NLM

Let $T$ be the tension in the rope, then for the block, we have
$T = M g\tag{... (1}$
If N is the normal reaction between the man and the ground, then for the man, we have
$N + T \sin (6 0 ^ {\circ}) = 6 0 g$
$\Rightarrow N = 6 0 g - T \sin (6 0 ^ {\circ})\tag{... (2}$
Also $T\cos (60^{\circ}) = \mu N$
... (3)
Solving these three equations we get
$M = 3 2. 1 5 \mathrm{kg} \approx 3 2 \mathrm{kg}$

Hence, the correct answer is (D).

$F \cos \phi - \mu N = M a\tag{... (1}$
$F \sin \phi + N = M g$
$\Rightarrow N = M g - F \sin \phi$........(2)
Substituting (2) in (1), we get
$F \cos \phi - \mu M g + \mu F \sin \phi = M a$
$\Rightarrow a = \frac {F}{M} (\cos \phi + \mu \sin \phi) - \mu g$
Hence, the correct answer is (D).

According to Newton's Third Law to every action there is equal and opposite reaction and both act on two different bodies. But in this case both are acting on the same body and hence they cancel so that the boat remains stationary.
Hence, the correct answer is (D).

B

$\begin{array}{l} m a _ {0} \cos \alpha = m g \sin \alpha \\ \Rightarrow \quad a _ {0} = g \tan \alpha \\ \Rightarrow \quad \alpha = \tan^ {- 1} \left(\frac {a _ {0}}{g}\right) \end{array}$
Hence, the correct answer is (A).

Thrust on the block = F = v $\frac{dm}{dt}$
$\Rightarrow F = (5) (1) = 5 N$
Acceleration of the block = $\frac{F}{M}$ = $\frac{5}{2}$ ms $^{-2}$
Hence, the correct answer is (B).

$F = \frac{mv_x^2}{R} = \frac{mv_0^2\cos^2\alpha}{R}$
Hence, the correct answer is (B).

$kx = 2mg$
By Law of Conservation of Energy
$\begin{array}{r l} & M g x = \frac {1}{2} k x ^ {2} \\ \Rightarrow & k x = 2 M g \\ \Rightarrow & M = m \end{array}$
Hence, the correct answer is (B).

$m g - U = m a _ {0}$
$\text {and} U - (m - \Delta m) g = (m - \Delta m) a _ {0}$
$\Rightarrow \Delta m = \left(\frac {2 a _ {0}}{g + a _ {0}}\right) m $
Hence, the correct answer is (C).

Since windows are closed and the balloon is filled with helium which is lighter than air.
The concept of pseudo force cannot be applied here as in this case the density of body is much less than the density of surrounding air. So when car takes a left turn the expected answer is that Helium balloon will be pushed to the right whereas actually it will be pushed to the left.
Hence, the correct answer is (B).

The laws of Physics are the same in all inertial frames or the frames moving with constant velocity with respect to an inertial frame.
Hence, the correct answer is (B).

$\begin{array}{r l} & F = \sqrt {T ^ {2} + T ^ {2} - 2 T ^ {2} \cos 2 \theta_ {0}} \\ \Rightarrow & F = \sqrt {2} T \sqrt {1 - \cos 2 \theta_ {0}} \\ \Rightarrow & F = \sqrt {2} T \sqrt {2} \sin \theta_ {0} \\ \Rightarrow & F = 2 T \sin \theta_ {0} \end{array}$
Hence, the correct answer is (C).

For the man not to slip i.e. for rotational equilibrium to exist, the algebraic sum of the moments of the forces acting on the man must be zero. Let $N$ be the normal reaction exerted by the earth on the man. The following forces are acting on the man.
(a) Weight $mg$ acting vertically downwards from the centre of gravity $G$ of the man.
(b) Normal reaction N exerted by earth on the man along OG for rotational equilibrium to exist.

For the man not to slip.
$\begin{array}{l l} & N \sin \theta_ {0} \leq f _ {\text { limiting }} \\ \text { But } & f _ {\text { limiting }} = \mu \big (N \cos \theta_ {0} \big) \\ \Rightarrow & N \sin \theta_ {0} \leq \mu N \cos \theta_ {0} \\ & \tan \theta_ {0} \leq \mu \end{array}\tag{... (1}$
Hence, the correct answer is (B).

The vertically hanging curtains bend forward thus indicating that the pseudo force acting on the curtain points in the forward direction. For this to happen the plane must be slowing down for landing.
Hence, the correct answer is (C).

