Physics Part Ii Chapter 9 Mechanical Properties Of Solids

Hooke’s law states that the extension produced in the wire is directly proportional to the load applied within the elastic limit i.e. Acc to Hooke’s low Stress α Strain Stress = E x Strain E = Modulus of elasticity

Modulus of elasticity = Stress / Strain, As Strain is a dimensionless quantity being a ratio between change in dimension to the original dimension, the dimensions of Modulus of Elasticity are the same as that of stress. That is kg/mm^2 are the dimensions of Modulus of Elasticity.

3 Types of Elastic Moduli. Young's modulus (E) describes tensile elasticity, or the tendency of an object to deform along an axis when opposing forces are applied along that axis; it is defined as the ratio of tensile stress to tensile strain. It is often referred to simply as the elastic modulus.

B.

5× 10⁴ J

Thermal energy corresponds to internal energy
Mass = 1 kg
density = 8 kg/m³

C.

9F

D.

B.

S²/2Y

C.

I_A < I_B

Let same mass and same outer radii of solid sphere and hollow sphere are M and R respectively. The moment of inertia of solid sphere A about its diameter
 .... (i)
Similarly the moment of inertia of hollow sphere (spherical shell) B about its diameter
  (ii)
It is clear from eqs. (i) and (ii). I_A < I_B

Work done in stretching the wire = potential energy stored

$$\begin{gathered} \frac{1}{2} \mathrm{x} \mathrm{stress} \mathrm{x} \mathrm{strain} \mathrm{x} \mathrm{volume} \\ = \frac{1}{2} \mathrm{x} \frac{\mathrm{F}}{\mathrm{A}} \mathrm{x} \frac{\mathrm{l}}{\mathrm{L}} \mathrm{x} \mathrm{AL} \\ = \frac{1}{2} \mathrm{Fl} \end{gathered}$$

B.

5 kHz

In solids, Velocity of wave

V = $$\begin{gathered} \sqrt{\frac{Y}{\rho }} = \sqrt{\frac{9.27 x {10}^{10}}{2.7 x {10}^3}} \\ v = 5.83 x {10}^3 m/sec \end{gathered}$$

Since rod is clamped at middle fundamental wave shape is as follow

$$\begin{gathered} \frac{\lambda }{2} = L \\ \Rightarrow \lambda = 2L \\ \lambda = 1.2 m \\ (\because L - 60 cm = 0.6 m ) \\ u\sin g v = f\lambda \\ \Rightarrow f = \frac{v}{\lambda } =\frac{5.85 x {10}^3}{1.2} \\ = 4.88 x {10}^3 Hz \cong 5 KHz \end{gathered}$$

D.

4.5%

Density = Mass/Volume

$$\begin{gathered} \frac{I∆d}{d} = \frac{I∆M}{M} + \frac{3∆L}{L} \\ = 1.5 + 3\left(1\right) \\ = 4.5\% \end{gathered}$$

D.

mg/3Ka

$$\begin{gathered} Bulk modulus, K = \frac{Volumetric stress}{Volumetric strain} \\ K= \frac{mg}{a\left({\displaystyle \frac{dV}{V}}\right)} \\ \Rightarrow \frac{dV}{V} = \frac{mg}{Ka} ....\left(i\right) \\ Volume of sphere, \\ V = \frac{4}{3}\pi R^3 \\ Fractional Change in volume \frac{dV}{V} = \frac{3dr}{r} .... \left(ii\right)\hspace{0.17em} \\ U\sin g equ \left(i\right) \& \left(ii\right) \frac{3dr}{r} = \frac{mg}{Ka} \\ \therefore \frac{dr}{r} = \frac{mg}{3Ka} (fractional decrement in radius) \end{gathered}$$

D.

p/3B

C.

25 rad/s²

τ = I α
F × R = MR²α
30 × 0.4 = 3 × (0.4)² α
12 = 3 × 0.16 α
400 = 16 α
α = 25 rad/s2

A.

9F

Wire 1 : 

$∆\mathcal{l} = \left(\frac{F}{AY}\right)3\mathcal{l} ..... \left(i\right)$

Wire 2:

$$\begin{gathered} ∆\mathcal{l} = \left(\frac{F\text{'}}{3AY}\right)\mathcal{l} .... \left(ii\right) \\ From equation \left(i\right) and \left(ii\right) \\ ∆\mathcal{l} = \left(\frac{F}{AY}\right)3\mathcal{l} = \left(\frac{F\text{'}}{3AY}\right)\mathcal{l} \\ Or, \\ F\text{'} = 9F \end{gathered}$$

A.

