MCQ Questions for Class 11 Engineering Physics Mechanical Properties Of Solids Quiz 3

When a tension

$F$ is applied, the elongation produced in uniform wire of length

$L$ , radius

$r$ is

$e$ . When tension

$2F$ is applied, the elongation produced in another uniform wire of length

$2L$ and radius

$2r$ made of same material is:

0%
$0.5e$
0%
$1.0e$
0%
$1.5e$
0%
$2.0e$

Explanation

$Y = \dfrac{F / \pi r^{2}}{\triangle l/l}$

$\triangle l = \dfrac{F l}{\pi r^{2} Y}$

$E = \dfrac{F l}{\pi r^{2} Y} - ------1$

$\triangle l = \dfrac{2 F 2l}{\pi 4 r^{2}Y}$

$\triangle l = \dfrac{F l}{\pi r^{2}Y}= e$ ---------- (By 1)

=> elongation is none wire is same as

$e$

A load of 1kg weight is attached to one end of a steel wire of cross sectional area 3mm

$^{2}$ and Youngs modulus 10

$^{11}$ N/m

$^{2}$ . The other end is suspended vertically from a hook on a wall, then the load is pulled horizontally and released.When the load passes through its lowest position the fractional change in length is: (g

$=$ 10m/ s

$^{2}$ )

0%
$0.3 \times 10^{-4}$
0%
$0.3 \times 10^{-3}$
0%
$10^{3}$
0%
$10^{4}$

Explanation

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

$\dfrac{\triangle l}{l} = \dfrac{F}{YA}$

$= \dfrac{1 \times 10}{10^{11} \times 3 \times 10^{-6}}$

$= \ 0.3 \times 10^{-4}$

Two wires A and B of the same dimensions are under loads of

$4$ and

$5.5 kg$ respectively. The ratio of Young's modulii of the materials of the wires for the same elongation is:

0%
$64 : 121$
0%
$\sqrt{11}:\sqrt{8}$
0%
$11:8$
0%
$8 : 11$

Explanation

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

$Y_{A} = \dfrac{F_A/A}{\triangle l/l} ........................... 1$

$Y_{B} = \dfrac{F_B/A}{\triangle l/l} ........................... 2$
Dividing 1 and 2

$\dfrac{Y_{A}}{Y_{B}} = \dfrac{F_{A}}{F_{B}} = \dfrac{4}{5.5} = \dfrac{8}{11}$

$3 cm$ long copper wire is stretched to increase its length by

$0.3cm.$ If poisson's ratio for copper is

$0.26$ , the lateral strain in the wire is

0%
0.26
0%
2.6
0%
0.026
0%
0.0026

Explanation

$\epsilon x=\dfrac{0.3}{3}$

$=0.1$ (standard result)

$\sigma =0.26$ (given)

$0.26=\dfrac{-\epsilon y}{0.1}$

$-0.026=\epsilon y$

A wire is subjected to a longitudinal strain of

$0.05.$ If its material has a Poisson's ratio

$0.25$ , the lateral strain experienced by it is

0%
0.00625
0%
0.125
0%
0.0125
0%
0.0625

Explanation

$\epsilon x=0.05$ (given)

$\sigma =0.25$

$\dfrac{\epsilon y}{0.05}=-0.25$ (standard result)

$=-0.0125$

A metallic ring of radius 'r', cross sectional area 'A' is fitted into a wooden circular disk of radius 'R' (R > r). If the Young's modulus of the material of the ring is 'Y', the force with which the metal ring expands is :

0%
$\dfrac{AYR}{r}$
0%
$\dfrac{AY(R-r)}{r}$
0%
$\dfrac{Y(R-r)}{Ar}$
0%
$\dfrac{YR}{Ar}$

Explanation

$Y = \dfrac{F/A}{\triangle l/l}$ (standard result)

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

$= \dfrac{Y (R-r) A}{r} (\because \triangle l = R-r)$

When a certain force is applied on a string it extends by

$0.01cm$ . When the same force is applied on another string of same material, twice the length and double the diameter, then the extension in second string is

0%
0.005 cm
0%
0.02 cm
0%
0.08 cm
0%
0.04 cm

Explanation

$\triangle l = \dfrac{F l}{\pi (\dfrac{d}{2})^{2}Y}$

$0.01 = \dfrac{F l}{\pi (\dfrac{d}{2})^{2}Y}$

$\triangle l_{1} = \dfrac{F 2l}{\pi 4 (\dfrac{d}{2})^{2}Y}$

$= \dfrac{1}{2} \dfrac{F l}{\pi (\dfrac{d}{2})^{2}Y}$ (also take the ratio in the next step)

$\dfrac{1}{2} (0.01)$

$= 0.005cm$

When load is applied to a wire, the extension is

$3mm.$ The extension in the wire of same material and length but half the radius extended by the same load is :

0%
$0.75mm$
0%
$6mm$
0%
$1.5mm$
0%
$12.0mm$

Explanation

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

$\triangle l = \dfrac{F l}{Y \pi R^{2}}$

$So, \triangle l \alpha \dfrac{1}{R^{2}}$

$\dfrac{\triangle l_{1}}{\triangle l_{2}} = \dfrac{r_{2}^{2}}{r_{1}^{2}} = (\dfrac{1}{2})^{2} = \dfrac{1}{4} (\because \dfrac{r_{1}}{r_{2}} = 2)$

