Q901
Advance
LINKED COMPREHENSION TYPE QUESTIONS
MCQ
Comprehension Passage: Comprehension - 26
A man of mass 50 kg is standing on a platform of mass 30 kg kept on a horizontal surface. Mass of block is 160 kg. Distance between block and pulley P is 6 m. There is no friction between block and platform. The man pulls the string with constant force 40 N.
Based on the above facts, answer the following questions.
Length of string pulled by man in 1 s is
Based on the above facts, answer the following questions.
Length of string pulled by man in 1 s is
A.
2 m
B.
6 m
C.
8 m
D.
12 m
Q902
Advance
LINKED COMPREHENSION TYPE QUESTIONS
MCQ
Comprehension Passage: Comprehension - 26
A man of mass 50 kg is standing on a platform of mass 30 kg kept on a horizontal surface. Mass of block is 160 kg. Distance between block and pulley P is 6 m. There is no friction between block and platform. The man pulls the string with constant force 40 N.
Based on the above facts, answer the following questions.
The minimum coefficient between platform and the surface so that it doesn't slip on the surface
Based on the above facts, answer the following questions.
The minimum coefficient between platform and the surface so that it doesn't slip on the surface
A.
0.05
B.
0.1
C.
0.53
D.
None of these
Q903
Advance
LINKED COMPREHENSION TYPE QUESTIONS
MCQ
Comprehension Passage: Comprehension - 27
It is observed that sometimes the frictional force opposing the motion of a body is not constant but increases with speed of the body. One of the familiar example is the air resistance on a freely falling body. For small speeds this resisting force is approximately proportional to the speed. The equations of motion are not directly applied in such cases if variable force. Based on the above facts, answer the following questions.
The variation of the position of a body falling freely through air can be best represented graphically by (consider y = 0 at the highest point)
A.
B.
C.
D.
Q904
Advance
LINKED COMPREHENSION TYPE QUESTIONS
MCQ
Comprehension Passage: Comprehension - 27
It is observed that sometimes the frictional force opposing the motion of a body is not constant but increases with speed of the body. One of the familiar example is the air resistance on a freely falling body. For small speeds this resisting force is approximately proportional to the speed. The equations of motion are not directly applied in such cases if variable force. Based on the above facts, answer the following questions.
A 5 kg mass freely falling experiences a retarding force of 5v where v is the speed in $ms^{-1}$ has
A.
5
B.
10
C.
15
D.
20
Q905
Advance
LINKED COMPREHENSION TYPE QUESTIONS
MCQ
Comprehension Passage: Comprehension - 27
It is observed that sometimes the frictional force opposing the motion of a body is not constant but increases with speed of the body. One of the familiar example is the air resistance on a freely falling body. For small speeds this resisting force is approximately proportional to the speed. The equations of motion are not directly applied in such cases if variable force. Based on the above facts, answer the following questions.
The variation of velocity with time is best represented by
A.
B.
C.
D.
Q906
Advance
LINKED COMPREHENSION TYPE QUESTIONS
MCQ
Comprehension Passage: Comprehension - 28
As shown two block of mass $m_{1}=2$ kg and $m_{2}=4$ kg are connected by an ideal spring of force constant $k=1000\ Nm^{-1}$ and the system is kept on a smooth inclined plane inclined at $\theta=30^{\circ}$ with horizontal. A third block of mass $m_{3}=1$ kg is attached to $m_{2}$ by an ideal string. A force F=15 N is applied on $m_{1}$ and the system is released from rest. Assume that the spring was initially unstretched. $(g=10\ ms^{-2})$ .
Based on the above facts, answer the following questions.
The common acceleration of blocks $m_{1}, m_{2}$ and $m_{3}$ is
Based on the above facts, answer the following questions.
The common acceleration of blocks $m_{1}, m_{2}$ and $m_{3}$ is
A.
$\frac{25}{7} \mathrm{~ms}^{-2}$
B.
$\frac{35}{7} \mathrm{~ms}^{-2}$
C.
$\frac{15}{7} \mathrm{~ms}^{-2}$
D.
$\frac{7}{25} \mathrm{~ms}^{-2}$
Q907
Advance
LINKED COMPREHENSION TYPE QUESTIONS
MCQ
Comprehension Passage: Comprehension - 28
As shown two block of mass $m_{1}=2$ kg and $m_{2}=4$ kg are connected by an ideal spring of force constant $k=1000\ Nm^{-1}$ and the system is kept on a smooth inclined plane inclined at $\theta=30^{\circ}$ with horizontal. A third block of mass $m_{3}=1$ kg is attached to $m_{2}$ by an ideal string. A force F=15 N is applied on $m_{1}$ and the system is released from rest. Assume that the spring was initially unstretched. $(g=10\ ms^{-2})$ .
Based on the above facts, answer the following questions.
Maximum elongation in spring is
Based on the above facts, answer the following questions.
Maximum elongation in spring is
A.
1 cm
B.
2.4 cm
C.
3.6 cm
D.
4.8 cm
Q908
Advance
LINKED COMPREHENSION TYPE QUESTIONS
MCQ
Comprehension Passage: Comprehension - 28
As shown two block of mass $m_{1}=2$ kg and $m_{2}=4$ kg are connected by an ideal spring of force constant $k=1000\ Nm^{-1}$ and the system is kept on a smooth inclined plane inclined at $\theta=30^{\circ}$ with horizontal. A third block of mass $m_{3}=1$ kg is attached to $m_{2}$ by an ideal string. A force F=15 N is applied on $m_{1}$ and the system is released from rest. Assume that the spring was initially unstretched. $(g=10\ ms^{-2})$ .
Based on the above facts, answer the following questions.
Acceleration of $m_{3}$ when spring has maximum elongation is (approx.)
Based on the above facts, answer the following questions.
Acceleration of $m_{3}$ when spring has maximum elongation is (approx.)
A.
$1.2 \, ms^{-2}$
B.
$2 \, ms^{-2}$
C.
$1 \, ms^{-2}$
D.
