AL\alpha}{2A}$

= \frac{ρ L α}{2}

$=\dfrac{\rho L\alpha}{2}$

Which of the following statements are correct?

0%
Poisson's ratio can be greater than $0.5$
0%
Poisson's ratio is a characteristic property of the material of the body
0%
Poisson's ratio of a body depends upon its shape and size
0%
None of these

Explanation

Suppose a uniform rod is pulled by a tension due to which a longitudinal stress

S

$S$ is developed in its material. Then tensile strain in the material of the rod,

ϵ_{1} = \frac{S}{Y}

$\epsilon_{1} = \dfrac {S}{Y}$ . If its Poisson's ratio of

σ

$\sigma$ , then lateral strain will be equal to

σ ϵ_{1}

$\sigma \epsilon_{1}$ . Lateral strain has opposite nature to that of the longitudinal strain. We also know that volumetric strain

ϵ_{v} = ϵ_{x} + ϵ_{y} + ϵ_{z}

$\epsilon_{v} = \epsilon_{x} + \epsilon_{y} + \epsilon_{z}$

∴ ϵ_{v} = ϵ_{1} + 2 (- σ ϵ_{1}) = (1 - 2 σ) ϵ_{1}

$\therefore \epsilon_{v} = \epsilon_{1} + 2(-\sigma \epsilon_{1}) = (1 - 2\sigma)\epsilon_{1}$
Since a tension is applied on the material, therefore, its volume cannot decrease. Hence,

ϵ_{v}

$\epsilon_{v}$ cannot be negative. Therefore

(1 - 2 σ)

$(1-2\sigma)$ cannot be less than

0

$0$ or

σ

$\sigma$ cannot be greater than

0.5

$0.5$ . Poisson's ratio is the property of the material of the body. (b) is correct.

The bulk modulus of water is

2.0 \times 10^{9} N / m^{2}

$2.0\times { 10 }^{ 9 }\ { N }/{ { m }^{ 2 } }$ . The pressure required to increase the density of water

0.1 %

$0.1 \%$ is

0%
$2.0\times { 10 }^{ 3 }\ { N }/{ { m }^{ 2 } }$
0%
$2.0\times { 10 }^{ 6 }\ { N }/{ { m }^{ 2 } }$
0%
$2.0\times { 10 }^{ 5 }\ { N }/{ { m }^{ 2 } }$
0%
$2.0\times { 10 }^{ 7 }\ { N }/{ { m }^{ 2 } }$

Explanation

ρ α V

$\rho \quad \alpha \quad V$

\frac{Δ V}{V} = \frac{0.1}{100}

$\dfrac{\Delta V}{V}=\dfrac{0.1}{100}$

B = \frac{- p}{Δ V / V}

$B=\dfrac{-p}{\Delta V/V}$

| B | = \frac{P}{Δ V / V}

$\left| B \right| =\dfrac { P }{ \Delta V/V } \\$

p = 2 \times 10^{9} \times 10^{- 3}

$p=2\times 10^9\times 10^{-3}$

p = 2 \times 10^{6} N / m^{2}

$p=2\times 10^6 N/m^2$

Figure shows the strain-stress curve for a given material. The Young's modulus of the material is

0%
$5\times { 10 }^{ 9 }N\quad { m }^{ -2 }$
0%
$5\times { 10 }^{ 10 }N\quad { m }^{ -2 }$
0%
$7.5\times { 10 }^{ 9 }N\quad { m }^{ -2 }$
0%
$7.5\times { 10 }^{ 10 }N\quad { m }^{ -2 }$

Explanation

From the given graph for a stress of

150 \times 10^{6} N

$150\times {10}^{6}N$

m^{- 2}

${m}^{-2}$ when strain is

0.002

$0.002$

∴

$\therefore$ Young's modulus

Y = \frac{s t r e s s}{s t r a i n}

$Y=\cfrac { stress }{ strain }$

Y = \frac{150 \times 10^{6}}{0.002} N m^{- 2} = 7.5 \times 10^{10} N m^{- 2}

$Y=\cfrac { 150\times { 10 }^{ 6 } }{ 0.002 } N\quad { m }^{ -2 }=7.5\times { 10 }^{ 10 }N\quad { m }^{ -2 }$

A uniform slender rod of length

L

$L$ , cross-sectional area

A

$A$ and Young's modulus

Y

$Y$ is acted upon by the forces shown in the figure. The elongation of the rod is:

0%
$\displaystyle \dfrac{3FL}{5AY}$
0%
$\displaystyle \dfrac{2FL}{5AY}$
0%
$\displaystyle \dfrac{3FL}{8AY}$
0%
$\displaystyle \dfrac{8FL}{3AY}$

Explanation

Divide the rod into 2 parts of length

2 l / m

$2l/m$ and

l / m

$l/m$ where the force F is acting.

