NLM

$F = kt$
$\Rightarrow m \frac {d v}{d t} = k t$
$\Rightarrow m v = \frac {k t ^ {2}}{2}\tag{... (1}$
$\mathrm{Since,} a = \frac {F}{m}$
$\Rightarrow a = \frac {k}{m} t\tag{... (2}$
From (1) and (2), we get
$m v = \frac {k}{2} \times \left(\frac {m a}{k}\right) ^ {2}$
Hence, (A), (B) and (C) are correct.

B,C,D

Free body diagram of $(A+B)$ system

$\begin{array}{r l} & f _ {B C} = F \neq 0 \\ \Rightarrow & f _ {C B} \neq 0 \end{array}$
Free body diagram of $(B + C)$ system

$f _ {C G} = F = f _ {B C}$
Free body diagram of $A$ is

No force acts on A parallel to the surface in contact, so $f_{AB} = 0$
Hence, (B) and (C) are correct.

(A) Magnitude of velocity is changing hence acceleration is present
C) Velocity is changing hence it can happen by change in direction also as in a uniform circular motion. Hence acceleration is present.
Hence, (A) and (C) are correct.

Since, v = 2t and if the radius of circular path is r, then we have
$\begin{array}{r l} & a _ {T} = \frac {d v}{d t} = 2 \\ & a _ {r} = \frac {v ^ {2}}{r} = \frac {4 t ^ {2}}{r} \\ \Rightarrow & a = \sqrt {a _ {T} ^ {2} + a _ {r} ^ {2}} \\ \Rightarrow & a = \sqrt {4 + \frac {1 6 t ^ {4}}{r ^ {2}}} \end{array}$
Hence, (B) and (D) are correct.

$N_{A}=mg\cos\theta$ and $N_{B}=mg\cos(53^{\circ})$
$\begin{array}{r l} & {\mathrm{Since,} \frac {N _ {A}}{N _ {B}} = \frac {4}{3}} \\ & {\Rightarrow \frac {\cos \theta}{\cos 5 3 ^ {\circ}} = \frac {4}{3}} \\ & {\Rightarrow \cos \theta = \frac {4}{3} \times \frac {3}{5}} \\ & {\Rightarrow \cos \theta = \frac {4}{5}} \\ & {\Rightarrow a = \frac {m g [ \sin (5 3 ^ {\circ}) - \sin \theta ]}{2 m} = \frac {g}{1 0}} \end{array}$
Hence, (A) and (C) are correct.

$R = T$
... (1)
$N = m g\tag{... (2}$
For equilibrium, we have
$\Sigma \vec {\tau} = \vec {0}$
So, lets calculate torque about centre of mass of ladder i.e., O.

$\begin{array}{r l} & N \times \frac {L}{2} \cos \theta = R \times \frac {L}{2} \sin \theta + T \times \frac {L}{2} \sin \theta \\ \Rightarrow & \frac {m g}{2} \cos \theta = T \sin \theta \\ \Rightarrow & T = \frac {m g}{2} \cot \theta \end{array}$
Hence, (A) and (B) are correct.

Free body diagram of block A:

Friction on A due to B, $f_{AB}$ acts downwards and friction on A due to Wall (W), $f_{AW}$ acts upwards. So, we have
$f _ {A W} = f _ {A B} + 1 0 \mathrm{g}$.... (1)
$N _ {A W} = N _ {A B}$.....(2)
Free body diagram of block B:
$c f _ {B A}$
B F $\leftarrow \text { N } _ {B A}$
5g
$f _ {B A} = 5 \mathrm{g}$
$N _ {B A} = F$.... (3)
From (1) and (3), we get
$f _ {A W} = 1 5 0 \mathrm{N}$
From (2) and (4), we get
$N _ {A W} = F$
$\Rightarrow \mu N _ {A W} \geq 1 5 0 \mathrm{N}$
$\Rightarrow 0.4 \times F \geq 1 5 0 \mathrm{N}$
$\Rightarrow F \geq 3 7 5 \mathrm{N}$
Hence, (A), (B) and (C) are correct.

Acceleration of block after application of force $a = \frac{2t}{m} -\mu g$ Hence, (B), (C) and (D) are correct.

