A particle of mass m1 is fastened to one end of a string and another particle of mass m2 is attached to the middle point, the other end of the string being fastened to a fixed point on a smooth horizontal table. If the particles are made to revolve in a horizontal circular path as shown in the figure, then the ratio of tension in the part of the string between the centre and m2 to the tension in the part of the string between m2 and m1 is

Question

A particle of mass m1 is fastened to one end of a string and another particle of mass m2 is attached to the middle point, the other end of the string being fastened to a fixed point on a smooth horizontal table. If the particles are made to revolve in a horizontal circular path as shown in the figure, then the ratio of tension in the part of the string between the centre and m2 to the tension in the part of the string between m2 and m1 is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-yw1wp4yh9wif.png" /> (A) (m1/m1 + m2) (B) (m1 + m2/m1) (C) (2 m1 + m2/2 m1) (D) (2 m1/m1 + m2)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-yw1wp4yhp3h3.png" /> for m1, mathrmT1= mathrmm1 ω2(2 mathrmr) ldots ldots ......(i) for m2, T2-T1=m2ω 2r ....(ii) Dividing both the equations: (T1/T2 - T1)=(2 m1/m2) ⇒ (T2/T1)=(2 m1 + m2/2 m1)

$\frac{m _{1}}{m _{1} + m _{2}}$

$\frac{m _{1} + m _{2}}{m _{1}}$

$\frac{2 m _{1} + m _{2}}{2 m _{1}}$

$\frac{2 m _{1}}{m _{1} + m _{2}}$

for $m_{1},$
$T_{1} = m_{1} ω^{2} (2 r) \dots\dots$ ......(i)
for $m_{2},$
$T_{2} - T_{1} = m_{2} ω^{2} r$ ....(ii)
Dividing both the equations:
$\frac{T _{1}}{T _{2} - T _{1}} = \frac{2 m _{1}}{m _{2}}$
$\Rightarrow \frac{T _{2}}{T _{1}} = \frac{2 m _{1} + m _{2}}{2 m _{1}}$

m1​+m2​m1​​

m1​m1​+m2​​

2m1​2m1​+m2​​

m1​+m2​2m1​​

for m1​, T1​=m1​ω2(2r)…… ......(i) for m2​, T2​−T1​=m2​ω2r ....(ii) Dividing both the equations: T2​−T1​T1​​=m2​2m1​​ ⇒T1​T2​​=2m1​2m1​+m2​​

$\frac{m _{1}}{m _{1} + m _{2}}$

$\frac{m _{1} + m _{2}}{m _{1}}$

$\frac{2 m _{1} + m _{2}}{2 m _{1}}$

$\frac{2 m _{1}}{m _{1} + m _{2}}$

for $m_{1},$
$T_{1} = m_{1} ω^{2} (2 r) \dots\dots$ ......(i)
for $m_{2},$
$T_{2} - T_{1} = m_{2} ω^{2} r$ ....(ii)
Dividing both the equations:
$\frac{T _{1}}{T _{2} - T _{1}} = \frac{2 m _{1}}{m _{2}}$
$\Rightarrow \frac{T _{2}}{T _{1}} = \frac{2 m _{1} + m _{2}}{2 m _{1}}$