Mechanical Properties of Solids Test

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Mechanical Properties of Solids Test - 27

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Weekly Quiz Competition

Question 1
1 / -0

What is the density of lead under a pressure of $$2.0\times 10^{8} N/m^{2}$$ , if the bulk modulus of lead is $$8.0\times10^{9} N/m^{2}$$. Also, the initial density of lead is $$11.4 g/cm$$$$^{3}$$.

A
$$12.89 g/cm^{3}$$

B
$$14 g/cm^{3}$$

C
$$11.69 g/cm^{3}$$

D
$$Zero$$

Solution

Using the definition of Bulk modulus we have
$$K=\rho\dfrac{dP}{d\rho}$$
or
$$\rho_2-\rho_1=\rho_1\dfrac{dP}{K}$$
or
$$\rho_2=\rho_1(1+\dfrac{dP}{K})$$
$$ = 11.4 (1+\dfrac{2 \times 10^{8}}{8 \times 10^{9}})$$
$$=11.69 g/cm^{3}$$
Question 2
1 / -0

Three blocks, each of same mass $$m$$, are connected with wire $$W_{1}$$ and $$W_{2}$$ of same cross sectional area 'a' and Young's modulus Y. Neglecting friction, the strain developed in wire $$W_{2}$$ is

A
$$\dfrac{2}{3}\dfrac{mg}{aY}$$

B
$$\dfrac{3}{2}\dfrac{mg}{aY}$$

C
$$\dfrac{1}{3}\dfrac{mg}{aY}$$

D
$$\dfrac{3mg}{aY}$$

Solution

Writing equations of motion for all three particles
$$mg -T =ma _ 1$$
$$ T-T_{1} = ma _ 2$$
$$ T_{1} = ma _ 3$$
$$\therefore T = 2ma$$
$$mg -2ma =ma$$
$$mg =3ma$$
$$a = g/3$$
$$ \therefore T = 2m \dfrac{g}{3}$$
$$Y = \dfrac{F/A}{\triangle L/L}$$
$$ \dfrac{\triangle L}{L} = strain = \dfrac{2mg/3}{Y} = \dfrac{2mg}{3aY}$$
Question 3
1 / -0

A thin rod of negligible mass and cross-sectional area $$4\times 10$$$$^{-6}m^{2}$$ , suspended vertically from one end, has a length of 0.5 m at 100$$^{o}$$C . The rod is cooled to 0$$^{o}$$C .Young's modulus $$=$$ 10$$^{11}$$Nm$$^{-2}$$, coefficient of linear expansion $$=$$ 10$$^{-5}$$K$$^{-1}$$ and g $$=10\ ms^{-2}$$. The total energy stored in the rod is :

A
$$0.1\ J$$

B
$$0.2\ J$$

C
$$0.3\ J$$

D
$$0.4\ J$$

Solution

Given:
$$A=4\times 10^{-6}\\L=.5m\\Y=10^{11}\\ \alpha=10^{-5}\\g=10ms^{-2}\\ \Delta T=100$$
Total energy $$=\dfrac{1}{2} k(\Delta l)^2$$
$$\Delta l$$ is due to temperature change.
$$l'=l(1+\alpha\Delta T)$$ where $$l'$$ is new length
$$l'=l+l\alpha\Delta T$$
$$l'-l=l\alpha\Delta T$$
$$\Delta l=l\alpha \Delta T$$
$$=\dfrac{1}{2}\times 10^{-5} \times 100=5\times 10^{-4}$$

$$k_{rod}=\dfrac{YA}{l}$$

Total Energy $$=\dfrac{1}{2} \dfrac{YA}{l}(\Delta l)^2=\dfrac{1}{2} \times \dfrac{10^{11} \times 4\times 10^{-6} }{0.5} \times (5\times 10^{-4})^2=0.1 J$$
Question 4
1 / -0

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

A
1.67x10$$^{-3}$$m

B
6.17x10$$^{-3}$$m

C
1.76x10$$^{-3}$$m

D
7.16x10$$^{-3}$$m

Solution

By equating force in vertical direction

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

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

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

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

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

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

$$ \simeq 1.67 \times 10^{-3}m$$
Question 5
1 / -0

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

A
decreases by$$\dfrac{Vl \rho g}{Y\pi r^{2}}$$

B
increases by$$\dfrac{Vr \rho g}{Y\pi l^{2}}$$

C
decreases by$$\dfrac{V \rho g}{Y\pi r}$$

D
increases by$$\dfrac{V \rho g}{\pi rl}$$

Solution

$$ Y = \dfrac{F/A}{\triangle l/l}$$
$$ \triangle l = \dfrac{Fl}{AY}$$
$$ \triangle l_{1} = \dfrac{(F -F_{B})l}{AY}$$
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.
Question 6
1 / -0

One end of a string of length L and cross-sectional area A is fixed to a support and the other end is fixed to a bob of mass m. The bob is revolved in a horizontal circle of radius r, with an angular velocity $$\omega $$, such that the string makes an angle $$\theta $$ with the vertical. The increase $$\Delta L$$ in length of the string is

A
$$\dfrac{ML}{AY}$$

B
$$\dfrac{MgL}{AY cos\theta }$$

C
$$\dfrac{MgL}{AY sin\theta }$$

D
$$\dfrac{MgL}{AY}$$

Solution

$$\Delta L=\dfrac{T L}{AY}$$
Substituting, $$T=mg/cos\theta $$
$$\Delta L=\dfrac{mg L}{AY cos\theta }$$
Question 7
1 / -0

