Q501
Advance
LINKED COMPREHENSION TYPE QUESTIONS
MCQ
Comprehension Passage: Comprehension - 14
A person standing on the roof of a building throws a ball vertically upward at an instant t=0. The ball leaves his hand with an upward speed $20 \, ms^{-1}$ and it is then in free fall. The ball rises to a certain height and then moves down. On its way down, the ball just misses to hit the roof of the building and keeps falling towards the earth. The ball hits earth at t = 5 s. Considering that
(i) the vertically upward direction is the positive $y$ -direction
(ii) the position of ball at t = 0 is the origin
(iii) the ball does not rebound and comes to rest at the same place where it hits earth and
(iv) air resistance is negligible, answer these questions. $\left(\text{Take } g = 10 \text{ ms}^{-2}\right)$
Based on the above facts, answer the following questions.
Acceleration of the ball will vary with time as
(i) the vertically upward direction is the positive $y$ -direction
(ii) the position of ball at t = 0 is the origin
(iii) the ball does not rebound and comes to rest at the same place where it hits earth and
(iv) air resistance is negligible, answer these questions. $\left(\text{Take } g = 10 \text{ ms}^{-2}\right)$
Based on the above facts, answer the following questions.
A.
B.
C.
D.
Q502
Advance
LINKED COMPREHENSION TYPE QUESTIONS
MCQ
Comprehension Passage: Comprehension - 15
The position of a particle is moving along the x-axis depends on the time according to the equation $x = 6t^{2} - t^{3}$ , where x is in metre and t in seconds.
Based on the above facts, answer the following questions.
Time at which velocity of the particle is maximum along positive direction of $x$ -axis is
A.
1 s
B.
2 s
C.
3 s
D.
4 s
Q503
Advance
LINKED COMPREHENSION TYPE QUESTIONS
MCQ
Comprehension Passage: Comprehension - 15
The position of a particle is moving along the x-axis depends on the time according to the equation $x = 6t^{2} - t^{3}$ , where x is in metre and t in seconds.
Based on the above facts, answer the following questions.
Distance travelled by the particle during time interval $t = 3 \mathrm{~s}$ to $t = 5 \mathrm{~s}$ is
A.
$2 \mathrm{~m}$
B.
$5 \mathrm{~m}$
C.
$12 \mathrm{~m}$
D.
$10 \mathrm{~m}$
Q504
Advance
LINKED COMPREHENSION TYPE QUESTIONS
MCQ
Comprehension Passage: Comprehension - 15
The position of a particle is moving along the x-axis depends on the time according to the equation $x = 6t^{2} - t^{3}$ , where x is in metre and t in seconds.
Based on the above facts, answer the following questions.
Average speed of the particle during time interval $t = 0$ s to $t = 6$ s is
A.
$2 \, \text{ms}^{-1}$
B.
Zero
C.
$4 \, \text{ms}^{-1}$
D.
None of these
Q505
Advance
LINKED COMPREHENSION TYPE QUESTIONS
MCQ
Comprehension Passage: Comprehension - 16
A particle is moving along x-axis and its initial velocity is $27 \, ms^{-1}$ . The acceleration of particle is given by the relation $a = (-6t) \, \text{ms}^{-2}$ , where t is in seconds. At t = 0 particle is at x = 0.
Based on the above facts, answer the following questions.
The velocity of particle, when it travels 26 m is
A.
$21 \, ms^{-1}$
B.
$15 \, ms^{-1}$
C.
$24 \, ms^{-1}$
D.
$18 \, ms^{-1}$
Q506
Advance
LINKED COMPREHENSION TYPE QUESTIONS
MCQ
Comprehension Passage: Comprehension - 16
A particle is moving along x-axis and its initial velocity is $27 \, ms^{-1}$ . The acceleration of particle is given by the relation $a = (-6t) \, \text{ms}^{-2}$ , where t is in seconds. At t = 0 particle is at x = 0.
Based on the above facts, answer the following questions.
Maximum value of velocity along positive x-direction is
A.
$35 \, ms^{-1}$
B.
$33 \, ms^{-1}$
C.
$27 \, ms^{-1}$
D.
$30 \, ms^{-1}$
Q507
Advance
LINKED COMPREHENSION TYPE QUESTIONS
MCQ
Comprehension Passage: Comprehension - 16
A particle is moving along x-axis and its initial velocity is $27 \, ms^{-1}$ . The acceleration of particle is given by the relation $a = (-6t) \, \text{ms}^{-2}$ , where t is in seconds. At t = 0 particle is at x = 0.