Given that, $a_{T} = a_{N}$
$\Rightarrow a _ {T} = \frac {v ^ {2}}{R}$
$\Rightarrow - \frac {d v}{d t} = \frac {v ^ {2}}{R}$
$\Rightarrow d v = - \frac {v ^ {2}}{R} d t$
$\Rightarrow \int_ {v _ {0}} ^ {v} \frac {d v}{v ^ {2}} = - \frac {1}{R} \int_ {0} ^ {t} d t$
$\Rightarrow \quad - \frac {1}{v} \bigg | _ {v _ {0}} ^ {v} = - \frac {1}{R} t$
$\Rightarrow \quad - \frac {1}{v} \bigg | _ {v _ {0}} ^ {v} = - \frac {1}{R} t$
$\Rightarrow \frac {1}{v} = \frac {1}{v _ {0}} + \frac {t}{R}$
$\Rightarrow v = \frac {v _ {0}}{1 + \frac {v _ {0} t}{R}}$
Since, $a_{T}=a_{N}=\frac{v^{2}}{R}$ and $a_{T}=v\frac{dv}{dS}$
$\Rightarrow v \frac {d v}{d S} = - \frac {v ^ {2}}{R}$
$\Rightarrow \int_ {v _ {0}} ^ {v} \frac {d v}{v} = - \frac {1}{R} \int_ {0} ^ {S} d S$
$\Rightarrow \log_ {e} v \big | _ {v _ {0}} ^ {v} = - \frac {S}{R}$
$\Rightarrow \log_ {e} \left(\frac {v}{v _ {0}}\right) = - \frac {S}{R}$
At $A$ and $C$ ; $N_A < mg$
$\Rightarrow \frac {v}{v _ {0}} = e ^ {- \frac {S}{R}}$
N _ {C} < m g
At $B$ and $D$ ; $N_B > mg$
$N _ {D} > m g$
$\Rightarrow v = v _ {0} e ^ {- \frac {S}{R}}$
Hence, (A), (B) and (C) are correct.
Hence, (A) and (B) are correct.

At A:

$m g - N _ {A} = \frac {m v ^ {2}}{r _ {A}}$
$N _ {A} = m g - \frac {m v ^ {2}}{r _ {A}}$
At B:
From figure $r_{B} < r_{D}$ , so $N_{B} > N_{D}$

At C:
$\Rightarrow 1 2 0 = 8 0 + T$
$N _ {D} = m g + \frac {m v ^ {2}}{r _ {D}}$
$N _ {B} - m g = \frac {m v ^ {2}}{r _ {B}}$
Hence $N_{B}$ is greatest
At D:
Since, $r_{C} < r_{A}$ , so $N_{C} < N_{A}$
$N _ {B} = m g + \frac {m v ^ {2}}{r _ {B}}$
$\Rightarrow 2 m \times 1 0 \times \frac {4}{5} = 4 0$
Hence $N_{C}$ is least
$N _ {C} = m g - \frac {m v ^ {2}}{r _ {C}}$
Also, force applied by 20 kg block on inclined is
$R = \sqrt {N ^ {2} + f ^ {2}} = 2 0 g \cos (3 7 ^ {\circ}) \sqrt {1 + \mu^ {2}} = 1 7 8. 8 \mathrm{N}$
Hence, (A) and (C) are correct.

For 20 kg block, we have
$20g\sin(37^\circ)=\mu[20g\cos(37^\circ)]+T$
$\Rightarrow 20\times10\times\frac{3}{5}=0.5\times20\times10\times\frac{4}{5}+T$
$\Rightarrow 120=80+T$
$\Rightarrow T=40\ \text{N}$
Also, $2mg\cos(37^\circ)=T$
$\Rightarrow 2m\times10\times\frac{4}{5}=40$
$\Rightarrow m=2.5\ \text{kg}$
Also, force applied by 20 kg block on inclined is
$R=\sqrt{N^2+f^2}=20g\cos(37^\circ)\sqrt{1+\mu^2}=178.8\ \text{N}$
Hence, (A) and (C) are correct.

