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Mechanical Properties of Solids Test - 69

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Mechanical Properties of Solids Test - 69
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  • Question 1
    1 / -0
    Two wires of different material and radius have their length in ratio of $$1:2.$$ if these were stretched by the same force$$,$$ the strain produced will be in the ratio$$.$$  
    Solution
    We know that the young's modulus is given by$$,$$
    $$Y = stress / strain = (F/A) / strain$$
    $$=> strain = F/AY$$
    where$$,$$
    F $$=$$ force applied
    A $$=$$ area of cross section
    since the wires are made of same material so there young's modulus will be same$$.$$
    Also the force applied and the area of cross section is same $$($$because radius is same$$)$$
    Hence$$,$$
    $$ strai{n_1}/strai{n_2}= F₁A₂Y₂/F₂A₁Y₁ = 1$$
    So$$,$$ the ratio of strain will be 1:1 despite of the ratio of length$$.$$
    Hence,
    option $$(B)$$ is correct answer.
  • Question 2
    1 / -0
    $$10^{22}$$ particle each of mass $$10^{-26}Kg$$ are striking perpendicularly on a wall of area $$1m^2$$ with speed $$10^4m/s$$ in $$1sec$$. The pressure on the wall if collisions are perfectly elastic is:
    Solution
    $$v=10^4m/s$$
    $$m=10^{-26}$$
    $$n=10^{22}$$
    $$A=1m^2$$
    $$\Delta p=2mnv$$
    $$\Delta p=2\times 10^{22}\times 10^{-26}\times 10^{4}=2$$
    $$P=\dfrac{F}{A}=2N/m^2$$
  • Question 3
    1 / -0
    Two elastic wire $$A \& B$$ having length $$\ell_A = 2 \,m$$ and $$\ell_B = 1.5 \,m$$ and the ratio of young's modules $$Y_A : Y_B$$ is $$7:4$$. If radius of wire $$B (r_B)$$ is $$2 mm$$ then choose the correct value of radius of wire $$A$$. Given that due to application of the same force charge in length in both $$A \& B$$ is same
    Solution
    $$Y = \dfrac{F\ell}{\pi r^2 \Delta \ell}$$

    $$\dfrac{Y_A}{Y_B} = \dfrac{F\ell_A}{\pi r^2_A \Delta \ell} \times \dfrac{\pi r^2_B \Delta \ell}{F \ell_B}$$

    $$\dfrac{7}{4} = \dfrac{2}{1.5} \times \dfrac{2^2}{r^2_A} \Rightarrow r^2_A = \dfrac{4 \times 2 \times 2^2}{1.5 \times 7}$$

    $$r_A = 1.7 \,mm$$
  • Question 4
    1 / -0
    For an equal stretching force F, the Young's modulus $$ (Y_s) $$ for steel and rubber $$ (Y_r) $$ are related as
    Solution

  • Question 5
    1 / -0
    If $$F$$ is the force applied to an elastic bar to produce an extension of $$\Delta\ell$$. The, the energy lost in the process is 
    Solution
    $$\begin{aligned}F_{S}=f & \\\Rightarrow \frac{A Y \Delta l}{l} &=F \\\text { Energy stored } &=\frac{A Y \Delta l^{2}}{2 l} \\&=\frac{A Y \Delta l}{\ell} \times \frac{\Delta l}{2} \\&=\frac{F \Delta l}{2}\end{aligned}$$
  • Question 6
    1 / -0
    Pick up the correct statement:
    Solution
    $$ \begin{array}{l} \text { Many of the crystals are anisotropic in nature. } \\ \text { because some properties like electrical resistance } \\ \text { or refractive index of crystalline sulids are. } \\ \text { different when measured along different direction. } \\ \text { This is called anisotropy. } \end{array} $$
  • Question 7
    1 / -0
    An elastic string of length $$42sm$$ and cross-sectional area $$10^{-4}m^2$$ is attached between two pegs ar distance of $$6mm$$ as shown in the figure. A particle of mass m is kept at midpoint of string and stretched as shown in figure by $$20cm$$ and release. As the string its natural length, the particle attains a speed of $$20m/s$$. Then young modulus $$Y$$ of string is of order

    Solution
    $$\dfrac{1}{2}\times Y\times {(\dfrac{\Delta l}{L}) ^2}= \dfrac{1}{2}mv^2$$$$Y\times \dfrac{(0.2)^2}{0.42}\times 10^{4}$$$$= 0.05 \times 400 =\dfrac{1}{2} mv^2$$$$Y=\dfrac{0.05\times 400 \times 0.42}{(0.2)^2\times 10^{4}}$$$$=2.1\times 10^{6 }N/m^2$$
  • Question 8
    1 / -0
    Select the correct alternative $$(s)$$
    a) Elastic forces are always conservative
    b) Elastic forces are not always conservative
    c) Elastic forces are conservative only when Hooke's law is obeyed
    d) Elastic forces may be conservative even when Hooke's law is not obeyed
    Solution
    $$ \begin{array}{l} \text { Elastic forces are not aluays conservative } \\ \text { gt is conservative only when Hooke's law is obeyed } \\ \text { because wol Hookes law is obeyed thpto elastic limit. } \\ \text { After elastic limit Elastic force is not conservative. } \\ \text { as it dcesn't regain its original shape. } \end{array} $$
  • Question 9
    1 / -0
    A wire of density 9$$gm/cm^2$$ is stretched between two clamps 1.00 m apart subjected extension of 0.05 cm. The lowest frequency of transverse vibrations in the wire is 
    (Assume Young's modulus Y = $$9* 10^10 N/m^2$$)
    Solution
    $$\begin{array}{l} The\, \, mass\, \, per\, \, unit\, \, length, \\ m=\dfrac { M }{ l } =\dfrac { { Al\rho  } }{ l } =A\rho  \\ ules\, \, of\, \, elasticity \\ y=\dfrac { { \dfrac { T }{ A }  } }{ { \dfrac { { \Delta l } }{ l }  } }  \\ T=\dfrac { { Y\Delta lA } }{ l }  \\ Hence\, \, lowest\, \, frequency\, \, of\, \, vibration \\ n=\dfrac { 1 }{ { 2l } } \sqrt { \dfrac { T }{ m }  }  \\ =\dfrac { 1 }{ { 2l } } \sqrt { \dfrac { { y\left( { \dfrac { { \Delta l } }{ l }  } \right) A } }{ { A\rho  } }  } =\dfrac { 1 }{ { 2l } } \sqrt { \dfrac { { y\Delta l } }{ { l\rho  } }  }  \\ n=\dfrac { 1 }{ { 2\times 1 } } \sqrt { \dfrac { { 9\times { { 10 }^{ 10 } }\times 4.9\times { { 10 }^{ -4 } } } }{ { 1\times 9\times { { 10 }^{ 3 } } } }  }  \\ =35Hz \end{array}$$
  • Question 10
    1 / -0
    The relation between Young's modulus $$(Y)$$, bulk modulus $$(K)$$ and modulus of elasticity $$(n)$$ is 
    Solution
    $$ \begin{aligned} \frac{3}{y} &=\frac{1}{n}+\frac{1}{3 \beta} \\ \text { where } & \text { Y- Ycoung modulus } \\ n \text { - rigidity modulus } \\ B &=\text { Bulk modulus } \end{aligned} $$
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