NLM

Since, the block is at rest under two forces
(i) weight of block.
(ii) contact force from the plane (resultant of force of friction and normal reaction).
Contact force should be equal to weight (or 30 N) in upward direction. Because under the action of two forces, a body remains in equilibrium when both the forces are equal in magnitude, but opposite in direction, i.e. 30 N.

When the system of two blocks of masses $M$ and $m$ is either at rest or moving with the same speed, no net force acts on them.

In this equilibrium state,
$N = m g - Q \cos \theta$

$P + Q \sin \theta = \mu N = \mu (m g - Q \cos \theta)$
$\mu = \frac {(P + Q \sin \theta)}{(m g - Q \cos \theta)}$

$m_{1}g > m_{2}g \sin\theta + \mu m_{2}g \cos\theta$
⇒
$\frac {m _ {1}}{m _ {2}} > \sin \theta + \mu \cos \theta$

$N = Mg - F\cos \theta = Mg - Mg\cos \theta = 2Mg\sin^2\frac{\theta}{2}$
Further, block can be pulled, if $F \sin\theta \geq \mu N$
$\begin{array}{l l} \Rightarrow & 2 M g \sin \frac {\theta}{2} \cdot \cos \frac {\theta}{2} \geq 2 \mu M g \sin^ {2} \frac {\theta}{2} \\ \Rightarrow & \cot \left(\frac {\theta}{2}\right) \geq \mu \end{array}$

$a = \frac{2F - F}{2m} = \frac{F}{2m}$
(towards left)
Writing equation of right hand side block,
$\begin{array}{r l} 2 F - N \sin 3 0 ^ {\circ} & = m a = \frac {F}{2} \\ \frac {N}{2} & = 2 F - \frac {F}{2} = \frac {3 F}{2} \Rightarrow N = 3 F \end{array}$

In critical case, weight of hanging part = force of friction of the part of rope lying on table.
$\frac {m}{l} \cdot l _ {1} g = \mu \frac {m}{l} (l - l _ {1}) g$
Solving above equation, we get
$l _ {1} = \left(\frac {\mu}{1 + \mu}\right) l$

Block $A$ moves due to friction. Maximum value of friction can be $\mu m_A g$ . Therefore, maximum acceleration of $A$ can be $\frac{\mu m_A g}{m_A}$ or $\mu g = 0.2g = \frac{g}{5}$ . When force $2mg$ is applied on lower block common acceleration (if both move together) will be
block, common acceleration (if both move together) will be
$\begin{array}{c} m \xrightarrow {0.2 \text { mg }} \\ 3 m \xrightarrow {2 \text { mg }} \end{array}$
$a = \frac {\mathrm{Netforce}}{\mathrm{Totalmass}} = \frac {2 m g}{4 m} = \frac {g}{2}$
Since, a = g/2 is greater than maximum acceleration of A which can be given to it by friction. Therefore, slipping will take place.
$\begin{array}{r l} & {a _ {A} = 0.2 g = \frac {g}{5}} \\ & {a _ {B} = \frac {2 m g - 0.2 m g}{3 m} = 0.6 g = \frac {3 g}{5}} \\ & {a _ {A B} = a _ {A} - a _ {B} = \left(\frac {g}{5} - \frac {3 g}{5}\right)} \\ & {\qquad = \frac {- 2 g}{5} (\mathrm{inbackwarddirection})} \end{array}$

$f_{1}$ = maximum value of friction between m and 2m = $\mu mg$ $f_{2}$ = maximum value of friction between 2m and ground $=\frac{\mu}{3}(3m)g=\mu mg$
Now, let us draw free body diagram of m and 2m in all four cases.
(i)
$\begin{array}{c} \frac {\mu}{3} m g \quad \boxed {m} \longrightarrow F = \frac {\mu}{3} m g \\ T = 0 \quad \boxed {2 m} \longrightarrow \frac {\mu}{3} m g \\ \frac {\mu}{3} m g \\ \mu m g \quad \boxed {m} \longrightarrow F = \mu m g \end{array}\tag{ii}$
$\begin{array}{c} T = 0 \\ \leftarrow \quad \boxed {2 m} \quad \mu m g \\ \mu m g \end{array}\tag{iii}$
$\mu m g \boxed {m} \longrightarrow F = 2 \mu m g$
$\begin{array}{c} T = 0 \\ \leftarrow \quad \boxed {2 m} \quad \mu m g \\ \mu m g \end{array}$
$\mu m g \boxed {m} \longrightarrow F = 3 \mu m g$
$\begin{array}{c} T = 0 \\ \leftarrow \quad \boxed {2 m} \quad \mu m g \\ \mu m g \end{array}$
As tension is zero in all four cases. So, option (c) is incorrect.

