Mechanical Properties of Solids Test

Result

Mechanical Properties of Solids Test - 29

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

Velocity of sound, (Y) youngs modules, and p density of solid are related as

A
$$v=\sqrt{Y p}$$

B
$$v=\sqrt{Y/p}$$

C
$$v^2 = Yp$$

D
Both (a) and (b)

Solution

$$\begin{array}{l}\text { The speed of sound in a solid the depends } \\\text { on the youngls modulus of the medium and Density.}\end{array}$$

$$V=\sqrt{\frac{Y}{\rho}}$$

$$\begin{aligned}Y &=\text { Young's modulus } \\\rho &=\text { density of } \\& \text { medium }\end{aligned}$$
Hence option B is correct
Question 2
1 / -0

(i) In Searle's experiment, after every step of loading, one waits for sometime (2 or 3 min) before taking reading.
(ii) In this duration, the wire becomes free from kinks.

A
Both (i) and (ii) are true and (ii) is reason for (i)

B
Both (i) and (ii) are true but (ii) is not reason for (i)

C
Only (i) is true

D
Only (ii) is true

Solution

In Searle's experiment, after every step of loading, one waits for sometime (2 or 3 min) before taking reading. This statement is true because in this time interval the wire becomes free from kinks, so that it can get desired length. The vertical oscillations in the wire also subsides in this time interval. So both statements are true and and (ii) is the correct reason for (i).
Question 3
1 / -0

The modulus of elasticity of a material does not depend upon

A
shape

B
temperature

C
nature of material

D
impurities mixed

Solution

Modulus of elasticity is an internal property of matter and hence is unaffected by shape, size etc.
Question 4
1 / -0

The ratio of change in dimension at right angles to applied force to the initial dimension is defined as

A
$$Y$$

B
$$\eta$$

C
$$\beta$$

D
$$K$$

Solution

This is a factual question. The ratio is labelled $$\beta$$.
Question 5
1 / -0

A wire is stretched through $$1 mm$$ by certain load. The extension produced in the wire of same material with double the length and radius will be

A
$$4 mm$$

B
$$3 mm$$

C
$$1 mm$$

D
$$0.5 mm$$

Solution

$$\Delta l=\dfrac { FL }{ AE } \\ Double\quad length\quad and\quad radius\quad means,\quad L'=2L\quad and\quad A'={ 2 }^{ 2 }A=4A\\ Thus\quad \Delta l'=\dfrac { F.2L }{ 4A.E } =\dfrac { \Delta l }{ 2 } =\dfrac { 1 }{ 2 } =0.5mm$$
Question 6
1 / -0

Equal weights are suspended from the wires of same material and same lengths but with radii in the ratio $$1 : 2$$. The ratio of extensions produced in them will be

A
$$4 : 1$$

B
$$1 : 4$$

C
$$1 : 2$$

D
$$2 : 1$$

Solution

We know that $$\Delta l=\dfrac{FL}{AY}$$. Thus, for same force, same material (Y) and same length, extension is inversely proportional to area, or $${r}^{2}$$
Thus, for radii in ratio $$1:2$$, extension will be in ratio $$4:1$$
Question 7
1 / -0

The modulus of elasticity at constant temperature is

A
$$\gamma P$$

B
$$\frac{P}{\gamma}$$

C
$$P$$

D
$$\frac{P}{V}$$

Solution

Bulk modulus of elasticity $$B=-V\frac { dP }{ dV } $$
Use Perfect gas equation, $$PV=RT$$
From this, we get $$\frac { dP }{ dV } =-\frac { RT }{ { V }^{ 2 } } $$
Put this in the expression for $$B$$,
$$B=-V\left( \frac { -RT }{ { V }^{ 2 } } \right) $$=$$\frac { RT }{ { V } } $$
At constant temperature,
$$B=\frac { RT }{ { V } }= P$$
Question 8
1 / -0

A stress of $$2\ kg/mm^{2}$$ is applied on a wire. If $$Y=10^{12}\ dyne/cm^2$$ then the percentage increase in its length will be

A
$$0.196\%$$

B
$$19.6\%$$

C
$$1.96\%$$

D
$$0.0196\%$$

Solution

Given$$:Stress=2\ kg/mm^2= 2\times\dfrac{9.8}{{ 10 }^{ -6 }} Pa= 19.6\times{10}^{6}Pa$$

