NLM

Net torque must be zero, so
$\begin{array}{r l} & {\vec {\tau} _ {N} + \vec {\tau} _ {f r} = \vec {0}} \\ {\Rightarrow} & {\vec {\tau} _ {N} = - \vec {\tau} _ {f r}} \end{array}$
Hence, the correct answer is (D).

When the entire incline is smooth, then both the blocks move down with an acceleration $g \sin(30^{\circ}) = 5 \, \text{ms}^{-2}$ . However, if A is smooth and B is rough then initial acceleration of A will be $g \sin \theta$ , whereas that of B is less than $g \sin \theta$ and hence there will be elongation in the spring.
Hence, (A) and (D) are correct.

Newton's Laws are valid in all inertial frames i.e., frames either at rest or frames moving with constant velocity. Further a frame moving with constant velocity with respect to an inertial frame is also Inertial. Earth is not an inertial frame due to rotation about its axis and revolution round the sun.
Hence, (B) and (C) are correct.

$F - fy - Mg = Ma$
{at a distance y}
$\begin{array}{r l} \Rightarrow & a = \frac {F - f y - M g}{M} \\ \Rightarrow & \frac {d v}{d t} = \frac {F - f y - M g}{M} \\ \Rightarrow & v \frac {d v}{d y} = \frac {F - f y - M g}{M} \end{array}$
$\begin{array}{l l} \Rightarrow & \int_ {0} ^ {v} v d v = \frac {1}{M} \int_ {0} ^ {s} (F - f y - M g) d y \\ \Rightarrow & v = \sqrt {\frac {2 s}{M} \left(F - M g - \frac {f s}{2}\right)} \end{array}$
Hence, (A) and (C) are correct.

When lift moves with uniform velocity no pseudo force exists and hence $f = \mu mg(A \& B)$ . When lift moves down with acceleration of $4.8 ms^{-2}$ then $N = m(9.8 - 4.8) = 5 \, \text{m}$ .
So, frictional force = $\mu N = 5 \mu m$
Hence, (A), (B), (C) and (D) are correct.

The block with lower value of $\mu$ will tend to have greater acceleration down the slope.
Hence, (A) and (B) are correct.

$\begin{array}{r l} 2 T \cos \theta & = \sqrt {2} M g \\ & \Rightarrow 2 M g \cos \theta = \sqrt {2} M g \\ & \Rightarrow \cos \theta = \frac {1}{\sqrt {2}} \\ & \Rightarrow \theta = 4 5 ^ {\circ} \end{array}\tag{... (1}$
Hence, (C) is correct.

For equilibrium, we get
$T\cos \theta = mg$ and $T\sin \theta = ma$

$\Rightarrow \tan \theta = \frac{a}{g}$
$\text{Also, } T = \frac{mg}{\cos \theta} = m\sqrt{a^2 + g^2}$
$\Rightarrow T = m\sqrt{a^2 + g^2}$
Hence, (B) and (D) are correct.

Since friction always has a tendency to stop the relative motion between the blocks, so when $m_{B} > m_{A}$ , then for a frictional force f between A and B, we have

$m _ {B} g \sin \theta = T + f$.... (1)
$m _ {A} g \sin \theta = T - f$.....(2)
Subtracting (2) from (1), we get
$ {\big (m _ {B} - m _ {A} \big) g \sin \theta = 2 f}$
$\Rightarrow f = \frac{1}{2} \big(m _ {B} - m _ {A} \big) g \sin \theta$
$Now, f = 0, when, m_A = m_B$
So, (B) is correct.
$Also, for f \leq f_{max}, the blocks will remain stationary.$
$\Rightarrow \frac {1}{2} (m _ {B} - m _ {A}) g \sin \theta \leq \mu (m _ {B} g \cos \theta)$
$\Rightarrow \mu \geq \left(\frac {m _ {B} - m _ {A}}{2 m _ {B}}\right) \tan \theta$
$Hence the acceleration of the system is zero if \mu \geq \left(\frac{m_B - m_A}{2m_B}\right)\tan \theta and m_B > m_A.$
Also, from (1) and (2), we get
$T = \left(\frac {m _ {A} + m _ {B}}{2}\right) g \sin \theta$

$a = \frac {m _ {B} g}{m _ {A} + m _ {B}} = \frac {2}{7} (9. 8) \mathrm{ms} ^ {- 2}$
$\Rightarrow a = 2. 8 \mathrm{ms} ^ {- 2} = 2 8 0 \mathrm{cms} ^ {- 2}$
Since the initial velocity of A is towards left whereas acceleration is towards the right. So this acceleration acts as retardation for motion of A towards left. (Take left to right as positive).
$\begin{array}{l l} & v \big | _ {t = 1 s} = - 5 0 + (2 8 0) (1) \\ \Rightarrow & v \big | _ {t = 1 s} = + 2 3 0 \mathrm{cms} ^ {- 1} (\text { towards right }) \text { and } \\ & s \big | _ {t = 1 s} = - (5 0) (1) + \frac {1}{2} (2 8 0) (1) \\ \Rightarrow & s \big | _ {t = 1 s} = + 9 0 \mathrm{cm} \quad \{\text { displacement towards right } \} \end{array}$
So distance from pulley = 200 - 90 = 110 cm
Hence, (B) and (D) are correct.