$4\triangle l_{1} = \triangle l_{2}$

$So, \triangle l_{2} = 4(3)$

$= 12mm$

A steel wire is 1m long and 1

$mm^2$ in the area of cross-section. If it takes

$200N$ to stretch the wire by

$1mm$ , the force that will be required to stretch the wire of the same material and cross-sectional area from a length of

$10m$ to

$1002 cm$

0%
$100N$
0%
$200 N$
0%
$400 N$
0%
$2000N$

Explanation

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

$Y = \dfrac{200 \times 1}{10^{-6}\times 10^{-3}}$

$Y = 2\times 10^{11}$

$\triangle l = 1002 - 1000 = 2cm , l = 10m$

$F = \dfrac{YA \triangle l}{l}$

$=\dfrac{2 \times 10^{11}\times 10^{-6} \times 2 \times 10^{-2}}{10}$

$= 400N$

When a uniform wire of radius r is stretched by a

$2 kg$ weight, the increase in its length is

$2.00 mm$ . If the radius of the wire is

$r/2$ and other conditions remain in the same, increase in its length is

0%
$2.00mm$
0%
$4.00 mm$
0%
$6.00 mm$
0%
$8.00mm$

Explanation

$Y = \dfrac{F / A}{\triangle l/l}$ (standard result)

$\triangle l = \dfrac{F l}{\pi r^{2} Y}$

$\dfrac{2}{1000} = \dfrac{F l}{\pi (\dfrac{r}{2})^{2}Y}$

$\triangle l_{1} = \dfrac{F l}{\pi (\dfrac{r}{2})^{2}Y}$

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

$= 4(\dfrac{2}{1000})$

$= \dfrac{8}{1000} m$

$= 8 mm$

The elongation of a steel wire stretched by a force is '

$e$ '. If a wire of the same material of double the length and half the diameter is subjected to double the force, its elongation will be

0%
$16e$
0%
$4e$
0%
$\left(\dfrac{1}{4}\right)e$
0%
$\left(\dfrac{1}{16}\right)e$

Explanation

$Y = \dfrac{F / \pi (d/2)^{2}}{\triangle l/l}$

$\triangle l = \dfrac{F l}{\pi (\dfrac{d}{2})^{2}Y}$

Now, according to problem

$e = \dfrac{F l}{\pi (\dfrac{d}{2})^{2}Y}$

$Y = \dfrac{F /A}{\triangle l_{1}/l }$

$\triangle l_{1} = \dfrac{F l}{Y A}$

$= \dfrac{2F 2l}{Y \pi (\dfrac{d}{2})^{2} \dfrac{1}{4}}$

$= 16 \dfrac{F l}{Y \pi (\dfrac{d}{2})^{2}}$

$=16 e$

So, new elongation is 16e

What percent of length of a wire will increase by applying a stress of

$1 kg$ . wt/mm

$^{2}$ on it.
[Y

$=$ 1x10

$^{11}$ Nm

$^{-2}$ and

$1 kg wt$

$=$ 9.8N]

0%
$0.0078$ %
0%
$0.0088$ %
0%
$0.0098$ %
0%
$0.0067$ %

Explanation

$Y = \dfrac{Stress}{Strain}$

$Strain = \dfrac{Stress}{Y}$

$= \dfrac{9.8 \times 10^{6}}{1 \times 10^{11}}$

$= 9.8 \times 10^{-3}$ %

$= 0.0098$ %

So, change in length is

$0.0098$ %

An elongation of

$0.1$ % in a wire of cross-section

$10^{-6}m^{2}$ causes a tension of

$100N$ .

$Y$ for the wire is

0%
$10^{12}N/m^{2}$
0%
$10^{11}N/m^{2}$
0%
$10^{10}N/m^{2}$
0%
$100\ N/m^{2}$

Explanation

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

$\triangle l = \dfrac{Fl}{AY}$

$So, \triangle l \alpha \dfrac{1}{\triangle}$

$\therefore \dfrac{\triangle l_{1}}{\triangle L_{2}} = \dfrac{A_{2}}{A_{1}} = 3$

$\dfrac{\triangle l_{1}}{3} = \triangle l_{2}$

$\triangle l_{2} = \dfrac{0.1}{3} = 0.033mm$

An iron wire of length

$4m$ and diameter

$2 mm$ is loaded with a weight of

$8kg$ . If the young's modulus '

$Y$ ' for iron is

$2$ x

$10$

$^{11}$ N/m

$^{2}$ , then the increase in the length of the wire is

0%
$0.2 mm$
0%
$0.5mm$
0%
$2mm$
0%
$1 mm$

Explanation

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

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

$\quad = \dfrac { 80 \times l }{ r { (\dfrac { 2 }{ 2 } ) }^{ 2 }\times { 10 }^{ -6 }\times 2\times { 10 }^{ 11 } }$

$\quad = 5.0 \times { 10 }^{ -4 }m$

$\quad = 0.5\ mm$

Four wires P,Q,R and S of same materials have diameters and stretching forces as shown below. Arrange their strains in the decreasing order.