Now solve the problem for the 2 rods separately.

Force acting on both sides of

2 l / 3

$2l/3$ is

3 F

$3F$ .

So extension

x_{1}

$x_1$ will be

x_{1} = \frac{[3 F \times \frac{2 l}{3}]}{A Y}

$x_1=\dfrac{\left[3F\times \dfrac{2l}{3}\right]}{AY}$

Force acting on both sides of

l / 3

$l/3$ is

2 F

$2F$

So extension

x 2

$x2$ will be

x_{2} = \frac{[2 F \times \frac{l}{3}]}{A Y}

$x_2=\dfrac{\left[2F\times \dfrac{l}{3}\right]}{AY}$

Now total extension x is

x = x_{1} + x_{2} = 8 F L / 3 A Y

$x=x_1+x_2=8FL/3AY$

Identical springs of steel and copper

(Y_{s t e e l} > Y_{c o p p e r})

$\left( { Y }_{ steel }>\quad { Y }_{ copper } \right)$ are equally stretched

0%
Less work is done on copper spring
0%
Less work is done on steel spring
0%
Equal work is done on both the springs
0%
Data is incomplete

Explanation

\begin{matrix} Since, young's modulus of steel is greater than that of copper. Hence, in order to produce same extension, lange force will have to be applied on the steel spring than that of copper spring ∴ W (steel) > W(copper ) \end{matrix}

$\begin{array}{l}\text { Since, young's modulus of steel is greater than } \\\text { that of copper. Hence, in order to produce } \\\text { same extension, lange force will have to be applied } \\\text { on the steel spring than that of copper spring } \\\therefore W\text { (steel)}>\text { W(copper )}\end{array}$

Hence option A is correct

Stress is a ______ quantity.

0%
scalar
0%
vector
0%
tensor
0%
dimensionless

Two wires of equal length and cross-sectional area are suspended as shown in figure. Their Young's modulii are Y

_{1}

$_1$ and Y

_{2}

$_2$ respectively. The equivalent Young's modulii will be:

0%
$\displaystyle Y_1 + Y_2$
0%
$\displaystyle \frac{Y_1 Y_2}{Y_1 + Y_2}$
0%
$\displaystyle \frac{Y_1 + Y_2}{2}$
0%
$\displaystyle \sqrt{Y_1 Y_2}$

Explanation

\begin{matrix} Y_{1} = \frac{F_{1} / A}{Δ L_{1} / L} Y_{2} = \frac{F_{2} / A}{Δ L_{2} / L} F_{1} = F_{2} = F Δ L_{1} = \frac{F L}{Y_{1} A} Δ L_{2} = \frac{F L}{Y_{2} A} ∴ Y_{12} = \frac{F / A}{Δ L_{12} / L} Δ L_{12} = Δ L_{2} + Δ L_{2} ∴ \frac{1}{Y} = \frac{1}{Y_{1}} + \frac{1}{Y_{2}} Y = \frac{Y_{1} Y_{2}}{Y_{1} + Y_{2}} \end{matrix}

$\begin{array}{l}{Y_1} = \dfrac{{{F_1}/A}}{{\Delta {L_1}/L}}\\{Y_2} = \dfrac{{{F_2}/A}}{{\Delta {L_2}/L}}\\{F_1} = {F_2} = F\\\Delta {L_1} = \dfrac{{FL}}{{{Y_1}A}}\\\Delta {L_2} = \dfrac{{FL}}{{{Y_2}A}}\\\therefore {Y_{12}} = \dfrac{{F/A}}{{\Delta {L_{12}}/L}}\\\Delta {L_{12}} = \Delta {L_2} + \Delta {L_2}\\\therefore \dfrac{1}{Y} = \frac{1}{{{Y_1}}} + \frac{1}{{{Y_2}}}\\Y = \dfrac{{{Y_1}{Y_2}}}{{{Y_1} + {Y_2}}}\end{array}$