One end of a string of length $$L$$ and cross-sectional area $$A$$ is fixed to a support and the other end is fixed to a bob of mass $$m$$. The bob is revolved in a horizontal circle of radius $$r$$ with an angular velocity $$\omega $$ such that the string makes an angle $$\theta $$ with the vertical. The stress in the string is :

A
$$\dfrac{mg}{A}$$

B
$$\dfrac{mg}{A} \left ( 1-\dfrac{r}{L} \right )$$

C
$$\dfrac{mg}{A} \left ( 1+\dfrac{r}{L} \right )$$

D
none of these

Solution

Stress$$=\dfrac{T}{A}$$
$$=\dfrac{mg}{A cos\theta }$$ (on taking the component in the string direction)
where $$cos \theta =\dfrac{\sqrt{L^{2}-r^{2}}}{L}$$
$$=\sqrt{1-\dfrac{r^{2}}{L^{2}}}$$
Stress$$=\dfrac{mg}{A\sqrt{1-\dfrac{r^{2}}{L^{2}}}}$$
Question 8
1 / -0

In Young's double slit experiment, the two equally bright slits are coherent, but of phase difference $$\dfrac{\pi}{3}$$ .If maximum intensity on the screen is$$I_0$$, the intensity at the point on the screen equidistant from the slits is____?

A
$$I_0$$

B
$$\dfrac{I_0}{2}$$

C
$$\dfrac{I_0}{4}$$

D
$$\dfrac{3 I_0}{4}$$

Solution

$$[I = I_o sin^2 \theta]$$

$$=I_o \left ( \dfrac{\sqrt{3}}{2} \right )^2$$

$$ = \dfrac{3 I_o}{4} $$
Question 9
1 / -0

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 :

A
$$\dfrac{mg}{A}$$

B
$$\dfrac{2mg}{3A}$$

C
$$\dfrac{3mg}{4A}$$

D
$$\dfrac{4mg}{3A}$$

Solution

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
Question 10
1 / -0

The average depth of Indian Ocean is about 3000 m. Calculate the fractional compression,$$\frac{\Delta V}{V}$$ of water at the bottom of the ocean, given that the bulk modulus of water is $$2.2 \times 10^9Nm^{-2}$$ (consider $$g=10 ms^{-2}$$)

A
0.82%

B
0.91%

C
1.24%

D
1.36%

Solution

We know one thing
P = P₀ + ρgh
Where P₀ is the atmospheric pressure , g is acceleration due to gravity, h is the height from the Earth surface and ρ is density of water
Here, P₀ = 10⁵ N/m² , g = 10m/s² , h = 3000m and ρ = 10³ Kg/m³
Now, P = 10⁵ + 10³ × 10 × 3000 = 3.01 × 10⁷ N/m²

Again, we have to use formula,
B = P/{-∆V/V}
Here, B is bulk modulus and { -∆V/V} is the fractional compression
So, -∆V/V = P/B
Put , P = 3.01 × 10⁷ N/m² and B= 2.2 × 10⁹ N/m²
∴ fractional compression = 3.01 × 10⁷/2.2 × 10⁹ = 1.368 × 10⁻²
Hence the ans is D

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Mechanical Properties of Solids Test - 27

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Mechanical Properties of Solids Test - 27

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Question 1

$$12.89 g/cm^{3}$$

$$14 g/cm^{3}$$

$$11.69 g/cm^{3}$$

$$Zero$$

Solution

Question 2

$$\dfrac{2}{3}\dfrac{mg}{aY}$$

$$\dfrac{3}{2}\dfrac{mg}{aY}$$

$$\dfrac{1}{3}\dfrac{mg}{aY}$$

$$\dfrac{3mg}{aY}$$

Solution

Question 3

$$0.1\ J$$

$$0.2\ J$$

$$0.3\ J$$

$$0.4\ J$$

Solution

Question 4

1.67x10$$^{-3}$$m

6.17x10$$^{-3}$$m

1.76x10$$^{-3}$$m

7.16x10$$^{-3}$$m

Solution

Question 5

decreases by$$\dfrac{Vl \rho g}{Y\pi r^{2}}$$

increases by$$\dfrac{Vr \rho g}{Y\pi l^{2}}$$

decreases by$$\dfrac{V \rho g}{Y\pi r}$$

increases by$$\dfrac{V \rho g}{\pi rl}$$

Solution

Question 6

$$\dfrac{ML}{AY}$$

$$\dfrac{MgL}{AY cos\theta }$$

$$\dfrac{MgL}{AY sin\theta }$$

$$\dfrac{MgL}{AY}$$

Solution

Question 7

$$\dfrac{mg}{A}$$

$$\dfrac{mg}{A} \left ( 1-\dfrac{r}{L} \right )$$

$$\dfrac{mg}{A} \left ( 1+\dfrac{r}{L} \right )$$

none of these

Solution

Question 8

$$I_0$$

$$\dfrac{I_0}{2}$$

$$\dfrac{I_0}{4}$$

$$\dfrac{3 I_0}{4}$$

Solution

Question 9

$$\dfrac{mg}{A}$$

$$\dfrac{2mg}{3A}$$

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

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

Solution

Question 10

0.82%

0.91%

1.24%

1.36%

Solution

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