Based on the above facts, answer the following questions.
Maximum value of displacement along positive x-direction is
A.
54 m
B.
27 m
C.
120 m
D.
None of these
Q508
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
From the v-t graph shown in figure, match the quantities in Column-I to their respective conclusions in Column-II.
| Column-I | Column-II | ||||||||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
|
Correct Answer: A → (s, u) B → (s, t) C → (s, u) D → (s, t) E → (q, r) F → (s, u) G → (p, s)
Explanation:
A → (s, u)
$\mathrm{B} \rightarrow (\mathrm{s}, \mathrm{t})$
$C \to (s, u)$
$\mathrm{D} \rightarrow (\mathrm{s}, t)$
$\mathrm{E} \rightarrow (\mathrm{q}, \mathrm{r})$
$\mathrm{F} \rightarrow (\mathrm{s}, \mathrm{u})$
$G \to (p, s)$
Between t=0 and t=1 s, motion is decelerated $a\neq0$ .
$(\mathrm{A}) \rightarrow (\mathrm{s}, \mathrm{u})$
Between t=1 s and t=2 s, motion is accelerated $a\neq0$ .
$\mathrm{So}, (\mathrm{B}) \rightarrow (\mathrm{s}, \mathrm{t})$
Between t = 2 s and t = 3 s, motion is decelerated ( $a \neq 0$ ).
$\mathrm{So}, (\mathrm{C}) \rightarrow (\mathrm{s}, \mathrm{u})$
Between $t = 3 \, \text{s}$ and $t = 4 \, \text{s}$ , motion is accelerated $a \neq 0$ .
$\mathrm{So}, (\mathrm{D}) \rightarrow (\mathrm{s}, \mathrm{t})$
Between $t = 4 \, \text{s}$ and $t = 5 \, \text{s}$ , motion is uniform i.e., $v \neq 0$ and $a = 0$ .
$\mathrm{So}, (\mathrm{E}) \rightarrow (\mathrm{q}, \mathrm{r})$
Between $t = 5 \, \text{s}$ and $t = 6 \, \text{s}$ , motion is decelerated ( $a \neq 0$ ).
$\mathrm{So}, (\mathrm{F}) \rightarrow (\mathrm{s}, \mathrm{u})$
At t=1 s and at t=3 s the motion reverses its direction, so v=0 and simultaneously $a\neq0$ .
$\mathrm{So}, (\mathrm{G}) \rightarrow (\mathrm{p}, \mathrm{s})$
$\mathrm{B} \rightarrow (\mathrm{s}, \mathrm{t})$
$C \to (s, u)$
$\mathrm{D} \rightarrow (\mathrm{s}, t)$
$\mathrm{E} \rightarrow (\mathrm{q}, \mathrm{r})$
$\mathrm{F} \rightarrow (\mathrm{s}, \mathrm{u})$
$G \to (p, s)$
Between t=0 and t=1 s, motion is decelerated $a\neq0$ .
$(\mathrm{A}) \rightarrow (\mathrm{s}, \mathrm{u})$
Between t=1 s and t=2 s, motion is accelerated $a\neq0$ .
$\mathrm{So}, (\mathrm{B}) \rightarrow (\mathrm{s}, \mathrm{t})$
Between t = 2 s and t = 3 s, motion is decelerated ( $a \neq 0$ ).
$\mathrm{So}, (\mathrm{C}) \rightarrow (\mathrm{s}, \mathrm{u})$
Between $t = 3 \, \text{s}$ and $t = 4 \, \text{s}$ , motion is accelerated $a \neq 0$ .
$\mathrm{So}, (\mathrm{D}) \rightarrow (\mathrm{s}, \mathrm{t})$
Between $t = 4 \, \text{s}$ and $t = 5 \, \text{s}$ , motion is uniform i.e., $v \neq 0$ and $a = 0$ .
$\mathrm{So}, (\mathrm{E}) \rightarrow (\mathrm{q}, \mathrm{r})$
Between $t = 5 \, \text{s}$ and $t = 6 \, \text{s}$ , motion is decelerated ( $a \neq 0$ ).
$\mathrm{So}, (\mathrm{F}) \rightarrow (\mathrm{s}, \mathrm{u})$
At t=1 s and at t=3 s the motion reverses its direction, so v=0 and simultaneously $a\neq0$ .
$\mathrm{So}, (\mathrm{G}) \rightarrow (\mathrm{p}, \mathrm{s})$
Q509
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
For a particle moving rectilinearly, the x varies with t as per the equation $x = -5t^{2} + 20t + 10$ , where x is in metre and t is in second.