To determine the Young's modulus of a wire, the formula is

Y = \frac{F}{A}, \frac{L}{△ l}

$Y = \dfrac {F}{A}, \dfrac {L}{\triangle l}$ ; where

L = l e n g t h, A =

$L = length, A =$ area of cross-section of the wire,

△ L =

$\triangle L =$ Change in length of the wire when stretched with a force

F

$F$ . The conversion factor to change it from

C G S

$CGS$ to

M K S

$MKS$ system is

0%
$1$
0%
$10$
0%
$0.1$
0%
$0.01$

Explanation

CGS unit of Y is dyn/

{c m}^{2}

${ cm }^{ 2 }$

MKS unit of Y is N/

m^{2}

${ m }^{ 2 }$

1 dyn/

{c m}^{2}

${ cm }^{ 2 }$ =

\frac{10^{- 5} N}{{(\frac{1}{100})}^{2} m^{2}}

$\dfrac { { 10 }^{ -5 }N }{ { \left( \dfrac { 1 }{ 100 } \right) }^{ 2 }{ m }^{ 2 } }$

[1 N = 10^{5} d y n 1 m = 100 c m]

$\left[ 1N={ 10 }^{ 5 }dyn\\ 1m=100cm \right]$

10^{- 5} \times 10^{4} N / m^{2}

${ 10 }^{ -5 }\times { 10 }^{ 4 }N/{ m }^{ 2 }$

0.1 N / m^{2}

$0.1N/{ m }^{ 2 }$

∴

$\therefore$ Conversion factor will be 0.1

Match the Column I with Column II

Column I	Column II
(A) A body which regains its original shape after the removal of external forces	(p) Elasticity
(B) A body which does not regain its original shape after the removal of external forces.	(q) Elastic body
(C) A body which does not show any deformation on applying external forces	(r) Plastic body
(D) The property of the body to regain its original configuration when the deforming forces are removed.	(s) Rigid body

0%
A-q, B-r, C-s, D-p
0%
A-p, B-q, C-r, D-s
0%
A-r, B-s, C-p, D-q
0%
A-s, B-p, C-q, D-r.

Which of the following statements is correct regarding Poisson's ratio?

0%
It is the ratio of the longitudinal strain to the lateral strain
0%
Its value is independent of the nature of the material
0%
It is unitless and dimensionless quantity
0%
The practical value of Poisson's ratio lies between $0$ and $1$

Explanation

The ratio of the lateral strain to longitudinal strain is called Poisson's ratio.
Hence, option (a) is an incorrect statement.
Its value depends only on the nature of the material.
Hence, option (b) is an incorrect statement.
It is the ratio of two like physical quantities.
Therefore, it is unitless and dimensionless quantity.
Hence option (c) is a correct statement
The practical value of Poisson's ratio lies between

0

$0$ and

0.5

$0.5$
hence option (d) is an incorrect statement.

The area of cross section of a steel wire (

y = 2 \times 10^{11} N / m^{2}

$y=2\times { 10 }^{ 11 }N/{ m }^{ 2 }$ ) is

0.1 {c m}^{2}

$0.1{cm}^{2}$ . The force required to double its length will be:

0%
$2\times { 10 }^{ 12 }N$
0%
$2\times { 10 }^{ 11 }N$
0%
$2\times { 10 }^{ 10 }N$
0%
$2\times { 10 }^{ 6 }N$

Explanation

Given,

Y = 2 \times 10^{11} N / m^{2}

$Y=2\times { 10 }^{ 11 }N/{ m }^{ 2 }$

A = 0.1 {c m}^{2} = 0.1 \times 10^{- 4} m^{2}

$A=0.1{ cm }^{ 2 }=0.1\times { 10 }^{ -4 }{ m }^{ 2 }$

Let, length of the wire is l. After applying force, length will be 2l.

∴

$\therefore$ increase in length =

(2 l - l) = l

$\left( 2l-l \right) =l$

we know,

Y = \frac{\frac{F}{A}}{\frac{△ l}{l}}

$Y=\dfrac { \dfrac { F }{ A } }{ \dfrac { \triangle l }{ l } }$

2 \times 10^{11} = \frac{\frac{F}{0.1 \times 10^{- 4}}}{\frac{l}{l}}

$2\times { 10 }^{ 11 }=\dfrac { \dfrac { F }{ 0.1\times { 10 }^{ -4 } } }{ \dfrac { l }{ l } }$

∴ F = 2 \times 10^{11} \times 0.1 \times 10^{- 4}

$\therefore \quad F=2\times { 10 }^{ 11 }\times 0.1\times { 10 }^{ -4 }$

∴ F = 2 \times 10^{6} N

$\therefore \quad F=2\times { 10 }^{ 6 }N$

Identical springs of steel and copper

(Y_{s t e e l} > Y_{c o p p e r})

