MCQ
A rectangular frame is to be suspended symmetrically by two strings of equal length on two supports. It can be done in one of the following three ways;

The tension in the strings will be
  • A
    The same in all cases.
  • B
    Least in $(a).$
  • Least in $(b)$.
  • D
    Least in $(c).$

Answer

Correct option: C.
Least in $(b)$.
According to the $\text{FBD}$ diagram of the rectangular frame.
Let $M$ be the mass of rectangular frame and $0$ be the angle which the tension $T$ in the string makes with the horizontal

Balancing vertical forces,
$2\text{T}\sin\theta-\text{mg}=0 [T$ is tension in the string$]$
$\Rightarrow2\text{T}\sin\theta=\text{mg}$
Total horizontal force will be zero because of equal and opposite forces balance each other $(\text{T}\cos\theta).$
Now from Eq. $(i), \text{T}=\frac{\text{mg}}{2\sin\theta}$
Here, we know $mg$ is constant.
$\Rightarrow\text{T}\propto\frac{1}{\sin\theta}$
$T$ is least if $\sin\theta$ has maximum value.
$\text{T}_\text{min}=\frac{\text{mg}}{2\sin\theta_\text{max}}\text{ (since,}\sin\theta_{\text{max}}=1)$
$\sin\theta_\text{max}=1\Rightarrow\theta=90^\circ$
So the correct option is option $(c).$
Hence, tension is least for the case $(b).$

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