
A body covered a distance of L metre along a curved path of a quarter circle. The ratio of distance to displacement.
(A) $\dfrac{\pi }{{2\sqrt 2 }}$
(B) $\dfrac{{2\sqrt 2 }}{\pi }$
(C) $\dfrac{\pi }{{\sqrt 2 }}$
(D) $\dfrac{{\sqrt 2 }}{\pi }$
Answer
161.1k+ views
Hint:In order to solve this question, we will first find the total distance covered by the body in terms of quarter circle length and radius and then we will find the displacement which is just the distance between initial and final point of the journey and therefore we will find the ratio of the distance to the displacement.
Complete answer:
Let us draw a quarter circle AOB right angled at O of radius R and AB represents the displacement of the body while the curved path of the quarter circle is the length L covered by the body as given in the question.

Now, the quarter circle has a perimeter of $\dfrac{C}{4}$ of total circle of radius R and as we know that perimeter of circle is $
C = 2\pi R \\
\Rightarrow \dfrac{C}{4} = \dfrac{{\pi R}}{2} \\
$ and this length is equal to length covered by the body which is given to us as L so,
so, Distance is represented by L and it’s given by $L = \dfrac{{\pi R}}{2} \to (i)$
Now, as displacement is the distance covered by the body between initial and final point which A and B in the diagram which is AB, so OAB is a right angled triangle hence, using Pythagorean theorem we have,
${(AB)^2} = {(OA)^2} + {(OB)^2}$
on putting the values we get,
$
{(AB)^2} = {(R)^2} + {(R)^2} \\
AB = \sqrt 2 R \to (ii) \\
$
as AB represent displacement covered by the body, so taking ratio of distance L from equation (i) to the displacement AB from equation (ii) we get,
$
\dfrac{L}{{AB}} = \dfrac{{\dfrac{{\pi R}}{2}}}{{\sqrt 2 R}} \\
\Rightarrow \dfrac{L}{{AB}} = \dfrac{\pi }{{2\sqrt 2 }} \\
$
So, the ratio of distance to displacement is $\dfrac{\pi }{{2\sqrt 2 }}$
Hence, the correct option is Option (A).
Note:While solving such questions, don’t confuse distance and displacement as the same quantity because distance is just total length covered by the body and displacement is actual length between initial and final point of the journey.
Complete answer:
Let us draw a quarter circle AOB right angled at O of radius R and AB represents the displacement of the body while the curved path of the quarter circle is the length L covered by the body as given in the question.

Now, the quarter circle has a perimeter of $\dfrac{C}{4}$ of total circle of radius R and as we know that perimeter of circle is $
C = 2\pi R \\
\Rightarrow \dfrac{C}{4} = \dfrac{{\pi R}}{2} \\
$ and this length is equal to length covered by the body which is given to us as L so,
so, Distance is represented by L and it’s given by $L = \dfrac{{\pi R}}{2} \to (i)$
Now, as displacement is the distance covered by the body between initial and final point which A and B in the diagram which is AB, so OAB is a right angled triangle hence, using Pythagorean theorem we have,
${(AB)^2} = {(OA)^2} + {(OB)^2}$
on putting the values we get,
$
{(AB)^2} = {(R)^2} + {(R)^2} \\
AB = \sqrt 2 R \to (ii) \\
$
as AB represent displacement covered by the body, so taking ratio of distance L from equation (i) to the displacement AB from equation (ii) we get,
$
\dfrac{L}{{AB}} = \dfrac{{\dfrac{{\pi R}}{2}}}{{\sqrt 2 R}} \\
\Rightarrow \dfrac{L}{{AB}} = \dfrac{\pi }{{2\sqrt 2 }} \\
$
So, the ratio of distance to displacement is $\dfrac{\pi }{{2\sqrt 2 }}$
Hence, the correct option is Option (A).
Note:While solving such questions, don’t confuse distance and displacement as the same quantity because distance is just total length covered by the body and displacement is actual length between initial and final point of the journey.
Recently Updated Pages
A steel rail of length 5m and area of cross section class 11 physics JEE_Main

At which height is gravity zero class 11 physics JEE_Main

A nucleus of mass m + Delta m is at rest and decays class 11 physics JEE_MAIN

A wave is travelling along a string At an instant the class 11 physics JEE_Main

The length of a conductor is halved its conductivity class 11 physics JEE_Main

Two billiard balls of the same size and mass are in class 11 physics JEE_Main

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Displacement-Time Graph and Velocity-Time Graph for JEE

Uniform Acceleration

Degree of Dissociation and Its Formula With Solved Example for JEE

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line
