Q401
Allen
COMPREHENSION TYPE QUESTIONS
MCQ
Comprehension Passage: PARAGRAPH FOR QUESTION NO. 76 TO 81
The maximum F which will not cause motion of any of the blocks is
A.
10 N
B.
15 N
C.
data insufficient
D.
None
Q402
Allen
COMPREHENSION TYPE QUESTIONS
MCQ
Comprehension Passage: PARAGRAPH FOR QUESTION NO. 76 TO 81
The maximum acceleration of 5 kg block is :-
A.
$1m / s^2$
B.
$3m / s^2$
C.
0
D.
None
Q403
Allen
COMPREHENSION TYPE QUESTIONS
MCQ
Comprehension Passage: PARAGRAPH FOR QUESTION NO. 76 TO 81
The acceleration of 10 kg block when F = 30 N is
A.
$2 \, m/s^{2}$
B.
$3 \, m/s^{2}$
C.
$1 \, m/s^{2}$
D.
None
Q404
Allen
COMPREHENSION TYPE QUESTIONS
MCQ
Comprehension Passage: PARAGRAPH FOR QUESTION NO. 76 TO 81
A block $A$ is placed over a long rough plank $B$ of same mass as shown in figure. The plank is placed over a smooth horizontal surface. At time $t = 0$ , block $A$ is given a velocity $v_{0}$ in horizontal direction. Let $v_{1}$ and $v_{2}$ be the velocities of $A$ and $B$ at time $t$ . Then choose the correct graph between $v_{1}$ or $v_{2}$ and $t$ .
A.
B.
C.
D.
Q405
Allen
COMPREHENSION TYPE QUESTIONS
MCQ
Comprehension Passage: PARAGRAPH FOR QUESTION NO. 82 AND 83
Given graph is a plot of magnitude of normal, limiting friction, kinetic friction and static friction versus angle of inclination $\theta$ for the coin lying on book. 17 points are labeled with numbers. The coin slips at a certain angle $\theta_{c}$ . The coefficient of static friction is 0.6, and the coefficient of kinetic friction is 0.4.
The magnitude of the normal force on the coin from the book,
The magnitude of the normal force on the coin from the book,
A.
6
B.
2
C.
7
D.
11
Q406
Allen
COMPREHENSION TYPE QUESTIONS
MCQ
Comprehension Passage: PARAGRAPH FOR QUESTION NO. 82 AND 83
Given graph is a plot of magnitude of normal, limiting friction, kinetic friction and static friction versus angle of inclination $\theta$ for the coin lying on book. 17 points are labeled with numbers. The coin slips at a certain angle $\theta_{c}$ . The coefficient of static friction is 0.6, and the coefficient of kinetic friction is 0.4.
$\theta_{c}$ is given by point
$\theta_{c}$ is given by point
A.
14
B.
15
C.
16
D.
17
Q407
Allen
COMPREHENSION TYPE QUESTIONS
Match the Columns
A sphere of mass 10 kg is placed in equilibrium in a V shaped groove plane made of two smooth surfaces 1 and 2 as shown in figure. $(g = 10 \, ms^{-2})$
| Column-I | Column-II | ||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
|
Correct Answer: (A-R), (B-Q), (C-S), (D-P)
Explanation:
Ans. (A-R), (B-Q), (C-S), (D-P)
$N_{2}\cos 37^{\circ} = N_{1}\cos 53\quad N_{2}\sin 37 + N_{1}\cos 53 = 100$
$N_{2} \times \frac{4}{5} = N_{1} \times \frac{3}{5}$
$N_{2} \times \frac{3}{5} + N_{1} \times \frac{4}{5} = 100$
$4N_{2} = 3N_{1}$
$3N_{2} + 4N_{1} = 500$
$N_{2}\cos 37^{\circ} = N_{1}\cos 53\quad N_{2}\sin 37 + N_{1}\cos 53 = 100$
$N_{2} \times \frac{4}{5} = N_{1} \times \frac{3}{5}$
$N_{2} \times \frac{3}{5} + N_{1} \times \frac{4}{5} = 100$
$4N_{2} = 3N_{1}$
$3N_{2} + 4N_{1} = 500$
Q408
Allen
COMPREHENSION TYPE QUESTIONS
Match the Columns
Match the situations in column I to the accelerations of blocks in the column II (acceleration due to gravity is g and F is an additional force applied to one of the blocks?
