A thin uniform rod of mass 5kg and length 1m is held in horizontal position with the help of string attached to ends of rod, other ends of strings are held by some external agent. Now end A is pulled down with speed ${\mathrm{V}}_{\mathrm{A}}=3\mathrm{t}$ and end B is pulled down with speed ${\mathrm{V}}_{\mathrm{B}}=\mathrm{t}$. Where ‘t’ is time in sec. Choose the correct choice(s) at time t = 0[Take  g = 10m/ s² ]

here a_A = 3m/s² and a_B = 1m/s²

Since the length of the string is constant, the acceleration of the left most point on the rod will be equal to a_A and the acceleration of the right most point will be a_B 

$\begin{array}{l}{\mathrm{a}}_{\mathrm{c}}+\mathrm{L\alpha }/2={\mathrm{a}}_1;{\mathrm{a}}_{\mathrm{c}}-\mathrm{L\alpha }/2={\mathrm{a}}_2\\ \mathrm{\alpha }=\frac{{\mathrm{a}}_1-{\mathrm{a}}_2}{\mathrm{L}}=2\mathrm{rad}/{\mathrm{s}}^2\\ {\mathrm{T}}_1+{\mathrm{T}}_2-\mathrm{mg}={\mathrm{ma}}_{\mathrm{c}}\\ \left({\mathrm{T}}_1-{\mathrm{T}}_2\right)\mathrm{L}/2=\frac{{\mathrm{ML}}^2}{12}\mathrm{\alpha }\end{array}$

$\left({\mathrm{T}}_1-{\mathrm{T}}_2\right)\frac{1}{2}=\frac{5{\left(1\right)}^2}{12}2\Rightarrow {\mathrm{T}}_1-{\mathrm{T}}_2=\frac{5}{3}...\left(1\right)$

$2{\mathrm{a}}_{\mathrm{c}}={\mathrm{a}}_1+{\mathrm{a}}_2\Rightarrow {\mathrm{a}}_{\mathrm{c}}=2\mathrm{m}/{\mathrm{s}}^2$

${\mathrm{T}}_1+{\mathrm{T}}_2=5(10+2)\Rightarrow {\mathrm{T}}_1+{\mathrm{T}}_2=60...\left(2\right)$

from (1) and (2) 

${\mathrm{T}}_1=\frac{185}{6}\mathrm{N}, {\mathrm{T}}_2=\frac{175}{6}\mathrm{N}$