Question

A particle of mass $1\text{kg}$ is fastened to one end of a string of length $2\text{m}$ and a mass $2\text{kg}$ is attached to the middle point, the other end of the string is fastened to a fixed point on a smooth horizontal table. The particles are then projected so that the two portions of the string are always in the same straight line and describes horizontal circles. Then the ratio of tensions $\left(\frac{T_1}{T_2}\right)$ in the two parts of the string as shown in diagram is

• A. $2$
• B. $\frac{2}{3}$
• C. $\frac{3}{2}$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $2$

Given $m_A=1\text{kg};m_B=2\text{kg}$
$r_A=$ length of $m_A$ from ${}^{\prime }{O}^{\prime }=2\text{m}$
$r_B=$ length of $m_B$ from ${}^{\prime }{O}^{\prime }=1\text{m}$

By the FBD of the mass $A$: let $T_2$ be the tension force acting on the string connected with mass $A$

By the force equation, we know that $(F_c=m{\omega }^{2}r)$

$T_2=m_A{r}_A{\omega }^2$, as $(\omega =2\text{rad/s})$ given

$T_2=1\times 2\times (2{)}^2=8\text{N}$

By the FBD of the mass $B$: let $T_1$ be the tension force acting on the left part of the string connected with mass $B$

$T_1-T_2=m_B{r}_B{\omega }^2$
$T_1-T_2=2\times 1\times (2{)}^2$
$=8\text{N}$
$\Rightarrow T_1=T_2+8=16\text{N}$
$\therefore \left(\frac{T_1}{T_2}\right)=2$

Hence option A is the correct answer