NLM

A

Taking upward direction as positive
Given $a_{A}=1\ ms^{-2}$ , $a_{B}=7\ ms^{-2}$ and $a_{C}=2\ ms^{-2}$
Let acceleration of pulley P is a upwards, then acceleration of pulley Q will be a downwards
$\begin{array}{r l} & {\Rightarrow a _ {P} = a} \\ & {\mathrm{and} a _ {Q} = - a} \\ & {\mathrm{Now} a _ {A P} = - a _ {B P}} \\ & {\Rightarrow a _ {A} - a _ {P} = a _ {P} - a _ {B}} \\ & {\Rightarrow 1 - a = a - 7} \\ & {\Rightarrow 2 a = 8} \\ & {\Rightarrow a = 4 \mathrm{ms} ^ {- 2}} \\ & {\mathrm{Further} a _ {C Q} = - a _ {D Q}} \\ & {\Rightarrow a _ {C} - a _ {Q} = a _ {Q} - a _ {D}} \\ & {\Rightarrow 2 - (- a) = - a - a _ {D}} \\ & {\Rightarrow a _ {D} = - 2 a - 2 = - 1 0 \mathrm{ms} ^ {- 2}} \end{array}$
So, acceleration of D is $10 \, ms^{-2}$ downwards.
Hence, the correct answer is (C).

Thrust = 2T = (4mM/M + m)g
⇒ Thrust = $\frac{4mg}{1+\frac{m}{M}}$ Since, $\frac{m}{M} \rightarrow 0$ ⇒ Thrust ≈ 4mg
Hence, the correct answer is (C).

From symmetrical behaviour of the arrangement, we observe that tension in both the branches BC and CD to be equal.
$\begin{array}{r l} \Rightarrow & 2 T \cos \beta = m g \\ \Rightarrow & T = \frac {m g}{2 \cos \beta} = \frac {m g}{2} \sec \beta \\ \Rightarrow & T = \frac {m g}{2} \sqrt {1 + \tan^ {2} \beta} \\ \Rightarrow & T = \frac {m g}{2} \sqrt {1 + \left(\frac {1 2}{5}\right) ^ {2}} \\ \Rightarrow & T = \frac {1 3}{1 0} m g \end{array}$
Hence, the correct answer is (B).

Block starts sliding over the surface when F is the maximum static force of friction
$\begin{array}{r l} \Rightarrow & K t = \mu_ {s} m g \\ \Rightarrow & t = \frac {\mu_ {s} m g}{K} \end{array}$
for $t > \frac{\mu_{s} mg}{K}$ , net acceleration of block is
$a = \frac {F - \mu_ {k} m g}{m} = \frac {K t}{m} - \mu_ {k} g$
which is a straight line with positive slope and negative intercept.
So, for $t \leq \frac{\mu_s mg}{K}$ , acceleration of particle is zero
and for $t > \frac{\mu_{s}mg}{K}$ , acceleration is $a = \frac{Kt}{m} - \mu_{k}g$ .
Hence, the correct answer is (D).

$\begin{array}{l} a = \frac {F}{m _ {1} + m _ {2} + m _ {3}} \\ \text {For equilibrium} m _ {1} a = m _ {2} g \\ \Rightarrow m _ {1} \left(\frac {F}{m _ {1} + m _ {2} + m _ {3}}\right) = m _ {2} g \\ \Rightarrow F = \frac {m _ {2} g}{m _ {1}} \big (m _ {1} + m _ {2} + m _ {3} \big) \end{array}$
Hence, the correct answer is (C).

$kx_{1}=2g$
$\begin{array}{l} k x _ {1} = 2 g \\ \Rightarrow x _ {1} = \frac {2 g}{k} \end{array}\tag{... (1}$
{ k = force constant of spring }
$\frac {3 g - k x _ {2}}{3} = \frac {3 g - 2 g}{5}$
$\Rightarrow x _ {2} = \frac {1 2}{5} \frac {g}{k}\tag{... (2}$
$\frac {k x _ {3} - g}{1} = \frac {2 g - 1 g}{3}$
$\Rightarrow x _ {3} = \frac {4 g}{3 k}\tag{... (3}$
From (1), (2) and (3) we see that
$x _ {2} > x _ {1} > x _ {3}$
Hence, the correct answer is (B).

Impulse = $\left(\begin{array}{l} \text{Area under force time graph} \\ \text{under the specified interval} \end{array}\right)$
Hence, the correct answer is (A).