| Column-I | Column-II | ||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
|
Correct Answer: (A-R), (B-T), (C-R), (D-P)
Explanation:
Ans. (A-R), (B-T), (C-R), (D-P)
A)
C)
D)
$mg + mg - T = ma$
3mg - T = 3ma
$T - mg = ma$
$T - mg = ma$
$mg - T_{1} = ma$
$T_{1} + 2mg - T_{2} = 2ma$
mg = 2ma
2mg = 4ma
$T_{2} - 2mg = 2ma$
$a = \frac{g}{2}$
$a = \frac{g}{2}$ mg = 5ma
$\frac {g}{5} = a$
A)
C)
D)
$mg + mg - T = ma$
3mg - T = 3ma
$T - mg = ma$
$T - mg = ma$
$mg - T_{1} = ma$
$T_{1} + 2mg - T_{2} = 2ma$
mg = 2ma
2mg = 4ma
$T_{2} - 2mg = 2ma$
$a = \frac{g}{2}$
$a = \frac{g}{2}$ mg = 5ma
$\frac {g}{5} = a$
Q409
Allen
COMPREHENSION TYPE QUESTIONS
Match the Columns
Velocity of three particles A, B and C varies with time t as, $\vec{v}_{A} = (2t\hat{i} + 6\hat{j})$ m/s; $\vec{v}_{B} = (3\hat{i} + 4\hat{j})$ m/s and $\vec{v}_{C} = (6\hat{i} - 4t\hat{j})$ m/s. Regarding the pseudo force match the following table
| Column-I | Column-II | ||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
|
Correct Answer: (A-T), (B-R), (C-R), (D-Q)
Explanation:
Ans. (A-T), (B-R), (C-R), (D-Q)
$\vec{a}_{A}=2\hat{i},\vec{a}_{B}=0,\vec{a}_{C}=-4\hat{j}$ and the direction of pseudo force on particle is opposite to the acceleration of non-inertial frame.
$\vec{a}_{A}=2\hat{i},\vec{a}_{B}=0,\vec{a}_{C}=-4\hat{j}$ and the direction of pseudo force on particle is opposite to the acceleration of non-inertial frame.
Q410
Allen
COMPREHENSION TYPE QUESTIONS
Match the Columns
In the figure shown, acceleration of 1 is x (upwards). Acceleration of pulley $P_{3}$ , w.r.t. pulley $P_{2}$ is y (downwards) and acceleration of 4 w.r.t. to pulley $P_{3}$ is z (upwards). Then
| Column-I | Column-II | ||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
|
Correct Answer: (A-S), (B-R), (C-Q)
Explanation:
Ans. (A-S), (B-R), (C-Q)
Taking upward direction as positive
$a _ {P _ {2}} = - a _ {1} = - x$
$a _ {P _ {3}} - a _ {P _ {2}} = - y$
$a _ {P _ {3}} = - x - y$
$a _ {4} - a _ {P _ {3}} = z$
$a _ {4} = a _ {P _ {3}} + z = (z - x - y) \uparrow$
$2 a _ {P _ {3}} = a _ {3} + a _ {4}$
$\Rightarrow a _ {3} = 2 a _ {P _ {3}} - a _ {4} = 2 (- x - y) - z + x + y = - (x + y + z) = (x + y + z) \downarrow$
$\Rightarrow 2 a _ {P _ {2}} = a _ {2} + a _ {P _ {3}}$
$a _ {2} = 2 a _ {P _ {2}} - a _ {P _ {3}} = - 2 x - (- x - y) = - x + y = (y - x) \uparrow$
Taking upward direction as positive
$a _ {P _ {2}} = - a _ {1} = - x$
$a _ {P _ {3}} - a _ {P _ {2}} = - y$
$a _ {P _ {3}} = - x - y$
$a _ {4} - a _ {P _ {3}} = z$
