Q1
Objective
Match the columns
Numerical
In the diagram shown in figure, match the following columns (Take, $g = 10 \, ms^{-2}$ )
| Column I | Column II | ||||||
| (A) | Normal reaction | (p) | 12 SI unit | ||||
| (B) | Force of friction | (q) | 20 SI unit | ||||
| (C) | Acceleration of block | (r) | zero | ||||
| (s) | 2 SI unit | ||||||
| Codes | |||||||
| A | B | C | A | B | C | ||
| (a) | p | q | r | (b) | q | p | r |
| (c) | q | p | s | (d) | p | s | r |
Correct Answer: C
Explanation:
$N = mg - 20\sqrt{2}\sin 45^{\circ} = 20\mathrm{N}, f = \mu_{k}N = 12\mathrm{N}$
Since, $20\sqrt{2}\cos45^{\circ}>\mu N$ block will move and its acceleration will be
$a = \frac {2 0 \sqrt {2} \cos 4 5 ^ {\circ} - \mu_ {k} N}{m} = \frac {2 0 - 1 2}{4} = 2 \mathrm{ms} ^ {- 2}$
Hence, A → q, B → p, C → s.
Since, $20\sqrt{2}\cos45^{\circ}>\mu N$ block will move and its acceleration will be
$a = \frac {2 0 \sqrt {2} \cos 4 5 ^ {\circ} - \mu_ {k} N}{m} = \frac {2 0 - 1 2}{4} = 2 \mathrm{ms} ^ {- 2}$
Hence, A → q, B → p, C → s.
Q2
Objective
Match the columns
Numerical
In the diagram shown in figure, match the following columns. (Take, $g = 10 \, ms^{-2}$ )
| Column I | Column II | |||||||
| (A) | Acceleration of 2 kg block | (p) 8 SI unit | ||||||
| (B) | Net force on 3 kg block | (q) 25 SI unit | ||||||
| (C) | Normal reaction between 2 kg and 1 kg | (r) 2 SI unit | ||||||
| (D) | Normal reaction between 3 kg and 2 kg | (s) 45 N | ||||||
| (t) None | ||||||||
| Codes | ||||||||
| A | B | C | D | A | B | C | ||
| (a) | q | r | t | t | (b) | p | s | r |
| (c) | s | s | p | t | (d) | r | t | q |
Correct Answer: D
Explanation:
Acceleration of system,
$a = \frac {6 0 - 1 8 - (m _ {1} + m _ {2} + m _ {3}) g \sin 3 0 ^ {\circ}}{(m _ {1} + m _ {2} + m _ {3})} = 2 \mathrm{ms} ^ {- 2}$
Net force on 3 kg block = $m_{3}a = 6N$
From free body diagram of 1 kg block, we have
$N _ {1 2} - m _ {1} g \sin 3 0 ^ {\circ} - 1 8 = m _ {1} a$
$N _ {1 2} = 2 5 \mathrm{N}$
From free body diagram of 3 kg block, we have
$a = \frac {6 0 - 1 8 - (m _ {1} + m _ {2} + m _ {3}) g \sin 3 0 ^ {\circ}}{(m _ {1} + m _ {2} + m _ {3})} = 2 \mathrm{ms} ^ {- 2}$
Net force on 3 kg block = $m_{3}a = 6N$
From free body diagram of 1 kg block, we have
$N _ {1 2} - m _ {1} g \sin 3 0 ^ {\circ} - 1 8 = m _ {1} a$
$N _ {1 2} = 2 5 \mathrm{N}$
From free body diagram of 3 kg block, we have
Q3
Objective
Match the columns
Numerical
In the diagram shown in figure. Match the following columns.
| Column I | Column II | ||||||
| (A) | Absolute acceleration of 1 kg block | (p) | $11 \text{ ms}^{-2}$ | ||||
| (B) | Absolute acceleration of 2 kg block | (q) | $6 \text{ ms}^{-2}$ | ||||
| (C) | Relative acceleration between the two | (r) | $17 \text{ ms}^{-2}$ | ||||
| (s) | None | ||||||
| Codes | |||||||
| A | B | C | A | B | C | ||
| (a) | p | q | r | (b) | s | p | s |
| (c) | r | q | p | (d) | p | p | r |
Correct Answer: B
Explanation:
Force diagram of both the blocks are as shown.
⇒
⇒
$a _ {2} = \frac {2 2}{2} = 1 1 \mathrm{m/s} ^ {2}$
⇒
$a _ {2 1} = a _ {2} - a _ {1} = 7 \mathrm{m/s} ^ {2}$
Hence, A → s, B → p, C → s.
⇒
⇒
$a _ {2} = \frac {2 2}{2} = 1 1 \mathrm{m/s} ^ {2}$
⇒
$a _ {2 1} = a _ {2} - a _ {1} = 7 \mathrm{m/s} ^ {2}$
Hence, A → s, B → p, C → s.
Q4
Objective
Match the columns
Numerical
In the diagram shown in figure, all pulleys are smooth and massless and strings are light. Match the following columns. $^{▲}$ $E = 30 \, N$
| Column I | Column II | ||||||||
| (A) | 1 kg block | (p) | will remain stationary | ||||||
| (B) | 2 kg block | (q) | will move down | ||||||
| (C) | 3 kg block | (r) | will move up | ||||||
| (D) | 4 kg block | (s) | $5 \text{ ms}^{-2}$ | ||||||
| (t) | $10 \text{ ms}^{-2}$ | ||||||||
| Codes | |||||||||
| A | B | C | D | A | B | C | D | ||
| (a) | r,t | p | q | q,s | (b) | p | r,s | q,t | s |
| (c) | p,s | r | q,s | t | (d) | p | t | r | s |
Correct Answer: A
Explanation:
Since, the pulleys are smooth, net force on each pulley should be zero. With this concept, tensions on all strings are shown below.
Now, we can draw free body diagrams of all the four blocks.
$\mathrm{Hence}, \mathrm{A} \rightarrow (\mathrm{r}, \mathrm{t}), \mathrm{B} \rightarrow \mathrm{p}, \mathrm{C} \rightarrow \mathrm{q}, \mathrm{D} \rightarrow (\mathrm{q}, \mathrm{s}).$
Now, we can draw free body diagrams of all the four blocks.
$\mathrm{Hence}, \mathrm{A} \rightarrow (\mathrm{r}, \mathrm{t}), \mathrm{B} \rightarrow \mathrm{p}, \mathrm{C} \rightarrow \mathrm{q}, \mathrm{D} \rightarrow (\mathrm{q}, \mathrm{s}).$