From constraint relations we can see that acceleration of block B is
$a _ {B} = \left(\frac {a _ {A} + a _ {C}}{2}\right) \mathrm{withpropersigns}$
Hence $a_{B} = \left(\frac{3 - 12t}{2}\right) = 1.5 - 6t$
$\Rightarrow \frac {d v _ {B}}{d t} = 1. 5 - 6 t$
$\Rightarrow \quad \int_ {0} ^ {v _ {B}} d v _ {B} = \int_ {0} ^ {t} (1. 5 - 6 t) d t$
$\Rightarrow v _ {B} = 1. 5 t - 3 t ^ {2}$
$\Rightarrow v _ {B} = 0 \mathrm{at} t = \frac {1}{2} \mathrm{s}$
Hence, the correct answer is (D).

Let $T$ be the tension in the string then
For body $A:F = T + \mu mg$
For body $B:T=\mu mg+\mu(m+M)g$
$\Rightarrow \quad F = \mu (3 m + M) g$
Hence, the correct answer is (D).

At the moment shown in figure net force on $P$ observed by $Q$ is zero.

Hence, the correct answer is (C).

$\frac {g}{2} = \frac {n - (1 6 - n) \left(\frac {1}{3}\right)}{1 6} g$
$ \Rightarrow = \frac {3 n - (1 6 - n)}{3} $ $\Rightarrow \quad 2 4 = 3 n - 1 6 + n$
$\Rightarrow n = 1 0$
Hence, the correct ans
Hence, the correct answer is (C).

Since the surfaces in contact are smooth, so force of friction between the block and the plane is zero. Hence, contact force is really the normal reaction between the two.

In the first case $F \sin 30^{\circ} = ma$
and $F \cos 30^{\circ} = mg \Rightarrow F = \frac{mg}{\cos 30^{\circ}}$
and in the second case
$\begin{array}{r l} & f = m g \cos 3 0 ^ {\circ} \\ \Rightarrow & \frac {F}{f} = \frac {1}{\cos^ {2} 3 0 ^ {\circ}} = \frac {4}{3} \end{array}$
Hence, the correct answer is (B).

$\left( \begin{array}{c} \text {Scale} \\ \text {reading} \\ \text {at time t} \end{array} \right) = \left( \begin{array}{c} \text {Average} \\ \text {force} \\ \text {exerted} \\ \text {by pebbles} \end{array} \right) + \left( \begin{array}{c} \text {Weight} \\ \text {of pebbles} \\ \text {collected} \\ \text {in box at time t} \end{array} \right)$
$\left( \begin{array}{c} \text { Scale } \\ \text { reading } \\ \text { at time } t \end{array} \right) = F _ {a v} + (m g) (n t) = m n \Big (\sqrt {2 g h} + g t \Big)$
Hence, the correct answer is (D).

Considering translational equilibrium of rod in horizontal direction.

$F _ {1} - F _ {2} = F \Rightarrow F _ {1} > F _ {2}$
Therefore, block B will tend to move first. Hence, the correct answer is (B).

Acceleration of body down the plane is
$\begin{array}{r l} & a = (g \sin 3 0 ^ {\circ}) (\cos 6 0 ^ {\circ}) \\ \Rightarrow & a = 1 0 \left(\frac {1}{2}\right) \left(\frac {1}{2}\right) = 2. 5 \mathrm{ms} ^ {- 2} \\ \Rightarrow & t = \sqrt {\frac {2 l}{a}} = \sqrt {\frac {2 \times 5}{2 . 5}} = 2 \mathrm{s} \end{array}$
$\left\{\because l = \frac {1}{2} a t ^ {2} \right\}$
Hence, the correct answer is (B).

Let us consider A and B as a system. Since, there is no vertical force in the upward direction to support their weight, hence the system cannot stay in equilibrium. Hence, the correct answer is (D).

Let a be the acceleration of the elevator. Drawing free body diagram of the lower portion.

Let the mass of lower portion of string be $m' = \frac{m}{L}(L - l)$ Equation of motion gives
$T - M g - m ^ {\prime} g = (M + m ^ {\prime}) a$
$\begin{array}{c} T \\ m ^ {\prime} \end{array} \quad \begin{array}{c} a \\ M \end{array}$
$\Rightarrow T - M g - \frac {m g}{L} (L - l) = \left(M + \frac {m}{L} (L - l)\right) a$
$\Rightarrow \quad a = \frac {T}{\left(M + m - \frac {m l}{L}\right)} - g$
Hence, the correct answer is (C).

