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Quiz 02 (Elasticity)

Total questions: 20

Worksheet time: 20mins

Name
Class
Date
1.

Unit of stress is

a)

N/m

b)

N.m

c)

N/m2

d)

N.m2

2.

Young's Modulus of material of a wire is that stress which

a)

does not change the length of the wire

b)

doubles the length of the wire

c)

increase the length of the wire by 50%

d)

decrease the area of cross section of the wire to half

3.

Steel is

a)

more elastic than rubber

b)

less elastic than rubber

c)

not elastic but is plastic

d)

same elastic as rubber

4.

A steel wire is of length 1 m and the area of cross-section 1 cm2. If Young's modulus of steel is 1011 N/m2 the force required to elongate the wire by 1 mm is

a)

103 N

b)

104 N

c)

105 N

d)

103 N

5.

The internal energy per unit volume of a stretched wire is

a)

12stress.strain\frac{1}{2}stress.strain

b)

12force.strain\frac{1}{2}force.strain

c)

stress.strainstress.strain

d)

force.strainforce.strain

6.

The elastic potential energy of a wire stretched by a force F is U. If the same wire is stretched by a force 2F, its potential energy will be

a)

U/2

b)

2U

c)

4U

d)

U2

7.

two wire A and B are of same metal. If the length of wire A is half of the length of the wire B and radius of wire A is twice of wire B, then on increasing their lengths by same amount, the force applied on A as compared to B should be

a)

3/8 times

b)

2 times

c)

4 times

d)

8 times

8.

A steel wire of length l increases by

  δl\delta l  due to its own weight. The fractional change in its volume will be

a)

 (1+σ) δll\left(1+\sigma\right)\ \frac{\delta l}{l}  

b)

 (1σ) δll\left(1-\sigma\right)\ \frac{\delta l}{l}  

c)

 (1+2σ) δll\left(1+2\sigma\right)\ \frac{\delta l}{l}  

d)

 (12σ) δll\left(1-2\sigma\right)\ \frac{\delta l}{l}  

9.

The increase in length of a unit cube on applying a force F at each face normally is

a)

FY(1+2σ)\frac{F}{Y}\left(1+2\sigma\right)

b)

FY(12σ)\frac{F}{Y}\left(1-2\sigma\right)

c)

F(1σ)YF\left(1-\sigma\right)Y

d)

F(1+σ)YF\left(1+\sigma\right)Y

10.

If Y, K and  \sigma  be the Young's modulus, Bulk Modulus and Poisson's ratio of a material then

a)

 Y=KσY=K\sigma  

b)

 Y=3K(12σ)Y=3K\left(1-2\sigma\right)  

c)

 Y=3(12σ)Y=3\left(1-2\sigma\right)  

d)

 Y=3σY=3\sigma  

11.

The correct relationship is

a)

Y>ηY>\eta

b)

σ<1\sigma<-1

c)

σ=Y2η\sigma=\frac{Y}{2\eta}

d)

σ=3KY\sigma=\frac{3K}{Y}

12.

Y, K and  \eta  are related as

a)

 Y(13η+1K)=3Y\left(\frac{1}{3\eta}+\frac{1}{K}\right)=3  

b)

 Y(1η+13K)=3Y\left(\frac{1}{\eta}+\frac{1}{3K}\right)=3  

c)

 Y(3η+1K)=3Y\left(\frac{3}{\eta}+\frac{1}{K}\right)=3  

d)

 Y(1η+3K)=3Y\left(\frac{1}{\eta}+\frac{3}{K}\right)=3  

13.

In relation

 Y=9KηAK+ηY=\frac{9K\eta}{AK+\eta}  the unknown A is

a)

3

b)

2

c)

6

d)

1

14.

The practical value of Poisson's ratio of a substance is

a)

between -1 and 0.5

b)

between 0 and 0.5

c)

more than 0, but less than 0.5

d)

all of the above

15.

the theoretical value of Poisson's ratio lies in between

a)

+1 and 0.5

b)

+0.5 and -1

c)

+0.5 and -0.5

d)

+1 and -1

16.

When a body is deformed, its internal energy

a)

increases

b)

decreases

c)

remains unchanged

d)

may increase or decrease

17.

Couple required per unit radian twist is

a)

πηr4 2l\frac{\pi\eta r^{4\ }}{2l}

b)

π2ηr4 4l\frac{\pi^2\eta r^{4\ }}{4l}

c)

πηr4 4l\frac{\pi\eta r^{4\ }}{4l}

d)

π2ηr4 2l\frac{\pi^2\eta r^{4\ }}{2l}

18.

The ratio of depressions of the two beams of same area of cross sections, but one of square sectionand the other of circular section, for a given load is

a)

9:π9:\pi

b)

4:π4:\pi

c)

3:π3:\pi

d)

16:π16:\pi

19.

The depressions of a mid point of a beam of rectangular cross-section of length l and breadth b and thickness t supported at the ends and loaded in the middle is

a)

δ=Wl34Ybt3\delta=\frac{Wl^3}{4Ybt^3}

b)

δ=Wb34Ylt3\delta=\frac{Wb^3}{4Ylt^3}

c)

δ=Wl34Y3bt3\delta=\frac{Wl^3}{4Y^3bt^3}

d)

δ=Wl34Y2b2t3\delta=\frac{Wl^3}{4Y^2b^2t^3}

20.

The depression at free end of a loaded cantilever of length l, breadth b and thickness t will be

a)

δ=Wl3Ybt3\delta=\frac{Wl^3}{Ybt^3}

b)

δ=2Wl3Ybt3\delta=\frac{2Wl^3}{Ybt^3}

c)

δ=Wl33Ybt3\delta=\frac{Wl^3}{3Ybt^3}

d)

δ=4Wl3Ybt3\delta=\frac{4Wl^3}{Ybt^3}