13.2 cm

For closed organ pipe, third harmonic

$$\begin{gathered} n = \frac{(2N-1)V}{4\mathcal{l}} = \frac{3V}{4\mathcal{l}} \\ (\because N =2) \end{gathered}$$

For open organ pipe, 

$$\begin{gathered} n = \frac{NV}{2\mathcal{l}}=\frac{V}{2\mathcal{l}\text{'}}(\because N= 1) \\ According to question, \\ \frac{3V}{4\mathcal{l}} = \frac{V}{2\mathcal{l}\text{'}} \\ \Rightarrow \mathcal{l}\text{'} = \frac{4\mathcal{l}}{3 x2} = \frac{2\mathcal{l}}{3} \\ =\frac{2 x 20}{3} = 13.33 cm \end{gathered}$$

A.

$\left[\mathrm{M} {\mathrm{L}}^{-1} {\mathrm{T}}^{-2}\right]$

The formula for the Young's modulus

$$\begin{gathered} \mathrm{Y} =\frac{\mathrm{stress}}{\mathrm{strain}} \\ \mathrm{Dimension} \mathrm{of} \mathrm{Y} = \frac{\left[\mathrm{M} {\mathrm{L}}^{-1} {\mathrm{T}}^{-2}\right]}{\left[{\mathrm{M}}^{\mathrm{o}} {\mathrm{L}}^{\mathrm{o}} {\mathrm{T}}^{\mathrm{o}}\right]} \\ =\left[\mathrm{M} {\mathrm{L}}^{-1}{\mathrm{T}}^{-2}\right] \end{gathered}$$

Young's modulus is the ratio of stress, which has units of pressure, to strain, which is dimensionless; therefore, Young's modulus has units of pressure.

B.

five

For possible interference maxima on the screen, the condition is

d sinθ = n λ

Given:- d = slit - width = 2 λ

∴  2 λ sinθ = n λ

⇒ 2 sinθ = n

The maximum value of sinθ is 1, hence,

n = 2 × 1 = 2

Thus, Eq. (i) must be satisfied by 5 integer values ie, -2, -1, 0, 1, 2. Hence, the maximum
number of possible interference maxima is 5.

C.

600 N

Young's modulus

       Y = $\frac{\mathrm{F} \mathrm{L}}{\mathrm{A} \mathrm{l}}$

        F = $\frac{\mathrm{Y} \mathrm{A} \mathrm{l} }{\mathrm{L}}$

         F ∝ A

⇒       F  ∝ r²

⇒       F  ∝ d²

∴      $\frac{{\mathrm{F}}_1}{{\mathrm{F}}_2 } = \frac{{\mathrm{d}}_1^2}{{\mathrm{d}}_2^2}$

Given:-   d₁ = d, d₂ = 2d and F₁ = 200N

∴       $\frac{200}{{\mathrm{F}}_2} = \frac{{\displaystyle {\left(\mathrm{d}\right)}^2}}{{\displaystyle {\left(2\mathrm{d}\right)}^2}}$

                 = $\frac{1}{4}$

⇒        F₂ = 4 × 200

           F₂  = 800 N

A.

2 × 10⁹ Nm^-2

Shearing modulus of cube

$\mathrm{\eta } = \frac{\mathrm{F} \mathrm{l}}{\mathrm{A}∆\mathrm{l}}$

       η = $\frac{8 \times {10}^3 \times 40 \times {10}^{-3}}{{\left(40 \times {10}^{-3}\right)}^2 \times \left(0.1 \times {10}^{-3} \right)}$

        η = 2 × 10⁹ N m^-2

B.

fringe width would decrease

Fridge width 

          β = $\frac{\mathrm{D\lambda }}{\mathrm{d}}$

⇒        β ∝ λ

If the experiment is performed in water, wavelength A decreases, so, fringe width also decreases.

C.

sin^-1 (0.001)

In Young's double slit experiment, half angular width is given by

          sin θ = $\frac{\mathrm{\lambda }}{\mathrm{d}}$

                 = $\frac{589 \times {10}^{-9}}{0.589 \times {10}^{-3}}$

                 = 10^-3

⇒           θ  = sin^-1 ( 0.001 )

C.

infinity

Bulk modulus:- The ratio of hydraulic stress to hydraulic strain called bulk modulus. It is denoted by B.

       B = $\frac{-\mathrm{p}}{{\displaystyle \raisebox{1ex}{$∆\mathrm{V}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{V}$}\right.}}$             ......(i)

The negative sign indicates that with an increase in pressure, a decrease in volume occurs. That is p is positive,  ΔV is negative

 For incompressible fluid, ΔV = 0

⇒ Equation(i) becomes

    B = $\frac{-\mathrm{p}}{0}$ = infinite

C.

sin^-1 (0.001)

In Young's double slit experiment half angular width is given by

             sin θ = $\frac{\mathrm{\lambda }}{\mathrm{d}}$

                      = $\frac{589 \times {10}^9 }{0.589 \times {10}^{-3}}$

                     = 10^-3

⇒            θ  = sin^-1 (0.001)

D.