Wire	Diameter	Stretching force
P	2 mm	10 N
Q	1 mm	20 N
R	4 mm	30 N
S	3 mm	40 N

0%
Q,S,P,R
0%
R,P,S,Q
0%
P,Q,R,S
0%
P,R,Q,S

Explanation

$Y = \dfrac{Stress}{Strain}$

$Strain = \dfrac{Stress}{Y}$

$= \dfrac{F/A}{Y}$

$Strain (P) = \dfrac{10/\pi (\dfrac{2}{2})^{2}}{Y} = \dfrac{10}{\pi Y} N/mm^{2}$

$Strain (Q) = \dfrac{20/\pi (\dfrac{1}{2})^{2}}{Y} = \dfrac{80}{\pi Y} N/mm^{2}$

$Strain (R) = \dfrac{30/\pi (\dfrac{4}{2})^{2}}{Y} = \dfrac{30}{4\pi Y} N/mm^{2}$

$Strain (S) = \dfrac{40/\pi (\dfrac{3}{2})^{2}}{Y} = \dfrac{160}{9\pi Y} N/mm^{2}$

Strain in order of decreasing

$Q>S>P>R$

An aluminium wire and steel wire of the same length and cross section are joined end to end.The composite wire is hung from a rigid support and a load is suspended from the free end. The young's modulus of steel is

$20/7$ times the aluminium. The ratio of increase of length of steel and aluminium is

0%
$20/7$
0%
$400/49$
0%
$7/20$
0%
$49/400$

Explanation

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

$Y \alpha \dfrac{1}{\triangle l}$

Now take the ratio,

$\dfrac{Y_{1}}{Y_{2}} = \dfrac{\triangle l_{2}}{\triangle l_{1}}$

$\therefore \dfrac{\triangle l_{2}}{\triangle l_{1}} = \dfrac{Y_{1}}{Y_{2}} = \dfrac{20}{7}$

$or \dfrac{\triangle l_{1}}{\triangle l_{2}} = \dfrac{7}{20}$

A wire elongates by l mm when a load W is hanged from it. If the wire goes over a pulley and two weights W each are hung at the two ends,the elongation of the wire will be (in mm)

0%
zero
0%
l / 2
0%
l
0%
2l

A ball falling in a lake to a depth

$200m$ shows a decrease of

$0.1$ % in its volume at the bottom. the bulk modulus of the ball is

0%
19.6 x 10 $^{8}$ N/m $^{2}$
0%
19.6 x 10 $^{-10}$ N/m $^{2}$
0%
19.6 x 10 $^{10}$ N/m $^{2}$
0%
19.6 x 10 $^{-8}$ N/m $^{2}$

Explanation

$B=\dfrac{\rho gh}{\Delta v/v}$ (standard result)

$=\dfrac{1000\times 200\times 9.8}{\dfrac{0.1}{100}}$

$=19.6\times 10^{8}N/m^{2}$

On taking a solid rubber ball from the surface to the bottom of a lake 200m deep, the reduction in volume is found to be 0.5%. If the density of water is 10

$^{3}$ kgm

$^{-3}$ and g

$=$ 10 ms

$^{-2}$ , find the bulk modulus of rubber.

0%
2 x10 $^{8}$ Pa
0%
4 x10 $^{8}$ Pa
0%
6 x10 $^{8}$ Pa
0%
8 x10 $^{8}$ Pa

Explanation

$B=\dfrac{\rho g\Delta h}{\Delta v/v}$

$=\dfrac{1000\times 10\times 200}{0.5/100}$

$=4\times 10^{8}pa$

A wire extends by

$1 mm$ when a force is applied. Double the force is applied to another wire of same material and length but of half the radius of cross section.The elongation of wire in mm will be

0%
$8$
0%
$4$
0%
$2$
0%
$1$

Explanation

$Y=\dfrac{Fl}{A\Delta l}$

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

$\dfrac{\Delta l_{1}}{\Delta l_{2}}=\dfrac{Fl}{\pi r^{2}Y}\times \dfrac{Y\pi \left ( \dfrac{r}{2} \right )^{2}}{2Fl}$

$\dfrac{\Delta l_{1}}{\Delta l_{2}}=\dfrac{1}{8}$

$8\Delta l_{1}=\Delta l_{2}$

So,

$\Delta l_{2}=8(1)$

$=8mm$

A sphere contracts in volume by

$0.01$ % when taken to the bottom of lake

$1km$ deep. If the density of water is

$1gm/cc$ , the bulk modulus of water is

0%
9.8 x 10 $^{5}$ N/m $^{2}$
0%
9.8 x 10 $^{8}$ N/m $^{2}$
0%
9.8 x 10 $^{10}$ N/m $^{2}$
0%
9.8 x 10 $^{6}$ N/m $^{2}$

Explanation

$B=\dfrac{\Delta p}{\Delta v/v}=\dfrac{\rho g\Delta h}{\Delta v/v}$

$=\dfrac{1000\times 9.8\times 1000}{0.00\times 10^{-2}}$

Bulk Modulus

$=9.8\times 10^{10}N/m^{2}$

The Y of a material having a cross sectional area of

$1 cm$

$^{2}$ is 2 x 10

$^{12}$ dynes/cm

$^{2}$ . The force required to double the length of the wire is:

0%
1 x 10 $^{12}$ dynes
0%
2 x 10 $^{12}$ dynes
0%
0.5 x 10 $^{12}$ dynes
0%
4 x 10 $^{12}$ dynes

Explanation

$Y=\dfrac{Fl}{A\Delta l}$

$F=\dfrac{YA\Delta l}{l}$

$=2\times 10^{12}\times 1$

$\left ( \therefore \dfrac{\Delta l}{l}=1 \right )$

$=2\times 10^{12}$ dynes.