For block,
$a = g \sin \theta = \frac {g}{\sqrt {2}}$

$\begin{array}{r} a _ {H} = a _ {V} = a \cos 4 5 ^ {\circ} \\ = \frac {g}{\sqrt {2}} \times \frac {1}{\sqrt {2}} = \frac {g}{2} \\ \frac {N}{\sqrt {2}} = \frac {m g}{2} \\ m g - \frac {N}{\sqrt {2}} = \frac {m g}{2} \end{array}\tag{... (i}$
... (ii)
For wedge,

$\begin{array}{c} {N ^ {\prime} = \frac {N}{\sqrt {2}} + 2 m g} \\ {\mu N ^ {\prime} = \frac {N}{\sqrt {2}}} \end{array}\tag{... (iii}$
... (iv)
Solving above four equations, we get
$\mu = 0.2$

2T = 250

Also, $T + 0.4F = 150$
$F = 6 2.5 \mathrm{N}$

$N_{1}\sin37^{\circ}=N_{2}\sin74^{\circ}$
$\Rightarrow N_{1} = 2N_{2}\cos 37^{\circ}$

Now, $N_{1}\cos 37^{\circ} - N_{2}\cos 74^{\circ} = mg$
$\Rightarrow 2N_{2}\cos^{2}37^{\circ} - N_{2}(2\cos^{2}37^{\circ} - 1) = mg$

$T \sin \theta - mg \sin \alpha = ma$

and $T\cos \theta = mg\cos \alpha$
From these two equations, we get
$\tan \theta = \frac {a + g \sin \alpha}{g \cos \alpha} \Rightarrow \theta = \tan^ {- 1} \left(\frac {a + g \sin \alpha}{g \cos \alpha}\right)$

A

$mg \sin \theta = 10 \times 10 \times \frac{1}{2} = 50 \, N$
$f_{\mathrm{friction}} = \mu mg\cos \theta = 0.7\times 10\times 10\times \frac{\sqrt{3}}{2} = 60.62\mathrm{N}$
Since, $\mu mg \cos \theta$ is more, block will remain stationary.

A

$(f _ {2}) _ {\mathrm{max}} = \mu_ {2} m g$
$(f _ {1}) _ {\mathrm{max}} = \mu_ {1} (2 m) g$
Since,
$(f _ {1}) _ {\mathrm{max}} > (f _ {2}) _ {\mathrm{max}}$
Lower block will not move at all.

$f - mg \sin \theta = ma \sin \theta$

$\therefore f = m(g + a)\sin \theta > mg\sin \theta$

$N _ {1} = m a _ {x} = 2 m$
$N _ {2} = m (g + a _ {y}) = 1 2 m$
$\frac {N _ {1}}{N _ {2}} = \frac {1}{6}$

D

The statement given in option (b) is incorrect, which can be corrected as,
Friction opposes the relative motion of the bodies in contact not the motion.

Only Statement II is correct while Statement I is incorrect, which can be corrected as,
Tension is non-uniform even, if string is not accelerated.

$F_{\mathrm{net}} = ma$
For $1\mathrm{kg}$ mass $\boxed{1\mathrm{kg}}\uparrow_{a}\Rightarrow T - g = a$
For 2 kg mass $\begin{array}{c}\uparrow T\\ \boxed{2\ kg}\\ \downarrow g\end{array}$ $\downarrow^{a}$ $\Rightarrow$ 2g - T = 2a
Solving, we get $a=\frac{10}{3}$ m/s $^{2}$
Force on $1\mathrm{kg}, F_{1} = \frac{10}{3}\mathrm{N}$ and on $2\mathrm{kg}, F_{2} = \frac{20}{3}\mathrm{N}$
Hence, Statement I is correct while Statement II is incorrect

As forces at two ends of string are different, so tension on it decreases from 2F to F in moving from A to B.
In case of massless string, $a = \frac{2F - F}{0} = \infty$
Hence, both statements are correct.