Match Column-I with Column-II
Match Column-I with Column-II
| Column-I | Column-II | ||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
|
Correct Answer: A → (q) B → (r) C → (s) D → (p)
Explanation:
A → (q)
$\mathrm{B} \rightarrow (\mathrm{r})$
$C \to (\mathrm{s})$
$\mathrm{D} \rightarrow (\mathrm{p})$
$x = - 5 t ^ {2} + 2 0 t + 1 0$
$\Rightarrow \frac {d x}{d t} = - 1 0 t + 2 0$
$\Rightarrow v = 2 0 - 1 0 t\tag{... (1}$
Now v=0 at t=2 s, so the distance travelled by the particle from t=0 to t=3 s is
$x = \left| \int_ {0} ^ {2} v d t \right| + \left| \int_ {2} ^ {3} v d t \right|$
$\mathrm{Now} \int v d t = (2 0 - 1 0 t) d t$
$\Rightarrow \int v d t = 2 0 t - 5 t ^ {2}$
$\Rightarrow x = \left| (2 0 t - 5 t ^ {2}) \right| _ {0} ^ {2} + \left| (2 0 t - 5 t ^ {2}) \right| _ {2} ^ {3}$
$\Rightarrow x = \vert 4 0 - 2 0 \vert + \vert 8 0 - 8 0 - 4 0 + 2 0 \vert$
$\Rightarrow x = | 2 0 | + | - 2 0 |$
$\Rightarrow x = 4 0 \mathrm{m}$
So, average speed is
$v _ {a v} = \frac {x}{t} = \frac {4 0}{4} = 1 0 \mathrm{ms} ^ {- 1}$
Displacement of the particle from t = 0 to t = 4 s is
$\Delta x = \int_ {0} ^ {4} v d t = \int_ {0} ^ {4} (2 0 - 1 0 t) d t$
$\Rightarrow \Delta x = (2 0 t - 5 t ^ {2}) \big | _ {0} ^ {4}$
$\Rightarrow \Delta x = 8 0 - 8 0 = 0$
So, average velocity is
$v _ {a v} = \frac {\Delta x}{\Delta t} = \frac {0}{4} = 0 \mathrm{ms} ^ {- 1}$
From (1), $a = \frac{dv}{dt} = -10\mathrm{ms}^{-2}$
$u = v | _ {t = 0} = 2 0 \mathrm{ms} ^ {- 1}$
Since, initially $u = \oplus$ and $a = \ominus$ , so a happens to be retardation for the motion from t = 0 to t = 2 s (till its velocity becomes zero momentarily i.e., the particle reverses its direction of motion).
Velocity at $t = 4$ s is
$v | _ {t = 4 \mathrm{s}} = 2 0 - 1 0 (4) = - 2 0 \mathrm{ms} ^ {- 1}$
Hence, speed at t = 4 s is $20 \, ms^{-1}$
$\mathrm{B} \rightarrow (\mathrm{r})$
$C \to (\mathrm{s})$
$\mathrm{D} \rightarrow (\mathrm{p})$
$x = - 5 t ^ {2} + 2 0 t + 1 0$
$\Rightarrow \frac {d x}{d t} = - 1 0 t + 2 0$
$\Rightarrow v = 2 0 - 1 0 t\tag{... (1}$
Now v=0 at t=2 s, so the distance travelled by the particle from t=0 to t=3 s is
$x = \left| \int_ {0} ^ {2} v d t \right| + \left| \int_ {2} ^ {3} v d t \right|$
$\mathrm{Now} \int v d t = (2 0 - 1 0 t) d t$
$\Rightarrow \int v d t = 2 0 t - 5 t ^ {2}$
$\Rightarrow x = \left| (2 0 t - 5 t ^ {2}) \right| _ {0} ^ {2} + \left| (2 0 t - 5 t ^ {2}) \right| _ {2} ^ {3}$
$\Rightarrow x = \vert 4 0 - 2 0 \vert + \vert 8 0 - 8 0 - 4 0 + 2 0 \vert$
$\Rightarrow x = | 2 0 | + | - 2 0 |$
$\Rightarrow x = 4 0 \mathrm{m}$
So, average speed is
$v _ {a v} = \frac {x}{t} = \frac {4 0}{4} = 1 0 \mathrm{ms} ^ {- 1}$
Displacement of the particle from t = 0 to t = 4 s is
$\Delta x = \int_ {0} ^ {4} v d t = \int_ {0} ^ {4} (2 0 - 1 0 t) d t$
$\Rightarrow \Delta x = (2 0 t - 5 t ^ {2}) \big | _ {0} ^ {4}$
$\Rightarrow \Delta x = 8 0 - 8 0 = 0$
So, average velocity is
$v _ {a v} = \frac {\Delta x}{\Delta t} = \frac {0}{4} = 0 \mathrm{ms} ^ {- 1}$
From (1), $a = \frac{dv}{dt} = -10\mathrm{ms}^{-2}$
$u = v | _ {t = 0} = 2 0 \mathrm{ms} ^ {- 1}$
Since, initially $u = \oplus$ and $a = \ominus$ , so a happens to be retardation for the motion from t = 0 to t = 2 s (till its velocity becomes zero momentarily i.e., the particle reverses its direction of motion).