One end of a wire

$2m$ long and

$0.2cm$

$^{2}$ in cross section is fixed to a ceiling and a load of

$4.8kg$ is attached to its free end. The elongation in the wire in mm is (if y

$=$ 2.0 x 10

$^{11}$ N.m

$^{-2}$ , g

$=$ 10ms

$^{-2}$ )

0%
$2.4$ x $10$ $^{-5}$
0%
$2.4$
0%
$0.024$
0%
$0.0024$

Explanation

$Y=\dfrac{Fl}{A\Delta l}$

$\Delta l=\dfrac{Fl}{\Delta Y}$

$=\dfrac{45\times 2}{0.2\times 10^{-4}\times 2\times 10^{11}}$

$=24\times 10^{-6}m$

$=24\times 10^{-3}mm$

$=0.024mm$

The increase in pressure required to decrease the

$200$ litres volume of a liquid by

$0.004$ % in kPa is : (bulk modulus of the liquid

$=$

$2100 MPa$ )

0%
$8.4$
0%
$84$
0%
$92.4$
0%
$168$

Explanation

$B=P\Delta v/v$

$P=B\times \dfrac{\Delta v}{v}$

$=2100\times 10^{3}\times \dfrac{0.004}{100}$

$=84k$ pa.

The extension of a wire by the application of a load is

$0.3 cm$ . The extension in the wire of the same material but of double the length and half the radius of cross section in cm is

0%
$3$
0%
$0.3$
0%
$1.2$
0%
$2.4$

Explanation

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

$\dfrac{\Delta l_{1}}{\Delta l_{2}}=\dfrac{Fl}{Y\pi r^{2}}\times \dfrac{Y\pi \left ( \dfrac{r}{2} \right )^{2}}{F2l}$

$\dfrac{\Delta l_{1}}{\Delta l_{2}}=\dfrac{1}{8}$

$8\Delta l_{1}=\Delta l_{2}$

So,

$\Delta l_{2}=8(0.3)$

$=2.4cm$

When a wire is subjected to a force along its length, its length increases by

$0.4$ % and its radius decreases by

$0.2$ %. Then the Poisons ratio of the material of the wire is

0%
$0.8$
0%
$0.5$
0%
$0.2$
0%
$0.1$

Explanation

Poison's ratio

$=\dfrac{\Delta d/d}{\Delta r/r}$

$=\dfrac{0.2/100}{0.4/100}$

$=\dfrac{1}{2}=0.5$

A 8m long string of rubber, having density 1.5 x 10

$^{3}$ kg/m

$^{3}$ and young's modulus

$5\times10^{6}$ N/m

$^{2}$ is suspended from the ceiling of a room. The increase in its length due to its own weight will be (g

$=$ 10m/s

$^{2}$ )

0%
$9.6\times 10^{-2}\ \text{m}$
0%
$19.2\times 10^{-5}\ \text{m}$
0%
$9.6\times 10^{-3}\ \text{m}$
0%
$9.6\ \text{m}$

Explanation

Mass of small dx element,

$\displaystyle dm = \lambda Adx$

Also

$\displaystyle \frac{stress}{strain}=Y$

$\displaystyle \frac{F/A}{\frac{\triangle L}{L}} = Y$

where

$F \rightarrow (x)(A)\lambda .g$ , gravitational force experienced by the part below to

$dm$ mass

Strain on

$dm$ mass =

$\dfrac{d(\triangle L)}{dx}$

$\displaystyle \frac{x\, A\lambda g}{A\lambda}=\frac{d(\triangle L)}{dx}$

$\displaystyle \Rightarrow\int_{o}^{\triangle L} d(\triangle L) = \frac{\lambda g}{y}\int_{o}^{L}xdx$

$\displaystyle \triangle L = \frac{\lambda gL^2}{2y}=9.6\times 10^{-2}$

The Poisson's ratio of a material is 0.If a force is applied to a wire of this material, there is a decrease of cross-sectional area by 2%. The percentage increase in its length is :

0%
$3$ %
0%
$2.5$ %
0%
$1$ %
0%
$0.5$ %

Explanation

Poisson's ratio,

$0.4=\cfrac{\Delta d/d}{\Delta l/l}\Rightarrow\cfrac{\Delta l}{l}=\cfrac{\Delta d/d}{0.4}$ -------------(1)

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

Taking log both sides:

$\log A=\log {(\frac{\pi}{4})}+2\log {d}$

Taking relative change both the sides:

$\cfrac{\Delta A}{A}=2\cfrac{\Delta d}{d}\\ \cfrac{\Delta d}{d}=\cfrac{1}{2}\cfrac{\Delta A}{A}=\dfrac{1}{2}\times 2\%=1\%$ ------------------------------(1)

From equation (1) and (2), the percentage increase length can be given as:

$\cfrac{\Delta l}{l}=\cfrac{1\%}{0.4}=2.5\%$

Option B is correct

$20 kg$ load is suspended from the lower end of a wire

$10 cm$ long and

$1 mm$

$^2$ in cross sectional area. The upper half of the wire is made of iron and the lower half with aluminium. The total elongation in the wire is
(

$Y_{iron} =$ 20 x 10

$^{10}$ N/m

$^{2}$ ,

$Y_{Al}$

$=$ 7 x 10

$^{10}$ N/m

$^{2}$ )

0%
1.92 x 10 $^{-4}$ m
0%
17.8 x 10 $^{-3}$ m
0%
1.78 x 10 $^{-3}$ m
0%
1.92 x 10 $^{-3}$ m