$a _ {4} = a _ {P _ {3}} + z = (z - x - y) \uparrow$
$2 a _ {P _ {3}} = a _ {3} + a _ {4}$
$\Rightarrow a _ {3} = 2 a _ {P _ {3}} - a _ {4} = 2 (- x - y) - z + x + y = - (x + y + z) = (x + y + z) \downarrow$
$\Rightarrow 2 a _ {P _ {2}} = a _ {2} + a _ {P _ {3}}$
$a _ {2} = 2 a _ {P _ {2}} - a _ {P _ {3}} = - 2 x - (- x - y) = - x + y = (y - x) \uparrow$
Q411
Allen
COMPREHENSION TYPE QUESTIONS
Match the Columns
A 50 kg box is held at rest on a slope of inclination $53^{\circ}$ above the horizontal. It is released and immediately pushed by a force F up the slope. Coefficients of static and kinetic friction between the slope and the box are 0.5 and 0.4 respectively. $g = 10 \, m/s^{2}$
| Column-I | Column-II | ||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
|
Correct Answer: (A-S), (B-S), (C-Q), (D-R)
Explanation:
Ans. (A-S), (B-S), (C-Q), (D-R)
$\text { Friction } = f _ {k} = \frac {4}{1 0} \times 5 0 0 \times \frac {3}{5} = 1 2 0 N\tag{A}$
(B) Friction $= f_{k} = \frac{4}{10}\times 500\times \frac{3}{5} = 120N$
(C) Friction $= f_{s} = 150N$
(D) Friction = 50 N
$\text { Friction } = f _ {k} = \frac {4}{1 0} \times 5 0 0 \times \frac {3}{5} = 1 2 0 N\tag{A}$
(B) Friction $= f_{k} = \frac{4}{10}\times 500\times \frac{3}{5} = 120N$
(C) Friction $= f_{s} = 150N$
(D) Friction = 50 N
Q412
Allen
COMPREHENSION TYPE QUESTIONS
Match the Columns
In the diagram shown in figure $(g = 10 \, m/s^{2})$
| Column-I | Column-II | ||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
|
Correct Answer: (A-R), (B-T), (C-Q), (D-T)
Explanation:
Ans. (A-R), (B-T), (C-Q), (D-T)
For system
$F _ {1} - F _ {2} - 6 g \sin 3 0 ^ {\circ} = 6 a$
$\Rightarrow a = 2 m / s ^ {2}$
For 3 kg
$F _ {1} - N _ {1} - 3 g \sin 3 0 ^ {\circ} = 3 a$
$\Rightarrow N _ {1} = 3 9 N$
For 2 kg
$N _ {1} - N _ {2} - 2 g \sin 3 0 ^ {\circ} = 2 a$
$\Rightarrow N _ {2} = 2 5 N$
For system
$F _ {1} - F _ {2} - 6 g \sin 3 0 ^ {\circ} = 6 a$
$\Rightarrow a = 2 m / s ^ {2}$
For 3 kg
$F _ {1} - N _ {1} - 3 g \sin 3 0 ^ {\circ} = 3 a$
$\Rightarrow N _ {1} = 3 9 N$
For 2 kg
$N _ {1} - N _ {2} - 2 g \sin 3 0 ^ {\circ} = 2 a$
$\Rightarrow N _ {2} = 2 5 N$
Q413
Allen
Newtons Laws of Motion
Numerical
Draw free body diagram of block of mass m.
(A)
(B)
(C)
(D)
(E)
(F)
(G)
(H)
(I)
(A)
(B)
(C)
(D)
(E)
(F)
(G)
(H)
(I)
Correct Answer: (a)
Explanation:
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
Q414
Allen
Newtons Laws of Motion
Numerical
In the given figure, what is the minimum possible tension $(N)$ in the string 2?