$\begin{array}{r l} & a = \left(\frac {M - m}{M + m}\right) g, \\ & T = \left(\frac {2 M m}{M + m}\right) g \\ & \mathrm{Thrust} = 2 T = \left(\frac {4 M m}{M + m}\right) g \text {and} \\ & (M + m) g - 2 T = (M + m) a _ {c m} \\ & \Rightarrow a _ {c m} = \left(\frac {M - m}{M + m}\right) ^ {2} g \\ & \mathrm{or} a _ {c m} = \frac {M a + m (- a)}{M + m} \qquad \left\{\because a _ {c m} = \frac {m _ {1} a _ {1} + m _ {2} a _ {2}}{m _ {1} + m _ {2}} \right\} \\ & \Rightarrow a _ {c m} = \left(\frac {M - m}{M + m}\right) a = \left(\frac {M - m}{M + m}\right) ^ {2} g \end{array}\tag{30.}$
Hence, (A), (B), (C) and (D) are correct.

Since the surfaces are smooth, so for any value of $\theta$ both A and B will have the same acceleration downwards, as a result of which neither of them exerts a force on each other.
Hence, (B) and (D) are correct.

According to Impulse-Momentum Theorem Impulse = Change in momentum
$\Rightarrow F (\Delta t) = m [ v - (- u) ]$
Hence, (A) and (B) are correct.

• Take downward as positive direction.
- The total length of the thread is a constant.
- Starting from centre of A and going to centre of B we have

$\left(x _ {B} - x _ {A}\right) + x _ {B} + 2 x _ {A} + x _ {C} + \left(x _ {C} - x _ {B}\right) = \text { constant }$
$\Rightarrow x _ {B} + x _ {A} + 2 x _ {C} = \mathrm{constant}$
$\Rightarrow \quad \ddot {x} _ {B} + \ddot {x} _ {A} + 2 \ddot {x} _ {C} = 0$
$\Rightarrow a _ {B} + a _ {A} + 2 a _ {C} = 0\tag{... (1}$
Applying cause-effect equations we get
$m g - 2 T = m a _ {C} (\mathrm{For} C)\tag{... (2}$
$m g + T - 2 T = m a _ {B} (\mathrm{For} B)$
$\Rightarrow m g - T = m a _ {B}\tag{... (3}$
$m g + T - 2 T = m a _ {A} (\text { For } A)$
$\Rightarrow m g - T = m a _ {A}\tag{... (4}$
From (3) and (4), $a_{A} = a_{B}$ so (1) gives
$\Rightarrow a _ {A} = a _ {B} = - a _ {C}$.....(5)
Adding (3) and (4), we get
$2 m g - 2 T = 2 m a _ {A}$ .... (6)
Subtract (2) from (6), we get
$m g = 2 m a _ {A} - m a _ {C}$
$\Rightarrow m g = 2 m a _ {A} + m a _ {A}$
$\left\{\because a _ {A} = - a _ {C} \right\}$
$\Rightarrow m g = 3 m a _ {A}$
$\Rightarrow a _ {A} = \frac {g}{3}$
So, $a_{A} = a_{B} = \frac{g}{3}$ ; Downwards
$a _ {C} = - \frac {g}{3} \mathrm{Upwards}$
Also we get T by putting value of $a_{A}$ in (4) i.e. $T=\frac{2}{3}mg$ .
Hence, (A) and (B) are correct.

$200 - kx = 20a_{1}$
and $kx = 10(12) = 120$
$\Rightarrow a _ {1} = 4 m s ^ {- 2}$
$a = \frac {F}{m _ {1}} + \frac {F}{m _ {2}} = \frac {F}{\left(\frac {m _ {1} m _ {2}}{m _ {1} + m _ {2}}\right)} = \frac {F}{\mu}$
$\Rightarrow \quad a = \frac {2 0 0}{(2 0 / 3)}$
$\Rightarrow a = 3 0 \mathrm{ms} ^ {- 2}$
Hence, (B), (C) and (D) are correct.

Centripetal force is a force which is directed radially inwards, but this does not imply that is a force constant in magnitude and direction.

So $|\vec{F}_C| = \frac{mv^2}{r}$
Also in uniform circular motion magnitude of velocity is constant but velocity is not constant as it changes its direction continuously. Further, $\vec{F} \perp \vec{v}$ implies work done is zero, so according to Work Energy Theorem Work done = Change in K.E.
$\Rightarrow \quad 0 = \Delta (K. E.)$
$\Rightarrow \quad \mathrm{K.E.} = \text { constant }$
Hence, (C) and (D) are correct.

