Question

Lift can move along both $y-$axis and $x-$axis. A ball of mass $m$ is attached to the ceiling of lift with inextensible light rope and box of mass $m$ is placed against a wall as shown in the figure. Neglect friction everywhere. $N$ is the net force acting on $‘m^{\prime }$.
Match the statements in List $1$ with results in List $2$

$\begin{array}{cc}\text{List 1}& \text{List 2}\\ \text{A) In the figure, lift starts moving along x -axis then value of T may be}& \text{p) Zero}\\ \text{B) Lift starts moving toward right along x -axis, then value of N may be}& \text{q)}>mg\\ \text{C) Lift starts moving in upward direction (y -axis) then the value of T may be}& \text{r)}<mg\\ \text{D) Lift starts moving in downward direction, then value of T may be}& \text{s)}=mg\end{array}$

• A. $A–q,s,B–q,r,s,C–q,s,D–p,r,s$
• B. $A–p,B–q,s,C–r,s,D–p,r,s$
• C. $A–q,s,B–q,r,s,C–q,r,D–p,r,s$
• D. $A–p,s,B–p,r,s,C–q,s,D–p,r,s$

Answer 1

The correct option is A $A–q,s,B–q,r,s,C–q,s,D–p,r,s$

A) If the lift is moving horizontally with an acceleration $‘a^{\prime }$, then $T$ will be the resultant of $mg$ and $ma$, will be greater than $mg$.
If it moves with constant velocity $T=mg$.
Hence $q,s$
B) If the lift is moving towards right with an acceleration $‘a^{\prime }$, then $N=ma$
As the lift can move with an acceleration $a>g$ or $a<g$ or $a=g$, then $N>mg$ or $N<mg$ or $N=mg$.
Hence $q,r,s$
C) If the lift moves up with an acceleration $a$, then $T=m(g+a)>mg$
If the lift moves with constant velocity $T=mg$
Hence $q,s$
D) If the lift moves down with an acceleration $a$, then $T=m(g-a)$
If $a=g\Rightarrow T=0$,
If $a<g\Rightarrow T<mg$,
If $a=0\text{(constant velocity)}\Rightarrow T=mg$,
Hence $p,r,s$