JEE Physics 2024 Question with Solution

Given: $m_A = 5 \, \text{kg}$, $m_B = 3 \, \text{kg}$, $m_C = 2 \, \text{kg}$, and applied force $F = 80 \, \text{N}$ on a smooth horizontal surface.

Find: The tensions $T_1$ and $T_2$.

Using Newton's second law for the entire system,

$$M_{\text{total}} = m_A + m_B + m_C = 5 + 3 + 2 = 10 \, \text{kg}$$ $$a = \frac{F}{M_{\text{total}}} = \frac{80}{10} = 8 \, \text{m/s}^2$$

Now apply Newton's second law to block $A$:

$$T_1 = m_A a = 5 \times 8 = 40 \, \text{N}$$

To find $T_2$, consider the subsystem consisting of blocks $A$ and $B$ together:

$$T_2 = (m_A + m_B)a = (5 + 3) \times 8 = 64 \, \text{N}$$

Therefore, the tensions are $40 \, \text{N}$ and $64 \, \text{N}$, respectively. The correct option is A.

Given: $m_A = 5 \, \text{kg}$, $m_B = 3 \, \text{kg}$, $m_C = 2 \, \text{kg}$, and $F = 80 \, \text{N}$.

Find: $T_1$ and $T_2$.

All three blocks move together on the smooth surface, so they have the same acceleration.

1. Calculate the total mass:

$$M_{\text{total}} = 5 + 3 + 2 = 10 \, \text{kg}$$

2. Calculate the common acceleration:

$$F = M_{\text{total}} a$$ $$80 = 10a$$ $$a = 8 \, \text{m/s}^2$$

3. For block $A$ alone, the only horizontal force is $T_1$:

$$T_1 = m_A a = 5 \times 8 = 40 \, \text{N}$$

4. For blocks $A + B$ together, the external horizontal pull on this subsystem is $T_2$:

$$T_2 = (m_A + m_B)a = (5 + 3) \times 8 = 64 \, \text{N}$$

Hence,

$$T_1 = 40 \, \text{N}, \qquad T_2 = 64 \, \text{N}$$

So the correct option is A.

The first solution snippet contains inconsistent intermediate statements for $T_2$, but the final answer and the second approach consistently give the correct values.