$1.5 \, ms^{-2}$
Q909
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
In the diagram shown in figure a light string is attached to the mass and is pulled by a constant force $F = 20\sqrt{2}$ N which makes an angle of $45^{\circ}$ with the horizontal. Taking $g = 10 \, ms^{-2}$ , match the quantities in COLUMN-I with their respective match(es) in COLUMN-II.
| Column-I | Column-II | ||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
|
Correct Answer: A → (q) B → (r) C → (s) D → (p)
Explanation:
$\mathrm{A} \rightarrow (\mathrm{q})$
$\mathrm{B} \rightarrow (\mathrm{r})$ $\mathrm{C} \rightarrow (\mathrm{s})$ $\mathrm{D} \rightarrow (\mathrm{p})$ $N + T \sin (4 5 ^ {\circ}) = m g, \text {where} T = 2 0 \sqrt {2} \mathrm{N}$ $\Rightarrow N = 4 0 - 2 0 = 2 0 \mathrm{N}$
So, $f_{s} = \mu_{s}N = 16\mathrm{N}$ and $f_{k} = \mu_{k}N = 12\mathrm{N}$
Since $T\cos(45^{\circ})=20\ N$ which happens to be greater than $f_{s}$ . So, the block will move and hence we have
$\begin{array}{r l} & T \cos (4 5 ^ {\circ}) - f _ {k} = m a \\ \Rightarrow & 2 0 - 1 2 = 4 a \\ \Rightarrow & a = 2 \mathrm{ms} ^ {- 2} = 2 0 0 \mathrm{cms} ^ {- 2} \end{array}$
$\mathrm{B} \rightarrow (\mathrm{r})$ $\mathrm{C} \rightarrow (\mathrm{s})$ $\mathrm{D} \rightarrow (\mathrm{p})$ $N + T \sin (4 5 ^ {\circ}) = m g, \text {where} T = 2 0 \sqrt {2} \mathrm{N}$ $\Rightarrow N = 4 0 - 2 0 = 2 0 \mathrm{N}$
So, $f_{s} = \mu_{s}N = 16\mathrm{N}$ and $f_{k} = \mu_{k}N = 12\mathrm{N}$
Since $T\cos(45^{\circ})=20\ N$ which happens to be greater than $f_{s}$ . So, the block will move and hence we have
$\begin{array}{r l} & T \cos (4 5 ^ {\circ}) - f _ {k} = m a \\ \Rightarrow & 2 0 - 1 2 = 4 a \\ \Rightarrow & a = 2 \mathrm{ms} ^ {- 2} = 2 0 0 \mathrm{cms} ^ {- 2} \end{array}$
Q910
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
In the diagram shown in figure, all pulleys are smooth and massless and strings are light. Match the following. (Take $g = 10 \, ms^{-2}$ ) Column-I and Column-II
| Column-I | Column-II | ||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
|
Correct Answer: A → (p, u) B → (t) C → (r, s) D → (q, r)
Explanation:
$\begin{array}{l} \mathrm {A\to(p,u)} \\ \mathrm {B\to(t)} \\ \mathrm {C\to(r,s)} \\ \mathrm {D\to(q,r)} \end{array}$
Since the pulleys are smooth and massless, so net force on each pulley (even when accelerating) is zero (because mass of each pulley tends to be zero).
Let acceleration of 1 kg, 2 kg, 3 kg and 4 kg blocks be $a_{1}$ , $a_{2}$ , $a_{3}$ and $a_{4}$ respectively. Then drawing the FBD's of all we get the values of $a_{1}$ , $a_{2}$ , $a_{3}$ and $a_{4}$ .
So, $a_1 = 10\mathrm{ms}^{-2}$ (up), $a_2 = 0$ ,
$a _ {3} = \frac {1 0}{3} \mathrm{ms} ^ {- 2} (\mathrm{down}), a _ {4} = 5 \mathrm{ms} ^ {- 2} (\mathrm{down})$
Since the pulleys are smooth and massless, so net force on each pulley (even when accelerating) is zero (because mass of each pulley tends to be zero).
Let acceleration of 1 kg, 2 kg, 3 kg and 4 kg blocks be $a_{1}$ , $a_{2}$ , $a_{3}$ and $a_{4}$ respectively. Then drawing the FBD's of all we get the values of $a_{1}$ , $a_{2}$ , $a_{3}$ and $a_{4}$ .
So, $a_1 = 10\mathrm{ms}^{-2}$ (up), $a_2 = 0$ ,
$a _ {3} = \frac {1 0}{3} \mathrm{ms} ^ {- 2} (\mathrm{down}), a _ {4} = 5 \mathrm{ms} ^ {- 2} (\mathrm{down})$
Q911
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
In the diagram shown in figure, match the following $\left(g = 10\mathrm{ms}^{-2}\right)$ .
and
and
| Column-I | Column-II | ||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
|
Correct Answer: A → (r) B → (s) C → (t) D → (q)
Explanation:
$\mathrm {A\to(r)}$ $\mathrm {B\to(s)}$ $\mathrm {C\to(t)}$ $\mathrm {D\to(q)}$
If a be the acceleration of the system, then
$\begin{array}{l l} & a = \frac {6 0 - 1 8 - (1 + 2 + 3) g \sin (3 0 ^ {\circ})}{1 + 2 + 3} \\ \Rightarrow & a = 2 \mathrm{ms} ^ {- 2} \end{array}$
Net force on the 3 kg block is $F = m_{3}a = 6$ N
$\mathrm{So}, (\mathrm{A}) \rightarrow (\mathrm{r}) \mathrm{and} (\mathrm{B}) \rightarrow (\mathrm{s})$
Let $N_{21}$ be the force on 2 due to 1 and $N_{32}$ be the force on 3 due to 2.
$N _ {2 3} - N _ {2 1} - m _ {2} g \sin \theta = m _ {2} a\tag{... (1}$
$6 0 - N _ {3 2} - m _ {3} g \sin \theta = m _ {3} a\tag{... (2}$
Since $a = 2 \, ms^{-2}$ , so from (2), we get
Substituting the value of $N_{32}$ in equation (1), we get
$\begin{array}{r l} & 3 9 - N _ {2 1} - (2) (1 0) \left(\frac {1}{2}\right) = (2) (2) \\ \Rightarrow & N _ {2 1} = 2 5 \mathrm{N} \end{array}$
If a be the acceleration of the system, then
$\begin{array}{l l} & a = \frac {6 0 - 1 8 - (1 + 2 + 3) g \sin (3 0 ^ {\circ})}{1 + 2 + 3} \\ \Rightarrow & a = 2 \mathrm{ms} ^ {- 2} \end{array}$
Net force on the 3 kg block is $F = m_{3}a = 6$ N
$\mathrm{So}, (\mathrm{A}) \rightarrow (\mathrm{r}) \mathrm{and} (\mathrm{B}) \rightarrow (\mathrm{s})$
Let $N_{21}$ be the force on 2 due to 1 and $N_{32}$ be the force on 3 due to 2.