Velocity at $t = 4$ s is
$v | _ {t = 4 \mathrm{s}} = 2 0 - 1 0 (4) = - 2 0 \mathrm{ms} ^ {- 1}$
Hence, speed at t = 4 s is $20 \, ms^{-1}$
Q510
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
Match the quantities in Column-I with the corresponding expressions in Column-II.
| Column-I | Column-II | ||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
|
Correct Answer: A → (q) B → (r) C → (p, s) D → (t)
Explanation:
A $\rightarrow$ (q)
$\mathrm{B} \rightarrow (\mathrm{r})$
$C \to (p, s)$
$\mathrm{D} \rightarrow (t)$
$\frac {d \vec {r}}{d t} = \mathrm{Velocity}$
$\frac {d | \vec {v} |}{d t} = \text { Tangential Acceleration }$
$\frac {d \vec {v}}{d t} = \frac {d ^ {2} \vec {r}}{d t ^ {2}} = \mathrm{Acceleration}$
$\left| \frac {d \vec {r}}{d t} \right| = \text { Instantaneous speed }$
$\mathrm{B} \rightarrow (\mathrm{r})$
$C \to (p, s)$
$\mathrm{D} \rightarrow (t)$
$\frac {d \vec {r}}{d t} = \mathrm{Velocity}$
$\frac {d | \vec {v} |}{d t} = \text { Tangential Acceleration }$
$\frac {d \vec {v}}{d t} = \frac {d ^ {2} \vec {r}}{d t ^ {2}} = \mathrm{Acceleration}$
$\left| \frac {d \vec {r}}{d t} \right| = \text { Instantaneous speed }$
Q511
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
A particle moves such that its x coordinate is related to the time t by the relation $t = \sqrt{x} + 3$ , where x is in metre, t is in second. Based on this information, match the values in COLUMN-I (in SI units) to their respective quantities for the particles motion given in COLUMN-II.
| Column-I | Column-II | ||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
|
Correct Answer: A → (r, t) B → (p) C → (q) D → (s)
Explanation:
A $\rightarrow (\mathbf{r},\mathbf{t})$
$\mathrm{B} \rightarrow (\mathrm{p})$
$\mathrm{C} \rightarrow (\mathrm{q})$
$\mathrm{D} \rightarrow (\mathrm{s})$
$x = (t - 3) ^ {2}$
$\Rightarrow x = t ^ {2} - 6 t + 9$
$\Rightarrow v = \frac {d x}{d t} = 2 t - 6$
2 t - 6 = 0
$\Rightarrow t = 3 \mathrm{s}$
Displacement of the particle from t=0 to t=6 s is zero.
Distance travelled by the particle from t = 0 to t = 6 is $9 + 9 = 18 \, m$
Average speed of the particle is
$v _ {a v} = \frac {\text { Total Distance Travelled }}{\text { Total Time Taken }}$
$\Rightarrow v _ {a v} = \frac {1 8}{6} = 3 \mathrm{ms} ^ {- 1}$
Average Velocity of the particle is
$\left| \vec {v} _ {a v} \right| = \frac {\text { Displacement }}{\text { Time }} = 0$
Acceleration of the particle over the entire duration of motion is $2 \, ms^{-2}$ .
$\mathrm{B} \rightarrow (\mathrm{p})$
$\mathrm{C} \rightarrow (\mathrm{q})$
$\mathrm{D} \rightarrow (\mathrm{s})$
$x = (t - 3) ^ {2}$
$\Rightarrow x = t ^ {2} - 6 t + 9$
$\Rightarrow v = \frac {d x}{d t} = 2 t - 6$
2 t - 6 = 0
$\Rightarrow t = 3 \mathrm{s}$
| t | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| x | 9 | 4 | 1 | 0 | 1 | 4 | 9 |
| v | -6 | -4 | -2 | 0 | 2 | 4 | 6 |
| a | 2 | 2 | 2 | 2 | 2 | 2 | 2 |
| Nature of Motion | Decelerating | Decelerating | Decelerating | Accelerating | Accelerating | Accelerating | Accelerating |
Displacement of the particle from t=0 to t=6 s is zero.