$\left( { Y }_{ steel }>{ Y }_{ copper } \right)$ are equally stretched then:

0%
Less work is done on copper spring
0%
Less work is done on steel spring
0%
Equal work is done on both the springs
0%
Data is incomplete

Explanation

Work done,

W = \frac{1}{2} \times F \times Δ L

$W=\cfrac { 1 }{ 2 } \times F\times \Delta L$
For a given

F

$F$

W \propto Δ L . . . (i)

$W\propto \Delta L...(i)\quad$
and

Δ L = \frac{F L}{A Y}

$\Delta L=\cfrac { FL }{ AY }$
As

F, A

$F, A$ and

L

$L$ are constants

∴ Δ L \propto \frac{1}{Y} . . . (i i)

$\therefore \Delta L\propto \cfrac { 1 }{ Y } ...(ii)$
From (i) and (ii), we get

W \propto \frac{1}{Y}

$\quad W\propto \cfrac { 1 }{ Y }$

∴ \frac{W_{s t e e l}}{W_{c o p p e r}} = \frac{Y_{c o p p e r}}{Y_{s t e e l}}

$\therefore \cfrac { { W }_{ steel } }{ { W }_{ copper } } =\cfrac { { Y }_{ copper } }{ { Y }_{ steel } }$

Y_{c o p p e r} < Y_{s t e e l}

${ Y }_{ copper }<{ Y }_{ steel }$

∴ W_{s t e e l} < W_{c o p p e r}

$\quad \therefore { W }_{ steel }<{ W }_{ copper }$
So, less work is done on steel spring

A wire stretches by a certain amount under a load. If the load and radius both are increased to four times. The stretch caused in the wire is then

0%
$l$
0%
$\cfrac { l }{ 2 }$
0%
$\cfrac { l }{ 3 }$
0%
$\cfrac { l }{ 4 }$

Explanation

Young's modulus,

Y = \frac{W}{A} \times \frac{L}{l}

$Y=\cfrac { W }{ A } \times \cfrac { L }{ l }$

∴

$\therefore$ Elongation,

l = \frac{W L}{A Y} = \frac{W L}{π r^{2} Y}

$l=\cfrac { WL }{ AY } =\cfrac { WL }{ \pi { r }^{ 2 }Y }$
When both load and radius are increased to four times, the elongation becomes

l^{'} = \frac{4 W \times L}{π {(4 r)}^{2} Y} = \frac{W L}{4 π r^{2} Y} = \frac{l}{4}

$l'=\cfrac { 4W\times L }{ \pi { (4r) }^{ 2 }Y } =\cfrac { WL }{ 4\pi { r }^{ 2 }Y } =\cfrac { l }{ 4 }$

When the load on a wire is increased from

3 k g

$3kg$

w t

$wt$ to

5 k g

$5kg$

w t

$wt$ the elongation increases from

0.61 m m

$0.61mm$ to

1.02 m m

$1.02mm$ . The required work done during the extension of the wire is:

0%
$16\times { 10 }^{ -3 }J$
0%
$8\times { 10 }^{ -2 }J$
0%
$20\times { 10 }^{ -2 }J\quad$
0%
$11\times { 10 }^{ -3 }J$

Explanation

Given:

Apply Load:

M_{1} = 3 k g Δ L = 0.61 m m M_{2} = 5 k g Δ L = 1.02 m m

$M_1=3kg \ \ \ \ \Delta L= 0.61mm \\ M_2=5kg \ \ \ \Delta L=1.02mm$

Work done is stretching the wire through

0.61 m m

$0.61mm$ under the load of

3 k g

$3kg$

w t

$wt$

W_{1} = \frac{1}{2} s t r e t c h i n g f o r c e \times e x t e n s i o n = \frac{1}{2} \times 3 \times 9.8 \times 0.61 \times 10^{- 3} = 8.967 \times 10^{- 3} J

${ W }_{ 1 }=\cfrac { 1 }{ 2 } stretching\quad force\times extension=\cfrac { 1 }{ 2 } \times 3\times 9.8\times 0.61\times { 10 }^{ -3 }=8.967\times { 10 }^{ -3 }J$
Work done in stretching the wire through

1.02 m m

$1.02mm$ under the load of

5 k g

$5kg$

w t

$wt$

W_{2} = \frac{1}{2} \times 5 \times 9.8 \times 1.02 \times 10^{- 3} = 24.99 \times 10^{- 3} J