$N _ {2 3} - N _ {2 1} - m _ {2} g \sin \theta = m _ {2} a\tag{... (1}$
$6 0 - N _ {3 2} - m _ {3} g \sin \theta = m _ {3} a\tag{... (2}$
Since $a = 2 \, ms^{-2}$ , so from (2), we get
Substituting the value of $N_{32}$ in equation (1), we get
$\begin{array}{r l} & 3 9 - N _ {2 1} - (2) (1 0) \left(\frac {1}{2}\right) = (2) (2) \\ \Rightarrow & N _ {2 1} = 2 5 \mathrm{N} \end{array}$
Q912
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
Velocity of five particles A, B, C, D and E varies with time t as, $\vec{v}_{A}=\left(8\hat{t}+\hat{6}\hat{j}\right)\mathrm{ms}^{-1}$ , $\vec{v}_{B}=\left(4\hat{i}+3\hat{j}\right)\mathrm{ms}^{-1}$ , $\vec{v}_{C}=\left(6\hat{i}-4\hat{t}\hat{j}\right)$ , $\vec{v}_{D}=\left(-3\hat{t}+\hat{5}\hat{j}\right)\mathrm{ms}^{-1}$ and $\vec{v}_{E}=\left(2\hat{i}+5\hat{t}\hat{j}\right)\mathrm{ms}^{-1}$ . Match the pseudo force as observed and specified in COLUMN-I to the respective directions given in COLUMN-II.
| Column-I | Column-II | ||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
|
Correct Answer: A → (t) B → (r) C → (p) D → (s) E → (q)
Explanation:
$\begin{array}{l}\mathrm{A} \rightarrow (\mathrm{t})\\\mathrm{B} \rightarrow (\mathrm{r})\\\mathrm{C} \rightarrow (\mathrm{p})\\\mathrm{D} \rightarrow (\mathrm{s})\\\mathrm{E} \rightarrow (\mathrm{q})\end{array}$
Pseudo force comes into play, when the frame of reference (from where the phenomenon is being observed) is accelerating. This pseudo force is then directed opposite to the acceleration of the observation frame. So for the observation frames A, B, C, D and E, respective values of accelerations are
$\begin{array}{l} \vec {a} _ {A} = 8 \hat {i}, \vec {a} _ {B} = \vec {0} \\ \vec {a} _ {C} = - 4 \hat {j}, \vec {a} _ {D} = - 3 \hat {i} \mathrm{and} \\ \vec {a} _ {E} = 5 \hat {j} \end{array}$
Hence, the pseudo force on A as observed by B is zero. The pseudo force on B as observed by C is along +y direction.
The pseudo force on C as observed by D is along +x direction.
The pseudo force on D as observed by E is along -y direction.
The pseudo force on E as observed by A is along -x direction.
Pseudo force comes into play, when the frame of reference (from where the phenomenon is being observed) is accelerating. This pseudo force is then directed opposite to the acceleration of the observation frame. So for the observation frames A, B, C, D and E, respective values of accelerations are
$\begin{array}{l} \vec {a} _ {A} = 8 \hat {i}, \vec {a} _ {B} = \vec {0} \\ \vec {a} _ {C} = - 4 \hat {j}, \vec {a} _ {D} = - 3 \hat {i} \mathrm{and} \\ \vec {a} _ {E} = 5 \hat {j} \end{array}$
Hence, the pseudo force on A as observed by B is zero. The pseudo force on B as observed by C is along +y direction.
The pseudo force on C as observed by D is along +x direction.
The pseudo force on D as observed by E is along -y direction.
The pseudo force on E as observed by A is along -x direction.
Q913
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
In the arrangement shown in figure the coefficient of friction between 1 kg and 2 kg block is $\mu=0.4$ , and that between 2 kg block and group is 0.6. Match the following.
Column-I
Column-I
| Column-I | Column-II | ||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
|
Correct Answer: A → (q) B → (s) C → (p, t) D → (r)
Explanation:
A → (q)
$\mathrm{B} \rightarrow (\mathrm{s})$
$C \to (p, t)$
$\mathrm{D} \rightarrow (\mathrm{r})$
$f _ {1} = \mu_ {1} N _ {1} = (0. 4) (1) (1 0) = 4 \mathrm{N}$
$f _ {2} = \mu_ {2} N _ {2} = (0. 6) (1 + 2) (1 0) = 1 8 \mathrm{N}$
So, net force acting on 1 is 4 N, hence acceleration of 1 is $a_{1}=\frac{f_{1}}{1}=4\ ms^{-2}$ , towards left.
Similarly, net force acting on 2 is $f_{1} + f_{2} = 22$ N.
So, acceleration of 2 is
$a _ {2} = \frac {f _ {1} + f _ {2}}{2} = 1 1 \mathrm{ms} ^ {- 2}, \text { towards right }$
Finally, relative acceleration of one w.r.t. other is $15 \, ms^{-2}$ .
$\mathrm{B} \rightarrow (\mathrm{s})$
$C \to (p, t)$
$\mathrm{D} \rightarrow (\mathrm{r})$
$f _ {1} = \mu_ {1} N _ {1} = (0. 4) (1) (1 0) = 4 \mathrm{N}$
$f _ {2} = \mu_ {2} N _ {2} = (0. 6) (1 + 2) (1 0) = 1 8 \mathrm{N}$
So, net force acting on 1 is 4 N, hence acceleration of 1 is $a_{1}=\frac{f_{1}}{1}=4\ ms^{-2}$ , towards left.
Similarly, net force acting on 2 is $f_{1} + f_{2} = 22$ N.
So, acceleration of 2 is
$a _ {2} = \frac {f _ {1} + f _ {2}}{2} = 1 1 \mathrm{ms} ^ {- 2}, \text { towards right }$
Finally, relative acceleration of one w.r.t. other is $15 \, ms^{-2}$ .
