NLM

Force of friction between the two is
$f _ {A} = \mu m g$
Retardation of $A$ is $a_{A} = \frac{\mu mg}{m} = \mu g$
and acceleration of B is $a_{B}=\frac{\mu mg}{2m}=\frac{\mu g}{2}$


Hence, acceleration of B relative to A is
$a _ {B A} = a _ {A} + a _ {B} = \frac {3 \mu g}{2}$
{∵ both are in opposite direction}
Since $\mu=\frac{1}{2}$
$\Rightarrow a _ {B A} = \frac {3 g}{4}$
Hence, the correct answer is (C).

From the free body diagram, we observe that the component $N\sin\theta$ provides the necessary centripetal force, where as the component $N\cos\theta$ equals mg.

So, $N\cos \theta = mg$
$N \sin \theta = \frac {m v ^ {2}}{R}\tag{... (1}$
... (2)
So, $N = mg\sec \theta$
$\{\because \mathrm{of}(1)\}$
Hence, the correct answer is (C).

Blocks A and C both move due to friction and less friction is available to A as compared to C because normal reaction between A and B is less. Maximum friction between A and B can be
$f _ {\mathrm{max}} = \mu m _ {A} g = \left(\frac {1}{2}\right) m g$
Maximum acceleration of A can be
$a _ {\mathrm{max}} = \frac {f _ {\mathrm{max}}}{m} = \frac {g}{2}$
Further $a_{max} = a_{D}$ , because no slipping of A and C has to be on B (which moves forward with an acceleration $a_{D}$ ).
$\Rightarrow a _ {\mathrm{max}} = \frac {m _ {D} g}{3 m + m _ {D}}$
$\Rightarrow \frac {g}{2} = \frac {m _ {D} g}{3 m + m _ {D}}$
$\Rightarrow m _ {D} = 3 m$
Hence, the correct answer is (A).

Drawing free body diagram of block, we have
$\Sigma F _ {y} = 0$

$\Rightarrow N + F \sin 3 0 ^ {\circ} = m g \cos 4 5 ^ {\circ}$
$\Rightarrow N = m g \cos 4 5 ^ {\circ} - F \sin 3 0 ^ {\circ}$
$\Rightarrow N = (4) (1 0) \left(\frac {1}{\sqrt {2}}\right) - (1 0) \left(\frac {1}{2}\right)$
$\Rightarrow N = 2 3. 2 8 \mathrm{N}\tag{... (1}$
$f _ {\max} = \mu N = (0. 6) (2 3. 2 8) = 1 3. 9 7 \mathrm{N}$
$m g \sin (4 5 ^ {\circ}) = (4) (1 0) \left(\frac {1}{\sqrt {2}}\right) = 2 8. 2 8 \mathrm{N}$
and $F\cos(30^{\circ})=(10)\left(\frac{\sqrt{3}}{2}\right)=8.66\ N$
Since, $mg \sin 45^{\circ} > f_{\max} + F \cos 30^{\circ}$ , so the block will slide down and friction will be maximum. Therefore, net contact force is
$\begin{array}{r l} & F _ {C} = \sqrt {N ^ {2} + \left(f _ {\max}\right) ^ {2}} \\ \Rightarrow & F _ {C} = \sqrt {(2 3 . 2 8) ^ {2} + (1 3 . 9 7) ^ {2}} \\ \Rightarrow & F _ {C} = 2 7. 1 5 \mathrm{N} \end{array}$
Hence, the correct answer is (C).

For no slipping, both A and B move together down the incline with an acceleration $a = g \sin \theta$ , down the plane as no slipping exists between them. Horizontal component of this acceleration is $a_{H} = a \cos \theta$ and vertical component is $a_{V} = a \sin \theta$
where $a = g\sin \theta$

$\Rightarrow a_{H}=a\cos\theta=g\sin\theta\cos\theta$ and
$a _ {V} = a \sin \theta = g \sin^ {2} \theta$
FBD for A

Let N be the normal reaction between A and B. Equations of motion in horizontal and vertical directions give
$\begin{array}{r l} & m g - N = m a _ {V} \\ \Rightarrow & N = m g - m a _ {V} = m g - m g \sin^ {2} \theta \\ \Rightarrow & N = m g \cos^ {2} \theta \end{array}$
If $f$ be the frictional force between A and B, then
$m a _ {H} \leq \mu N$
$\Rightarrow m g \sin \theta \cos \theta \leq \mu m g \cos^ {2} \theta$
$\Rightarrow \quad \theta \leq \tan^ {- 1} (\mu)$
Hence, the correct answer is (D).