Correct Answer: 8
Explanation:
Ans. 8
$T_{1}\sin 37^{\circ} + T_{2}\cos \theta = 10$
$T_{1}\cos 37^{\circ} = T_{2}\sin \theta$
$\sin 37^{\circ}\left(\frac{T_{2}\sin\theta}{\cos 37^{\circ}}\right) + T_{2}\cos \theta = 10$
$T _ {2} = \frac {4 0}{3 \sin \theta + 4 \cos \theta}$
$T_{2\ min}$ occurs if $(3\sin\theta+4\cos\theta)$ is maximum
$T _ {2 \min} = \frac {4 0}{5} = 8$
$T_{1}\sin 37^{\circ} + T_{2}\cos \theta = 10$
$T_{1}\cos 37^{\circ} = T_{2}\sin \theta$
$\sin 37^{\circ}\left(\frac{T_{2}\sin\theta}{\cos 37^{\circ}}\right) + T_{2}\cos \theta = 10$
$T _ {2} = \frac {4 0}{3 \sin \theta + 4 \cos \theta}$
$T_{2\ min}$ occurs if $(3\sin\theta+4\cos\theta)$ is maximum
$T _ {2 \min} = \frac {4 0}{5} = 8$
Q415
Allen
Newtons Laws of Motion
Numerical
A ladder is hanging from ceiling as shown in figure. Three men A, B and C of masses 40 kg, 60 kg, and 50 kg are climbing the ladder. Man A is going up with retardation $2 \, m/s^{2}$ , C is going up with an acceleration of $1 \, m/s^{2}$ and man B is going up with a constant speed of 0.5 m/s. Find the tension in the string supporting the ladder. $[g = 9.8 \, m/s^{2}]$
Correct Answer: $1440N$
Explanation:
Ans. 1440 N
For man B,
$T _ {B} - 6 0 \times 9. 8 - 5 4 0 = 0$
$T _ {B} = 5 4 0 + 5 8 8 = 1 1 2 8 N$
For man A,
$4 0 \mathrm{g} + 1 1 2 8 - T _ {A} = 4 0 \times 2$
$T _ {A} = 3 9 2 + 3 9 2 + 1 1 2 8 - 1 6 0$
$\boxed {T _ {A} = 1 4 4 0 N}$
For man C,
$T _ {C} - 5 0 g = 5 0 \times 1$
$T _ {C} = 5 0 g = (1 + g)$
$T _ {C} = 5 0 (1 + 9. 8)$
$T _ {C} = 1 0. 8 \times 5 0$
$T _ {C} = 4 5 0 N$
For man B,
$T _ {B} - 6 0 \times 9. 8 - 5 4 0 = 0$
$T _ {B} = 5 4 0 + 5 8 8 = 1 1 2 8 N$
For man A,
$4 0 \mathrm{g} + 1 1 2 8 - T _ {A} = 4 0 \times 2$
$T _ {A} = 3 9 2 + 3 9 2 + 1 1 2 8 - 1 6 0$
$\boxed {T _ {A} = 1 4 4 0 N}$
For man C,
$T _ {C} - 5 0 g = 5 0 \times 1$
$T _ {C} = 5 0 g = (1 + g)$
$T _ {C} = 5 0 (1 + 9. 8)$
$T _ {C} = 1 0. 8 \times 5 0$
$T _ {C} = 4 5 0 N$
Q416
Allen
Newtons Laws of Motion
Numerical
A 40 kg monkey climbs a rope that passes over an ideal pulley. The other end of the rope is attached to a 60 kg weight placed on the ground. What is the maximum upward acceleration the monkey can have without lifting the weight? If he climbs the rope with upward acceleration 2g, with what acceleration the weight will rise up?
Correct Answer: $0.5g,g$
Explanation:
Ans. 0.5 g, g
T = 600 N (Block is on the verge of getting lifted)
T - 400 = 40a
600 - 400 = 40a $a = \frac{200}{40} = 5$ ; $\frac{g}{2} = 0.5g$ If boy move with 2g acceleration $T - 40\ g = 40 \times 2g \Rightarrow T = 120\ g$ For block 120 g - 60 g = 60 a $q = q$
a = g
T = 600 N (Block is on the verge of getting lifted)
T - 400 = 40a
600 - 400 = 40a $a = \frac{200}{40} = 5$ ; $\frac{g}{2} = 0.5g$ If boy move with 2g acceleration $T - 40\ g = 40 \times 2g \Rightarrow T = 120\ g$ For block 120 g - 60 g = 60 a $q = q$
a = g
Q417
Allen
Newtons Laws of Motion
Numerical
In the system shown, pulley and strings are ideal. The vertically upward pull F is being increased gradually, find magnitude of F and acceleration of the 5 kg block at the moment the 10 kg block leaves the floor.