Q914
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
One or more of the entries of COLUMN-II may be the cause of linear motions described in COLUMN-I. Give the appropriate match(es) for the entries of COLUMN-I from COLUMN-II.
| Column-I | Column-II | ||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
|
Correct Answer: A → (r, s) B → (p, s) C → (p, s) D → (p, r, s)
Explanation:
A → (r, s)
$\mathrm{B} \rightarrow (\mathrm{p}, \mathrm{s})$
$\mathrm{C} \rightarrow (\mathrm{p}, \mathrm{s})$
$\mathrm{D} \rightarrow (\mathrm{p}, \mathrm{r}, \mathrm{s})$
For an object moving along +x axis with constant velocity, there may be no net force acting on the object or there may not be any force acting along +x axis. So, (A) → (r, s).
For both the accelerated motions expressed in (B) and (C), we have either the net force on the object to act along +x axis or also no net force along +x axis such that components of force along +x axis happen to be zero. Hence (B) → (p, s) and (C) → (p, s).
Finally, for the motion of object along +x axis (accelerated or non-uniform) all $(p, r, s)$ satisfy the condition. So $(D) \rightarrow (p, r, s)$ .
$\mathrm{B} \rightarrow (\mathrm{p}, \mathrm{s})$
$\mathrm{C} \rightarrow (\mathrm{p}, \mathrm{s})$
$\mathrm{D} \rightarrow (\mathrm{p}, \mathrm{r}, \mathrm{s})$
For an object moving along +x axis with constant velocity, there may be no net force acting on the object or there may not be any force acting along +x axis. So, (A) → (r, s).
For both the accelerated motions expressed in (B) and (C), we have either the net force on the object to act along +x axis or also no net force along +x axis such that components of force along +x axis happen to be zero. Hence (B) → (p, s) and (C) → (p, s).
Finally, for the motion of object along +x axis (accelerated or non-uniform) all $(p, r, s)$ satisfy the condition. So $(D) \rightarrow (p, r, s)$ .
Q915
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
Two blocks having masses $m_{1}$ and $m_{2}$ are placed on smooth horizontal surfaces (as shown) for the different horizontal combination of the respective forces $F_{1}$ and $F_{2}$ applied on $m_{1}$ and $m_{2}$ . Further it is given that $m_{1}F_{2} > m_{2}F_{1}$ , $a_{1}$ and $a_{2}$ are the accelerations of $m_{1}$ and $m_{2}$ if $F_{1}$ and $F_{2}$ would have acted on them separately.
CASE-I
CASE-II
CASE-III
CASE-IV
If $T_{1}$ , $T_{2}$ , $N_{1}$ and $N_{2}$ be the respective tensions and the normal reactions in Case-I, II, III and IV, then match $T_{1}$ , $T_{2}$ , $N_{1}$ and $N_{2}$ from COLUMN-I to their values in COLUMN-II.
CASE-I
CASE-II
CASE-III
CASE-IV
If $T_{1}$ , $T_{2}$ , $N_{1}$ and $N_{2}$ be the respective tensions and the normal reactions in Case-I, II, III and IV, then match $T_{1}$ , $T_{2}$ , $N_{1}$ and $N_{2}$ from COLUMN-I to their values in COLUMN-II.
| Column-I | Column-II | ||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
|
Correct Answer: A → (q, r) B → (p, s) C → (p, s) D → (q, r)
Explanation:
A → (q, r)
$\mathrm{B} \rightarrow (\mathrm{p}, \mathrm{s})$
$C \to (p, s)$
$\mathrm{D} \rightarrow (\mathrm{q}, \mathrm{r})$
As per the question, we are given that
$a _ {1} = \frac {F _ {1}}{m _ {1}} \mathrm{and} a _ {2} = \frac {F _ {2}}{m _ {2}}$
For CASE-1 and 4:
If a be the combined acceleration of the system, then
$T _ {1} - F _ {1} = m _ {1} a \mathrm{and} F _ {2} - T _ {1} = m _ {2} a$
$\Rightarrow a = \frac {F _ {2} - F _ {1}}{m _ {1} + m _ {2}}$
Substituting in either of the equations, we get
$T _ {1} = \frac {m _ {1} m _ {2}}{m _ {1} + m _ {2}} \left(\frac {F _ {1}}{m _ {1}} + \frac {F _ {2}}{m _ {2}}\right) = \frac {m _ {1} m _ {2}}{m _ {1} + m _ {2}} \left(a _ {1} + a _ {2}\right)$
$\mathrm{So}, (\mathrm{A}) \rightarrow (\mathrm{q}, \mathrm{r})$
Also,
$N _ {2} = \frac {m _ {1} m _ {2}}{m _ {1} + m _ {2}} \left(\frac {F _ {1}}{m _ {1}} + \frac {F _ {2}}{m _ {2}}\right) = \frac {m _ {1} m _ {2}}{m _ {1} + m _ {2}} \left(a _ {1} + a _ {2}\right)$
$\mathrm{So}, (\mathrm{D}) \rightarrow (\mathrm{q}, \mathrm{r})$
For CASE-2 and 3:
If $a$ be the combined acceleration of the system, then
$F_{1} + T_{2} = m_{1}a$ and $F_{2} - T_{2} = m_{2}a$
$\Rightarrow a = \frac {F _ {1} + F _ {2}}{m _ {1} + m _ {2}}$
Substituting in either of the equations, we get
$T _ {2} = \frac {m _ {1} m _ {2}}{m _ {1} + m _ {2}} \left(\frac {F _ {2}}{m _ {2}} - \frac {F _ {1}}{m _ {1}}\right) = \frac {m _ {1} m _ {2}}{m _ {1} + m _ {2}} \left(a _ {2} - a _ {1}\right)$
