For the system shown in figure, all the pulleys are light and frictionless and strings are ideal. Initially the system is held at rest by some external means. Now, the system is released i.e., externa

Question

For the system shown in figure, all the pulleys are light and frictionless and strings are ideal. Initially the system is held at rest by some external means. Now, the system is released i.e., external effect has been removed, but still it has been observed that 10 kg and 5kg blocks are not moving. This is possible when

TestHub · Accepted Answer

Since the 10 kg block does not move, the tension in the top string is

$T_{1} = 10 g$

This string holds the second pulley, so the tension in the second string is

$T_{2} = \frac{T _{1}}{2} = 5 g$

which also keeps the 5 kg block stationary, as observed.

Similarly the third string (over the lowest pulley, which stays fixed) carries

$T_{3} = \frac{T _{2}}{2} = \frac{5}{2} g$

The lowest pulley is an Atwood pair m_1, m_2 on a fixed pulley; its string tension is

$T_{3} = \frac{2 m _{1} m _{2}}{m _{1} + m _{2}} g$

$\Rightarrow \frac{2 m _{1} m _{2}}{m _{1} + m _{2}} = \frac{5}{2} kg$

i.e. the harmonic mean of m_1 and m_2 is 5/2 kg - (D) correct.

Infinitely many pairs (m_1, m_2) satisfy this relation, e.g. (5, 5/3), so they have many possible values - (C) correct - while their sum is not fixed (it is 5 only for the symmetric pair), so (B) is wrong.

One valid choice is m_1 = m_2 = 5/2 kg (harmonic mean of equals equals each) - (A) correct as a possible value.

Correct options: (A), (C), (D).

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