Correct Answer: $200 \mathrm{~N}, 10 \mathrm{~m} / \mathrm{s}^{2}$
Explanation:
Ans. 200 N, 10 m/s $^{2}$ When 10 kg leave contact $\frac{F}{2}=100\Rightarrow F=200N$ 100 - 50 = 5a
a = 10 m/s $^{2}$
a = 10 m/s $^{2}$
Q418
Allen
Newtons Laws of Motion
Numerical
Force F is applied on upper pulley. If $F = 30t N$ where t is time in second. Find the time when $m_{1}$ loses contact with floor.
Correct Answer: 2 sec
Explanation:
Ans. 2 sec
If 4 kg is leaving ground,
20t = 40
If 4 kg is leaving ground,
20t = 40
Q419
Allen
Newtons Laws of Motion
Numerical
In the figure shown, all surfaces are smooth and block A and wedge B have mass 10 kg and 20 kg respectively. Find normal reaction between block A and B, spring force and normal reaction of ground on block B. $(g = 10 \, m/s^{2})$ .
Correct Answer: 80 N, 48 N, 264 N
Explanation:
Ans. 80 N, 48 N, 264 N
$\begin{array}{c} {N = 1 0 0 \cos 3 7 = 1 0 0 \times \frac {4}{5}} \\ {= 8 0 N} \end{array}$
$\begin{array}{r l} N _ {B} & = 2 0 0 + 8 0 \cos 3 7 \\ & = 2 0 0 + 8 0 \times \frac {4}{5} \\ & = 2 6 4 N \end{array}$
Spring force $kx = 16 \times 3 = 48 \, N$ .
$\begin{array}{c} {N = 1 0 0 \cos 3 7 = 1 0 0 \times \frac {4}{5}} \\ {= 8 0 N} \end{array}$
$\begin{array}{r l} N _ {B} & = 2 0 0 + 8 0 \cos 3 7 \\ & = 2 0 0 + 8 0 \times \frac {4}{5} \\ & = 2 6 4 N \end{array}$
Spring force $kx = 16 \times 3 = 48 \, N$ .
Q420
Allen
Newtons Laws of Motion
Numerical
Find the initial acceleration of blocks A and B if the
(A) Upper spring is cut
(B) Lower spring is cut
(A) Upper spring is cut
(B) Lower spring is cut
Correct Answer: (a) $3g \downarrow, 0$ (b) $2g \uparrow, g \downarrow$
Explanation:
Ans. (a) $3g \downarrow, 0$ (b) $2g \uparrow, g \downarrow$
Initially
(a)
$a _ {m} = \frac {3 m g}{m} = 3 g \downarrow$
(b)
$a _ {m} = \frac {2 m g}{m} = 2 g \uparrow$
$a _ {2 m} = \frac {2 m g}{2 m} = g \downarrow$
Initially
(a)
$a _ {m} = \frac {3 m g}{m} = 3 g \downarrow$
(b)
$a _ {m} = \frac {2 m g}{m} = 2 g \uparrow$
$a _ {2 m} = \frac {2 m g}{2 m} = g \downarrow$
Q421
Allen
Newtons Laws of Motion
Numerical
The system shown in the figure is initially in equilibrium. A is of mass 2m and B, C, D and E are of mass m. Certain actions are performed on the system. Every action has been taken individually when the system is intact. Find the direction and magnitude of acceleration of the blocks after each of the following actions has been taken:
(I) Spring 1 is cut
(II) Spring 2 is cut
(III) String between C and D is cut.
(IV) String between $B$ and $C$ is cut.
(I) Spring 1 is cut
(II) Spring 2 is cut
(III) String between C and D is cut.
