Question

A rope of 1 cm in diameter breaks if tension in it exceeds 500N. The maximum tension that may be given to a similar rope of diameter 2cm is
A. 2000N
B. 1000N
C. 500N
D. 250N

Answer 1

Hint: The tension responsible for breaking down a rope is equal to the amount of force that acts on it. The Young’s modulus remains the same for two ropes of similar material. So the tension on the second rope can be found using Young’s modulus mathematical expression for both the ropes and equating them with each other.

Formula Used:
Young’s modulus is given by:
$\eqalign{
  & Y = \dfrac{{F/A}}{{\Delta L/L}} = \dfrac{{FL}}{{A\Delta L}} \cdots \cdots \cdots \cdots \left( 1 \right) \cr
  & {\text{where }}F{\text{ is the force acting on the rope,}} \cr
  & \Delta L{\text{ is the change in its length,}} \cr
  & L{\text{ is the original length of the rope,}} \cr
  & A{\text{ is the area of cross section of rope}}{\text{.}} \cr} $

Complete step by step answer:
Consider a rope of length, l and cross section area A being pulled by equal and opposite forces. The length of the rope increases from its natural length $L{\text{ to }}L + \Delta L$. The fractional change $\Delta L/L$ is called the longitudinal strain.
Now, Young’s modulus is derived from Hooke’s law for small deformations i.e., if the deformation is small, the stress in a body is proportional to the corresponding strain.
Mathematically,
 $\dfrac{{{\text{Tensile stress}}}}{{{\text{Tensile strain}}}} = Y$
where Y is a constant for a given material. So, Young’s modulus for the material is the ratio of tensile stress over tensile strain.
For the above described situation, the Young’s modulus is
$\eqalign{
  & Y = \dfrac{{F/A}}{{\Delta L/L}} = \dfrac{{FL}}{{A\Delta L}} \cdots \cdots \cdots \cdots \left( 1 \right) \cr
  & {\text{where }}F{\text{ is the force acting on the rope,}} \cr
  & \Delta L{\text{ is the change in its length,}} \cr
  & L{\text{ is the original length of the rope,}} \cr
  & A{\text{ is the area of cross section of rope}}{\text{.}} \cr} $
Given:
Tension acting on rope 1, ${T_1} = 500N$
Let tension acting on rope 2 be ${T_2}$
The diameter of rope 1, ${D_1} = 1cm = 0.01m$
The Radius of rope 1, ${R_1} = \dfrac{1}{2}cm = \dfrac{{0.01}}{2}m$
The diameter of rope 2, ${D_2} = 2cm = 0.02m$
The radius of rope 1, ${R_2} = \dfrac{2}{2}cm = \dfrac{{0.02}}{2}m = 0.01m$
We know that the area of cross section of a rope is given by: $A = \pi {R^2}$
Rearranging equation (1) we get:
$\eqalign{
  & Y = \dfrac{{TL}}{{A\Delta L}}{\text{ }}\left[ {\because {\text{ here }}F = T} \right] \cr
  & \Rightarrow T = Y \times \dfrac{{\Delta L}}{L} \times \pi {R^2} \cr
  & \Rightarrow T \propto {R^2} \cr} $
Taking ratio of tension in rope 1 and in rope 2 and their respective radii, we get:
$\eqalign{
  & \dfrac{{{T_2}}}{{{T_1}}} = \dfrac{{R_2^2}}{{R_1^2}} \cr
  & \Rightarrow \dfrac{{{T_2}}}{{500N}} = \dfrac{{{{\left( {0.01} \right)}^2}}}{{{{\left( {\dfrac{{0.01}}{2}} \right)}^2}}} \cr
  & \Rightarrow \dfrac{{{T_2}}}{{500N}} = 4\dfrac{{{{\left( {0.01} \right)}^2}}}{{{{\left( {0.01} \right)}^2}}} \cr
  & \Rightarrow {T_2} = 4 \times 500N \cr
  & \therefore {T_2} = 2000N \cr} $
Therefore, the correct option is A. i.e., 2000N is the maximum tension for a similar rope of diameter 2cm.

Note: Another way to solve this same question is to form 2 equations using the formula that stress acting on the rope is given by ${\text{Stress, }}S = \dfrac{T}{A}$. Then equate the two equations because the breaking stress must be the same for two different ropes of the same material. Thus find the required tension for rope 2.