Two blocks of masses m and M are connected by means of a metal wire of cross sectional area A

Answer

warning Report Error

Page 4

• Correct Answer: A

  Solution :

  [a]
  
  \[\frac{r-R}{x}=\frac{3R-R}{L}\Rightarrow r=R\left( 1+\frac{2x}{L} \right);Y=\frac{Mg}{\pi {{R}^{2}}\frac{dL}{dx}}\]             \[dL\frac{Mg}{\pi {{R}^{2}}};\frac{dx}{{{\left( 1+\frac{2x}{L} \right)}^{2}}}\]             \[\Delta L=\frac{Mg}{Y\pi {{R}^{2}}}\int\limits_{0}^{L}{\frac{dx}{{{\left( 1+\frac{2x}{L} \right)}^{2}}}=\frac{MgL}{3\pi {{R}^{2}}Y};}\] \[L'=L+\Delta K=L\left( 1+\frac{1}{3}\frac{Mg}{\pi {{R}^{2}}Y} \right)\]

warning Report Error

Page 5

• Correct Answer: D

  Solution :

  [d] \[\beta \frac{\Delta V}{V}=-\Delta P\]                         \[\Delta V\simeq -2\pi ab\Delta a\] \[{{V}_{f}}=\pi {{a}^{2}}b\] \[\frac{\Delta V}{V}=\frac{-2\pi ab\Delta a}{\pi {{a}^{2}}b}=\frac{-2\Delta a}{a}\] \[\Rightarrow \,\Delta P=\frac{2\beta \Delta a}{a}\] Normal force = \[=\frac{2\beta \Delta a}{a}2\pi ab\] \[=4\pi \beta \,\,b\,\Delta a\] friction \[=\mu N\] \[=4\pi \mu \beta \,\,b\,\,a\]

warning Report Error

Page 6

• Correct Answer: A

  Solution :

  [a] \[\ell =1m\]             \[r=5\times {{10}^{-3}}m\]             \[F=50\pi \,\times {{10}^{3}}N\]                       \[\gamma =\frac{F/A}{\frac{\Delta \ell }{\ell }}\]             \[\gamma =\frac{\Delta \ell }{\ell }=\frac{F}{A}\]             \[\gamma =\frac{50\pi \times {{10}^{3}}}{\pi \times {{(5\times {{10}^{-3}})}^{2}}}\times \frac{\ell }{\Delta \ell }\]             \[\gamma =\frac{50\times {{10}^{3}}}{25\times {{10}^{-6}}}\times \frac{1}{\Delta \ell }\Rightarrow \gamma =\frac{2\times {{10}^{9}}}{\Delta \ell }\]             \[\gamma =\frac{2\times {{10}^{9}}}{\varepsilon }\] \[\gamma \max .=2\times {{10}^{9}}\]

warning Report Error

Page 7

• Correct Answer: C

  Solution :

  [c] \[Stress=\frac{Force}{area}=\frac{mg}{A}\] \[=\frac{volume\times density\times g}{Ara}\]            \[Stress=\frac{{{L}^{3}}\rho g}{{{L}^{2}}}\]
  

warning Report Error

Page 8

• Correct Answer: C

  Solution :

  [c] Due to thermal expansion            
  
  Due to External pressure
  
  Equating both
  
  get
  
  

warning Report Error

Page 9

• Correct Answer: A

  Solution :

  [a] \[K=\frac{mg/a}{\frac{\Delta V}{V}}\] and \[V=\frac{4}{3}\pi {{r}^{3}}\] \[\therefore \frac{\Delta V}{V}=3\frac{dr}{r}\] So, \[K=\frac{mg/a}{3\frac{dr}{r}}\Rightarrow \frac{dr}{r}=\frac{mg}{3Ka}\]

warning Report Error

Page 10

• Correct Answer: B

  Solution :

  [b] For same material the ration of stress to strain is same For first cube \[Stres{{s}_{1}}=\frac{{{10}^{5}}}{({{0.1}^{2}})}\] \[strai{{n}_{1}}=\frac{0.5\times {{10}^{-2}}}{0.1}\] For second block, \[stres{{s}_{2}}=\frac{{{10}^{5}}}{{{(0.2)}^{2}}}\] \[strai{{n}_{2}}=\frac{x}{0.2}\] where\[x\] is the displacement for second block. For same material, \[\frac{stres{{s}_{1}}}{strai{{n}_{1}}}=\frac{stres{{s}_{2}}}{strai{{n}_{2}}}\] From this \[x=0.25cm\]

warning Report Error

Page 11

Page 12

Page 13

• Correct Answer: D

  Solution :

  [d] Let \[\rho \]and \[\sigma \]be the density of the liquid and material of the load respectively. In first case, the extension in the wire is \[x=Mg/k=V\rho g/k\]                             ….(i) When the load is immersed in the liquid, up thrust + internal force due to extension in wire = weight of the load \[\Rightarrow V\sigma g+k{{x}_{1}}=V\rho g\Rightarrow {{x}_{1}}=Vg(\rho -\sigma )/k\] Using (i) and (ii), \[{{x}_{1}}=\frac{Vgx}{Vg\rho }(\rho -\sigma )=x\left( 1-\frac{\sigma }{\rho } \right)=4\times \left( 1-\frac{2}{8} \right)=3mm\]

warning Report Error

Page 14

• Correct Answer: C

  Solution :