(IV) String between $B$ and $C$ is cut.
Correct Answer: (i) $a_{A} = g\downarrow ,a_{B} = \frac{2g}{3}\uparrow ,a_{c} = \frac{2g}{3}\downarrow ,a_{D} = \frac{2g}{3}\downarrow ,a_{E} = 0$
(ii) $a_{A} = 0$ , $a_{B} = \frac{g}{3}\downarrow$ , $a_{C} = \frac{g}{3}\uparrow$ , $a_{D} = \frac{g}{3}\uparrow$ , $a_{E} = g\downarrow$ ,
(iii) $a_{A} = 0, a_{B} = g \downarrow, a_{C} = g \uparrow, a_{D} = 2g \downarrow, a_{E} = 0$
(iv) $a_{A} = 0, a_{B} = 3g \downarrow, a_{C} = \frac{3g}{2} \downarrow, a_{D} = \frac{3g}{2} \downarrow, a_{E} = 0$
Explanation:
Ans. (i) $a_{A} = g \downarrow, a_{B} = \frac{2g}{3} \uparrow, a_{c} = \frac{2g}{3} \downarrow, a_{D} = \frac{2g}{3} \downarrow, a_{E} = 0$
(ii) $a_{A} = 0$ , $a_{B} = \frac{g}{3}\downarrow$ , $a_{C} = \frac{g}{3}\uparrow$ , $a_{D} = \frac{g}{3}\uparrow$ , $a_{E} = g\downarrow$ ,
(iii) $a_{A} = 0, a_{B} = g \downarrow, a_{C} = g \uparrow, a_{D} = 2g \downarrow, a_{E} = 0$
(iv) $a_{A} = 0, a_{B} = 3g \downarrow, a_{C} = \frac{3g}{2} \downarrow, a_{D} = \frac{3g}{2} \downarrow, a_{E} = 0$
(i) Spring 1 is cut
$\vec {a} _ {A} = g \downarrow$
Acceleration of B, C, D are a.
$m g - T _ {1} = m a$
$T _ {1} - m g - T _ {2} = m a$
$T _ {2} - 2 m g = m a$
- 2 m g = 3 m a
$a = \frac {2 g}{3}$
$a _ {B} = \frac {2 g}{3} \uparrow a _ {C} = \frac {2 g}{3} \downarrow a _ {D} = \frac {2 g}{3} \downarrow$
$a_{E} = 0$
(ii) Spring (2) is cut $a_{A} = 0$
Let acceleration on B, C, D be a
$3 m g - T _ {1} = m a \quad a _ {B} = \frac {g}{3} \downarrow$
$a _ {c} = \frac {g}{3} \uparrow$
$T _ {1} - T _ {2} - m g = m a$
$a _ {D} = g / 3 \uparrow$
$T _ {2} - m g = m a \qquad \vec {a} _ {E} = g \downarrow$
$a = \frac {g}{3}$
(iii) String between $C$ and $D$ is cut
$a _ {A} = 0$
$a _ {E} = 0$
$a _ {D} = 2 g \downarrow$
Let acceleration of $B$ and $C$ be $a$
$3 m g - T _ {1} = m a$
$a _ {B} = g \downarrow$
$T _ {1} - m g = m a$
$a _ {C} = g \uparrow$
(iv) Between $B$ and $C$ is cut
$a _ {A} = 0$
$a _ {E} = 0$
$\mathrm{For} a _ {B}$
$3 m g = m a _ {B}$
$a _ {B} = 3 g \downarrow$
Let acceleration of $C$ and $D$ be $a$
2 m g - T = m a
m g + T = m a
$a = \frac {3 g}{2}$
$a _ {\mathrm{c}} = \frac {3 g}{2} \downarrow$
(ii) $a_{A} = 0$ , $a_{B} = \frac{g}{3}\downarrow$ , $a_{C} = \frac{g}{3}\uparrow$ , $a_{D} = \frac{g}{3}\uparrow$ , $a_{E} = g\downarrow$ ,
(iii) $a_{A} = 0, a_{B} = g \downarrow, a_{C} = g \uparrow, a_{D} = 2g \downarrow, a_{E} = 0$
(iv) $a_{A} = 0, a_{B} = 3g \downarrow, a_{C} = \frac{3g}{2} \downarrow, a_{D} = \frac{3g}{2} \downarrow, a_{E} = 0$
(i) Spring 1 is cut
$\vec {a} _ {A} = g \downarrow$
Acceleration of B, C, D are a.
