A 2.0 cm
2)0 cm steel rod of length 80 cm has its left end clamped and a force of
Applied perpendicular to its length at its right end.By how much does the right end shift from its original position? For steel Young's modulus is
,the bulk modulus is
,and the shear modulus is
)
D) More information is needed.

Question

A 2.0 cm 
2)0 cm steel rod of length 80 cm has its left end clamped and a force of 
Applied perpendicular to its length at its right end.By how much does the right end shift from its original position? For steel Young's modulus is 
,the bulk modulus is 
,and the shear modulus is 
)
D) More information is needed.

Quizplus · Accepted Answer

The right end of the steel rod will experience a displacement given by the formula:
$$\Delta L = \frac{F\cdot L}{A\cdot E}$$
where $F$ is the applied force, $L$ is the length of the rod, $A$ is the cross-sectional area of the rod, and $E$ is the Young's modulus of the steel. 
Substituting the given values, we have:
$$\Delta L = \frac{(40\ 	ext{N})\cdot (80\ 	ext{cm})}{(\pi\cdot (1\ 	ext{cm})^2)\cdot (2	imes 10^{11}\ 	ext{N/m}^2)} \approx 0.101\ 	ext{mm}$$
Therefore, the right end of the rod shifts by about 0.101 mm. 
The best choice is B, which matches this result.

A 2.0 Cm