The tension is maximum in the string when it is in the lowest position. So,
$h = l \left(1 - \cos \theta_ {0}\right)$
Speed of mass m in its lowest position is
$v ^ {2} = 2 g h = 2 g l \big (1 - \cos \theta_ {0} \big)$

Further since, $T_{max} - mg = \frac{mv^{2}}{l}$
$\begin{array}{r l} \Rightarrow & T _ {\max} = m g + \frac {m v ^ {2}}{l} \\ \Rightarrow & T _ {\max} = m g + 2 m g (1 - \cos \theta_ {0}) \\ \Rightarrow & T _ {\max} = m g (3 - 2 \cos \theta_ {0}) \end{array}$
The block of mass 4m will not move, when
$\begin{array}{r l} & T _ {\max} \leq \mu (4 m) g \\ \Rightarrow & m g (3 - 2 \cos \theta_ {0}) \leq 4 \mu \mathrm{mg} \\ \Rightarrow & \mu \geq \frac {3 - 2 \cos \theta_ {0}}{4} \\ \Rightarrow & \mu_ {\mathrm{MIN}} = \frac {3 - 2 \cos \theta_ {0}}{4} \end{array}$
Hence, the correct answer is (D).

Net pulling force on the system is $Mg + mg - mg$ or simply Mg. Total mass being pulled is $M + 2m$ . Hence acceleration of system is
$a = \frac {M g}{M + 2 m}$

Now since a < g, there should be an upward force on M so that its acceleration becomes less than g. Hence for any value of M spring will be elongated.
Hence, the correct answer is (D).

Net pulling force = 2g - 1g = 10 N
Mass being pulled = 2 + 1 = 3 kg
So, acceleration of the system is $a=\frac{10}{3}$ ms $^{-2}$
Velocity of both the blocks at t = 1 s will be
$v _ {0} = a t = \left(\frac {1 0}{3}\right) (1) = \frac {1 0}{3} \mathrm{ms} ^ {- 1}$
Now, at this moment velocity of 2 kg block becomes zero, while that of 1 kg block is $\frac{10}{3}$ ms $^{-1}$ upwards. Hence, string becomes tight again when, displacement of 1 kg block = displacement of 2 kg block
$\Rightarrow v _ {0} t - \frac {1}{2} g t ^ {2} = \frac {1}{2} g t ^ {2}$
$\Rightarrow t = \frac {v _ {0}}{g} = \frac {\frac {1 0}{3}}{1 0} = \frac {1}{3} s$
Hence, the correct answer is (D).

Limiting force of friction between $A$ and $B$ is
$f _ {1} = \mu_ {1} m _ {A} g = (0. 3) (3 0) (1 0) = 9 0 \mathrm{N}$
Limiting force of friction between $B$ and $C$ is
$\begin{array}{l} {f _ {2} = \mu_ {2} (m _ {A} + m _ {B}) g} \\ {f _ {2} = (0.2) (3 0 + 1 0) (1 0) = 8 0 \mathrm{N}} \end{array}$
and limiting force of friction between C and ground is
$f _ {3} = \mu_ {3} (m _ {A} + m _ {B} + m _ {C}) g$
$f _ {3} = (0.1) (3 0 + 1 0 + 2 0) (1 0) = 6 0 \mathrm{N}$
As F is gradually increased the force of friction between A and B will increase.
When F = 60 N block A will exert a horizontal force of 60 N on B. There will be no relative motion between A and B since this force is less than 90 N. B will also exert a horizontal force of 60 N on C. Hence C will be on the point of motion.
Hence, the least value of F is 60 N. All the three blocks will be on the point of motion as a single body.
Hence, the correct answer is (A).

Since $a_{\text{down}} = g(\sin \theta - \mu \cos \theta)$
$\Rightarrow a _ {\mathrm{down}} = a = 5 (1 - 0.5 \sqrt {3}) = 0.6 7 \mathrm{ms} ^ {- 2}$
So, minimum speed while reaching the bottom can be
$v ^ {2} = u ^ {2} + 2 a s$
$\Rightarrow v = (6) ^ {2} + 2 (0.6 7) (1 5) = 5 6.1$
$\Rightarrow v = 7.4 9 \mathrm{ms} ^ {- 1} \cong 7. 5 \mathrm{ms} ^ {- 1}$
Hence, the correct answer is (B).