$m g - T _ {1} = m a$
$T _ {1} - m g - T _ {2} = m a$
$T _ {2} - 2 m g = m a$
- 2 m g = 3 m a
$a = \frac {2 g}{3}$
$a _ {B} = \frac {2 g}{3} \uparrow a _ {C} = \frac {2 g}{3} \downarrow a _ {D} = \frac {2 g}{3} \downarrow$
$a_{E} = 0$
(ii) Spring (2) is cut $a_{A} = 0$
Let acceleration on B, C, D be a
$3 m g - T _ {1} = m a \quad a _ {B} = \frac {g}{3} \downarrow$
$a _ {c} = \frac {g}{3} \uparrow$
$T _ {1} - T _ {2} - m g = m a$
$a _ {D} = g / 3 \uparrow$
$T _ {2} - m g = m a \qquad \vec {a} _ {E} = g \downarrow$
$a = \frac {g}{3}$
(iii) String between $C$ and $D$ is cut
$a _ {A} = 0$
$a _ {E} = 0$
$a _ {D} = 2 g \downarrow$
Let acceleration of $B$ and $C$ be $a$
$3 m g - T _ {1} = m a$
$a _ {B} = g \downarrow$
$T _ {1} - m g = m a$
$a _ {C} = g \uparrow$
(iv) Between $B$ and $C$ is cut
$a _ {A} = 0$
$a _ {E} = 0$
$\mathrm{For} a _ {B}$
$3 m g = m a _ {B}$
$a _ {B} = 3 g \downarrow$
Let acceleration of $C$ and $D$ be $a$
2 m g - T = m a
m g + T = m a
$a = \frac {3 g}{2}$
$a _ {\mathrm{c}} = \frac {3 g}{2} \downarrow$
Q422
Allen
Newtons Laws of Motion
Numerical
Two trolley A and B are moving with accelerations a and 2a respectively in the same direction. To an observer in trolley A, the magnitude of pseudo force acting on a block of mass m on the trolley B is
Correct Answer: ma
Explanation:
Ans. ma
$a _ {D} = \frac {3 g}{2} \downarrow$
Pseudo force = $m_{object} \times a_{observer}$
$a _ {D} = \frac {3 g}{2} \downarrow$
Pseudo force = $m_{object} \times a_{observer}$
Q423
Allen
Newtons Laws of Motion
Numerical
A block of mass m lies on wedge of mass M as shown in figure.
With what minimum acceleration must the wedge be moved towards right horizontally so that block m falls freely.
With what minimum acceleration must the wedge be moved towards right horizontally so that block m falls freely.
Correct Answer: $a = g\cot \theta$
Explanation:
Ans. $a = g \cot \theta$
$a_{2}$ : Acceleration w.r.t. wedge, $a_{1}$ : Acceleration of wedge
$a _ {\mathrm{x}} = a _ {2} \cos \theta - a _ {1} = 0$
$a _ {1} = a _ {2} \cos \theta$
Acceleration B along y-axis
$a _ {\mathrm{y}} = a _ {2} \sin \theta = g$
$a _ {2} = \frac {g}{\sin \theta}$
$a _ {1} = \frac {g}{\sin \theta} \cos \theta = g \cot \theta$
$a_{2}$ : Acceleration w.r.t. wedge, $a_{1}$ : Acceleration of wedge
$a _ {\mathrm{x}} = a _ {2} \cos \theta - a _ {1} = 0$
$a _ {1} = a _ {2} \cos \theta$
Acceleration B along y-axis
$a _ {\mathrm{y}} = a _ {2} \sin \theta = g$
$a _ {2} = \frac {g}{\sin \theta}$
$a _ {1} = \frac {g}{\sin \theta} \cos \theta = g \cot \theta$
Q424
Allen
Newtons Laws of Motion
Numerical
All the surfaces shown in figure are assumed to be frictionless. The block of mass m slides on the prism which in turn slides backward on the horizontal surface. Find the acceleration of the smaller block with respect to the prism.