$\frac{\mathrm{a}}{2 {\mathrm{b}}^2 \mathrm{c}}$

Free body diagram of the two blocks are

      $\frac{{\mathrm{l}}_1}{{\mathrm{l}}_2} = \mathrm{a} , \frac{{\mathrm{r}}_1}{{\displaystyle {\mathrm{r}}_2}} = \mathrm{b} , \frac{{\mathrm{Y}}_1}{{\mathrm{Y}}_2} = \mathrm{c}$

Let Young's modulus of steel is Y₁  and of brass is Y₂

          Y₁ = $\frac{{\mathrm{F}}_1 . {\mathrm{l}}_1}{{\mathrm{A}}_1 ∆{\mathrm{l}}_1}$                                 .... (i)

 and    Y₂ = $\frac{{\mathrm{F}}_2 {\mathrm{l}}_2}{{\mathrm{A}}_2 ∆{\mathrm{l}}_2}$                                 .... (ii)

Dividing Eq. (i) by Eq. (ii), we get

           $\frac{{\mathrm{Y}}_1}{{\mathrm{Y}}_2} = \frac{{\displaystyle \frac{{\mathrm{F}}_1 . {\mathrm{l}}_1}{{\mathrm{A}}_1 . ∆{\mathrm{l}}_1}}}{{\displaystyle \frac{{\mathrm{F}}_2 . {\mathrm{l}}_2}{{\mathrm{A}}_2 . ∆{\mathrm{l}}_2}}}$

⇒        $\frac{{\mathrm{Y}}_1}{{\mathrm{Y}}_2} = \frac{{\mathrm{F}}_1 . {\mathrm{A}}_2 . {\mathrm{l}}_1 . ∆{\mathrm{l}}_2}{{\mathrm{F}}_2 . {\mathrm{A}}_1 . {\mathrm{l}}_2 .∆{\mathrm{l}}_1}$                     .....(iii)

Force on steel wire from free body diagram 

    T = F₁ = 2 g N

Force on brass wire from free body diagram

     F₂ = T' = T + 2g

          = 4 g N

Now, putting the values of F₁ , F₂   Eq. (iii) we get

     $\frac{Y_1}{Y_2} = \left(\frac{2 g}{4 g}\right) . \left(\frac{\pi r_2^2}{\pi r_1^2}\right) .\left(\frac{l_1}{l_2}\right). \left(\frac{∆l{ }_2}{∆l_1}\right)$

      $\mathrm{c} = \frac{1}{2} \left(\frac{1}{{\mathrm{b}}^2}\right). \mathrm{a} \left(\frac{∆{\mathrm{l}}_2}{∆{\mathrm{l}}_1}\right)$

⇒ $\frac{∆ {\mathrm{l}}_1}{∆ {\mathrm{l}}_2} = \left(\frac{\mathrm{a}}{2 {\mathrm{b}}^2 \mathrm{c}}\right)$

C.

0.05 mm

Young's coefficient of elasticity

 Given m = 1 kg = 1000 g

    Area of cross section 1 mm² = 10^-6 m²

G          Y = $\frac{\mathrm{m} \mathrm{g} \mathrm{L}}{\mathrm{\pi } {\mathrm{r}}^2 ∆\mathrm{L}}$

            ΔL = $\frac{\mathrm{m} \mathrm{g} \mathrm{L} }{\mathrm{\pi } {\mathrm{r}}^2 \mathrm{Y}}$

                = $\frac{{10}^3 \times 10 \times 1}{{10}^{-6} \times 2\times {10}^{11}}$     

                 = 5 × 10^-2      

          Δ L = 0.05 mm

C.

0.15 cm^-3

As, we know that

         v = $\sqrt{\frac{\mathrm{R}}{\mathrm{\rho }}}$

where, k = bulk modulus of elasticity

⇒    k = v² ρ

          = (1330)² × 800 N/m²

Also   k = $\frac{{\displaystyle \raisebox{1ex}{$\mathrm{F}$}\!\left/ \!\raisebox{-1ex}{$ \mathrm{A}$}\right.}}{{\displaystyle \raisebox{1ex}{$∆\mathrm{V}$}\!\left/ \!\raisebox{-1ex}{$ \mathrm{V}$}\right.}}$

       ΔV = $\frac{\mathrm{pressure}}{\mathrm{k}}$

            = $\frac{2 \times {10}^5}{1330 \times 1330 \times 800}$

        ΔV  = 0.15 cm³

C.

2.1 × 10⁵ N/m²

Given:-  Bulk modulus of water B = 2.1 × 10⁹ N/m²

Also given that

    $\frac{∆\mathrm{V}}{\mathrm{V}}$= 0.01 %

            = $\frac{0.01}{100}$

     $\frac{∆\mathrm{V}}{\mathrm{V}}$= 10^-4

We know that,

The ratio of change in pressure to the fractional volume compression is called the bulk modulus of the material.