$F \cos \theta = \mu N$ and
$N + F \sin \theta = m g\tag{... (1}$
$\Rightarrow N = m g - F \sin \theta\tag{... (2}$
Substituting (2) in (1), we get
$F \cos \theta = \mu m g - \mu F \sin \theta$
$\Rightarrow F = \frac {\mu m g}{\cos \theta + \mu \sin \theta}\tag{... (3}$
Let $Z = \text{Denominator} = \cos \theta + \mu \sin \theta$

For F to be MINIMUM, Z must be MAXIMUM
$\begin{array}{l l} \Rightarrow & \frac {d Z}{d \theta} = 0 \\ \Rightarrow & - \sin \theta + \mu \cos \theta = 0 \\ \Rightarrow & \tan \theta = \mu \\ \Rightarrow & \theta = \tan^ {- 1} \mu \end{array}$
and $F_{\min}=\frac{\mu mg}{\sqrt{1+\mu^{2}}}$
Hence, the correct answer is (C).

Let a be the acceleration of the block with respect to prism. Let the prism move backward with acceleration $a_{0}$ , then a fictitious (pseudo) force acts on the mass (see figure).

For the mass to move down with acceleration a, we have
$\begin{array}{r l} & m a _ {0} \cos \theta + m g \sin \theta = m a \\ \Rightarrow & a = a _ {0} \cos \theta + g \sin \theta \end{array}\tag{... (1}$
Also the block is in equilibrium perpendicular to the incline i.e.
$N _ {1} + m a _ {0} \sin \theta = m g \cos \theta$
For the prism, we have that it is placed on the ground (an inertial frame) with no fictitious force acting on the prism. Following forces are acting on the prism.

(a) Weight of the prism Mg.
(b) Normal reaction $N_{2}$ exerted by the surface on the prism.
(c) Normal reaction $N_{1}$ exerted by the block placed on the prism.
$N_{1}\sin \theta$ is the component responsible for horizontal motion of the prism. So,
$N _ {1} \sin \theta = M a _ {0}$
$\Rightarrow N _ {1} = \frac {M a _ {0}}{\sin \theta}\tag{... (3}$
Substituting (3) in (2), we get
$\begin{array}{l} \frac {M a _ {0}}{\sin \theta} + m a _ {0} \sin \theta = m g \cos \theta \\ \Rightarrow \quad a _ {0} = \left(\frac {m \cos \theta}{M + m \sin^ {2} \theta}\right) g \sin \theta \end{array}\tag{... (4}$
Substituting $a_{0}$ in (1) we get
$a = \left(\frac {M + m}{M + m \sin^ {2} \theta}\right) g \sin \theta$
Hence, the correct answer is (C).
OBJECTIVE TRICK
if $\theta \to \frac{\pi}{2}$ , then $m$ (i.e. the block) will fall freely with
acceleration g. This is satisfied by both (A) and (C). To check which one of the two we can further think that if M is very heavy as compared to m, then M must not move and acceleration of m w.r.t. M is g sin $\theta$ . This is satisfied by (C) only. If we can directly think the second way first, then we can even solve the problem without any cumbersome steps.

Since $x = k t ^ {2} = 1. 7 3 2 t ^ {2}$
$ \Rightarrow x = \sqrt {3} t ^ {2}$
$\Rightarrow \dot {x} = 2 \sqrt {3} t$
$\Rightarrow \ddot {x} = a = 2 \sqrt {3} \mathrm{ms} ^ {- 2} $
$\Rightarrow a = g (\sin 4 5 - \mu \cos 4 5) = 2 \sqrt {3}$
$ \Rightarrow g \frac {1}{\sqrt {2}} (1 - \mu) = 2 \sqrt {3}$
$\Rightarrow 1 - \mu = \frac {2 \sqrt {6}}{g} $
$\Rightarrow \quad \mu = 1 - \frac {2 \sqrt {6}}{g} = 1 - \frac {2 (2 . 4 5)}{9.8}$
$\Rightarrow \mu = 0. 5$
Hence, the correct answer is (A).

