Question

Match the following:

A particle suspended from a string of length is given a horizo ntal speed $u=3\sqrt{gl}$ at the bottom. Then for the particle match the following column

$\begin{array}{cccc}& \text{Column -I}& & \text{Column-II}\\ \left(a\right)& \text{Speed at B}& \left(p\right)& \text{7 mg}\\ \left(b\right)& \text{Speed at C}& \left(q\right)& \sqrt{5gl}\\ \left(c\right)& \text{Tension in string at B}& \left(r\right)& \sqrt{7gl}\\ \left(d\right)& \text{Tension in string at C}& \left(s\right)& \text{4 mg}\end{array}$

• A. (a) – (r); (b) – (q); (c) – (p); (d) – (s)
• B. (a) – (q); (b) – (r); (c) – (p); (d) – (s)
• C. (a) – (r); (b) – (q); (c) – (s); (d) – (q)
• D. (a) – (q); (b) – (r); (c) – (s); (d) – (p)

Answer 1

The correct option is A (a) – (r); (b) – (q); (c) – (p); (d) – (s)

$\therefore v^2=u^2-2gh$

${v_b}^2=9gl-2gl=7gl$

and ${v_c}^2=9gl-4gl=5gl$

$T_B=\frac{m{v}_{B^2}}{l}=7mg$

$T_c+mg=\frac{m{v}^{2}c}{l}$

$\Rightarrow T_c=4mg$