${ W }_{ 2 }=\cfrac { 1 }{ 2 } \times 5\times 9.8\times 1.02\times { 10 }^{ -3 }=24.99\times { 10 }^{ -3 }J$
Hence work done in stretching the wire from

0.61 m m

$0.61mm$ to

1.02 m m

$1.02mm$

Δ W = W_{2} - W_{1} = (24.99 - 8.961) \times 10^{- 3} ≃ 16 \times 10^{- 3} J

$\Delta W={ W }_{ 2 }-{ W }_{ 1 }=\left( 24.99-8.961 \right) \times { 10 }^{ -3 }\simeq 16\times { 10 }^{ -3 }J$
1031875_936989_ans_e85dccce7c664c61b4a64f10f29eb961.PNG

Let

Y_{S}

${Y}_{S}$ and

Y_{A}

${Y}_{A}$ represent Young's modulus for steel and aluminium respectively It is said that steel is more elastic than aluminium. Therefore, it follows that

0%
${ Y }_{ S }={ Y }_{ A }$
0%
${ Y }_{ S }< { Y }_{ A }$
0%
${ Y }_{ S }> { Y }_{ A }$
0%
$\cfrac { { Y }_{ S } }{ { Y }_{ A } } =0$

Explanation

Steel is more elastic than aluminium
It means

Y_{S} > Y_{A}

${ Y }_{ S }> { Y }_{ A }$

15 k g

$15kg$ mass fastened to the end of a steel wire of unstretched length

1.0 m

$1.0m$ is whirled in a vertical circle with an angular velocity of

2 r e v

$2rev$

s^{- 1}

${s}^{-1}$ at the bottom of the circle. The cross-section of the wire is

0.05 {c m}^{2}

$0.05{cm}^{2}$ . The elongation of the wire when the mass is at the lowest point of its path is
(Take

g = 10 m s^{- 2}, Y_{s t e e l} = 2 \times 10^{11} N m^{- 2}

$g=10m{ s }^{ -2 },{ Y }_{ steel }=2\times { 10 }^{ 11 }N\quad { m }^{ -2 }\quad$ )

0%
$0.52mm$
0%
$1.52mm$
0%
$2.52mm$
0%
$3.52mm$

Explanation

Mass

(m) = 15 k g

$(m) = 15\ kg$

Length of wire

(L) = 1 m

$( L) =1\ m$

Area of cross section of wire

(A) = 0.05 c m ² = 5 \times 10^{- 6} m^{2}

$(A) = 0.05 cm² = 5 × 10^{-6}\ m^2$

Angular frequency

(ν) = 2 r e v / s

$(\nu) = 2\ rev/s$

So, Angular velocity,

(ω) = 2 π ν = 2 π (2) = 4 π r a d / s

$(\omega) = 2π\nu= 2π(2) = 4π\ rad/s$

Young's modulus for steel

(Y) = 2 \times 10^{11} N / m^{2}

$(Y) = 2 × 10^{11}\ N/m^2$

At the lowest point of the vertical circle,

T - m g = m ω^{2} L

$T - mg = m\omega^2\ L$

T = m g + m ω^{2} L

$T = mg + m\omega^2 L$

= 15 [9.8 + (4 π) ² \times 1]

$\ \ \ = 15[{ 9.8 + (4π)²×1 }]$

= 15 (9.8 + 16 π ²)

$\ \ \ = 15( { 9.8 + 16π²})$

= 15 \times 167.72 N

$\ \ \ = 15 × 167.72\ N$

= 2515.8 N

$\ \ \ = 2515.8\ N$

Now,

Youngs modulus

= \frac{stress}{strain}

$=\dfrac{\text{stress}}{\text{strain}}$

Y = \frac{T L}{A Δ L}

$Y= \dfrac{TL}{A∆L}$

Δ L = \frac{F L}{A Y}

$∆L =\dfrac{FL}{AY}$

= \frac{2515.8 \times 1}{(5 \times 10^{- 6}) \times (2 \times 10 ¹ ¹)}

$\ \ \ =\dfrac{ 2515.8 × 1}{(5×10^{-6})×(2×10¹¹)}$

= 2.52 \times 10^{-} 3 m

$\ \ \ = 2.52 × 10^-{3}\ m$

Δ L = 2.52 m m

$\Delta L= 2.52\ mm$

Hence, option

(C)

$(C)$ is correct.