Distance travelled by the particle from t = 0 to t = 6 is $9 + 9 = 18 \, m$
Average speed of the particle is
$v _ {a v} = \frac {\text { Total Distance Travelled }}{\text { Total Time Taken }}$
$\Rightarrow v _ {a v} = \frac {1 8}{6} = 3 \mathrm{ms} ^ {- 1}$
Average Velocity of the particle is
$\left| \vec {v} _ {a v} \right| = \frac {\text { Displacement }}{\text { Time }} = 0$
Acceleration of the particle over the entire duration of motion is $2 \, ms^{-2}$ .
Q512
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
A man can row a boat with $4 \, kmh^{-1}$ in still water. The man wishes to cross the river of width 4 km having a water current of $2 \, kmhr^{-1}$ . To cross the river with zero drift he swims making at an angle $\alpha$ degree with the current flow taking a time $t_{1}$ minutes to cross the river.
Now he wishes to cross the river in the shortest time $t_{2}$ minutes making an angle $\beta$ degree with the river flow.
Further he takes a time $t_{3}$ minutes to row 2 km upstream and then downstream back to the start point. Assuming all the cases to be independent of each other, the man to start from the river bank from the same point in the first two cases and from the midpoint of the river in the third case, match the quantities in COLUMN-I to the values in COLUMN-II.
Now he wishes to cross the river in the shortest time $t_{2}$ minutes making an angle $\beta$ degree with the river flow.
Further he takes a time $t_{3}$ minutes to row 2 km upstream and then downstream back to the start point. Assuming all the cases to be independent of each other, the man to start from the river bank from the same point in the first two cases and from the midpoint of the river in the third case, match the quantities in COLUMN-I to the values in COLUMN-II.
| Column-I | Column-II | ||||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
|
Correct Answer: A → (t) B → (s) C → (p) D → (q) E → (r)
Explanation:
A → (t)
$\mathrm{B} \rightarrow (\mathrm{s})$
$C \rightarrow (p)$
$\mathrm{D} \rightarrow (\mathrm{q})$
$\mathrm{E} \rightarrow (\mathrm{r})$
Given, that $v_{br} = 4 \, kmhr^{-1}$ and $v_{r} = 2 \, kmhr^{-1}$
$\Rightarrow \quad \alpha = \sin^ {- 1} \left(\frac {v _ {r}}{v _ {b r}}\right) = \sin^ {- 1} \left(\frac {2}{4}\right) = \sin^ {- 1} \left(\frac {1}{2}\right) = 3 0 ^ {\circ}$
Hence, to reach the point directly opposite to starting point he should head the boat at an angle of $30^{\circ}$ with AB or $90^{\circ} + 30^{\circ} = 120^{\circ}$ with the river flow.
Time taken by the boatman to cross the river for zero drift condition is
$t _ {1} = \frac {l}{v _ {b r} \cos \alpha}$
where $l = 4 \, km$ , $v_{br} = 4 \, kmh^{-1}$ and $\alpha = 30^{\circ}$
$\Rightarrow t _ {1} = \frac {4}{4 \cos (3 0 ^ {\circ})} = \frac {2}{\sqrt {3}} \mathrm{hr}$
$\Rightarrow t_{1} = \frac{120}{\sqrt{3}} = 40\sqrt{3}$ minute
For shortest time $\theta=0^{\circ}$ i.e., the man must row the boat perpendicular to the river flow. So $\beta=90^{\circ}$
and $t_{\min}=t_{2}=\frac{l}{v_{br}\cos(0^{\circ})}=\frac{4}{4}=1\mathrm{hr}=60\mathrm{minute}$
Hence, he should head his boat perpendicular to the river current for crossing the river in shortest time and this shortest time $t_{2}$ of 60 minute.
