Question

The edge of an aluminium cube is 10 cm long. One face of the cube is firmly fixed to a vertical wall. A mass of 100 kg is then attached to the opposite face of the cube. The shear modulus of aluminium is 25 GPa. What is the vertical deflection of this face?

Answer 1

$\huge{\red{\underline{\underline{\sf{......Answer......}}}}}$

$\star{\blue{\underline{\sf{Given:-}}}}$

$\tt{\implies Length\;(l)=10\;cm=0.10\;m}$

$\tt{\implies Mass\;(m)=100\;kg}$

$\tt{\implies Shear\;modulus\;(\eta)=25\;GPa=25\times 10^{9}\;Pa}$

$\tt{\implies Acceleration\;due\;to\;gravity\;(g)=10\;m/s^{2}}$

$\star{\blue{\underline{\sf{To\;Find:-}}}}$

⇒ Vertical Deflection.

$\star{\blue{\underline{\sf{Formula\;used:-}}}}$

$\star{\boxed{\blue{\tt{\eta = \dfrac{Fl}{A.\theta}}}}}$

$\textsf{Now, first we have to find area,}$

$\tt{\implies Area=(length)^{2}}$

$\tt{\implies Area=0.01\;m^{2}}$

$\textsf{Now, put the values in the formula, we get,}$

$\tt{\implies \eta=\dfrac{Fl}{A.\theta}}$

$\tt{\implies \theta=\dfrac{mg.l}{l^{2}\times \eta}}$  

$\tt{\implies \dfrac{100\times 10\times 0.1}{0.01\times 25\times 10^{9}}}$

$\tt{\implies \dfrac{100}{0.25\times 10^{9}}}$

$\tt{\implies \dfrac{1\times 10^{2}}{0.25\times 10^{9}}}$

$\tt{\implies 4\times 10^{-7}\;m}$

Hence, vertical deflection = 4 × 10⁻⁷

$\huge{\underline{\underline{\sf{Extra\;Information:}}}}$

• Shear modulus of elasticity (η): In an elastic limit the ratio of tangential stress to tangential strain is called shear modulus of elasticity.

• $\sf{\eta=\dfrac{Tangential\;stress}{Tangential\;strain}}$

• $\sf{\eta=\dfrac{F}{A\times \theta}}$

Answer 2

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