Question

The twisting couple per unit twist for a solid cylinder of radius 4.9cm is 0.1N/m. The twisting couple per unit twist for a hollow cylinder of same material with outer and inner radii of 5cm and 4cm respectively, will be
A. $0.64N/m$
B. $0.64\times {{10}^{-1}}N/m$
C. $0.64\times {{10}^{-2}}N/m$
D. $0.64\times {{10}^{-3}}N/m$

Answer 1

Hint: Twisting a couple per unit twist is the torque required to deform the cylinder for a unit rotation for a unit length of the cylinder, when a couple acts on it. A couple is a moment (torque) that is caused due to two forces equal in magnitude but opposite in the directions, acting on a body at the same distance from the hinge point or axis of rotation.

Complete step by step answer:
A couple is a moment (torque) that is caused due to two forces equal in magnitude but opposite in the directions, acting on a body at the same distance from the hinge point or axis of rotation. Twisting couple per unit twist of a cylinder is also called the torsional rigidity of the modulus of torsion of the cylinder. It is the amount of resistance of a body against torsional deformation. It is also defined as the torque required to deform the cylinder for a unit rotation for a unit length of the cylinder. Torsional rigidity (C) is similar to shear modulus ($G$) of a body. Torsional rigidity is the ratio of the shear stress to the angle of deformation of the cylinder and is given by $C=\dfrac{\pi \eta {{r}^{4}}}{2l}$, where $\eta$ is the coefficient of rigidity, r is the radius of the cylinder and l is the length of the cylinder. This is for a solid cylinder. If the cylinder is hollow with some thickness then the torsional rigidity is given by $C=\dfrac{\pi \eta ({{r}_{2}}^{4}-{{r}_{1}}^{4})}{2l}$, $r_1$ and $r_2$ are the inner and outer radii of the cylinder respectively.
Let the torsional rigidity in the first be $C_1$ and let it be $C_2$ in the second case.
In the first case, ${{C}_{1}}=0.1N/m$ and $r=4.9cm=4.9\times {{10}^{-2}}m$.
In the second case, ${{r}_{1}}=4cm=4\times {{10}^{-2}}m$ and ${{r}_{2}}=5cm=5\times {{10}^{-2}}m$. We have to find $C_2$.
Since the material is the same in both the cases, $\eta$ will be the same in both cases. Moreover, let the length also remain the same.
Therefore, ${{C}_{1}}=\dfrac{\pi \eta {{(4.9\times {{10}^{-2}})}^{4}}}{2l}=0.1$ and ${{C}_{2}}=\dfrac{\pi \eta \left( {{\left( 5\times {{10}^{-2}} \right)}^{4}}-{{\left( 4\times {{10}^{-2}} \right)}^{4}} \right)}{2l}$
Divide $C_2$ by $C_1$.
$\dfrac{{{C}_{2}}}{{{C}_{1}}}=\dfrac{{{C}_{2}}}{0.1}=\dfrac{\dfrac{\pi \eta \left( {{\left( 5\times {{10}^{-2}} \right)}^{4}}-{{\left( 4\times {{10}^{-2}} \right)}^{4}} \right)}{2l}}{\dfrac{\pi \eta {{(4.9\times {{10}^{-2}})}^{4}}}{2l}}=\dfrac{{{\left( 5\times {{10}^{-2}} \right)}^{4}}-{{\left( 4\times {{10}^{-2}} \right)}^{4}}}{{{(4.9\times {{10}^{-2}})}^{4}}}$
$\Rightarrow \dfrac{{{C}_{2}}}{0.1}=\dfrac{{{\left( 5\times {{10}^{-2}} \right)}^{4}}-{{\left( 4\times {{10}^{-2}} \right)}^{4}}}{{{(4.9\times {{10}^{-2}})}^{4}}}=\dfrac{\left( {{5}^{4}}-{{4}^{4}} \right)\times {{10}^{-8}}}{{{4.9}^{4}}\times {{10}^{-8}}}=\dfrac{\left( {{5}^{4}}-{{4}^{4}} \right)}{{{4.9}^{4}}}$
$\Rightarrow {{C}_{2}}=\dfrac{\left( {{5}^{4}}-{{4}^{4}} \right)}{{{4.9}^{4}}}\times 0.1=0.64\times {{10}^{-1}}N/m$
Therefore, the twisting couple per unit twist if the hollow cylinder is $0.64\times {{10}^{-1}}N/m$.
Hence, the correct option is (a) $0.64\times {{10}^{-1}}N/m$.

Note: To cause a torsional strain or to deform a cylinder with a couple, one end of the cylinder must be fixed or apply the two same couple at both the ends of the cylinder. If this is not done, then there will be no deformation in the cylinder instead, the cylinder will rotate about its axis.