$t = t _ {P \to Q} + t _ {Q \to P}$
$\Rightarrow t = \frac {P Q}{v _ {b r} + v _ {r}} + \frac {Q P}{v _ {b r} + v _ {r}}$
$\Rightarrow t = \frac {2}{4 + 2} + \frac {2}{4 - 2} = \frac {4}{3} \mathrm{hr}$
$\Rightarrow$ t = 80 minute
$\mathrm{B} \rightarrow (\mathrm{s})$
$C \rightarrow (p)$
$\mathrm{D} \rightarrow (\mathrm{q})$
$\mathrm{E} \rightarrow (\mathrm{r})$
Given, that $v_{br} = 4 \, kmhr^{-1}$ and $v_{r} = 2 \, kmhr^{-1}$
$\Rightarrow \quad \alpha = \sin^ {- 1} \left(\frac {v _ {r}}{v _ {b r}}\right) = \sin^ {- 1} \left(\frac {2}{4}\right) = \sin^ {- 1} \left(\frac {1}{2}\right) = 3 0 ^ {\circ}$
Hence, to reach the point directly opposite to starting point he should head the boat at an angle of $30^{\circ}$ with AB or $90^{\circ} + 30^{\circ} = 120^{\circ}$ with the river flow.
Time taken by the boatman to cross the river for zero drift condition is
$t _ {1} = \frac {l}{v _ {b r} \cos \alpha}$
where $l = 4 \, km$ , $v_{br} = 4 \, kmh^{-1}$ and $\alpha = 30^{\circ}$
$\Rightarrow t _ {1} = \frac {4}{4 \cos (3 0 ^ {\circ})} = \frac {2}{\sqrt {3}} \mathrm{hr}$
$\Rightarrow t_{1} = \frac{120}{\sqrt{3}} = 40\sqrt{3}$ minute
For shortest time $\theta=0^{\circ}$ i.e., the man must row the boat perpendicular to the river flow. So $\beta=90^{\circ}$
and $t_{\min}=t_{2}=\frac{l}{v_{br}\cos(0^{\circ})}=\frac{4}{4}=1\mathrm{hr}=60\mathrm{minute}$
Hence, he should head his boat perpendicular to the river current for crossing the river in shortest time and this shortest time $t_{2}$ of 60 minute.
$t = t _ {P \to Q} + t _ {Q \to P}$
$\Rightarrow t = \frac {P Q}{v _ {b r} + v _ {r}} + \frac {Q P}{v _ {b r} + v _ {r}}$
$\Rightarrow t = \frac {2}{4 + 2} + \frac {2}{4 - 2} = \frac {4}{3} \mathrm{hr}$
$\Rightarrow$ t = 80 minute
Q513
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
For one dimensional motion if $v_{av}$ be the average speed, $\vec{v}_{av}$ be the average velocity, $v_{inst}$ be the instantaneous speed, $\vec{v}_{inst}$ be the instantaneous velocity and v be the speed, then match the following
Column-I
Column-I
| Column-I | Column-II | ||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
|
Correct Answer: A → (q) B → (r) C → (p, q) D → (s)
Explanation:
A → (q) B → (r) C → (p, q) D → (s)
Q514
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
Match the following Column-I with Column-II:
| Column-I | Column-II | ||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
|
Correct Answer: A → (r) B → (p) C → (q) D → (p)
Explanation:
A → (r) B → (p) C → (q) D → (p)
Q515
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
The displacement-time graph of a body moving on a straight line is given by
Match Column-I with Column-II:
:
Match Column-I with Column-II:
:
| Column-I | Column-II | ||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
|
Correct Answer: A → (s) B → (r) C → (p) D → (q)
Explanation:
A → (s) B → (r) C → (p) D → (q)
Q516
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
For the velocity-time graph shown in figure, in a time interval from t=0 to t=6 s, match the following
Column-I
Column-I
| Column-I | Column-II | ||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
|
Correct Answer: A → (r) B → (p) C → (r) D → (s)
Explanation:
$\begin{array}{l} \mathrm{A} \to (\mathrm{r}) \\ \mathrm{B} \to (\mathrm{p}) \\ \mathrm{C} \to (\mathrm{r}) \\ \mathrm{D} \to (\mathrm{s}) \end{array}$
$v _ {i} = + 1 0 \mathrm{ms} ^ {- 1} \text {and} v _ {f} = 0$
$\Rightarrow \Delta v = v _ {f} - v _ {f} = - 1 0 \mathrm{ms} ^ {- 1}$
$\Rightarrow a _ {a v} = \frac {\Delta v}{\Delta t} = \frac {- 1 0}{6} = \frac {- 5}{3} \mathrm{ms} ^ {- 2}$
Total displacement = area under v-t graph (with sign) and acceleration = slope of v-t graph.