$\mathrm{So}, (\mathrm{C}) \rightarrow (\mathrm{p}, \mathrm{s})$
Also,
$N _ {1} = \frac {m _ {1} m _ {2}}{m _ {1} + m _ {2}} \left(\frac {F _ {2}}{m _ {2}} - \frac {F _ {1}}{m _ {1}}\right) = \frac {m _ {1} m _ {2}}{m _ {1} + m _ {2}} \left(a _ {2} - a _ {1}\right)$
$\mathrm{So}, (\mathrm{B}) \rightarrow (\mathrm{p}, \mathrm{s})$
$\mathrm{B} \rightarrow (\mathrm{p}, \mathrm{s})$
$C \to (p, s)$
$\mathrm{D} \rightarrow (\mathrm{q}, \mathrm{r})$
As per the question, we are given that
$a _ {1} = \frac {F _ {1}}{m _ {1}} \mathrm{and} a _ {2} = \frac {F _ {2}}{m _ {2}}$
For CASE-1 and 4:
If a be the combined acceleration of the system, then
$T _ {1} - F _ {1} = m _ {1} a \mathrm{and} F _ {2} - T _ {1} = m _ {2} a$
$\Rightarrow a = \frac {F _ {2} - F _ {1}}{m _ {1} + m _ {2}}$
Substituting in either of the equations, we get
$T _ {1} = \frac {m _ {1} m _ {2}}{m _ {1} + m _ {2}} \left(\frac {F _ {1}}{m _ {1}} + \frac {F _ {2}}{m _ {2}}\right) = \frac {m _ {1} m _ {2}}{m _ {1} + m _ {2}} \left(a _ {1} + a _ {2}\right)$
$\mathrm{So}, (\mathrm{A}) \rightarrow (\mathrm{q}, \mathrm{r})$
Also,
$N _ {2} = \frac {m _ {1} m _ {2}}{m _ {1} + m _ {2}} \left(\frac {F _ {1}}{m _ {1}} + \frac {F _ {2}}{m _ {2}}\right) = \frac {m _ {1} m _ {2}}{m _ {1} + m _ {2}} \left(a _ {1} + a _ {2}\right)$
$\mathrm{So}, (\mathrm{D}) \rightarrow (\mathrm{q}, \mathrm{r})$
For CASE-2 and 3:
If $a$ be the combined acceleration of the system, then
$F_{1} + T_{2} = m_{1}a$ and $F_{2} - T_{2} = m_{2}a$
$\Rightarrow a = \frac {F _ {1} + F _ {2}}{m _ {1} + m _ {2}}$
Substituting in either of the equations, we get
$T _ {2} = \frac {m _ {1} m _ {2}}{m _ {1} + m _ {2}} \left(\frac {F _ {2}}{m _ {2}} - \frac {F _ {1}}{m _ {1}}\right) = \frac {m _ {1} m _ {2}}{m _ {1} + m _ {2}} \left(a _ {2} - a _ {1}\right)$
$\mathrm{So}, (\mathrm{C}) \rightarrow (\mathrm{p}, \mathrm{s})$
Also,
$N _ {1} = \frac {m _ {1} m _ {2}}{m _ {1} + m _ {2}} \left(\frac {F _ {2}}{m _ {2}} - \frac {F _ {1}}{m _ {1}}\right) = \frac {m _ {1} m _ {2}}{m _ {1} + m _ {2}} \left(a _ {2} - a _ {1}\right)$
$\mathrm{So}, (\mathrm{B}) \rightarrow (\mathrm{p}, \mathrm{s})$
Q916
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
A block of mass $m = 5 \, \text{kg}$ is moving right with velocity $v_0 = 33 \, \text{ms}^{-1}$ on a rough surface. A force $F = 20\sqrt{2}$ is applied as shown ( $g = 10 \, \text{ms}^{-2}$ )
Column-I
Column-I
| Column-I | Column-II | ||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
|
Correct Answer: A → (r) B → (t) C → (q) D → (p)
Explanation:
A → (r)
B → (t)
C → (q)
D → (p)
$N=mg+F\sin(45^\circ)$
$\Rightarrow N=50+(20\sqrt{2})\frac{1}{\sqrt{2}}=70\ \text{N}$
$\Rightarrow f_{\max}=\mu N=(0.5)(70)=35\ \text{N}$
So, retardation given to the block is
$a=\frac{F\cos(45^\circ)+f}{m}=\frac{35+20}{5}=11\ \text{m s}^{-2}$
Since $v=u+at$, so the block attains zero velocity at time $t_0$, we have
$0=33-11t_0$
$\Rightarrow t_0=3\ \text{s}$
Till $t < t_0,(=3\ \text{s})$, we have the following two facts:
a) The force of friction acting on the block is
$f_{\max}=\mu N_1=35\ \text{N}$
b) The net retarding force acting on the block is
$F_{\text{net, retarding}}=F\cos(45^\circ)+f_{\max}=55\ \text{N}$
So, we have A) $\rightarrow$ r) and C) $\rightarrow$ q).
Now, for $t=4.5\ \text{s}$, i.e., $t > t_0$, since the block has zero velocity at $t_0=3\ \text{s}$. At this very instant the friction will oppose the applied force and happens to be more than the applied force, so the block will remain at rest after $t_0$.
At $t=4.5\ \text{s}$, the applied force is
$F\cos(45^\circ)=20\ \text{N}$
Hence,
$f_{\text{static}}=F_{\text{applied}}=F\cos(45^\circ)=20\ \text{N}$
Hence, (B) $\rightarrow$ (t).
Also, at $t=4.5\ \text{s}$, the net force on the block is zero as it happens to be in static mode. Hence, (D) $\rightarrow$ (p).
B → (t)
C → (q)
D → (p)
$N=mg+F\sin(45^\circ)$
$\Rightarrow N=50+(20\sqrt{2})\frac{1}{\sqrt{2}}=70\ \text{N}$
$\Rightarrow f_{\max}=\mu N=(0.5)(70)=35\ \text{N}$
So, retardation given to the block is
$a=\frac{F\cos(45^\circ)+f}{m}=\frac{35+20}{5}=11\ \text{m s}^{-2}$
Since $v=u+at$, so the block attains zero velocity at time $t_0$, we have
$0=33-11t_0$
$\Rightarrow t_0=3\ \text{s}$
Till $t < t_0,(=3\ \text{s})$, we have the following two facts:
a) The force of friction acting on the block is
$f_{\max}=\mu N_1=35\ \text{N}$
b) The net retarding force acting on the block is
$F_{\text{net, retarding}}=F\cos(45^\circ)+f_{\max}=55\ \text{N}$
So, we have A) $\rightarrow$ r) and C) $\rightarrow$ q).