$T\cos \theta = mg\cos \alpha$ and
... (1)
$T \sin \theta = m a _ {0} + m g \sin \alpha$.......(2)
$\Rightarrow \tan \theta = \frac {a _ {0} + g \sin \alpha}{g \cos \alpha}$
$\Rightarrow \theta = \tan^ {- 1} \left(\frac {a _ {0} + g \sin \alpha}{g \cos \alpha}\right)$
$\Rightarrow \theta = \tan^ {- 1} \left(\frac {\frac {g}{2} + g \sin (3 0 ^ {\circ})}{g \cos (3 0 ^ {\circ})}\right)$
$\Rightarrow \theta = \tan^ {- 1} \left(\frac {2}{\sqrt {3}}\right)$
Hence, the correct answer is (B).

When centre of gravity lies vertically below its centre, the restoring torque while displaced from its equilibrium position, brings it back to its equilibrium position as shown below.

C = centre of sphere
$G =$ centre of gravity
and $\tau =$ restoring torque
Hence, the correct answer is (B).

mg = 2T sin 45
$\begin{array}{l l} \Rightarrow & m g = \sqrt {2} T \\ & T _ {1} \cos \theta = T \cos 4 5 \\ \Rightarrow & T _ {1} \cos \theta = \frac {T}{\sqrt {2}} = \frac {m g}{2} \end{array}\tag{... (1}$
$\left\{\because T = \frac {m g}{\sqrt {2}} \right\}$
Further, $Mg + T\cos 45 = T_{1}\sin \theta$
$\Rightarrow T _ {1} \sin \theta = M g + \frac {m g}{\sqrt {2}} \frac {1}{\sqrt {2}}$
$\Rightarrow T _ {1} \sin \theta = M g + \frac {m g}{2}\tag{..(2}$
$\Rightarrow \quad \tan \theta = \frac {M g + \frac {m g}{2}}{\frac {m g}{2}} = 1 + \frac {2 M}{m} \quad \{\text { Divide (2) by (1) } \}$
Hence, the correct answer is (A).

From constraint relations we can see that
$a _ {A} = 2 a _ {B}$
so, let $a_{B} = a$ then $a_{A} = 2a$
Free body diagram of B gives
$4 0 - T = 4 a\tag{... (1}$
Free body diagram of $A$ is
$\frac {T}{2} - 1 0 = 4 a$
$\Rightarrow T - 2 0 = 8 a\tag{... (2}$
Solving equations (1) and (2) we get

$2 a = \frac {1 0}{3} \mathrm{ms} ^ {- 2}$
$\Rightarrow a _ {A} = \frac {1 0}{3} \mathrm{ms} ^ {- 2}$
Hence, the correct answer is (A).

Let acceleration of $A$ be $a_1$ downwards and that of $B$ is $a_2$ upwards and let $T$ be the tension in the string. Then equations of motion of $A$ and $B$ gives
$m _ {1} g - T = m _ {1} a _ {1}$
$2 T - m _ {2} g = m _ {2} a _ {2}\tag{... (1}$
$\ldots(2)$
Equilibrium of pulley $C$ gives
2 T - T = 0
$\Rightarrow T = 0$
Substituting in (1) and (2)
$a _ {1} = g \text { and } a _ {2} = - g$
So, both the blocks are falling freely. Hence, the correct answer is (D).

$T \sin 45 = ma$ (For A)
... (1)
$m g - T \sin 4 5 = m a (\mathrm{For} B)$....(2)
$\Rightarrow T = \frac {m g}{2 \sin 4 5} = \frac {m g}{\sqrt {2}}$

Hence, the correct answer is (C).

For the first 3 second, the acceleration of the combined system is
$a = \left(\frac {5 - m}{5 + m}\right) g\tag{... (1}$
Velocity gained by m (starting from rest) at the end of 3 second is
v = 0 + a t
$\Rightarrow = 3 \left(\frac {5 - m}{5 + m}\right) g$.....(2)
At t = 3s, the string is cut (i.e., at position B).
As soon as the string is cut the mass m starts moving under the retardation g and covers a further distance of $\frac{49}{90}$ m before its velocity becomes zero (at c) and then it begins to descend.

$\Rightarrow 0 - v ^ {2} = 2 (- g) \frac {4 9}{9 0}$
$\Rightarrow 9 \left(\frac {5 - m}{5 + m}\right) ^ {2} g ^ {2} = - 2 g \left(\frac {4 9}{9 0}\right)$
$\Rightarrow \left(\frac {5 - m}{5 + m}\right) ^ {2} = \frac {1}{8 1}$
$\Rightarrow \frac {5 - m}{5 + m} = \pm \frac {1}{9}$
$\Rightarrow \quad m = 4 \text{ kg or } m = 6.25 \text{ kg}$
But $m = 6.25 \, kg$ is not acceptable, as the value of lighter mass cannot exceed 5 kg.
Hence, the correct answer is (A).