Let $T$ be the tension in the rope and $a$ the acceleration of rope. The absolute acceleration of man is therefore $\left(\frac{5g}{4} - a\right)$ . Equations of motion for mass and man gives

$T - 1 0 0 g = 1 0 0 a$...(1)
$T - 6 0 g = 6 0 \left(\frac {5 g}{4} -a\right)$..(2)
Solving (1) and (2) we get
$T = 1 2 1 8 \mathrm{N}$
Hence, the correct answer is (C).

Let $T$ be the tension in the string. The upward force exerted on the clamp is $T \sin 30^{\circ} = \frac{T}{2}$ .

$\begin{array}{l} \mathrm{Given} \frac {T}{2} = 4 0 \mathrm{N} \\ \Rightarrow T = 8 0 \mathrm{N} \end{array}$...(i)
If a is the acceleration of monkey in upward direction. Equation of monkey in upward direction is
$a = \frac {T - m g}{m}$
$a = \frac {8 0 - (5) (1 0)}{5} = 6 \mathrm{ms} ^ {- 2}$
Hence, the correct answer is (C).

The string is under tension, hence there is limiting friction between the block and the plane. Drawing free body diagram of the block.

$\begin{array}{r l} & {\Sigma F _ {x} = 0} \\ {\Rightarrow} & {\mu N + 5 0 \cos 4 5 ^ {\circ} = 1 5 0 \sin 4 5 ^ {\circ}} \\ & {\Sigma F _ {y} = 0} \\ {\Rightarrow} & {N = 5 0 \sin 4 5 ^ {\circ} + 1 5 0 \cos 4 5 ^ {\circ}} \end{array}\tag{... (1}$
$\ldots(2)$
Solving (1) and (2) we get
$\mu = \frac {1}{2}$
Hence, the correct answer is (A).

Acceleration of the system and hence tension (T) on the string attached with the pulley is same in both the cases. Hence
$F _ {1} = F _ {2} = 2 T$
Hence, the correct answer is (C).

As long as $mg \sin \theta < \mu mg \cos \theta$ , the block will remain stationary.
So, $x = 0$ , till $\tan \theta < \mu$ .
As soon as $\tan\theta > \mu$ , then the spring of force constant k gets extended by say x.
So, we have, for equilibrium
$k x + \mu m g \cos \theta = m g \sin \theta$
$\Rightarrow x = \frac {m g \sin \theta - \mu m g \cos \theta}{k}$

$\theta \rightarrow \frac {\pi}{2}, x = \frac {m g}{k}$
$\left\{\because N = 0 \right\}$
Hence, the correct answer is (A).

Maximum friction that can be obtained between $A$ and $B$ is

$f _ {1} = \mu m _ {A} g = (0. 3) (1 0 0) (1 0) = 3 0 0 \mathrm{N}$
and maximum friction between B and ground is
$\begin{array}{l} {f _ {2} = \mu (m _ {A} + m _ {B}) g} \\ {f _ {2} = (0. 3) (1 0 0 + 1 4 0) (1 0) = 7 2 0 \mathrm{N}} \end{array}$
Drawing free body diagrams of A, B and C in limiting case
Equilibrium of $A$ gives
$T _ {1} = f _ {1} = 3 0 0 \mathrm{N}\tag{... (1}$
Equilibrium of $B$ gives
$\begin{array}{r l} & 2 T _ {1} + f _ {1} + f _ {2} = T _ {2} \\ \Rightarrow & T _ {2} = 2 (3 0 0) + 3 0 0 + 7 2 0 = 1 6 2 0 \mathrm{N} \end{array}\tag{... (2}$
and equilibrium of $C$ gives
$\begin{array}{r l} & m _ {C} g = T _ {2} \\ \Rightarrow & 1 0 m _ {C} = 1 6 2 0 \\ \Rightarrow & m _ {C} = 1 6 2 \mathrm{kg} \end{array}$
Hence, the correct answer is (D).

For equilibrium
$\left( \begin{array}{c} \text {Weight of} \\ \text {hanging part} \\ \text {of chain} \end{array} \right) = \left( \begin{array}{c} \text {Force of friction} \\ \text {on the part resting} \\ \text {on table} \end{array} \right)$
Let $l$ be the length of the chain hanging
$\Rightarrow \left(\frac {M}{L} l\right) g = \mu \left[ \frac {M}{L} (L - l) g \right]$
$\Rightarrow l = \mu (L - l)$
$\Rightarrow \quad (\mu + 1) l = \mu L$
$\Rightarrow l = \frac {\mu L}{\mu + 1}$
Hence, the correct answer is (A).

The directions of friction forces at different surfaces are shown in figure.

Here, $f_{1}$ is the force of friction between A and B and $f_{2}$ is the force of friction between A and vertical wall.
Hence, the correct answer is (B).

