Mechanical Properties of Solids Test

Result

Mechanical Properties of Solids Test - 22

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

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

A
0.005 cm

B
0.02 cm

C
0.08 cm

D
0.04 cm

Solution

$$ \triangle l = \dfrac{F l}{\pi (\dfrac{d}{2})^{2}Y}$$
$$ 0.01 = \dfrac{F l}{\pi (\dfrac{d}{2})^{2}Y} $$
$$ \triangle l_{1} = \dfrac{F 2l}{\pi 4 (\dfrac{d}{2})^{2}Y}$$
$$ = \dfrac{1}{2} \dfrac{F l}{\pi (\dfrac{d}{2})^{2}Y} $$ (also take the ratio in the next step)
$$\dfrac{1}{2} (0.01)$$
$$ = 0.005cm $$
Question 2
1 / -0

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 :

A
$$0.75mm$$

B
$$6mm$$

C
$$1.5mm$$

D
$$12.0mm$$

Solution

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

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

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

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

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

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

$$ = 12mm$$
Question 3
1 / -0

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 $$1002 cm$$

A
$$100N$$

B
$$200 N$$

C
$$400 N$$

D
$$2000N$$

Solution

$$ Y = \dfrac{F l}{A \triangle l}$$
$$ Y = \dfrac{200 \times 1}{10^{-6}\times 10^{-3}}$$
$$ Y = 2\times 10^{11}$$
$$ \triangle l = 1002 - 1000 = 2cm , l = 10m$$
$$F = \dfrac{YA \triangle l}{l}$$
$$ =\dfrac{2 \times 10^{11}\times 10^{-6} \times 2 \times 10^{-2}}{10}$$
$$ = 400N$$
Question 4
1 / -0

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:

A
$$0.5e$$

B
$$1.0e$$

C
$$1.5e$$

D
$$2.0e$$

Solution

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

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

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

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

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

=> elongation is none wire is same as $$e$$
Question 5
1 / -0

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

A
$$2.00mm$$

B
$$4.00 mm$$

C
$$6.00 mm$$

D
$$8.00mm$$

Solution

$$ Y = \dfrac{F / A}{\triangle l/l}$$ (standard result)
$$ \triangle l = \dfrac{F l}{\pi r^{2} Y}$$
$$\dfrac{2}{1000} = \dfrac{F l}{\pi (\dfrac{r}{2})^{2}Y}$$
$$\triangle l_{1} = \dfrac{F l}{\pi (\dfrac{r}{2})^{2}Y}$$
$$= \dfrac { F l }{ \pi { r }^{ 2 }Y } $$
$$ = 4(\dfrac{2}{1000})$$
$$ = \dfrac{8}{1000} m $$
$$ = 8 mm$$
Question 6
1 / -0

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

A
$$16e$$

B
$$4e$$

C
$$\left(\dfrac{1}{4}\right)e$$

D
$$\left(\dfrac{1}{16}\right)e$$

Solution

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

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

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
Question 7
1 / -0

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]

A
$$0.0078$$%

B
$$0.0088$$%

C
$$0.0098$$%

D
$$0.0067$$%

Solution

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

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

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

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

$$ = 0.0098$$ %

So, change in length is $$0.0098$$%
Question 8
1 / -0

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 :

A
$$\dfrac{AYR}{r}$$

B
$$\dfrac{AY(R-r)}{r}$$

C
$$\dfrac{Y(R-r)}{Ar}$$

D
$$\dfrac{YR}{Ar}$$

Solution

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

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

$$ = \dfrac{Y (R-r) A}{r} (\because \triangle l = R-r)$$
Question 9
1 / -0

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

A
Q,S,P,R

B
R,P,S,Q

C
P,Q,R,S

D
P,R,Q,S

Solution

$$ Y = \dfrac{Stress}{Strain}$$
$$ Strain = \dfrac{Stress}{Y}$$
$$ = \dfrac{F/A}{Y}$$
$$ Strain (P) = \dfrac{10/\pi (\dfrac{2}{2})^{2}}{Y} = \dfrac{10}{\pi Y} N/mm^{2}$$
$$Strain (Q) = \dfrac{20/\pi (\dfrac{1}{2})^{2}}{Y} = \dfrac{80}{\pi Y} N/mm^{2}$$
$$ Strain (R) = \dfrac{30/\pi (\dfrac{4}{2})^{2}}{Y} = \dfrac{30}{4\pi Y} N/mm^{2}$$
$$ Strain (S) = \dfrac{40/\pi (\dfrac{3}{2})^{2}}{Y} = \dfrac{160}{9\pi Y} N/mm^{2}$$
Strain in order of decreasing
$$Q>S>P>R$$
Question 10
1 / -0

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

A
$$20/7$$

B
$$400/49$$

C
$$7/20$$

D
$$49/400$$

Solution

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

$$ Y \alpha \dfrac{1}{\triangle l}$$
Now take the ratio,
$$\dfrac{Y_{1}}{Y_{2}} = \dfrac{\triangle l_{2}}{\triangle l_{1}}$$

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

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

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

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

Mechanical Properties of Solids Test - 22

Result

Mechanical Properties of Solids Test - 22

Score

-

Rank

-

SHARING IS CARING

Question 1

0.005 cm

0.02 cm

0.08 cm

0.04 cm

Solution

Question 2

$$0.75mm$$

$$6mm$$

$$1.5mm$$

$$12.0mm$$

Solution

Question 3

$$100N$$

$$200 N$$

$$400 N$$

$$2000N$$

Solution

Question 4

$$0.5e$$

$$1.0e$$

$$1.5e$$

$$2.0e$$

Solution

Question 5

$$2.00mm$$

$$4.00 mm$$

$$6.00 mm$$

$$8.00mm$$

Solution

Question 6

$$16e$$

$$4e$$

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

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

Solution

Question 7

$$0.0078$$%

$$0.0088$$%

$$0.0098$$%

$$0.0067$$%

Solution

Question 8

$$\dfrac{AYR}{r}$$

$$\dfrac{AY(R-r)}{r}$$

$$\dfrac{Y(R-r)}{Ar}$$

$$\dfrac{YR}{Ar}$$

Solution

Question 9

Q,S,P,R

R,P,S,Q

P,Q,R,S

P,R,Q,S

Solution

Question 10

$$20/7$$

$$400/49$$

$$7/20$$

$$49/400$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.