Explanation

$Y=\dfrac{F/A}{\Delta L/L} \Rightarrow \Delta L=\dfrac{FL}{AY}$

Therefore,

$\Delta L_i=\dfrac{FL_i}{Y_i A}$ and

$\Delta L_A =\dfrac{FL_A}{Y_A A}$

Total elongation

$=\Delta L_i+\Delta L_A$

$=\dfrac{FL}{A}\left ( \dfrac{1}{Y_i}+\dfrac{1}{Y_A} \right )$

$=\dfrac{200\times 5}{10^{-6}\times 100}\left ( \dfrac{1}{20\times 10^{10}}+\dfrac{1}{7\times 10^{10}} \right )$

$=1.92\times 10^{-4}m$

A piece of copper wire has twice the radius of steel wire. One end of the copper wire is joined to one end of steel wire so that both of them can be subjected to the same longitudinal force.

$Y$ for steel is twice that of copper. When the length of copper wire is increased by

$1$ %, the steel wire will be stretched by

0%
$2$ % of its original length
0%
$1$ % of its original length
0%
$4$ % of its original length
0%
$0.5$ % of its original length

Explanation

$Y=\dfrac{F/A}{\Delta l/l}$

$\dfrac{\Delta l}{l}=\dfrac{F}{YA}$

$\dfrac{\Delta l_c}{l_c}=\dfrac{F}{Y_cA_c}=\dfrac{F}{Y_c \pi (r_c)^2}$ ------(1)

$\dfrac{\Delta l_s}{l_s}=\dfrac{F}{Y_sA_s}=\dfrac{F}{Y_s \pi (r_s)^2}$ ------(2)

Dividing (1) by (2)

$\dfrac{\Delta l_c/l_c}{\Delta l_s/l_s}=\dfrac{Y_s(r_s)^2}{Y_c(r_c)^2}=\dfrac{Y_s}{Y_c}\times \left ( \dfrac{r_s}{r_c} \right )^2=(2)\times (\dfrac{1}{2})^2$

$\Rightarrow \dfrac{\Delta l_c/l_c}{\Delta l_s/l_s}=\dfrac{1}{2}$

$\therefore \Delta l_s/l_s=2\Delta l_c/l_c$

$=2$ %

A steel wire of mass

$3.16 Kg$ is stretched to a tensile strain of 1 x 10

$^{-3}$ . What is the elastic deformation energy if density

$\rho =7.9 g/cc$ and Y

$=$ 2x10

$^{11}$ N/m

$^{2}$ ?

0%
$4 KJ$
0%
$0.4 KJ$
0%
$0.04KJ$
0%
$4J$

Explanation

$Volume=\dfrac{mass}{\rho }$

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

$stress=Y\times strain$

$energy=\dfrac{1}{2}\times stress\times strain\times volume$

$=\dfrac{1}{2}\times Y\times strain\times strain\times \dfrac{mass}{\rho }$

$=\dfrac{1}{2}\times 2\times 10^{11}\times 1\times 10^{-3}\times 1\times 10^{-3}\times \dfrac{3.16}{7.9\times 1000}$

$=0.04KJ.$

The length of a metal wire is

$l_{1}$ when the tension in it is

$F_{1}$ and

$l_{2}$ when the tension in it is

$F_{2}$ . The natural length of the wire is

0%
$\dfrac{l_{1}F_{1}+l_{2}F_{2}}{F_{1}+F_{2}}$
0%
$\dfrac{l_{2}-l_{1}}{F_{2}-F_{1}}$
0%
$\dfrac{l_{1}F_{2}-l_{2}F_{1}}{F_{2}-F_{1}}$
0%
$\dfrac{l_{1}F_{1}-l_{2}F_{2}}{F_{2}-F_{1}}$

Explanation

We have

$l_1-l=\dfrac{F_1l}{AY}$
and

$l_2-l=\dfrac{F_2l}{AY}$
Thus we get

$\dfrac{l_1-l}{F_1}=\dfrac{l_2-l}{F_2}$
or

$F_2l_1-F_2l=F_1l_2-F_1l$
or

$l=\dfrac{F_2l_1-F_1l_2}{F_2-F_1}$

When a wire of length 10m is subjected to a force of

$100 N$ along its length , the lateral strain produced is

$0.01$ x

$10$

$^{-3}$ . The Poisson's ratio was found to be

$0.4$ . If the area of cross-section of wire is

$0.025m$

$^{2}$ , its Young's modulus is :

0%
$1.6 \times 10^{8} N /m^{2}$
0%
$2.5 \times 10^{10} N /m^{2}$
0%
$12.5 \times 10^{11} N /m^{2}$
0%
$16 \times 10^{10} N /m^{2}$

Explanation

$\text{Poisson's ratio} =\dfrac{\Delta d/d}{\Delta d_0/d_0}$

$\dfrac{\Delta l}{l}=\dfrac{\Delta d/d}{0.4}$

$=\dfrac{0.01\times 10^{3}}{0.4}$

$Y=\dfrac{F/A}{\Delta l/l}$

$=\dfrac{100/0.025}{0.01\times 10^{3}/0.4}$

$=\dfrac{100\times 0.4}{0.025\times 0.01\times 10^{3}}$

$=1.6\times 10^{8}N/m^{2}$

The increase in length of a wire on stretching is

$0.025$ %. If its Poisson's ratio is

$0.4$ , then the percentage decrease in the diameter is :