Concept of Pseudo-Force.
Hence, (A), (C) and (D) are correct.

$m_{1}g\sin \theta -T - \mu N_{1} = m_{1}a$
$\Rightarrow m _ {1} g \sin \theta - T - \mu m _ {1} g \cos \theta = m _ {1} a\tag{... (1}$

$T - m _ {2} g \sin \theta - \mu N _ {2} = m _ {2} a$
$\Rightarrow T - m _ {2} g \sin \theta - \mu m _ {2} g \cos \theta = m _ {2} a\tag{... (2}$
Adding (1) and (2), we get
$a = \left[ \frac {(m _ {1} - m _ {2}) \sin \theta - \mu (m _ {1} + m _ {2}) \cos \theta}{m _ {1} + m _ {2}} \right] g$
for $m_{1}$ downwards along the incline.
So for $m_{2}$ acceleration is directed up the incline i.e.
$a _ {2} = - a$
Hence, (A) and (B) are correct.

$\left( \begin{array}{c} T i m e \text { taken by } \\ \mathrm{m} _ {2} \text { to reach } \\ \text { maximum height } \end{array} \right) = \frac {1}{2} \left( \begin{array}{c} \text { Time taken by } \\ \mathrm{m} _ {1} \text { to reach } \\ \text { maximum height } \end{array} \right)$
$a _ {1} = \left(\frac {m _ {1} - m _ {2} \sin 3 0}{m _ {1} + m _ {2}}\right) g \mathrm{and}$
$a _ {2} = \left(\frac {m _ {2} - m _ {1} \sin 3 0}{m _ {1} + m _ {2}}\right) g$
Since $l=\frac{1}{2}a_{1}t_{1}^{2}=\frac{1}{2}a_{2}t_{2}^{2}$
$\Rightarrow \frac {1}{2} \left(\frac {2 m _ {1} - m _ {2}}{m _ {1} + m _ {2}}\right) \frac {t _ {2} ^ {2}}{4} = \frac {1}{2} \left(\frac {2 m _ {2} - m _ {1}}{m _ {1} + m _ {2}}\right) t _ {2} ^ {2}$
$\Rightarrow \left(\frac {2 m _ {1} - m _ {2}}{m _ {1} + m _ {2}}\right) \frac {1}{4} = \left(\frac {2 m _ {2} - m _ {1}}{m _ {1} + m _ {2}}\right)$
$\Rightarrow 8 m _ {2} - 4 m _ {1} = 2 m _ {1} - m _ {2}$
$\Rightarrow 9 m _ {2} = 6 m _ {1}$
$\Rightarrow \quad \frac {m _ {1}}{m _ {2}} = \frac {3}{2}$
Hence, (A) and (C) are correct.

A,D

$T_{\mathrm{upper}} = (2.9 + 0.2 + 1.9)(9.8 + 0.2)$
$\Rightarrow T _ {\mathrm{upper}} = 5 0 \mathrm{N} \text { and }$
$T _ {\mathrm{lower}} = (1. 9 + 0. 1) (9. 8 + 0. 2)$
$\Rightarrow T _ {\mathrm{lower}} = 2 0 \mathrm{N}$
Hence, (A) and (D) are correct.

$T _ {2} \cos \phi = M g$
$T _ {2} \sin \phi = M g$ .... (1)
$T _ {1} \sin \theta = T _ {2} \sin \phi$....(2)
$T _ {1} \cos \theta = T _ {2} \cos \phi + M g$.... (3)
From (1) and (2), we get
$\tan \phi = 1$3 OPTION (A)<br /> $T _ {2} = \sqrt {2} M g$ OPTION (C)
From (3), we get
$T _ {1} \sin \theta = \sqrt {2} M g \frac {1}{\sqrt {2}}$
$T _ {1} \sin \theta = M g$
From (4), we get
$T _ {1} \cos \theta = \sqrt {2} M g \left(\frac {1}{\sqrt {2}}\right) + M g$ ....(5)
$\Rightarrow T _ {1} \cos \theta = 2 M g$ .... (6)
From (5) and (6), we get
$\tan \theta = \frac {1}{2} = 0. 5$ OPTION (B)
and $T_{1}=\sqrt{5}Mg$
{OPTION (D)}
Hence, (A), (B), (C) and (D) are correct.

Concept of Pseudo Force.
Hence, (A) and (C) are correct.

$a_{BA}=3a_{0}$
{using constraint relations}
$a _ {B G} = \sqrt {a _ {0} ^ {2} + 9 a _ {0} ^ {2} + 6 a _ {0} ^ {2} \cos \theta}$
$a _ {0} \sqrt {1 0 + 6 \cos \theta}$
Hence, (A) and (D) are correct.

With respect to wedge, the acceleration of block is same as that of wedge with respect to ground.
Hence, (A) and (B) are correct.