$v _ {i} = + 1 0 \mathrm{ms} ^ {- 1} \text {and} v _ {f} = 0$
$\Rightarrow \Delta v = v _ {f} - v _ {f} = - 1 0 \mathrm{ms} ^ {- 1}$
$\Rightarrow a _ {a v} = \frac {\Delta v}{\Delta t} = \frac {- 1 0}{6} = \frac {- 5}{3} \mathrm{ms} ^ {- 2}$
Total displacement = area under v-t graph (with sign) and acceleration = slope of v-t graph.
Q517
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
A balloon rises up with constant net acceleration of $10 \, ms^{-2}$ . After 2 s a particle drops from the balloon. After further 2 s match the following (Take $g = 10 \, ms^{-2}$ )
Column-I
Column-I
| Column-I | Column-II | ||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
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Correct Answer: A → (r) B → (p) C → (s) D → (q)
Explanation:
A → (r)
$\begin{array}{l} \mathrm{A} \to (\mathrm{r}) \\ \mathrm{B} \to (\mathrm{p}) \\ \mathrm{C} \to (\mathrm{s}) \\ \mathrm{D} \to (\mathrm{q}) \end{array}$
After 2 s velocity of balloon and hence the velocity of the particle will be $20 \, \text{ms}^{-1} (= \text{at})$ and its height from the ground will be $20 \, m \left( = \frac{1}{2} at^{2} \right)$ . Now g, will start acting on the particle.
$\begin{array}{l} \mathrm{A} \to (\mathrm{r}) \\ \mathrm{B} \to (\mathrm{p}) \\ \mathrm{C} \to (\mathrm{s}) \\ \mathrm{D} \to (\mathrm{q}) \end{array}$
After 2 s velocity of balloon and hence the velocity of the particle will be $20 \, \text{ms}^{-1} (= \text{at})$ and its height from the ground will be $20 \, m \left( = \frac{1}{2} at^{2} \right)$ . Now g, will start acting on the particle.
Q518
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
A body accelerates from rest for time $t_{1}$ at a constant rate $\alpha$ for distance x then it decelerates at constant rate $\beta$ for time $t_{2}$ and covers distance y in this time and come at rest. If all quantities are in SI units, then match the following columns.
Column-I
Column-I
| Column-I | Column-II | ||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
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Correct Answer: A → (p) B → (q) C → (s) D → (r)
Explanation:
A → (p)
$\mathrm{B} \rightarrow (\mathrm{q})$
$C \to (s)$
$\mathrm{D} \rightarrow (\mathrm{r})$
$\frac {v _ {m}}{t _ {1}} = \alpha \mathrm{and} \frac {v _ {m}}{t _ {2}} = \beta$
$\text { Also } x + y = \frac {1}{2} \times v _ {m} \times \left(t _ {1} + t _ {2}\right)$
$\Rightarrow \quad (x + y) = \frac {1}{2} \times v _ {m} \times \left(\frac {v _ {m}}{\alpha} + \frac {v _ {m}}{\beta}\right)$
$\Rightarrow v _ {m} = \sqrt {\frac {2 (x + y) \alpha \beta}{(\alpha + \beta)}}$
Now, $v_{av}=\frac{x+y}{t_{1}+t_{2}}=\frac{x+y}{\frac{v_{m}}{\alpha}+\frac{v_{m}}{\beta}}=\frac{\alpha\beta}{\alpha+\beta}\cdot\frac{x+y}{v_{m}}$
$\mathrm{B} \rightarrow (\mathrm{q})$
$C \to (s)$
$\mathrm{D} \rightarrow (\mathrm{r})$
$\frac {v _ {m}}{t _ {1}} = \alpha \mathrm{and} \frac {v _ {m}}{t _ {2}} = \beta$
$\text { Also } x + y = \frac {1}{2} \times v _ {m} \times \left(t _ {1} + t _ {2}\right)$
$\Rightarrow \quad (x + y) = \frac {1}{2} \times v _ {m} \times \left(\frac {v _ {m}}{\alpha} + \frac {v _ {m}}{\beta}\right)$
$\Rightarrow v _ {m} = \sqrt {\frac {2 (x + y) \alpha \beta}{(\alpha + \beta)}}$
Now, $v_{av}=\frac{x+y}{t_{1}+t_{2}}=\frac{x+y}{\frac{v_{m}}{\alpha}+\frac{v_{m}}{\beta}}=\frac{\alpha\beta}{\alpha+\beta}\cdot\frac{x+y}{v_{m}}$
Q519
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
The equation of one dimensional motion of particle is described in COLUMN-I. At t = 0, particle is at origin and at rest. Match the COLUMN-I with the statements in COLUMN-II.