Young's Modulus of elasticity Y$$=10^{ 12 }\ dyne/cm^2$$
Since$$,\ 1\ dyne= 10^{-5}\ N$$
Y$$= 10^{ 11 }\ N/m^2$$
Now, from Hooke's law, we know that
$$Stress=Y Strain$$
Percentage change in length is strain $$\%$$
Strain$$=\dfrac { \Delta L }{ L } $$
Strain$$\%=\dfrac { Stress }{ Y } \times 100$$
So, percentage change in length $$=\dfrac{ 19.6\times { 10 }^{ 6 }\times 100 }{ { 10 }^{ 11 } }=0.0196\%$$
Question 9
1 / -0

The expression for the determination of Poisson's ratio for rubber is

A
$$\displaystyle \sigma=\frac{1}{2}\left[1-\frac{dV}{AdL}\right]$$

B
$$\displaystyle \sigma=\frac{1}{2}\left[1+\frac{dV}{AdL}\right]$$

C
$$\displaystyle \sigma=\frac{1}{2}\frac{dV}{AdL}$$

D
$$\displaystyle \sigma=\frac{dV}{AdL}$$

Solution

Volume of rubber, $$V=AL$$
$$\dfrac{dV}{dL}=- L\dfrac{dA}{dL}+A$$, negative sign because the are decreases as length increases
Area, $$A=\pi {R}^{2}$$
So, $$\dfrac{dA}{A}=2\dfrac{dR}{R}$$
$$\therefore \quad \dfrac { dV }{ dL } = \dfrac { -2A\dfrac { dR }{ R } }{ \dfrac { dL }{ L } } +A=(-2\sigma +1)A$$
Hence,
$$\sigma =\dfrac { 1 }{ 2 } \left( 1-\dfrac { dV }{ AdL } \right) $$
Question 10
1 / -0

A wire of length L and cross-sectional area A is constructed of a material whose Young's modulus of elasticity is Y. The energy density stored in the wire when stretched by $$x$$ meter will be

A
$$Y\times x^2$$

B
$$\displaystyle \frac{Y\times x^2}{L}$$

C
$$\displaystyle \frac{Y\times x^2}{2L^2}$$

D
$$\displaystyle \frac{Y\times x^2}{L^2}$$

Solution

$$Energy\quad Density=\dfrac { 1 }{ 2 } \times Stress\times Strain=\dfrac { 1 }{ 2 } \times \dfrac { xY }{ l } \times \dfrac { x }{ l } \quad (\because Stress=Y\times Strain)\\ or,\quad \dfrac { { x }^{ 2 }Y }{ 2{ L }^{ 2 } } $$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Mechanical Properties of Solids Test - 29

Result

Mechanical Properties of Solids Test - 29

Score

-

Rank

-

SHARING IS CARING

Question 1

$$v=\sqrt{Y p}$$

$$v=\sqrt{Y/p}$$

$$v^2 = Yp$$

Both (a) and (b)

Solution

Question 2

Both (i) and (ii) are true and (ii) is reason for (i)

Both (i) and (ii) are true but (ii) is not reason for (i)

Only (i) is true

Only (ii) is true

Solution

Question 3

shape

temperature

nature of material

impurities mixed

Solution

Question 4

$$Y$$

$$\eta$$

$$\beta$$

$$K$$

Solution

Question 5

$$4 mm$$

$$3 mm$$

$$1 mm$$

$$0.5 mm$$

Solution

Question 6

$$4 : 1$$

$$1 : 4$$

$$1 : 2$$

$$2 : 1$$

Solution

Question 7

$$\gamma P$$

$$\frac{P}{\gamma}$$

$$P$$

$$\frac{P}{V}$$

Solution

Question 8

$$0.196\%$$

$$19.6\%$$

$$1.96\%$$

$$0.0196\%$$

Solution

Question 9

$$\displaystyle \sigma=\frac{1}{2}\left[1-\frac{dV}{AdL}\right]$$

$$\displaystyle \sigma=\frac{1}{2}\left[1+\frac{dV}{AdL}\right]$$

$$\displaystyle \sigma=\frac{1}{2}\frac{dV}{AdL}$$

$$\displaystyle \sigma=\frac{dV}{AdL}$$

Solution

Question 10

$$Y\times x^2$$

$$\displaystyle \frac{Y\times x^2}{L}$$

$$\displaystyle \frac{Y\times x^2}{2L^2}$$

$$\displaystyle \frac{Y\times x^2}{L^2}$$

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.