A steel rod of length

1 m

$1m$ and radius

10 m m

$10mm$ is stretched by a force

100 k N

$100kN$ along its length. The percentage strain in the rod is then

(Y_{s t e e l} = 2 \times 10^{11} N m^{- 2})

$\left( { Y }_{ steel }=2\times { 10 }^{ 11 }N\quad { m }^{ -2 } \right)$

0%
$0.04$ %
0%
$0.08$ %
0%
$0.16$ %
0%
$0.24$ %

Explanation

Given:

force,

F = 100 K N

$F =100KN$

radius,

L = 10 m m

$L= 10mm$

length,

L = 1 m

$L= 1m$

Area

A = π r^{2}

$A=\pi r^2$

Stress

σ = \frac{F}{A} = \frac{10^{5}}{π (10^{- 2})^{2}}

$\sigma =\dfrac{F}{A}=\dfrac{10^5}{\pi (10^{-2})^2}$

Elongation

Δ L = \frac{(F / A) L}{Y} = \frac{(\frac{10^{5}}{π (10^{- 2})^{2}}) (1 m)}{2 \times 10^{11} N m^{- 2}} = 1.59 \times 10^{- 3} m = 1.59 m m

$\quad \Delta L=\cfrac { (F/A)L }{ Y } =\cfrac { \left( \dfrac{10^5}{\pi (10^{-2})^2}\right) \left( 1m \right) }{ 2\times { 10 }^{ 11 }N\quad { m }^{ -2 } } =1.59\times { 10 }^{ -3 }m=1.59mm\quad$

Strain produced in the rod is

S t r a i n = \frac{Δ L}{L} = \frac{1.59 \times 10^{- 3} m}{1 m} = 1.59 \times 10^{- 3} = 0.16

$Strain=\cfrac { \Delta L }{ L } =\cfrac { 1.59\times { 10 }^{ -3 }m }{ 1m } =1.59\times { 10 }^{ -3 }=0.16\quad$

If the work done in stretching a wire by

1 m m

$1mm$ is

2 J

$2J$ , the work necessary for stretching another wire of same material but with double radius of cross-section and half the length by

1 m m

$1mm$ is:

0%
$16J$
0%
$8J$
0%
$4J$
0%
$\cfrac { 1 }{ 4 } J$

Explanation

Stretching force,

F = \frac{Y π r^{2} Δ L}{L}

$F=\cfrac { Y\pi { r }^{ 2 }\Delta L }{ L }$
where the symbols have their usual meanings.
Both the wires are of same material, so

Y

$Y$ will be equal, extension in both the wires is same, so

Δ L

$\Delta L$ will be equal

∴ F \propto \frac{r^{2}}{L}

$\therefore F\propto \cfrac { { r }^{ 2 } }{ L }$

∴ \frac{F^{'}}{F} = \frac{{(2 r)}^{2}}{(L / 2)} \times \frac{L}{r^{2}} = 8

$\quad \therefore \cfrac { F' }{ F } =\cfrac { { (2r) }^{ 2 } }{ (L/2) } \times \cfrac { L }{ { r }^{ 2 } } =8$
or

F^{'} = 8 F . . . (i)

$F'=8F...(i)\quad$
Work done is stretching a wire,

W = \frac{1}{2} \times F \times Δ L

$W=\cfrac { 1 }{ 2 } \times F\times \Delta L\quad$
For sam extension

W \propto F

$W\propto F$

∴ \frac{W^{'}}{W} = \frac{F^{'}}{F} = 8

$\therefore \cfrac { W' }{ W } =\cfrac { F' }{ F } =8$ [Using (i)]

W^{'} = 8 W = 8 \times 2 J = 16 J

$W'=8W=8\times 2J=16J$

A wire of length

L

$L$ has a linear mass density

μ

$\mu$ and area of cross-section

A

$A$ and Young's modulus

Y

$Y$ is suspended vertically from a rigid support. The extension produced in the wire due to its own weight is:

0%
$\cfrac { \mu g{ L }^{ 2 } }{ YA }$
0%
$\cfrac { \mu g{ L }^{ 2 } }{ 2YA }$
0%
$\cfrac { 2\mu g{ L }^{ 2 } }{ YA }$
0%
$\cfrac { 2\mu g{ L }^{ 2 } }{ 3YA }$

Explanation

Consider a small element of length

d x

$dx$ at a distance

x

$x$ from the free end of wire as shown in the figure. Tension in the wire at distance

x

$x$ from the lower end is

T (x) = μ g x . . . . (i)

