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  • I have a doubt, kindly clarify. As shown in the figure, forces of 105 N each are applied in opposite directions, on the upper and lower faces of a cube of side 10 cm, shifting the upper face parallel to itself by 0.5 cm. If the side of another cube o

As shown in the figure, forces of 105 N each are applied in opposite directions, on the upper and lower faces of a cube of side 10 cm, shifting the upper face parallel to itself by 0.5 cm. If the side of another cube of the same material is 20 cm, then under similar conditions as above, the displacement will be :

  • Option 1)

    0.25cm

  • Option 2)

    0.37cm

  • Option 3)

    0.75cm

  • Option 4)

    1.00cm

 

Answers (2)

best_answer

As we learnt

Modulus of Rigidity -

G = \frac{F}{A\phi }
 

- wherein

F=force

A=area

\theta =shear\: strain

 

 shearing modulus =\frac{shearing\: stress}{shearing\: strain}=\frac{F/A_{1}}{\left ( \Delta x/L_{1} \right )}

\Delta x=\left ( \frac{FL_{1}}{A_{1}.n} \right )

\therefore \frac{\Delta x_{1}}{\Delta x_{2}}=\frac{L_{1}}{A_{1}}.\frac{A_{2}}{L_{2}}=\frac{10cm*(20cm)^{2}}{\left ( 10cm \right )^{2}*20cm}\therefore \frac{\Delta x_{1}}{\Delta x_{2}}=\frac{L_{1}}{A_{1}}.\frac{A_{2}}{L_{2}}=\frac{10cm*(20cm)^{2}}{\left ( 10cm \right )^{2}*20cm}=2

\therefore \Delta x_{2}= \frac{\Delta x_{1}}{2}=0.25cm

 


Option 1)

0.25cm

Option 2)

0.37cm

Option 3)

0.75cm

Option 4)

1.00cm

Posted by

divya.saini

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