     B = $\frac{\mathrm{p} \mathrm{V}}{∆\mathrm{V}}$

     p = $\mathrm{B} \left(\frac{∆\mathrm{V}}{\mathrm{V}}\right)$

         = 2.1 × 10⁹ × 10^-4

    p = 2.1 × 10⁵ N/m²

C.

$\frac{\mathrm{b}}{12}$

We know that, intensity, 

       I = 4 l₀ cos² $\frac{\mathrm{\varphi }}{2}$

Case (i)

     2 I₀ = 4 I₀ cos² $\frac{\mathrm{\varphi }}{2}$

     $\mathrm{\varphi } = \frac{2 \mathrm{\pi }}{\mathrm{\lambda }} ∆\mathrm{x}$

     $\frac{\mathrm{\pi }}{2} = \frac{2 \mathrm{\pi }}{\mathrm{\lambda }}$ Δx

      $\frac{\mathrm{\pi }}{2} = \frac{2\mathrm{\pi }}{\mathrm{\lambda }} \frac{{\mathrm{Y}}_1}{\mathrm{D}}$                    $\left[∆ \mathrm{x} = \frac{{\mathrm{Y}}_1}{\mathrm{D}} \mathrm{and} \mathrm{\varphi } = \frac{\mathrm{\pi }}{2}\right]$

⇒     ${\mathrm{Y}}_1 = \frac{\mathrm{\lambda } \mathrm{D}}{4 \mathrm{d}}$                            .....(i)

Case (ii)

      I₀ = 4I₀ ${\cos }^2 \frac{\mathrm{\varphi }}{2}$ 

∴      $\mathrm{\varphi } = \frac{2 \mathrm{\pi }}{\mathrm{\lambda }}$ Δx

       $\frac{2\mathrm{\pi }}{3} = \frac{2 \mathrm{\pi }}{\mathrm{\lambda }} \frac{{\mathrm{Y}}_2 \mathrm{d}}{\mathrm{D}}$                 $\left[\because \mathrm{\varphi } = \frac{2\mathrm{\pi }}{3} \mathrm{and} ∆ \mathrm{x} = \frac{{\mathrm{y}}_2 \mathrm{d}}{\mathrm{D}}\right]$

              Y₂ = $\frac{\mathrm{\lambda } \mathrm{D}}{3 \mathrm{d}}$

∴     ΔY = Y₂ $-$ Y₁

            = $\frac{\mathrm{\lambda } \mathrm{D}}{3\mathrm{d}} - \frac{\mathrm{\lambda } \mathrm{D}}{4\mathrm{d}}$

             = $\frac{\mathrm{\lambda } \mathrm{D}}{12 \mathrm{d}}$

$\because \frac{\mathrm{\lambda D}}{\mathrm{d}} = \mathrm{b}$

     ΔY = $\frac{\mathrm{b}}{12}$

D.

If both assertion and reason are false.

An ideal rigid body is a body with perfectly definite and unchanging shape.

A.

$\frac{\mathrm{P}}{\mathrm{B}}$

Bulk modulus:-  It is the ratio of hydraulic strain to the corresponding hydraulic strain is called Bulk modulus. 

It is denoted by symbol B

          B = $\frac{-∆\mathrm{P}}{{\displaystyle \raisebox{1ex}{$∆\mathrm{V} $}\!\left/ \!\raisebox{-1ex}{$ \mathrm{V}$}\right.}}$

⇒         $\left|\frac{∆\mathrm{V}}{\mathrm{V}}\right| = \frac{\mathrm{P}}{\mathrm{B}}$ 

A.

3.75  mm

The cross-sectional area of a wire of radius r = l mm= 10^-3m is  A = $\mathrm{\pi } {\mathrm{r}}^2$ = 3.14 x 10^-6 m².
The maximum force that can be applied without exceeding the elastic limit is therefore

          F = ${\left(\frac{\mathrm{F}}{\mathrm{A}}\right)}_{\max }\times \left(\mathrm{A}\right)$

          F = (2.5 × 10⁸ N m^-2 ) ( 3.14 × 10^-6 m² )

           F = 785 N

When this force is applied, the wire will stretch by

          ΔL = $\frac{\mathrm{L}}{\mathrm{Y}} \frac{\mathrm{F}}{\mathrm{A}}$

                = $\frac{\left(3 \mathrm{m}\right) \left(785 \mathrm{N}\right)}{\left(2 \times {10}^{11} {\displaystyle \raisebox{1ex}{$\mathrm{N}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{m}}^2$}\right.}\right) \left(3.14 \times {10}^{-6} {\mathrm{m}}^2\right)}$

                 = 3.75 × 10^-3 m

            ΔL = 3.75 mm