0%
$0.01$
0%
$0.02$
0%
$0.03$
0%
$0.04$

Explanation

$\text{Poisson's ratio} =\dfrac{\Delta d/d}{\Delta l/l}$

$0.4=\dfrac{\Delta d/d}{0.025/100}$

$\dfrac{0.4\times 0.025}{100}=\Delta d/d$

$\Delta d/d=0.01$ %

One end of uniform wire of length

$L$ and of weight

$W$ is attached rigidly to a point in the roof and a weight

$W_{1}$ is suspended from the lower end. If

$A$ is the area of cross-section of the wire, the stress in the wire at a height

$\dfrac{3L}{4}$ from its lower end is:

0%
$\dfrac{W_{1}}{A}$
0%
$\dfrac{\left ( W_{1}+\dfrac{W}{4} \right )}{A}$
0%
$\dfrac{\left ( W_{1}+\dfrac{3W}{4} \right )}{A}$
0%
$\dfrac{W_{1}+W}{A}$

Explanation

$stress=\dfrac{Tension}{Area}$

Tension at height

$\dfrac{3L}{4}$ from lower end

is

$\dfrac{3}{4}w+w_1$

So, stress

$=\dfrac{\dfrac{3}{4}w+w_1}{S}$

The length of an elastic spring is '

$a$ ' when tension is

$4 N$ and '

$b$ ' when the tension is

$5N$ . The length of the spring when tensionis 9N is

0%
4a - 5b
0%
5b - 4a
0%
5b + 4a
0%
5(b-a)

Explanation

$Y=\dfrac{F/A}{\Delta l/l}$

$\Delta l=\dfrac{Fl}{YA}$

$(a-l)=\dfrac{4l}{YA}$ -------(1)

$(b-l)=\dfrac{5l}{YA}$ -------(2)

Dividing (1) by (2)

$\dfrac{a-l}{b-l}=\dfrac{4}{5}$

$5a-5l=4b-bl$

$l=5a-4b$ ------(3)

$\Delta l=\dfrac{9l}{YA}$ (when force

$=9N$ )

$x-l=9\dfrac{(a-l)}{4}$ ( BY (1))

$x-l=9\left [ a-(5a-4b) \right ]/4$ ( BY (3))

$x=9(b-a)+l$

$=5b-4a$

A wire suspended from one end carries a sphere at its other end. The elongation in the wire reduces from

$2 mm$ to

$1.6 mm$ on completely immersing the sphere in water. The density of the material of the sphere is

0%
3200 kg/m $^{3}$
0%
800 kg/m $^{3}$
0%
1250 kg/m $^{3}$
0%
5000 kg/m $^{3}$

Explanation

$Y = \dfrac{F/A}{\triangle L/L}$

$Y = \dfrac{F_{1}/A}{2/L} = \dfrac{F_{2}/A}{1.6/L}$

$\dfrac { 4 }{ 5 } F_{1} = F_{2}$
Bouyant force

$= F_{1} -F_{2} = \dfrac{F_{1}}{5}$

$F_{1} = \rho_{s} V_{s} g$
Bouyant force

$= \rho_{w} V_{s} g = \dfrac{F_{1}}{5}$

$\Rightarrow 5\rho_{w} V_{s} g = \rho_{s} V_{s} g$

$\Rightarrow \rho_{s} = 5\rho_{w}$

$\Rightarrow \rho_{s} = 5000 kg / m^{3} (\because \rho_{w} = 1000 kg/m^{3})$
So, density of material is

$5000 kg / m^{3}$

Two wires of different materials each of length

$l$ and cross sectional area ‘

$A$ ’ are joined in series to form a composite wire. If their Young’s modulii are

$Y$ and

$2Y$ , the total elongation produced by applying a force

$F$ to stretch the composite wire.

0%
$3FA/2Yl$
0%
$2FA/3Yl$
0%
$2FA/3AY$
0%
$3Fl/2AY$

Explanation

Applying a force F to stretch is applied on both wires simultaneously so we can directly add the elongation of each wire with force F, so,

$Y = \dfrac{F/A}{\triangle L_{1} / L }$ for wire with youngs modulus

$Y$

$2Y = \dfrac{F/A}{\triangle L_{2} / L }$ for wire with youngs modulus

$2Y$

$\triangle L_{1} = \dfrac{F L}{Y A }$

$\triangle L_{2} = \dfrac{F L}{2YA}$

$\triangle L_{1} + \triangle L_{2} = \dfrac{F L}{Y A} + \dfrac{F L}{2YA} = \dfrac{3F L}{2YA}$

A force of

$15N$ increases the length of a wire by 1mm. The additional force required to increase the length by

$2.5mm$ in N is

0%
$22.5$
0%
$37.5$
0%
$52.5$
0%
$75$

Explanation

$Y=\dfrac{F/A}{\Delta l/l}$

$\Delta l=\dfrac{Fl}{AY}$

$\Delta l_1=\dfrac{F_1l}{AY}$ -----(1)

$\Delta l_2=\dfrac{F_2l}{AY}$ -----(2)

Dividing (1) by (2)

$\dfrac{\Delta l_1}{\Delta l_2}=\dfrac{F-1}{F-2}$

$\dfrac{1}{2.5}=\dfrac{15}{F_2}$

$F_2=15\times 2.5=37.5$ N

So, additional force

$=F_2-F_1$

$=37.5-15$

$=22.5$ N

When a mass is suspended from the end of a wire the top end of which is attached to the roof of the lift, the extension is