Now, for $t=4.5\ \text{s}$, i.e., $t > t_0$, since the block has zero velocity at $t_0=3\ \text{s}$. At this very instant the friction will oppose the applied force and happens to be more than the applied force, so the block will remain at rest after $t_0$.
At $t=4.5\ \text{s}$, the applied force is
$F\cos(45^\circ)=20\ \text{N}$
Hence,
$f_{\text{static}}=F_{\text{applied}}=F\cos(45^\circ)=20\ \text{N}$
Hence, (B) $\rightarrow$ (t).
Also, at $t=4.5\ \text{s}$, the net force on the block is zero as it happens to be in static mode. Hence, (D) $\rightarrow$ (p).
Q917
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
In the arrangement shown, a block A of mass $m_{A} = 1 \, kg$ is sliding on a block B of mass $m_{B} = 4 \, kg$ . Friction is absent between all surfaces in contact. Based on this information and $g = 10 \, ms^{-2}$ , match the items of COLUMN-I with their respective values in COLUMN-II.
| Column-I | Column-II | ||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
|
Correct Answer: A → (r) B → (t) C → (p) D → (q)
Explanation:
A → R
B → T
C → P
D → Q
For A, $T-m_Ag\sin\theta=m_Aa$ … (1)
For B, $m_Bg\sin\theta-T=m_Ba$ … (2)
Add 1 and 2, we get
$(m_B-m_A)g\sin\theta=(m_A+m_B)a$
$\Rightarrow a=\left(\frac{m_B-m_A}{m_A+m_B}\right)\sin\theta$
$\Rightarrow a=\left(\frac{4-1}{4+1}\right)(10)\left(\frac{1}{2}\right)$
$\Rightarrow a=3\ \text{m s}^{-2}$
So, A → P
Substitute the value of $a$ in (1), we get
$T-(10)\left(\frac{1}{2}\right)=(1)(3)$
$\Rightarrow T=8\ \text{N}$
So, Thrust $=2T=16\ \text{N}$, i.e., B → (T)
Force exerted by A on B is
$N_A=m_Ag\cos\theta=5\sqrt{3}\ \text{N}$
and force exerted by B on incline is
$N_B=(m_A+m_B)g\cos\theta=25\sqrt{3}\ \text{N}$
Hence,C → P and D → Q
B → T
C → P
D → Q
For A, $T-m_Ag\sin\theta=m_Aa$ … (1)
For B, $m_Bg\sin\theta-T=m_Ba$ … (2)
Add 1 and 2, we get
$(m_B-m_A)g\sin\theta=(m_A+m_B)a$
$\Rightarrow a=\left(\frac{m_B-m_A}{m_A+m_B}\right)\sin\theta$
$\Rightarrow a=\left(\frac{4-1}{4+1}\right)(10)\left(\frac{1}{2}\right)$
$\Rightarrow a=3\ \text{m s}^{-2}$
So, A → P
Substitute the value of $a$ in (1), we get
$T-(10)\left(\frac{1}{2}\right)=(1)(3)$
$\Rightarrow T=8\ \text{N}$
So, Thrust $=2T=16\ \text{N}$, i.e., B → (T)
Force exerted by A on B is
$N_A=m_Ag\cos\theta=5\sqrt{3}\ \text{N}$
and force exerted by B on incline is
$N_B=(m_A+m_B)g\cos\theta=25\sqrt{3}\ \text{N}$
Hence,C → P and D → Q
Q918
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
Configuration of L shaped block B of mass 6 kg and two other blocks are shown. System is set free to move. At any moment, match the following parameters. $\left(g = 10 \, \text{ms}^{-2}\right)$ .
Column-I
Column-I
| Column-I | Column-II | ||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
|
A.
B.
C.
D.
Q919
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
A block A of mass 10 kg lies on a block B of mass 25 kg resting on a smooth horizontal surface. The coefficient of static friction between A and B is $\mu = 0.5$ . A force F is applied on the block A, as shown.
Column-I
Column-I
| Column-I | Column-II | ||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
|
Correct Answer: A → (t, u) B → (p, u) C → (q, u) D → (s, u)
Explanation:
A → (t, u)
$\mathrm{B} \rightarrow (\mathrm{p}, \mathrm{u})$
$C \to (q, u)$
$\mathrm{D} \rightarrow (\mathrm{s}, \mathrm{u})$
Since $f_{A}=\mu m_{A}g=(0.5)(10)(10)=50\ N$
So, friction on B due to A is 50 N, towards right Hence (A) → (t, u)
Now, when $F = 35 \, N < f_{A}$ , so no slipping of A on B and since B lies on a frictionless surface, hence both A and B move together with a common acceleration a (say). Then
$a = \frac{F}{m_A + m_B} = \frac{35}{35} = 1\mathrm{ms}^{-2}$ , towards right
So, we get (B) → (p, u)
Now, when $F = 250 \mathrm{~N} > 50 \mathrm{~N}$ , then for block $A$
$F - \mu m _ {A} g = m _ {A} a _ {A}$
$\Rightarrow 2 5 0 - 5 0 = (1 0) a _ {A}$
$\Rightarrow a_{A}=20\ ms^{-2}$ , towards right
$\mathrm{So}, (\mathrm{C}) \rightarrow (\mathrm{q}, \mathrm{u})$
Also, for $B$ , we get
$f = m _ {B} a _ {B}$
$\Rightarrow 5 0 = 2 5 a _ {B}$
$\Rightarrow a_{B}=2\ ms^{-2}$ , towards right
Hence, finally (D) → (s, u)
$\mathrm{B} \rightarrow (\mathrm{p}, \mathrm{u})$
$C \to (q, u)$
$\mathrm{D} \rightarrow (\mathrm{s}, \mathrm{u})$
Since $f_{A}=\mu m_{A}g=(0.5)(10)(10)=50\ N$
So, friction on B due to A is 50 N, towards right Hence (A) → (t, u)
Now, when $F = 35 \, N < f_{A}$ , so no slipping of A on B and since B lies on a frictionless surface, hence both A and B move together with a common acceleration a (say). Then
$a = \frac{F}{m_A + m_B} = \frac{35}{35} = 1\mathrm{ms}^{-2}$ , towards right
So, we get (B) → (p, u)
Now, when $F = 250 \mathrm{~N} > 50 \mathrm{~N}$ , then for block $A$
$F - \mu m _ {A} g = m _ {A} a _ {A}$
$\Rightarrow 2 5 0 - 5 0 = (1 0) a _ {A}$
$\Rightarrow a_{A}=20\ ms^{-2}$ , towards right
$\mathrm{So}, (\mathrm{C}) \rightarrow (\mathrm{q}, \mathrm{u})$
Also, for $B$ , we get
$f = m _ {B} a _ {B}$
$\Rightarrow 5 0 = 2 5 a _ {B}$
$\Rightarrow a_{B}=2\ ms^{-2}$ , towards right
Hence, finally (D) → (s, u)
Q920
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
If the net force acting on a system is represented by $\vec{F}$ and its momentum is $\vec{p}$ , then match the quantities in COLUMN-I with the quantities in COLUMN-II.