$T(x)=\mu gx....(i)$
Let

d l

$dl$ be increase in length of the element. Then

T = \frac{T (x) / A}{d l / d x}

$T=\cfrac { T(x)/A }{ dl/dx }$

d l = \frac{T (x) d x}{Y A} = \frac{μ g x d x}{Y A}

$dl=\cfrac { T(x)dx }{ YA } =\cfrac { \mu gxdx }{ YA }$ [Using (i)]

Total extension produced in the wire is

l = \int_{0}^{L} \frac{μ g x}{Y A} = \frac{μ g}{Y A} {[\frac{x^{2}}{2}]}_{0}^{L} = \frac{μ g L^{2}}{2 Y A}

$\quad l=\displaystyle \int _{ 0 }^{ L }{ \cfrac { \mu gx }{ YA } } =\cfrac { \mu g }{ YA } { \left[ \cfrac { { x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ L }=\cfrac { \mu g{ L }^{ 2 } }{ 2YA }$

1032758_936941_ans_4688ef5f66034df7ae24bbb30dff23f7.PNG

The following four wires of length

L

$L$ and radius

r

$r$ are made of the same material. Which of these will have the largest extension, when the same tension is applied?

0%
$L=100cm, r=0.2mm$
0%
$L=200cm, r=0.4mm$
0%
$L=300cm, r=0.6mm$
0%
$L=400cm, r=0.8mm$

Explanation

Young's modulus

Y = \frac{F}{A} \frac{L}{Δ L} = \frac{T}{π r^{2}} \frac{L}{Δ L}

$Y=\cfrac { F }{ A } \cfrac { L }{ \Delta L } =\cfrac { T }{ \pi { r }^{ 2 } } \cfrac { L }{ \Delta L }$

Δ L = \frac{T}{π r^{2}} \frac{L}{Y}

$\Delta L=\cfrac { T }{ \pi { r }^{ 2 } } \cfrac { L }{ Y }$ \
where the symbols have their ususal meanings
As the four wires are made of the same material, therefore Young's modulus is the same for four wires.
As the

T

$T$ and

Y

$Y$ are the same for the four wires.

∴ Δ L \propto \frac{L}{r^{2}}

$\therefore \Delta L\propto \cfrac { L }{ { r }^{ 2 } }$

\frac{L}{r^{2}}

$\cfrac { L }{ { r }^{ 2 } }$
is maximum for wire length

L = 100 c m

$L=100cm$ and radius

r = 0.2 m m

$r=0.2mm$

Two wires of the same material and length but diameter in the ratio

1 : 2

$1:2$ are stretched by the same load. The ratio of elastic potential energy per unit volume for the two wires is:

0%
$1:1$
0%
$2:1$
0%
$4:1$
0%
$16:1$

Explanation

Hint: Elastic energy per unit volume: $\quad \mathrm{E}=\dfrac{1}{2} \times$ Stress $\times$ Strain

Correct Answer = Option (D)

Step 1: Using Hooke's law: $\quad$ Strain $=\dfrac{\text { Stress }}{\mathrm{Y}}$

$\Longrightarrow \quad \mathrm{E}=\dfrac{(\text { Stress })^{2}}{2 \mathrm{Y}}=\dfrac{\mathrm{F}^{2}}{2 \mathrm{YA}^{2}} \quad$ where $\mathrm{A}=\dfrac{\pi \mathrm{D}^{2}}{4}$

$\Longrightarrow \quad \mathrm{E}=\dfrac{8 \mathrm{~F}^{2}}{\pi^{2} \mathrm{D}^{4} \mathrm{Y}}$

$\therefore \dfrac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\dfrac{\mathrm{D}_{2}}{\mathrm{D}_{1}}\right)^{4}$

$(\because \mathrm{F}$ and $\mathrm{Y}$ are same for both)

Given : $\dfrac{\mathrm{D}_{1}}{\mathrm{D}_{2}}=\dfrac{1}{2}$

on substituing the value of $\dfrac{\mathrm{D}_{1}}{\mathrm{D}_{2}}=\dfrac{1}{2}$

$\dfrac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\dfrac{2}{1}\right)^{4}=\dfrac{16}{1}$

Among solids, liquids and gases, which posses the greatest bulk modulus?

0%
Solids
0%
Liquids
0%
Gases
0%
Both solids and liquids

Explanation

Bulk modulus is defining the ability of a material to resist deformation in terms of volume change, under pressure.
Bulk modulus $=\cfrac { P }{ \Delta V/V }$

As solids have long-range order by its nature. Because the atoms of a crystal are arranged in an ordered pattern that stretches all the way across its material because of which if the same amount of pressure will be applied on solids, liquid, and gases then minimum fractional change in volume will be for solid.
Thus solids are having the greatest Bulk Modulus.
Option A is correct.

