
Two particles of equal mass are connected to a rope AB of negligible mass, such that one is at end A and other dividing the length of the rope in the ratio $1:2$ from A. The rope is rotated about end B in a horizontal plane. Ratio of the tensions in the smaller part to the other is (ignore effect of gravity)
A. $4:3$
B. $1:4$
C. $1:2$
D. $1:3$
Answer
489k+ views
Hint: In order to solve this question we need to understand Newton’s laws of motion and pseudo forces. So according to Newton’s third law of motion, it states that for every action there is equal and opposite reaction, so if mass is rotated in a circle then it creates tension in rope so for this action there must be some reaction and that is why a centripetal force acts in opposite direction to tension, Centripetal force is a pseudo force which only acts to conserve Newton’s universal law for all inertial frames moving with non-relativistic velocity.
Complete step by step answer:
For outer mass $m$, let tension be ${T_1}$ and it rotates in radius (shorter length) $\dfrac{L}{3}$
Also let the speed of mass be $v$. So Upon rotation centripetal force acts which is given by, $F = \dfrac{{m{v^2}}}{r}$
$\Rightarrow F = \dfrac{{m{v^2}}}{{\dfrac{L}{3}}}$
$\Rightarrow F = \dfrac{{3m{v^2}}}{L}$
So in case of equilibrium, ${T_1} = \dfrac{{3m{v^2}}}{L}$
For the inner part of length $\dfrac{{2L}}{3}$ and mass $2m$ let the tension be, ${T_2}$.
So tension in inner part is, ${T_2} = {T_1} + {F_2}$
Where, ${F_2}$ is centripetal force due to mass $2m$ and length $\dfrac{{2L}}{3}$ is given by, ${F_2} = \dfrac{{2m{v^2}}}{{\dfrac{{2L}}{3}}}$
${F_2} = \dfrac{{3m{v^2}}}{L}$
Putting values we get,
${T_2} = \dfrac{{3m{v^2}}}{L} + \dfrac{{3m{v^2}}}{L}$
$\Rightarrow {T_2} = \dfrac{{6m{v^2}}}{L}$
$\Rightarrow \dfrac{{{T_1}}}{{{T_2}}} = \dfrac{{\dfrac{{3m{v^2}}}{L}}}{{\dfrac{{6m{v^2}}}{L}}}$
$\therefore \dfrac{{{T_1}}}{{{T_2}}} = \dfrac{1}{2}$
So the correct option is C.
Note: It should be remembered that velocity with which masses are rotating is non relativistic so it does not involve relativistic properties like time dilation and length contraction. Also centripetal force is a pseudo force which acts in the opposite direction of real force and with the same acceleration, but it only appears in a rotating frame of reference.
Complete step by step answer:
For outer mass $m$, let tension be ${T_1}$ and it rotates in radius (shorter length) $\dfrac{L}{3}$
Also let the speed of mass be $v$. So Upon rotation centripetal force acts which is given by, $F = \dfrac{{m{v^2}}}{r}$
$\Rightarrow F = \dfrac{{m{v^2}}}{{\dfrac{L}{3}}}$
$\Rightarrow F = \dfrac{{3m{v^2}}}{L}$
So in case of equilibrium, ${T_1} = \dfrac{{3m{v^2}}}{L}$
For the inner part of length $\dfrac{{2L}}{3}$ and mass $2m$ let the tension be, ${T_2}$.
So tension in inner part is, ${T_2} = {T_1} + {F_2}$
Where, ${F_2}$ is centripetal force due to mass $2m$ and length $\dfrac{{2L}}{3}$ is given by, ${F_2} = \dfrac{{2m{v^2}}}{{\dfrac{{2L}}{3}}}$
${F_2} = \dfrac{{3m{v^2}}}{L}$
Putting values we get,
${T_2} = \dfrac{{3m{v^2}}}{L} + \dfrac{{3m{v^2}}}{L}$
$\Rightarrow {T_2} = \dfrac{{6m{v^2}}}{L}$
$\Rightarrow \dfrac{{{T_1}}}{{{T_2}}} = \dfrac{{\dfrac{{3m{v^2}}}{L}}}{{\dfrac{{6m{v^2}}}{L}}}$
$\therefore \dfrac{{{T_1}}}{{{T_2}}} = \dfrac{1}{2}$
So the correct option is C.
Note: It should be remembered that velocity with which masses are rotating is non relativistic so it does not involve relativistic properties like time dilation and length contraction. Also centripetal force is a pseudo force which acts in the opposite direction of real force and with the same acceleration, but it only appears in a rotating frame of reference.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

