Question

Three identical particles are joined together by a thread as shown in figure. All the three particles moving in a horizontal plane. If the velocity of the outermost particle is $v_0$ then the ratio of tensions in the three sections of the string is then

• A. 3: 5: 7
• B. 3: 4: 5
• C. 7: 11: 6
• D. 3: 5: 6

Answer 1

The correct option is D 3: 5: 6

Assume mass of each ball is $m$

$V_0=3wL$

$w=\frac{V_0}{3L}$

For Ball $A$ force due to ratio on $1L$

$m{w}^{2}L=\frac{m{V}_0}{aL}$

For $B=m{w}^2\left(2L\right)=\frac{m{V}_0}{9L}$

For $C=m{w}^{2}3L=\frac{m{V}_0}{3L}$

From digname

Let $=$ Tension defined as $\left(T\right)$.

$T$ Between $B$ and $C=m{V}_0/3C=T_1$

$T$ Between $A$ and $B=\frac{2m{V}_0}{aL}+\frac{m{V}_0}{3L}=T_2$

$T$ Between $U$ and $A=\frac{m{V}_0}{aL}+\frac{2m{V}_0}{aL}+\frac{m{V}_0}{3L}$

$=\frac{2m{V}_0}{3L}=T_3$

$T_1:T_2:T_3=\frac{1}{3}:\frac{5}{9}:\frac{2}{3}$

$=\frac{1}{3}\times 9:\frac{5}{9}\times 9:\frac{2}{3}\times 9$

$=3:5:6$.