NLM

$\begin{array}{r l} & a = \left(\frac {1 0 - 4 \sin 3 0 - (0.2) (4) \cos 3 0}{1 0 + 4}\right) g \\ & \Rightarrow \quad a = \frac {(1 0 - 2 - 0.7)}{1 4} g \\ & \Rightarrow \quad a = \left(\frac {7.3}{1 4}\right) 9.8 \\ & \Rightarrow \quad a = 5.1 1 \mathrm{ms} ^ {- 2} \\ & \text {Since} v ^ {2} - u ^ {2} = 2 a s \\ & \Rightarrow \quad v ^ {2} - 0 ^ {2} = 2 (5.1 1) (4) \\ & \Rightarrow \quad v \simeq 6.5 \mathrm{ms} ^ {- 1} \end{array}$
Hence, the correct answer is (A).

Friction force between A and $B(=\mu mg)$ will accelerate B and retard A till slipping stops between A and B. Since mass of both are equal, So we have acceleration of B = retardation of $A = \mu g$ .
$\Rightarrow v _ {A} = v _ {0} - (\mu g) t \text { and } v _ {B} = (\mu g) t$

After slipping ceases, the common velocity of both becomes $\frac{v_0}{2}$ .
This can also be obtained from conservation of linear momentum also (to be studied later).
Hence, the correct answer is (B).

D

$\sin \theta = \frac{h}{l}$
$\begin{array}{r l} & M g - T = M a \text {and} \\ \Rightarrow & T - m g \sin \theta = m a \\ \Rightarrow & a = \left(\frac {M - m \sin \theta}{M + m}\right) g \end{array}$
Since $v^{2}-0^{2}=2as$ , where s is the distance travelled by m before M is detached. After detaching M, m will move the remaining distance of $(l-s)$ up the incline, under a retardation $g\sin\theta$ .
$\begin{array}{l l} \Rightarrow & 0 ^ {2} = v ^ {2} - 2 (g \sin \theta) (l - s) \\ \Rightarrow & 0 ^ {2} = 2 a s - 2 g \sin \theta (l - s) \\ \Rightarrow & s = \frac {M + m}{M} \left(\frac {h l}{h + l}\right) \end{array}$
Hence, the correct answer is (D).

Let the initial lengths of the respective segments of strings be $l_{1}$ , $l_{2}$ , $l_{3}$ and $l_{4}$ . Then
$\binom{\text {Total length of the}}{\text {string at t = 0}} = \binom{\text {Total length of the}}{\text {string at time t}}$

Let A move forward by x, B move backwards (towards left) by y and correspondingly C moves down, say by z. Then
$\begin{array}{r l} \not {l} _ {1} + \not {l} _ {2} + \not {l} _ {3} + \not {l} _ {4} & = \left[ \not {l} _ {1} - (x + y) \right] + \not {l} _ {2} + \\ & \left[ \not {l} _ {3} - (x + y) \right] + \left(\not {l} _ {4} + z\right) \end{array}$
z = 2 (x + y)
$\Rightarrow \quad \ddot {z} = 2 (\ddot {x} + \ddot {y})$
$\Rightarrow a_{C} = 2(a + b)$ , downwards
Also, as A moves forward, C also moves forward, hence
$\begin{array}{r l} & {\vec {a} _ {C} = a \hat {i} - 2 (a + b) \hat {j}} \\ {\Rightarrow} & {\left| \vec {a} _ {C} \right| = \sqrt {a ^ {2} + 4 (a + b) ^ {2}}} \\ {\Rightarrow} & {\left| \vec {a} _ {C} \right| = \sqrt {5 a ^ {2} + 4 b ^ {2} + 8 a b}} \end{array}$
Hence, the correct answer is (D).

$a = \mu g$
$\begin{array}{r l} & a = \mu g \\ & s = 0 + \frac {1}{2} (\mu g) t _ {M I N} ^ {2} \\ & \Rightarrow t _ {\mathrm{MIN}} = \sqrt {\frac {2 s}{\mu g}} \\ & \Rightarrow t _ {\mathrm{MIN}} \propto \frac {1}{\sqrt {\mu}} \end{array}$
Hence, the correct answer is (B).

FBD of blocks is shown
$\begin{array}{c} 1 \mathrm{N} \\ \xleftarrow {2 \mathrm{kg}} T \\ f _ {\max} = 2 \mathrm{N} \end{array} \quad \begin{array}{c} 3 \mathrm{kg} \\ \xleftarrow {} f _ {\max} = 6 \mathrm{N} \end{array}$
Net force without friction on system is 7 N towards right side, so maximum friction will first come on 3 kg block
$\begin{array}{c c} 1 \mathrm{N} & 2 \mathrm{N} \\ \leftarrow & \\ 1 \mathrm{N} & \\ 2 \mathrm{kg} & \\ \leftarrow & \\ 6 \mathrm{N} & \end{array} \quad \begin{array}{c c} 2 \mathrm{N} & 3 \mathrm{kg} \\ \leftarrow & \\ 8 \mathrm{N} & \\ 6 \mathrm{N} & \end{array}$
$\Rightarrow f _ {1} = 1 \mathrm{N}, f _ {2} = 6 \mathrm{N}, T = 2 \mathrm{N}$
Hence, the correct answer is (C).