Free body diagram of blocks are drawn.
For $m_{1}$ :

$\begin{array}{r l} & a _ {1} = \frac {m _ {1} g - T}{m _ {1}} \\ \Rightarrow & a _ {1} = g - \frac {T}{m _ {1}} \end{array}$
For $m_{2}$ :

$a _ {2} = \frac {m _ {2} g - T}{m _ {2}}$
$\Rightarrow a _ {2} = g - \frac {T}{m _ {2}}$
$\mathrm{For} t = 0, T = 0$
$\left\{\because x = 0 \right\}$
$\Rightarrow a _ {1} = a _ {2}$
$\mathrm{For} t > 0, T > 0$
$\Rightarrow a _ {1} < a _ {2}$
Hence, (A) and (D) are correct.

$\mu m_{3}g = \sqrt{\left(m_{1}g\right)^{2} + \left(m_{2}g\right)^{2}}$
Hence, (A), (C) and (D) are correct.

$\left(f_{1}\right)$ maximum $= 0.2\times 10 = 2\mathrm{N}$
$\left(f _ {2}\right) \text { maximum } = 0. 8 \times 2 0 = 1 6 \mathrm{N}$
$\left(f _ {3}\right) \text { maximum } = 0. 3 \times 3 0 = 9 \mathrm{N}$
$\begin{array}{c}\boxed {A} \rightarrow f _ {1}\\f _ {1} \leftarrow \boxed {B} \rightarrow 2 0 N\\f _ {2} \leftarrow \boxed {C} \rightarrow f _ {2}\\f _ {3} \leftarrow\end{array}$
So, acceleration of the lowest block is
$a _ {3} = \frac {f _ {2} - f _ {3}}{1} = 4. 5 \mathrm{ms} ^ {- 2}$
Acceleration of upper most block is
$a _ {1} = \frac {f _ {1}}{1} = 2 \mathrm{ms} ^ {- 2}$
Hence, (A), (B) and (D) are correct.

B,C

B,D

Applying Lami's Theorem, we get
$\frac {m g}{\sin 9 0 ^ {\circ}} = \frac {T _ {1}}{\sin 1 5 0 ^ {\circ}} = \frac {T _ {2}}{\sin 1 2 0 ^ {\circ}}$
$\Rightarrow T _ {1} = m g \sin (1 5 0 ^ {\circ}) = 5 0 \times \frac {1}{2} = 2 5 \mathrm{N}$
$\Rightarrow T _ {2} = m g \sin 1 2 0 ^ {\circ} = 5 0 \times \frac {\sqrt {3}}{2} = 2 5 \sqrt {3} \mathrm{N}$
Hence, (A) and (D) are correct.

Since friction is absent everywhere, so
$a = \frac {F}{M}$
$\mathrm{Since} s = \frac {1}{2} a t ^ {2}$
$\Rightarrow \quad l = \frac {1}{2} \left(\frac {F}{M}\right) t ^ {2}$
$\Rightarrow t = \sqrt {\frac {2 l M}{F}}$
Hence, (B) and (D) are correct.

$f = m _ {A} a = 4 0 \mathrm{N}$
$f \leftarrow \boxed {B} \rightarrow F$
$F - f = m _ {B} a = 5 0 \times 3 \mathrm{N} = 1 5 0 \mathrm{N}$
$\Rightarrow F = 1 5 0 \mathrm{N} + 4 0 \mathrm{N} = 1 9 0 \mathrm{N}$
Hence, (A), (B) and (C) are correct.

The normal reaction on the wedge can be due to the wall, ground and the block

From free body diagrams of $m_{1}$ and $m_{2}$ , we get
$N _ {3} = m _ {2} g \cos \alpha$
$N _ {3} \sin \alpha = N _ {1}$
$N _ {2} = m _ {1} g + N _ {3} \cos \alpha$
$\Rightarrow N _ {1} = m _ {2} g \sin \alpha \cos \alpha$
$\Rightarrow N _ {2} = m _ {1} g + m _ {2} g \cos^ {2} \alpha$
Hence, (A), (B) and (C) are correct.

(i) $F = \frac{dp}{dt}$ (Newton's Second Law)
(ii) Newton's Third Law
Hence, the correct answer is (B).

$f = mg\sin \theta$
... (1)
$M = mg\cos \theta$
$R = \sqrt {N ^ {2} + f ^ {2}} = m g$

Hence, the correct answer is (A).

Conceptual
Hence, the correct answer is (D).

Law of inertia.
Hence, the correct answer is (A).