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A square lead slab of side 50 cm and thickness 10 cm is subject to a shearing force on its narrow face) of 9.0 xx 10 ^(4) N.The lower edge is riveted to the floor. How much will the upper edge be displaced ? Shear modulus of lead G=5.8 xx 10 ^(9) N//m^(2)

Answer» <html><body><p></p>Solution :<img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/KPK_AIO_PHY_XI_P2_C09_E02_008_S01.png" width="80%"/> <br/> The load slab is <a href="https://interviewquestions.tuteehub.com/tag/fixed-457597" style="font-weight:bold;" target="_blank" title="Click to know more about FIXED">FIXED</a> and the force is applied parallel to the narrow face. The <a href="https://interviewquestions.tuteehub.com/tag/area-13372" style="font-weight:bold;" target="_blank" title="Click to know more about AREA">AREA</a> of the face parallel to which this force is applied is, <br/> `A=l xx <a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>` <br/> `= 50 cm xx 10 cm = 0.5 m xx 0.1 m` <br/> `therefore A = 0.05 m ^(2)` <br/> Stress = `("Force")/("Area")= (9 xx 10 ^(4))/(0.05) =1.8 xx 10 ^(<a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a>) Nm ^(-2)`<br/> Shear modulus `(G) =("Stress")/("Shear strain")` <br/> `therefore` Shear strain`= ("Stress")/(G)` <br/> ` (Delta x)/(L) = ("Stress")/(G)` <br/> `therefore Deleta x = ("Stress")/(G) xx l` <br/> `therefore Delta x = (1.8 xx 10 ^(6))/(5.6 xx 10 ^(9)) xx 0.5` <br/> `therefore Delta x =0.1607 xx 10 ^(-<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)` <br/> `therefore Delta x ~~ 1.6 xx 10 ^(-4) m = 0.16 mm`</body></html>


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