| Column-I | Column-II | ||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
|
Correct Answer: A → (p, q, r) B → (p, q, r) C → (p, q) D → (s)
Explanation:
A → (p, q, r)
$\mathrm{B} \rightarrow (\mathrm{p}, \mathrm{q}, \mathrm{r})$
$C \to (p, q)$
$\mathrm{D} \rightarrow (\mathrm{s})$
The directions of $\vec{F}$ and $\vec{p}$ may be different, so $\vec{p}$ may change its direction.
If direction of $\vec{F}$ and $\vec{p}$ are the same, then direction of $\vec{p}$ would not change.
If $\vec{F}$ is non-zero in magnitude then magnitude of $\vec{p}$ will change.
$\mathrm{B} \rightarrow (\mathrm{p}, \mathrm{q}, \mathrm{r})$
$C \to (p, q)$
$\mathrm{D} \rightarrow (\mathrm{s})$
The directions of $\vec{F}$ and $\vec{p}$ may be different, so $\vec{p}$ may change its direction.
If direction of $\vec{F}$ and $\vec{p}$ are the same, then direction of $\vec{p}$ would not change.
If $\vec{F}$ is non-zero in magnitude then magnitude of $\vec{p}$ will change.
Q921
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
ABCD is a horizontal surface on which a block of mass m = 20 kg is placed and coefficient of friction between this body and surface is $\mu = 0.5$ as shown. y-axis is vertical axis and x-z plane is horizontal. The surface starts moving with an acceleration which depends on time as $\vec{a} = (3t\hat{i} + 10\hat{j} + 4t\hat{k}) \, \text{ms}^{-2}$ .
Assuming that at t=0, the block is at rest with respect to the surface. If, t is time when block starts slipping, a is the acceleration of surface when block starts slipping. $f_{l}$ is limiting friction when block starts slipping, F is total force applied by surface on block at t=1 s, then match the quantities in COLUMN-I with their respective values in COLUMN-II.
Assuming that at t=0, the block is at rest with respect to the surface. If, t is time when block starts slipping, a is the acceleration of surface when block starts slipping. $f_{l}$ is limiting friction when block starts slipping, F is total force applied by surface on block at t=1 s, then match the quantities in COLUMN-I with their respective values in COLUMN-II.
| Column-I | Column-II | ||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
|
Correct Answer: A → (s) B → (r) C → (p) D → (q)
Explanation:
A → (s)
$\mathrm{B} \rightarrow (\mathrm{r})$
$\mathrm{C} \rightarrow (\mathrm{p})$
$\mathrm{D} \rightarrow (\mathrm{q})$
Since $N = m(g + a_y) = 20(10 + 10) = 400 \, \mathrm{N}$
and $F_{\mathrm{pseudo}} = m\sqrt{(3t)^2 + (4t)^2} = 100t$
$\Rightarrow f _ {\ell} = \mu N = 0. 5 \times 4 0 0 = 2 0 0 \mathrm{N}$
Slipping starts when, 100t = 200
$\Rightarrow t = 2 \mathrm{s}$
So, $\vec{a}$ at $t = 2$ s is $\vec{a} = (6\hat{i} +10\hat{j} +8\hat{k})\mathrm{ms}^{-2}$
$\Rightarrow | \vec {a} | = 1 0 \sqrt {2} \mathrm{ms} ^ {- 2}$
Total force F applied by the surface on the block at $t = 1 \, s$ is $\sqrt{N^{2} + f_{l}^{2}}$ (also called as Contact Force (C.F.)) $\Rightarrow$ Contact Force = $\sqrt{(400)^{2} + (100)^{2}}$ $\Rightarrow$ C.F. = $100\sqrt{17}$ N
$\mathrm{B} \rightarrow (\mathrm{r})$
$\mathrm{C} \rightarrow (\mathrm{p})$
$\mathrm{D} \rightarrow (\mathrm{q})$
Since $N = m(g + a_y) = 20(10 + 10) = 400 \, \mathrm{N}$
and $F_{\mathrm{pseudo}} = m\sqrt{(3t)^2 + (4t)^2} = 100t$
$\Rightarrow f _ {\ell} = \mu N = 0. 5 \times 4 0 0 = 2 0 0 \mathrm{N}$
Slipping starts when, 100t = 200
$\Rightarrow t = 2 \mathrm{s}$
So, $\vec{a}$ at $t = 2$ s is $\vec{a} = (6\hat{i} +10\hat{j} +8\hat{k})\mathrm{ms}^{-2}$
$\Rightarrow | \vec {a} | = 1 0 \sqrt {2} \mathrm{ms} ^ {- 2}$
Total force F applied by the surface on the block at $t = 1 \, s$ is $\sqrt{N^{2} + f_{l}^{2}}$ (also called as Contact Force (C.F.)) $\Rightarrow$ Contact Force = $\sqrt{(400)^{2} + (100)^{2}}$ $\Rightarrow$ C.F. = $100\sqrt{17}$ N
Q922
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
Match the force or set of forces in the left hand column to a characteristic of force or set of forces in the right-hand column.
