Mechanical Properties of Solids Test

Result

Mechanical Properties of Solids Test - 4

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

For a rope of yield strength S_y loaded in tension with weight Mg the minimum area A of the rope should be

A
A ≥ Mg S_y

B
A ≥ Mg/3 S_y

C
A ≥ Mg/ S_y

D
A ≥ Mg/2 S_y

Solution

We know that Yield strength times the Area give the weight of the body.

As Yield strength = weight of body/area

For safety purposes

Yield strength ≥ weight of body/area

S_y ≥ Mg/A

A ≥ Mg/ S_y
Question 2
1 / -0

Rectangular section is rarely used in beams because

A
stresses are always plastic in rectangular section

B
stresses are uniform in a rectangular section

C
less material in kilos is required

D
more material in kilos is required

Solution

A pillar with rounded ends supports less load than that with a distributed shape at the ends.
Question 3
1 / -0

Columns are loaded in

A
shear

B
tension

C
compression

D
hydraulic stress

Solution

Stress on the columns compress it.
Question 4
1 / -0

Material is said to be ductile if

A
material breaks suddenly at little elongation

B
fracture occurs soon after the elastic limit is passed

C
material cross section is not significantly reduced at failure

D
a large amount of plastic deformation takes place between the elastic limit and the fracture point

Solution

If The ultimate strength and fracture points are close in stress-strain curve, the material is said to be brittle. If they are far apart, the material is said to be ductile and show large plastic range.
Question 5
1 / -0

Material is said to be brittle if

A
material cross section is significantly reduced at failure

B
material elongates a lot before finally breaking

C
fracture occurs soon after the elastic limit is passed

D
a large amount of plastic deformation takes place between the elastic limit and the fracture point

Solution

If the ultimate strength and fracture points are close, it means very small plastic range beyond elastic limit, the material is said to be brittle.
Question 6
1 / -0

A steel wire of length 4.7 m and cross-sectional area 3.0 × 10⁻⁵ m² stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 × 10⁻⁵ m² under a given load. What is the ratio of the Young’s modulus of steel to that of copper?

A
1.8

B
1.6

C
1.2

D
2.0

Solution

Length of the steel wire, L1 = 4.7 m
Area of cross-section of the steel wire, A1 = 3.0 × 10^-5 m²
Length of the copper wire, L2 = 3.5 m
Area of cross-section of the copper wire, A2 = 4.0 × 10^-5 m²
Change in length = ΔL1 = ΔL2 = ΔL
Force applied in both the cases = F
Young’s modulus of the steel wire:
Y1 = (F1 / A1) (L1 / ΔL1)
= (F / 3 X 10^-5) (4.7 / ΔL) ….(i)
Young’s modulus of the copper wire:
Y2 = (F2 / A2) (L2 / ΔL2)
= (F / 4 × 10^-5) (3.5 / ΔL) ….(ii)
Dividing (i) by (ii), we get:
Y1 / Y2 = (4.7 × 4 × 10^-5) / (3 × 10^-5 × 3.5)
= 1.79 : 1
The ratio of Young’s modulus of steel to that of copper is 1.8 : 1.
Question 7
1 / -0

A steel rod 2.0 m long has a cross-sectional area of 0.30 cm². It is hung by one end from a support, and a 550-kg milling machine is hung from its other end. Determine the stress on the rod.

A
1.8 × 10⁸ Pa

B
1.6 × 10⁸ Pa

C
1.2 × 10⁸ Pa

D
1.4 × 10⁸ Pa

Solution

Stress=F/A

F=m×a

F=550Kg×9.8m/s²=5390N

A=0.30cm²=0.30×10^-4

Stress=F/A

Stress=5390/0.30×10^-4=1.8×10⁸Pa

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Mechanical Properties of Solids Test - 4

Result

Mechanical Properties of Solids Test - 4

Score

-

Rank

-

SHARING IS CARING

Question 1

A ≥ Mg Sy

A ≥ Mg/3 Sy

A ≥ Mg/ Sy

A ≥ Mg/2 Sy

Solution

Question 2

stresses are always plastic in rectangular section

stresses are uniform in a rectangular section

less material in kilos is required

more material in kilos is required

Solution

Question 3

shear

tension

compression

hydraulic stress

Solution

Question 4

material breaks suddenly at little elongation

fracture occurs soon after the elastic limit is passed

material cross section is not significantly reduced at failure

a large amount of plastic deformation takes place between the elastic limit and the fracture point

Solution

Question 5

material cross section is significantly reduced at failure

material elongates a lot before finally breaking

fracture occurs soon after the elastic limit is passed

a large amount of plastic deformation takes place between the elastic limit and the fracture point

Solution

Question 6

1.8

1.6

1.2

2.0

Solution

Question 7

1.8 × 108 Pa

1.6 × 108 Pa

1.2 × 108 Pa

1.4 × 108 Pa

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.

A ≥ Mg S_y

A ≥ Mg/3 S_y

A ≥ Mg/ S_y

A ≥ Mg/2 S_y

1.8 × 10⁸ Pa

1.6 × 10⁸ Pa

1.2 × 10⁸ Pa

1.4 × 10⁸ Pa