JEE MAIN - Physics (2022 - 26th July Morning Shift - No. 1)
Three masses $$M=100 \mathrm{~kg}, \mathrm{~m}_{1}=10 \mathrm{~kg}$$ and $$\mathrm{m}_{2}=20 \mathrm{~kg}$$ are arranged in a system as shown in figure. All the surfaces are frictionless and strings are inextensible and weightless. The pulleys are also weightless and frictionless. A force $$\mathrm{F}$$ is applied on the system so that the mass $$\mathrm{m}_{2}$$ moves upward with an acceleration of $$2 \mathrm{~ms}^{-2}$$. The value of $$\mathrm{F}$$ is :
( Take $$\mathrm{g}=10 \mathrm{~ms}^{-2}$$ )
_26th_July_Morning_Shift_en_1_1.png)
Explanation
_26th_July_Morning_Shift_en_1_2.png)
Here we will use the Newton's laws of motion.
Let acceleration of 100 kg block w.r.t. ground is a.
Now, Let's draw FBD of 100 kg block w.r.t. ground,
_26th_July_Morning_Shift_en_1_3.png)
By NLM,
$F-T-N_1=100a\quad \text{... (1)}$
FBD of 20 kg block w.r.t. 100 kg block,
_26th_July_Morning_Shift_en_1_4.png)
By NLM,
$${N_1} = {m_2}a = 20a \Rightarrow {N_1} = 20a\quad$$ ... (2)
$$T - {m_2}g = {m_2}(2)$$
$$ \Rightarrow T - 20g = 20(2)$$
$$ \Rightarrow T = 40 + 200 = 240$$
$$ \Rightarrow T = 240N\quad$$ .... (3)
Now, FBD of 10 kg block w.r.t. 100 kg block,
_26th_July_Morning_Shift_en_1_5.png)
By NLM,
$${m_1}a - T = {m_1}(2)$$
$$ \Rightarrow 10a - T = 20$$
$$ \Rightarrow 10a - 240 = 20\quad$$ [from (3)]
$$ \Rightarrow 10a = 260$$
$$ \Rightarrow a = 26\,m{s^{ - 2}}$$
So, from (2), $${N_1} = 20(26) = 520N$$
Now, from (1), $$F - 240 - 520 = 100(26)$$
$$ \Rightarrow F = 2600 + 240 + 520$$
$$ \Rightarrow F = 3360\,N$$
Comments (0)