Block slides down with constant velocity. Hence, net force on the block is zero.

So, net contact force $F_{C}$ on the body is
$F _ {C} = \sqrt {f ^ {2} + N ^ {2}}$
where $f = mg\sin \theta$
$\left\{\because a = 0 \Rightarrow f = m g \sin \theta \right\}$
and $N = mg\cos \theta$
$\Rightarrow F _ {C} = \sqrt {m ^ {2} g ^ {2} (\cos^ {2} \theta + \sin^ {2} \theta)} = m g$
Hence, the correct answer is (D).

Since, we observe that $mg \sin(30^{\circ}) > \mu mg \cos(30^{\circ})$

So, the block has a tendency to slip downwards. Let F be the minimum force applied on it, such that it does not slip. Then
$N = F + m g \cos (3 0 ^ {\circ})$
Also, $mg \sin (30^{\circ}) = \mu N = \mu (F + mg \cos 30^{\circ})$
$\Rightarrow F = \frac {m g \sin (3 0 ^ {\circ})}{\mu} - m g \cos (3 0 ^ {\circ})$
$\Rightarrow F = \frac {(2) (1 0) \left(\frac {1}{2}\right)}{0.5} - (2) (1 0) \left(\frac {\sqrt {3}}{2}\right)$
$\Rightarrow F = 2 0 - 1 7.3 2$
$\Rightarrow F = 2.6 8 \mathrm{N}$
Hence, the correct answer is (C).

Maximum frictional force between the block and the plank is
$f _ {\mathrm{max}} = \mu m g = (0.5) (1) (1 0) = 5 \mathrm{N}$
The free body diagrams of the block and the plank are shown for convenience to solve the problem
$5 \mathrm{N}$

Acceleration of block is
$a _ {B} = \frac {5}{1} = 5 \mathrm{ms} ^ {- 2}$
Acceleration of plank is
$a _ {P} = \frac {3 5 - 5}{2} = 1 5 \mathrm{ms} ^ {- 2}$
So, relative acceleration of block with respect to the plank is
$a _ {B P} = a _ {B} - a _ {P} = - 1 0 \mathrm{ms} ^ {- 2}$
$\Rightarrow a _ {B P} = 1 0 \mathrm{ms} ^ {- 2}, \mathrm{backwards}$
Since, $s = \frac{1}{2} at^{2}$ , where $s = 1.25 \, m$ , backwards
$\Rightarrow t = \sqrt {\frac {2 s}{a}} = \sqrt {\frac {(2) (- 1.2 5)}{(- 1 0)}}$
$\Rightarrow t = \sqrt {\frac {2.5}{1 0}} = 0.5 \mathrm{s}$
Hence, the correct answer is (B).

$ a = g (\sin \theta - \mu \cos \theta) $
$\Rightarrow a = g (\sin \theta - k x \cos \theta)$
At $x = 0$ , $a = g \sin \theta$ . As $x$ increases a decreases and becomes zero at $x_0 = \frac{1}{k} \tan \theta$ . For $x > x_0$ , the acceleration becomes negative.
Hence, the correct answer is (D).

$\begin{array}{c} T = M g \\ \text { mg } \end{array} \quad \text { Mg } = T$
Three forces are acting on the pulley T (horizontally), T (vertically downward) and the weight of the pulley (also acting vertically downwards) i.e. Force Mg acts horizontally and $(M + m)g$ acts vertically downwards. So, total force acting on the pulley is
$\begin{array}{r l} & F _ {\mathrm{net}} = \sqrt {M ^ {2} g ^ {2} + (M + m) ^ {2} g ^ {2}} \\ \Rightarrow & F _ {\mathrm{net}} = \left[ \sqrt {M ^ {2} + (M + m) ^ {2}} \right] g \end{array}$
Hence, the correct answer is (D).

$h = R\sin \theta$
Speed of particle at point B is
$\begin{array}{r l} & v ^ {2} = 2 g h \\ \Rightarrow & v ^ {2} = 2 g R \sin \theta \end{array}$
The centripetal force is given by
$F = \frac {m v ^ {2}}{R} = 2 m g \sin \theta\tag{... (1}$
Further since,
$\begin{array}{l l} \Rightarrow & N - m g \sin \theta = \frac {m v ^ {2}}{R} = 2 m g \sin \theta \\ \Rightarrow & N = 3 m g \sin \theta \end{array}\tag{.(2}$

The desired ratio is
$x = \frac {F}{N} = \frac {2}{3} = \mathrm{constant}$
Hence, the correct answer is (D).

Hence, the correct answer is (A).