Question

Good morning.
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plz solve question no 36
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Answer 1

Answer:-

$\mathsf{a = \dfrac{g}{7}}$

$\mathsf{T_2 = \dfrac{6g}{7}}$

Option → C

Given :-

$m_3 = 1 kg \\ m_2 = 3kg \\ m_1 = 3 kg$

Where,

$m_3 = a, m_2 = b, m_1 = c$

To find :-

The acceleration of the block of mass 1 kg and Tension force between A and B.

Solution:-

Let $T_1$ be the Tension force between Block B and Block C

and $T_2$ be the tension force between B and A.

Now,

• Consider Block C as system.

The force acting on it :-

• Mg force downward.
• Tension force upward.

Acceleration is upward.

$\mathsf{m_1 g - T_1 = m_1 a}$

$\mathsf{T_1 = m_1g - m_1 a}----1)$

• For Block B.

$\mathsf{m_2g + T_2 - T_1 = m_2a}----2)$

• Take Block A as system.

$\mathsf{ m_3g - T_2 = m_3a}$

$\mathsf{T_2 = m_3g - m_3 a}-------3)$

• Put the value of$T_1, T_2$in equation 2.

→$\mathsf{m_2g + m_3g - m_3a -(m_1g -m_1a) = m_2a}$

→$\mathsf{m_2g + m_3g - m_3a - m_1g - m_1a = m_2a }$

→$\mathsf{m_2g + m_3g - m_1 g = m_2a + m_3a + m_1a}$

• Put the masses of each one.

→$\mathsf{3g + g - 3g = 3a + a + 3a }$

→$\mathsf{ g = 7a }$

→$\mathsf{a = \dfrac{g}{7}}$

• put the value of a in eq. 3

→$\mathsf{T_2 = m_3g - m_3a}$

→$\mathsf{T_2 = g - a}$

→$\mathsf{T_2 = g - \dfrac{g}{7}}$

→$\mathsf{T_2 = \dfrac{7g-g}{7}}$

→$\mathsf{T_2 = \dfrac{6g}{7}}$

hence,

The acceleration is $\dfrac{g}{7}$ and Tension force between A and B is $\dfrac{6g}{7}$

Answer 2

$\huge{\fbox{\fbox{\mathfrak{\bigstar{\red{Answer}}}}}}$

$<marquee direction="left">$ Hope it helps