  [c] Energy of catapult\[=\frac{1}{2}\times {{\left( \frac{\Delta \ell }{\ell } \right)}^{2}}\times Y\times A\times \ell \] = Kinetic energy of the ball\[=\frac{1}{2}m{{\text{v}}^{2}}\] therefore, \[\frac{1}{2}\times {{\left( \frac{20}{42} \right)}^{2}}\times Y\times \pi \times {{3}^{2}}\times {{10}^{-6}}\times 42\times {{10}^{-2}}=\] \[\frac{1}{2}\times 2\times {{10}^{-2}}\times {{(20)}^{2}}\] \[Y\simeq 3\times {{10}^{6}}N{{m}^{-2}}\]

warning Report Error

Page 15

• Correct Answer: D

  Solution :

  [d] Tensile stresss in wire will be \[=\frac{\text{Tensile}\,\,\text{force}}{\text{Cross}\,\,\text{section}\,\,\text{Area}}\] \[=\frac{mg}{\pi {{R}^{2}}}=\frac{4\times 3.1\pi }{\pi \times 4\times {{10}^{-6}}}N{{m}^{-2}}=3.1\times {{10}^{6}}N{{m}^{-2}}\]

warning Report Error

Page 16

• Correct Answer: B

  Solution :

  [b] Given \[\frac{{{Y}_{A}}}{{{Y}_{B}}}=\frac{7}{4}\,\,\,\,\,\,\,\,\,\,{{L}_{A}}=2m\,\,\,\,\,\,\,\,\,\,{{A}_{A}}=\pi {{R}^{2}}\]     \[{{L}_{B}}=1.5m\]   \[{{A}_{B}}=\pi {{(2mm)}^{2}}\] \[\frac{F}{A}=Y\left( \frac{\ell }{L} \right)\] given F and \[\ell \]are same \[\Rightarrow \frac{AY}{L}\] is same \[\frac{{{A}_{A}}{{Y}_{A}}}{{{L}_{A}}}=\frac{{{A}_{B}}{{Y}_{B}}}{{{L}_{B}}}\] \[\Rightarrow \]\[\frac{\left( \pi {{R}^{2}} \right)\left( \frac{7}{4}{{Y}_{B}} \right)}{2}=\frac{\pi {{\left( 2mm \right)}^{2}}.{{Y}_{B}}}{1.5}\] \[R=1.74mm\]

warning Report Error

Page 17

• Correct Answer: A

  Solution :

  [a] \[\frac{F}{A}=stress\]             \[\frac{400\times 4}{\pi {{d}^{2}}}=379\times {{10}^{6}}\]             \[{{d}^{2}}=\frac{1600}{\pi \times 379\times {{10}^{6}}}=1.34\times {{10}^{-6}}\]             \[d=\sqrt{1.34}\times {{10}^{-3}}=1.15\times {{10}^{-3}}m\]

warning Report Error

Page 18

• Correct Answer: B

  Solution :

  [b]
  
  \[{{k}_{1}}=\frac{{{y}_{1}}{{A}_{1}}}{{{\ell }_{1}}}=\frac{120\times {{10}^{9}}\times A}{1}\] \[{{k}_{2}}=\frac{{{y}_{2}}{{A}_{2}}}{{{\ell }_{2}}}=\frac{60\times {{10}^{9}}\times A}{1}\] \[{{k}_{eq}}=\frac{{{k}_{1}}{{k}_{2}}}{{{k}_{1}}\times {{k}_{2}}}=\frac{120\times 60}{180}\times {{10}^{9}}\times A\] \[{{k}_{eq}}=40\times {{10}^{9}}\times A\] \[F={{k}_{eq}}(x)\] \[F=(40\times {{10}^{9}})A.(0.2\times {{10}^{-3}})\] \[\frac{F}{A}=8\times {{10}^{6}}N/{{m}^{2}}\]         No option is matching. Hence question must be bonus.

warning Report Error

Page 19

• Correct Answer: B

  Solution :

  [b] \[{{I}_{1}}=4{{I}_{0}}\]                 \[{{I}_{2}}={{I}_{0}}\]                \[{{I}_{\max }}={{(\sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}})}^{2}}\]             \[={{(2\sqrt{{{I}_{0}}}+\sqrt{{{I}_{0}}})}^{2}}=9{{I}_{0}}\]             \[{{I}_{\min }}={{(\sqrt{{{I}_{1}}}-\sqrt{{{I}_{2}}})}^{2}}\]             \[={{(2\sqrt{{{I}_{0}}}-\sqrt{{{I}_{0}}})}^{2}}={{I}_{0}}\]             \[\therefore \]\[\frac{{{I}_{\max }}}{{{\operatorname{I}}_{\min }}}=\frac{9}{1}\]

warning Report Error

Page 20

• Correct Answer: B

  Solution :

  [b]            
  
  \[\therefore \] Length of cylinder remains unchanged so\[{{\left( \frac{F}{A} \right)}_{Compressive}}={{\left( \frac{F}{A} \right)}_{Thermal}}\] \[\frac{F}{\pi {{r}^{2}}}=Y\alpha T\] (\[\alpha \]is linear coefficient of expansion) \[\therefore \]\[\alpha =\frac{F}{YT\pi {{r}^{2}}}\] \[\therefore \]The coefficient of volume expansion\[\gamma =3\alpha \] \[\therefore \]\[\gamma =3\frac{F}{YT\pi {{r}^{2}}}\]

warning Report Error

Page 21

warning Report Error

Page 22

• Correct Answer: A

  Solution :

  [a] Energy density \[=\frac{1}{2}\,stress\times strain\] Energy density\[=\frac{1}{2}\frac{F}{A}\times \frac{F}{AY}\] \[\frac{{{u}_{1}}}{{{u}_{2}}}={{\left( \frac{{{d}_{2}}}{d{{}_{1}}} \right)}^{4}}\] \[\frac{{{d}_{1}}}{{{d}_{2}}}={{\left( 4 \right)}^{1/4}}\] \[\frac{{{d}_{1}}}{{{d}_{2}}}=\sqrt{2}:1\]

warning Report Error