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Question: The twisting couple for unit twist for a solid cylinder of radius \(3\,cm\) is.\(0.1\,N/m\). What is...

The twisting couple for unit twist for a solid cylinder of radius 3cm3\,cm is.0.1N/m0.1\,N/m. What is the twisting couple per unit twist, for a hollow cylinder of the same material with outer and inner radius 5cm5\,cm and 4cm\,4\,cm respectively?
A. 0.1N/m0.1\,N/m
B. 0.455N/m0.455\,N/m
C. 0.91N/m0.91\,N/m
D. 1.82N/m1.82\,N/m

Explanation

Solution

We know that for a solid cylinder the twisting couple per unit twist, Cs{C_s}is given by the equation Cs=πηR42l{C_s} = \dfrac{{\pi \eta {R^4}}}{{2l}}, where η\eta is the coefficient of rigidity, RRis the radius of the cylinder and llis the length of the cylinder. For a hollow cylinder the twisting couple per unit twist, Ch{C_h} is given by equation
Ch=πη(r24r14)2l{C_h} = \dfrac{{\pi \eta \left( {r_2^4 - r_1^4} \right)}}{{2l}} ,Where r2{r_2} is the outer radius and r1{r_1} is the inner radius of the hollow cylinder. By taking the ratio of these two equations and substituting the given values, we can arrive at the final answer.

Complete step by step answer:
Radius of the solid cylinder is given as
R=3cm=0.03mR = 3\,cm = 0.03\,m
Twisting couple per unit twist of solid cylinder,Cs=0.1N/m{C_s} = 0.1\,N/m
Inner radius of the hollow cylinder,
r1=4cm=0.04m{r_1} = \,4\,cm = 0.04\,m
Outer radius of the hollow cylinder,
r2=5cm=0.05m{r_2} = 5cm = 0.05\,m
We are asked to find the couple per unit twist of a hollow cylinder.
We know that the twisting couple GG of a solid cylinder is given by the equation,
Gs=Csθs\,{G_s} = {C_s}{\theta _s}
Thus, Cs=Gsθs{C_s} = \dfrac{{{G_s}}}{{{\theta _s}}}
Cs{C_s} is the twisting couple per unit twist. is given by the equation
Cs=πηR42l{C_s} = \dfrac{{\pi \eta {R^4}}}{{2l}} (1)
Where η\eta is the coefficient of rigidity, RRis the radius of the cylinder and llis the length of the cylinder.
The twisting couple GG of a hollow cylinder is given by the equation,
Gh=Chθh\,{G_h} = {C_h}{\theta _h}
Thus, Ch=Ghθh{C_h} = \dfrac{{{G_h}}}{{{\theta _h}}}
Ch{C_h} is the twisting couple per unit twist in a hollow cylinder. It is given by equation
Ch=πη(r24r14)2l{C_h} = \dfrac{{\pi \eta \left( {r_2^4 - r_1^4} \right)}}{{2l}} (2)
Where r2{r_2} is the outer radius and r1{r_1} is the inner radius of the hollow cylinder.
Now, let us divide equation (1) by (2) then we get ,
CsCh=πηR42lπη(r24r14)2l\dfrac{{{C_s}}}{{{C_h}}} = \dfrac{{\dfrac{{\pi \eta {R^4}}}{{2l}}}}{{\dfrac{{\pi \eta \left( {r_2^4 - r_1^4} \right)}}{{2l}}}}
CsCh=R4(r24r14)\Rightarrow \dfrac{{{C_s}}}{{{C_h}}} = \dfrac{{{R^4}}}{{\left( {r_2^4 - r_1^4} \right)}}
Now, let us substitute the given values, in this equation.
CsCh=R4(r24r14)\dfrac{{{C_s}}}{{{C_h}}} = \dfrac{{{R^4}}}{{\left( {r_2^4 - r_1^4} \right)}}
CsCh=0.034(0.0540.044)\Rightarrow \dfrac{{{C_s}}}{{{C_h}}} = \dfrac{{{{0.03}^4}}}{{\left( {{{0.05}^4} - {{0.04}^4}} \right)}}
CsCh=0.2195\therefore \dfrac{{{C_s}}}{{{C_h}}} = 0.2195\,
Value of Cs{C_s} is already given in the question. On substituting the value, we get
0.1N/mCh=0.219\dfrac{{0.1\,N/m}}{{{C_h}}} = 0.219
Ch=0.1N/m0.219\Rightarrow {C_h} = \dfrac{{0.1\,N/m}}{{0.219}}
Ch=0.455N/m\therefore {C_h} = 0.455\,N/m
This is the twisting couple per unit twist, for a hollow cylinder of the same material.
Thus, the correct answer is option B.

Note: In the question it is given that both the solid cylinder and hollow cylinder are made up of the same material. Therefore, we can take the value of coefficient of rigidity to be the same for both solid and hollow cylinders. But nothing is mentioned about the length of the cylinders. So, we assumed that both the cylinders have the same length. In cases where the length of the cylinders are different, we should take that into consideration.