Match COLUMN-I (Force) with COLUMN-II (Characteristic)
Match COLUMN-I (Force) with COLUMN-II (Characteristic)
| Column-I | Column-II | ||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
|
Correct Answer: A → (p, s, t) B → (p, q, t) C → (p, t) D → (r)
Explanation:
A → (p, s, t) B → (p, q, t) C → (p, t) D → (r)
Q923
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
In the arrangement shown pulley is ideal and string is massless. 5 kg block is moving on the horizontal which is smooth. When $\theta$ is $37^{\circ}$ acceleration of 5 kg and 10 kg blocks are $a_{1}$ and $a_{2}$ . Tension in string connecting blocks is T and the total force applied by this string on pulley is F. Taking, $g = 10 \, ms^{-2}$ , match the quantities in COLUMN-I to their respective values (in SI units) in COLUMN-II.
| Column-I | Column-II | ||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
|
Correct Answer: A → (r) B → (p) C → (q) D → (s)
Explanation:
$\mathrm{A}\rightarrow (\mathrm{r})$
$\mathrm{B} \rightarrow (\mathrm{p})$
$\mathrm{C} \rightarrow (\mathrm{q})$
$\mathrm{D} \rightarrow (\mathrm{s})$
$a _ {1} \cos (3 7 ^ {\circ}) = a _ {2}$
$\Rightarrow 4 a _ {1} = 5 a _ {2}\tag{... (1}$
$\mathrm{Since,} 1 0 0 - T = 1 0 a _ {2}\tag{... (2}$
and $T \cos(37^{\circ}) = 5a_{1}$
$\Rightarrow T = \frac {2 5}{4} a _ {1}\tag{... (3}$
Force on pulley by the string is
$F = \sqrt {T ^ {2} + T ^ {2} + 2 T ^ {2} \cos (5 3 ^ {\circ})}$
$\mathrm{B} \rightarrow (\mathrm{p})$
$\mathrm{C} \rightarrow (\mathrm{q})$
$\mathrm{D} \rightarrow (\mathrm{s})$
$a _ {1} \cos (3 7 ^ {\circ}) = a _ {2}$
$\Rightarrow 4 a _ {1} = 5 a _ {2}\tag{... (1}$
$\mathrm{Since,} 1 0 0 - T = 1 0 a _ {2}\tag{... (2}$
and $T \cos(37^{\circ}) = 5a_{1}$
$\Rightarrow T = \frac {2 5}{4} a _ {1}\tag{... (3}$
Force on pulley by the string is
$F = \sqrt {T ^ {2} + T ^ {2} + 2 T ^ {2} \cos (5 3 ^ {\circ})}$
Q924
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
A metal sphere is hung by a string fixed to a wall. The forces acting on the sphere are shown. Match the following:
Column-I
Column-I
| Column-I | Column-II | ||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
|
Correct Answer: A → (p) B → (r) C → (q) D → (s)
Explanation:
A → (p)
$\mathrm{B} \rightarrow (\mathrm{r})$
$\mathrm{C} \rightarrow (\mathrm{q})$
$\mathrm{D} \rightarrow (\mathrm{s})$
Since the system is in equilibrium, so
$\vec {T} + \vec {W} + \vec {N} = 0$
$\Rightarrow \vec {T} = - \vec {W} - \vec {N}$
Also, $T\cos \theta = W$ and $T\sin \theta = N$
$\Rightarrow T^{2} = N^{2} + W^{2}$ and $N = W\tan \theta$
$\mathrm{B} \rightarrow (\mathrm{r})$
$\mathrm{C} \rightarrow (\mathrm{q})$
$\mathrm{D} \rightarrow (\mathrm{s})$
Since the system is in equilibrium, so
$\vec {T} + \vec {W} + \vec {N} = 0$
$\Rightarrow \vec {T} = - \vec {W} - \vec {N}$
Also, $T\cos \theta = W$ and $T\sin \theta = N$
$\Rightarrow T^{2} = N^{2} + W^{2}$ and $N = W\tan \theta$
Q925
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
A block A of mass 5 kg is placed on another block B of mass 10 kg placed on a horizontal rough surface. Horizontal forces $F_{1}$ and $F_{2}$ act on blocks A and B, respectively as shown in figure. Coefficient of friction between two blocks is 0.8 while between block B and horizontal surface is 0.2. Let $f_{1}$ denote friction force acting on block B due to A and $f_{2}$ denote friction force on horizontal surface. Then match the following.
Column-I
Column-I
| Column-I | Column-II | ||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
|
Correct Answer: A → (q, s) B → (p, s) C → (r) D → (r)
Explanation:
A → (q, s)
B → (p, s)
C → (r)
D → (r) $\left( f_1 \right)_{\text{max}} = \left( f_1 \right)_{\text{lim}} = (0.8)(50) = 40 \text{ N}$ $\left( f_2 \right)_{\text{max}} = \left( f_2 \right)_{\text{lim}} = (0.2)(150) = 30 \text{ N}$ $F_1 = 35 \text{ N}, F_2 = 0: \text{Block } A \text{ will not slip over block } B, \text{ because } F_{\text{app}} < \left( f_1 \right)_{\text{lim}}$ $F_1 = 50 \text{ N}, F_2 = 0: \text{Block } B \text{ will slip over horizontal surface, because } F_{\text{app}} > \left( f_1 \right)_{\text{lim}}$ $F_1 = 0, F_2 = 30 \text{ N}: \text{Block } B \text{ will be at rest, because } F_{\text{app}} = \left( f_2 \right)_{\text{lim}}$ $F_1 = 10 \text{ N}, F_2 = 50 \text{ N}: \text{Block } A \text{ have no tendency of slipping over block } B.$
B → (p, s)
C → (r)
D → (r) $\left( f_1 \right)_{\text{max}} = \left( f_1 \right)_{\text{lim}} = (0.8)(50) = 40 \text{ N}$ $\left( f_2 \right)_{\text{max}} = \left( f_2 \right)_{\text{lim}} = (0.2)(150) = 30 \text{ N}$ $F_1 = 35 \text{ N}, F_2 = 0: \text{Block } A \text{ will not slip over block } B, \text{ because } F_{\text{app}} < \left( f_1 \right)_{\text{lim}}$ $F_1 = 50 \text{ N}, F_2 = 0: \text{Block } B \text{ will slip over horizontal surface, because } F_{\text{app}} > \left( f_1 \right)_{\text{lim}}$ $F_1 = 0, F_2 = 30 \text{ N}: \text{Block } B \text{ will be at rest, because } F_{\text{app}} = \left( f_2 \right)_{\text{lim}}$ $F_1 = 10 \text{ N}, F_2 = 50 \text{ N}: \text{Block } A \text{ have no tendency of slipping over block } B.$