NLM

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.

$F = kt$
$\Rightarrow m \frac {d v}{d t} = k t$
$\Rightarrow m v = \frac {k t ^ {2}}{2}\tag{... (1}$
$\mathrm{Since,} a = \frac {F}{m}$
$\Rightarrow a = \frac {k}{m} t\tag{... (2}$
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.