As no force is acting on $A$ ,
$T = 0 \text { and } a _ {B} = \frac {F}{m _ {B}}$
Hence, the correct answer is (A).

As block moves to the right, N decreases. Hence, the correct answer is (C).

B

Only magnitude remains constant and direction changes
Hence, the correct answer is (A).

$x = 2t$
$\Rightarrow v _ {x} = \frac {d x}{d t} = 2$
$y = 2 t ^ {2}$
$\Rightarrow v _ {y} = \frac {d y}{d x} = 4 t$
$\Rightarrow \tan \theta = \frac {v _ {y}}{v _ {x}} = \frac {4 t}{2} = 2 t$
Differentiating with respect to time we get,
$\left(\sec^ {2} \theta\right) \frac {d \theta}{d t} = 2$
$\Rightarrow \left(1 + \tan^ {2} \theta\right) \frac {d \theta}{d t} = 2$
$\Rightarrow \left(1 + 4 t ^ {2}\right) \frac {d \theta}{d t} = 2$
$\Rightarrow \frac {d \theta}{d t} = \frac {2}{1 + 4 t ^ {2}}$
So, $\frac{d\theta}{dt}$ at $t = 2$ s is
$\frac {d \theta}{d t} = \frac {2}{1 + 4 (2) ^ {2}} = \frac {2}{1 7} \mathrm{rads} ^ {- 1}$
Hence, the correct answer is (A).

$2kx\cos(30^{\circ})=\left(\frac{4m_{1}m_{2}}{m_{1}+m_{2}}\right)g$
Hence, the correct answer is (A).

Using Lami's Theorem, we get

$\frac {m g}{\sin (1 8 0 ^ {\circ} - 3 7 ^ {\circ})} = \frac {N _ {2}}{\sin (1 8 0 ^ {\circ} - 3 7 ^ {\circ})}$
$\Rightarrow N _ {2} = m g$
Hence, the correct answer is (A).

Limiting friction on block B is
$f _ {l} = (0. 5) (2 + 8) (1 0) = 5 0 \mathrm{N}$
So, block B will not move.
Hence, the correct answer is (A).

Retardation $a = g(\sin\theta + \mu\cos\theta) = 5(\sqrt{3} + \mu)$ Since v = u - at $\Rightarrow \quad 0 = u - at$ $\Rightarrow \quad a = \frac{u}{t}$ $\Rightarrow \quad 5(\sqrt{3} + \mu) = 10$ $\Rightarrow \quad \mu = 0.27$ Hence, the correct answer is (C).

$\tan\theta=\frac{a_{N}}{a_{t}}$

Hence, the correct answer is (C).

B

Drawing F.B.D of m, we get
$N\sin \theta = mg$ and $N\cos \theta = ma$
$\Rightarrow \tan \theta = \frac {g}{a}$
$\Rightarrow a = g \cot \theta$
So, $F = (m + M)g\cot \theta$
Hence, the correct answer is (C).

C

$T = f$.... (1)
$N = m g$....(2)
Taking torque about A, we get
$T \times \frac {L}{\sqrt {2}} = m g \times \frac {L / 2}{\sqrt {2}}$
$\Rightarrow T = \frac {m g}{2} = 5 0 \mathrm{N}$
Hence, the correct answer is (A).

$F_{net} = F - f_{k} = 2t - \mu mg$
Hence, the correct answer is (C).

$R = u \sqrt {\frac {2 h}{g}} = 1 2 \sqrt {\frac {2 \times 5}{1 0}} = 1 2 \mathrm{m}\tag{138.}$
So, distance from origin is
$r = \sqrt {5 ^ {2} + 1 2 ^ {2}} = 1 3 \mathrm{m}$
Hence, the correct answer is (C).

$x = v _ {0} t \mathrm{and} h = \frac {1}{2} g t ^ {2}$
$\Rightarrow t = \sqrt {\frac {2 h}{g}} = \sqrt {\frac {2 \times 3 2}{1 0}} = \frac {8}{\sqrt {1 0}}$
$\mathrm{Since}, v _ {0} = \omega r = \frac {2 \pi}{T} \times 1 2 = \frac {2 \pi}{1 . 2} \times 1 2 = 2 0 \pi$
$\Rightarrow x = 2 0 \pi \left(\frac {8}{\sqrt {1 0}}\right) = \frac {1 6 0 \times 3 . 1 4}{3 . 1 6 2}$
$\Rightarrow x = 1 6 0 \mathrm{m}$
Hence, the correct answer is (D).

In this case spring force is zero initially. F.B.D of $A$ and $B$ are shown below.
$\begin{array}{c} m \\ A \\ \downarrow \\ m g \\ a _ {A} = g \end{array} \quad \begin{array}{c} 2 m \\ B \\ \downarrow \\ 2 m g \\ a _ {B} = g \end{array}$
Hence, the correct answer is (A).