$e$ when the lift is stationary. If the lift moves up with a constant acceleration

$g/2$ , the extension of the wire would be

0%
$\dfrac{2e}{3}$
0%
$\dfrac{3e}{2}$
0%
$2e$
0%
$3e$

Explanation

$Y = \dfrac{mg/A}{e/L} = \dfrac{(3/2) mg/A}{x/L} = Y$

$x = \dfrac{3e}{2}$
Hence, new elongation is

$\dfrac{3e}{2}$
In this problem Young's modulus will be same because were is same. No change in area, natural length will be also same. But tension will change. In the frame of lift when it is moving up by acceleration g/2 a psuedo force of mg/2 units will add in the tension so elongation will increase to

$\dfrac{3e}{2}$

The density of water at the surface of the ocean is

$\rho$ . If the bulk modulus of water is B, what is the density of ocean water at a depth where the pressure is nP

$_{o}$ , where P

$_{o}$ is the atmosphreic pressure:

0%
$\dfrac{\rho B}{B-(n-1)P_{o}}$
0%
$\dfrac{\rho B}{B+(n-1)P_{o}}$
0%
$\dfrac{\rho B}{B-nP_{o}}$
0%
$\dfrac{\rho B}{B+nP_{o}}$

Explanation

Let the volume of certain mass (M) of water at surface be

$V$

$\implies \rho =\dfrac{M}{V}$

Pressure at a depth

$P' = n P_o$

Thus change in pressure

$\Delta P = nP_o - P_o =(n-1)P_o$

Bulk modulus

$B = \dfrac{\Delta P}{\Delta V/V}$

$\implies \Delta V = \dfrac{(n-1) P_o V}{B}$

As at a depth, the pressure increases. Thus the volume of water will decrease at that depth.

$\therefore$ Volume of water of mass (M) at that depth

$V' = V - \Delta V$

$\therefore$

$V' = V - \dfrac{(n-1) P_o V}{B} = (B - (n-1) P_o) \dfrac{V}{B}$ ..............(1)

Density at that depth

$\rho' = \dfrac{M}{V'} = \rho \dfrac{V}{V'}$

$\therefore$

$\rho' = \dfrac{\rho B}{B - (n-1) P_o}$ (using 1)

A stone of

$0.5 kg$ mass is attached to one end of a

$0.8 m$ long aluminium wire of

$0.7 mm$ diameter and suspended vertically. The stone is now rotated in a horizontal plane at a rate such that wire makes an angle of 85

$^{o}$ with the vertical. If Y

$=$ 7x10

$^{10}$ Nm

$^{-2}$ , sin 85

$^{o}=$ 0.9962 and cos85

$^{o}=$ 0.0872 , the increase in length of wire is

0%
1.67x10 $^{-3}$ m
0%
6.17x10 $^{-3}$ m
0%
1.76x10 $^{-3}$ m
0%
7.16x10 $^{-3}$ m

Explanation

By equating force in vertical direction

$T cos 85^{\circ} = mg$

$T = \dfrac{mg}{Cos 85^{\circ}}$

$y = \dfrac{F/A}{\triangle l/l}$

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

$= \dfrac{mg l}{A Y Cos 85^{\circ}}$

$= 1.7 \times 10^{-3} m$

$\simeq 1.67 \times 10^{-3}m$

When a sphere of radius

$2 cm$ is suspended at the end of a wire, elongation is

$'e$ '. When the same wire is loaded with a sphere of radius

$3cm$ and made of the same material, the elongation would be :

0%
$\dfrac{8}{27}e$
0%
$\dfrac{27}{8}e$
0%
$\dfrac{4}{9}e$
0%
$\dfrac{9}{4}e$

Explanation

$Y=\dfrac{F/A}{\Delta l/l}$

$\Delta l=\dfrac{Fl}{YA}$

So,

$\dfrac{\Delta l_1}{\Delta l_2}=\dfrac{p\frac{4}{3}\pi r{_{1}}^{3}g}{p\dfrac{4}{3}\pi r{_{2}}^{3}g}$

$\dfrac{\Delta l_1}{\Delta l_2}=\dfrac{(2)^3}{(3)^3}$

$\dfrac{\Delta l_1}{\Delta l_2}=\dfrac{8}{27}$

$\Delta l_2=\dfrac{27}{8}\Delta l_1$

$=\dfrac{27}{8}e$ .

Two bars

$A$ and

$B$ of circular section and of the same volume and made of the same material are subjected to tension. If the diameter of

$A$ is half that of

$B$ and if the force supplied to both the rods is the same and is within the elastic limit, the ratio of the extension of

$A$ to that of

$B$ will be

0%
$16$
0%
$8$
0%
$4$
0%
$2$

Explanation

Using

$Y = \dfrac{F/A}{\triangle L/L }$

$Y = \dfrac{F/ \triangle_{1}}{\triangle L_{1} / L_{1}} - (1)\text{For bar A}$

$Y = \dfrac{F/ \triangle_{2}}{\triangle L_{2} / L_{2}} - (2)\text{For bar B}$
Volume is same for A and B

