Q35 / 181 · NTA Abhyas (JEE Main) 2020 · Physics · Laws of Motion

Two masses $m$ and $M$ are attached with strings as shown in the figure. For the system to be in equilibrium, we must have

Correct answer
A
$tan θ = 1 + \frac{2 M}{m}$
B
$tan θ = 1 - \frac{2 m}{M}$
C
$tan θ = 1 - \frac{M}{2 m}$
D
$tan θ = 1 + \frac{m}{2 M}$

Solution

$m g = 2 T \sin 45 °$

$\Rightarrow m g = \sqrt{2} T$

   $T_{1} cos θ = T cos 4 5^{\circ}$

$\Rightarrow T_{1} cos θ = \frac{T}{\sqrt{2}} = \frac{mg}{2}$               ...(i)

Further, $Mg + T cos 4 5^{\circ} = T_{1} sin θ \{∵ T = \frac{mg}{\sqrt{2}}\}$

$\Rightarrow T_{1} sin θ = Mg + \frac{mg}{\sqrt{2}} \frac{1}{\sqrt{2}}$               ...(ii)

$\Rightarrow tan θ = \frac{Mg + \frac{mg}{2}}{\frac{mg}{2}} = 1 + \frac{2 M}{m}$

{dividing Eq. (ii) by Eq. (i)}

Hence, the correct answer is option: 'A'.

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Two masses m and M are attached with strings as shown in the figure. For the system to be in equilibrium, we must have

Solution

More Laws of Motion Questions

Two masses $m$ and $M$ are attached with strings as shown in the figure. For the system to be in equilibrium, we must have