The average depth of Indian Ocean is about

3000 m

$3000m$ . The fractional compression,

\frac{Δ V}{V}

$\cfrac { \Delta V }{ V }$ of water at the bottom of the ocean is then
(Given: Bulk modulus of the water

= 2.2 \times 10^{9} N m^{- 2}

$=2.2\times { 10 }^{ 9 }N{ m }^{ -2 }$ and

g = 10 m s^{- 2}

$g=10m{ s }^{ -2 }$ )

0%
$0.82$ %
0%
$0.91$ %
0%
$1.36$ %
0%
$1.24$ %

Explanation

The pressure exerted by a

3000 m

$3000m$ column of water on the bottom layer is

P = h ρ g = 3000 m \times 1000 k g m^{- 3} \times 10 m s^{- 2} = 3 \times 10^{7} N m^{- 2}

$P=h\rho g=3000m\times 1000kg\quad { m }^{ -3 }\times 10m\quad { s }^{ -2 }=3\times { 10 }^{ 7 }N\quad { m }^{ -2 }\quad$
Fractional compression

\frac{Δ V}{V}

$\cfrac { \Delta V }{ V }$ is

\frac{Δ V}{V} = \frac{P}{B} = \frac{3 \times 10^{7} N m^{- 2}}{2.2 \times 10^{9} N m^{- 2}} = 1.36 \times 10^{- 2} = 1.36

$\cfrac { \Delta V }{ V } =\cfrac { P }{ B } =\cfrac { 3\times { 10 }^{ 7 }N\quad { m }^{ -2 } }{ 2.2\times { 10 }^{ 9 }N\quad { m }^{ -2 } } =1.36\times { 10 }^{ -2 }=1.36\quad$ %

Match the column I with column II

Column I	Column II
(A) The of shape rubber heel changes under stress	(p) Young's modulus of elasticity is involved
(B) In a suspended bridge, there is a strain in the ropes by the load of the bridge	(B) Bulk modulus of elasticity is involved
(C) In an automobile tyre, when air is compressed, the shape of tyre changes	(r) Modulus of rigidity is involved
(D) A solid body is subjected to a deforming force	(s) All the moduli of elasticity are involved

0%
A-q; B-r; C-s; D-p
0%
A-p; B-q; C-r; D-s
0%
A-r; B-q; C-p; D-s
0%
A-r; B-p; C-q; D-s

The length of a rubber cord is

l_{1}

${l}_{1}$ when the tension is

4 N

$4N$ and

l_{2} m

${l}_{2}m$ when the tension is

$6N$ . The length when the tension is

$9N$ , is:

0%
$\left( 2.5{ l }_{ 2 }-1.5{ l }_{ 1 } \right) m$
0%
$\left( 6{ l }_{ 2 }-1.5{ l }_{ 1 } \right) m$
0%
$\left( 3{ l }_{ 2 }-2{ l }_{ 1 } \right) m$
0%
$\left( 3.5{ l }_{ 2 }-2.5{ l }_{ 1 } \right) m$

Explanation

Let the original unstretched length be

$l$ .

$\therefore Y=\cfrac { T }{ A } \cfrac { l }{ \Delta l }$
Now

$Y=\cfrac { 4 }{ A } \cfrac { l }{ \left( { l }_{ 1 }-l \right) } =\cfrac { 6 }{ A } \cfrac { l }{ \left( { l }_{ 2 }-l \right) } =\cfrac { 9 }{ A } \cfrac { l }{ \left( { l }_{ 3 }-l \right) }$

$\quad \therefore 4\left( { l }_{ 3 }-l \right) =9\left( { l }_{ 1 }-l \right) \quad or\quad 4{ l }_{ 3 }+5l=9{ l }_{ 1 }...(i)\quad$

Again

$6\left( { l }_{ 3 }-l \right) =9\left( { l }_{ 2 }-l \right) \quad or\quad 2{ l }_{ 3 }+l=3{ l }_{ 2 }...(ii)$

To obtain

${l}_{3}$ , solve (i) and (ii) simulatneously

$\therefore { l }_{ 3 }=\left( 2.5{ l }_{ 2 }-1.5{ l }_{ 1 } \right) m$

CBSE Questions for Class 11 Engineering Physics Mechanical Properties Of Solids Quiz 7 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Mechanical Properties Of Solids Quiz 7 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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