Correct Answer: $\frac{(M + m)g\sin\theta}{M + m\sin^2\theta}$
Explanation:
Ans. $\frac{(M+m)g\sin\theta}{M+m\sin^{2}\theta}$
Let the acceleration of the prism be $a_{0}$ in the backward direction. Consider the motion of the smaller block the frame of the prism.
The forces on the block arc (figure).
i) $N$ normal force
ii) mg downward (gravity)
iii) $ma_{0}$ forward (pseudo)
a)
$\mathrm{Mg}\downarrow \mathbf{(b)}$
The block slides down the plane. Components of the force parallel to the incline to the incline give $ma_{0} \cos \theta + mg \sin \theta = ma$
or $a = a_0\cos \theta +g\sin \theta$
Components of the force perpendicular to the incline give
$N + m a _ {0} \sin \theta = m g \cos \theta$
Now consider the motion of the prism from the lab frame. No pseudo force is needed as the frame used is inertial. The forces are (figure).
i) Mg downward, ii) N normal to the incline (by the block) iii) N upward (by the horizontal surface) Horizontal components given,
$N \sin \theta = m a _ {0} \text { or, } N = m a _ {0} / \sin \theta .\tag{... (ii}$
Putting in ii)
$\frac {M a _ {0}}{\sin \theta} + m a _ {0} \sin \theta = m g \cos \theta$
or $a_0 = \frac{mg\sin\theta\cos\theta}{M + m\sin^2\theta}$
$\text { From i), } a = \frac {m g \sin \theta \cos^ {2} \theta}{M + m \sin^ {2} \theta} + g \sin \theta = \frac {(M + m) g \sin \theta}{M + m \sin^ {2} \theta}$
Let the acceleration of the prism be $a_{0}$ in the backward direction. Consider the motion of the smaller block the frame of the prism.
The forces on the block arc (figure).
i) $N$ normal force
ii) mg downward (gravity)
iii) $ma_{0}$ forward (pseudo)
a)
$\mathrm{Mg}\downarrow \mathbf{(b)}$
The block slides down the plane. Components of the force parallel to the incline to the incline give $ma_{0} \cos \theta + mg \sin \theta = ma$
or $a = a_0\cos \theta +g\sin \theta$
Components of the force perpendicular to the incline give
$N + m a _ {0} \sin \theta = m g \cos \theta$
Now consider the motion of the prism from the lab frame. No pseudo force is needed as the frame used is inertial. The forces are (figure).
i) Mg downward, ii) N normal to the incline (by the block) iii) N upward (by the horizontal surface) Horizontal components given,
$N \sin \theta = m a _ {0} \text { or, } N = m a _ {0} / \sin \theta .\tag{... (ii}$
Putting in ii)
$\frac {M a _ {0}}{\sin \theta} + m a _ {0} \sin \theta = m g \cos \theta$
or $a_0 = \frac{mg\sin\theta\cos\theta}{M + m\sin^2\theta}$
$\text { From i), } a = \frac {m g \sin \theta \cos^ {2} \theta}{M + m \sin^ {2} \theta} + g \sin \theta = \frac {(M + m) g \sin \theta}{M + m \sin^ {2} \theta}$
Q425
Allen
Newtons Laws of Motion
Numerical
The block A is moving downward with constant velocity $v_{0}$ . Find the velocity of the block B, when the string makes an angle $\theta$ with the horizontal
Correct Answer: $v_{0}/\cos\theta$
Explanation:
Ans. $v_{0} / \cos \theta$
Length of string remains constant
$v \cos \theta = v _ {0}$
$v = \frac {v _ {0}}{\cos \theta}$
Length of string remains constant
$v \cos \theta = v _ {0}$
$v = \frac {v _ {0}}{\cos \theta}$