$So, R(\dfrac{d_{1}}{2})^{2} L_{1} = R (\dfrac{d_{2}}{2})^{2} L_{2}$

$d_{1}^{2} L_{1} = 4d_{1}^{2} L_{2} (given d_{1} = \dfrac{1}{2}d_{2})$

$\dfrac{L_{1}}{L_{2}} = 4$

Dividing 1 and 2

$1 = \dfrac{L_{1} \triangle L_{2} A_{2}}{L_{2} \triangle L_{1} A_{1}}$

$\dfrac{\triangle L_{1} }{\triangle L_{2}} = 4\times 4 (\because \dfrac{A_{2}}{A_{1}} = (\dfrac{d_{2}}{d_{1}})^{2} = 4)$

$\dfrac{\triangle L_{1} }{\triangle L_{2}} = 16$

A wire of radius

$r$ , Youngs modulus

$Y$ and length

$l$ is hung from a fixed point and supports a heavy metal cylinder of volume

$V$ at its lower end. The change in length of wire when cylinder is immersed in a liquid of density

$\rho$ is in fact

0%
decreases by $\dfrac{Vl \rho g}{Y\pi r^{2}}$
0%
increases by $\dfrac{Vr \rho g}{Y\pi l^{2}}$
0%
decreases by $\dfrac{V \rho g}{Y\pi r}$
0%
increases by $\dfrac{V \rho g}{\pi rl}$

Explanation

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

$\triangle l = \dfrac{Fl}{AY}$

$\triangle l_{1} = \dfrac{(F -F_{B})l}{AY}$

So, change in length is

$\triangle l - \triangle l_{1} = \dfrac{F_{B} l}{AY}$

$= \dfrac{P Vgl}{AY} = \dfrac{P Vgl}{Y\pi r^{2}} (\because F_{B}= VPg)$

length decreases because bouyant force acts in upward direction, So, tension decreases, so length also decreases.

The density of a metal at normal pressure is

$\rho$ . It's density when it is subjected to an excess pressure p is

$\rho$ '. If B is the bulk modulus of the metal, the ratio

$\rho$ '/

$\rho$ is :

0%
$\dfrac{1}{1-p/B}$
0%
$1+\dfrac{P}{B}$
0%
$\dfrac{1}{1-B/p}$
0%
$1+B/p$

Explanation

$B=-V\dfrac{dP}{dV}$

Thus, $\Delta V=-\dfrac{pV}{B}$ or $V'-V=-\dfrac{pV}{B}$

Or, $V'=V(1-\dfrac{p}{B})$

Now, $\rho '=\dfrac{m}{V'}=\dfrac{m}{V(1-\dfrac{p}{B})}=\dfrac{\rho}{(1-\dfrac{p}{B})}$

Choose the correct statements from the following:

0%
steel is more elastic than rubber
0%
the stretching of a coil spring is determined by the Youngs modulus of the wire of the spring.
0%
the frequency of a tuning fork is determined by the shear modulus of the material of the fork
0%
when a material is subjected to a tensile (stretching) stress, the restoring forces are caused by interatomic attraction

Two blocks of masses

$m$ and

$M$ =

$2m$ are connected by means of a metal wire of cross sectional area A, passing over a frictionless fixed pulley as shown in figure. The system is then released.
The stress produced in the wire is :

0%
$\dfrac{mg}{A}$
0%
$\dfrac{2mg}{3A}$
0%
$\dfrac{3mg}{4A}$
0%
$\dfrac{4mg}{3A}$

Explanation

Let tension in string

$=T$ , acceleration of the system

$=a$

For

$M \rightarrow$

$2mg$ -

$T=2ma$

For

$m \rightarrow$

$T$ -

$\ mg = ma$

Solving both equations

$\Rightarrow$

$T=\dfrac{4mg}{3}$

Stress produced in wire

$=\dfrac{T}{A}=\dfrac{4mg}{3A}$

Option D

Two wire

$A$ and

$B$ have equal diameter and are made of the same material, but the length of

$A$ is twice that of wire

$B$ . Then, for a given load

0%
The extension of $A$ will be twice that of $B$
0%
The extensions of $A$ and $B$ will be equal
0%
The strain in $A$ will be half that is $B$
0%
The strains in $A$ and $B$ will be equal

Explanation

$Y=\dfrac{F/A}{\Delta l/l}$

$Y=\dfrac{F/A}{\Delta l_A/l_A}$ ----(1)

$Y=\dfrac{F/A}{\Delta l_B/l_B}$ -----(2)

dividing (1) by (2)

$1=\dfrac{l_A}{\Delta l_A}\times \dfrac{\Delta l_B}{l_B}$ ------(3)

$2=\dfrac{l_A}{l_B}=\dfrac{\Delta l_A}{\Delta l_B}$

$\therefore \Delta l_A=2\Delta l_B$

So, extension in A is twice and strain is same in A and B by (3) equation.

A uniform pressure P is exerted by an external agent on all sides of a solid cube at a temperature t

$^{0}C$ .By what amount should the temperature of the cube be raised in order to bring its volume back to its original volume before the pressure was applied if the bulk modulus is B and coefficient of volumetric expansion is

$\gamma$

0%
P $\gamma / B$
0%
$P / B\gamma$
0%
$B / \gamma P$
0%
$1 / BP\gamma$

Explanation

$B = - \dfrac{p}{[\dfrac{dv}{v}]} \implies dv = - \dfrac{pv}{B} \implies v\gamma \Delta T = \dfrac{pv}{B} \implies \Delta T = \dfrac{p}{\gamma B}$

CBSE Questions for Class 11 Engineering Physics Mechanical Properties Of Solids Quiz 3 - 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 3 - MCQExams.com

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

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