| Column-I | Column-II | ||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
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Correct Answer: A → (q) B → (p, q) C → (p, r) D → (r, s)
Explanation:
A → (q)
B → (p, q)
C → (p, r)
D → (r, s) $v = \frac{dx}{dt}$ $\Rightarrow a = \frac{dv}{dt}$ $\Rightarrow v = \int_{0}^{t} ad t$ $\Rightarrow x = \int_{0}^{t} v dt$
B → (p, q)
C → (p, r)
D → (r, s) $v = \frac{dx}{dt}$ $\Rightarrow a = \frac{dv}{dt}$ $\Rightarrow v = \int_{0}^{t} ad t$ $\Rightarrow x = \int_{0}^{t} v dt$
Q520
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
v-t graph of a particle moving along positive direction x is shown in figure. Match the items in COLUMN-I with the respective answers in COLUMN-II.
| Column-I | Column-II | ||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
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Correct Answer: A → (r) B → (p) C → (r) D → (r)
Explanation:
A → (r) B → (p) C → (r) D → (r)
Q521
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
Match the v-t graphs in COLUMN-I with the respective a-t graphs in COLUMN-II.
| Column-I | Column-II | ||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
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Correct Answer: A → (r) B → (q) C → (s) D → (p)
Explanation:
A → (r)
B → (q)
C → (s)
D → (p)
A → v = Kt
$⇒ a = \frac{dv}{dt} = K$
$B → v = Kt^2$
$⇒ a = \frac{dv}{dt} = 2Kt$
$C → K_1t → First part of graph$
$⇒ a = \frac{dv}{dt} = K_1$
$⇒ v = K_2 → Second part of graph$
$⇒ a = \frac{dv}{dt} = 0$
$D → v = v_0e^{-Kt}$
$⇒ a = \frac{dv}{dt} = (-v_0K)e^{-Kt}$
$⇒ a = -a_0e^{-Kt}$
$⇒ a_0 = v_0K$
B → (q)
C → (s)
D → (p)
A → v = Kt
$⇒ a = \frac{dv}{dt} = K$
$B → v = Kt^2$
$⇒ a = \frac{dv}{dt} = 2Kt$
$C → K_1t → First part of graph$
$⇒ a = \frac{dv}{dt} = K_1$
$⇒ v = K_2 → Second part of graph$
$⇒ a = \frac{dv}{dt} = 0$
$D → v = v_0e^{-Kt}$
$⇒ a = \frac{dv}{dt} = (-v_0K)e^{-Kt}$
$⇒ a = -a_0e^{-Kt}$
$⇒ a_0 = v_0K$
Q522
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
Let us call a motion, A when velocity is positive and increasing. $A^{-1}$ when velocity is negative and increasing. R when velocity is positive and decreasing and $R^{-1}$ when velocity is negative and decreasing. Now match the following two tables for the given s-t graph
Column-I
Column-I
| Column-I | Column-II | ||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
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Correct Answer: A → (r) B → (s) C → (p) D → (q)
Explanation:
A → (r)
B → (s)
C → (p)
D → (q)
In motion M : slope of s-t graph is positive and increasing. Therefore, velocity of the particle is positive and increasing. Hence, it is A type motion. Similarly, N, P and Q can be observed from the slope.
B → (s)
C → (p)
D → (q)
In motion M : slope of s-t graph is positive and increasing. Therefore, velocity of the particle is positive and increasing. Hence, it is A type motion. Similarly, N, P and Q can be observed from the slope.
Q523
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
In the $s-t$ equation $(s=10+20t-5t^{2})$ match the following
Column-I
Column-I
| Column-I | Column-II | ||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
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Correct Answer: A → (r) B → (q) C → (s) D → (p)
Explanation:
A → (r) B → (q) C → (s) D → (p)
Q524
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
The velocity time graphs for a particle moving along a straight line is given in each situation of COLUMN-I. Match the graph in COLUMN-I with corresponding statements in COLUMN-II.
| Column-I | Column-II | ||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
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Correct Answer: A → (r, s) B → (r, s) C → (p, q, r, t) D → (p, q, r, t)
Explanation:
A → (r, s) B → (r, s) C → (p, q, r, t) D → (p, q, r, t)
Q525
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
Match the statements in COLUMN-I with corresponding graphs in COLUMN-II.
| Column-I | Column-II | ||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
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Correct Answer: A → (q, s) B → (r) C → (p) D → (s)
Explanation:
A → (q, s) B → (r) C → (p) D → (s)