Consider the block not to slide, then it must have an acceleration same as that of the plane. Then

$f \cos \alpha - N \sin \alpha = m a_{0}$
$\text{and } f \sin \alpha + N \cos \alpha = mg \dots (2)$
$\text{From (1) and (2), we get}$
$f = m(a_{0} \cos \alpha + g \sin \alpha) \dots (3)$
$\text{and } N = m(g \cos \alpha - a_{0} \sin \alpha) \dots (4)$
$\Rightarrow \frac{f}{N} = \frac{a_{0} \cos \alpha + g \sin \alpha}{g \cos \alpha - a_{0} \sin \alpha}$
$\Rightarrow \frac{f}{N} = \frac{a_{0} + g \tan \alpha}{g - a_{0} \tan \alpha}$
$\text{Further } \frac{f}{N} = \mu_{s} = \tan \theta \text{ in the absence of slipping}$
$\Rightarrow \frac{a_{0} + g \tan \alpha}{g - a_{0} \tan \alpha} \leq \tan \theta$
$\Rightarrow a_{0} \leq g \left[ \frac{\tan \theta - \tan \alpha}{1 + \tan \alpha \tan \theta} \right]$
$\Rightarrow a_{0} \leq g \tan(\theta - \alpha) \tag{for no slipping}$
$\text{So for the block to slide } a_{0} > g \tan(\theta - \alpha)\text{.}$
Hence, the correct answer is (C).

Let $N_{1}$ be normal reaction between the man and the board
Let $N_{2}$ be normal reaction between the board and the floor.
For man: $F + N_{1} = Mg$
For board: $mg + N_{1} = N_{2}$
... (1)
$F = \mu N _ {2}\tag{... (2}$
... (3)
Hence, the correct answer is (D).

A

$mg(\sin\theta + \mu\cos\theta) = 2mg(\sin\theta - \mu\cos\theta)$
$\Rightarrow \tan \theta = 3 \mu$
$\Rightarrow \quad \tan \theta = 3 \tan \phi$
$\left\{\because \mu = \tan \phi \right\}$
Hence, the correct answer is (C).

Horizontal acceleration of the system is
$a = \frac {F}{2 m + m + 2 m} = \frac {F}{5 m}$

FBD A
$F - N _ {1} = (2 m) a$
$\Rightarrow N _ {1} = F - \frac {2 F}{5} = \frac {3 F}{5}$
$\longrightarrow a = \frac {F}{5 \mathrm{m}}$

FBD B
$N _ {1} - N _ {2} = m a$
$\Rightarrow N _ {2} = \frac {3 F}{5} - \frac {F}{5} = \frac {2 F}{5}$
Now B will not slide downwards if

$\begin{array}{r l} & {\mu N _ {2} \geq m _ {B} g} \\ {\Rightarrow} & {(\mu) \bigg (\frac {2}{5} F \bigg) \geq m g} \\ {\Rightarrow} & {F \geq \frac {5}{2 \mu} m g} \end{array}$
$\mathrm{So}, F _ {\mathrm{min}} = \frac {5 m g}{2 \mu}$
Hence, the correct answer is (B).

For $A$ , we have
$y _ {1} ^ {2} = l ^ {2} + x _ {1} ^ {2}$
$\Rightarrow 2 y _ {1} \frac {d y _ {1}}{d t} = 2 x _ {1} \frac {d x _ {1}}{d t}$
$\Rightarrow \quad \frac {d y _ {1}}{d t} = \frac {x _ {1}}{y _ {1}} \left(\frac {d x _ {1}}{d t}\right)$

$\begin{array}{l} \text {Since} \frac {x _ {1}}{y _ {1}} = \cos \alpha \\ \Rightarrow \frac {d y _ {1}}{d t} = v _ {1} \cos \alpha \end{array}$
... (1)
Similarly for $B$ , we have
$\begin{array}{r l} & y _ {2} ^ {2} = l ^ {2} + x _ {2} ^ {2} \\ \Rightarrow & 2 y _ {2} \frac {d y _ {2}}{d t} = 2 x _ {2} \frac {d x _ {2}}{d t} \\ \Rightarrow & \frac {d y _ {2}}{d t} = \frac {x _ {2}}{y _ {2}} \left(\frac {d x _ {2}}{d t}\right) \\ \text {Since} & \frac {x _ {2}}{y _ {2}} = \cos \beta \\ \Rightarrow & \frac {d y _ {2}}{d t} = v _ {2} \cos \beta \end{array}\tag{... (2}$
Also from our knowledge of constrained motion, we observe that
$\begin{array}{l l} & y _ {1} + y _ {2} = \text { constant } \\ \Rightarrow & \frac {d y _ {1}}{d t} + \frac {d y _ {2}}{d t} = 0 \\ \Rightarrow & \left| \frac {d y _ {1}}{d t} \right| = \left| \frac {d y _ {2}}{d t} \right| \\ \Rightarrow & v _ {1} \cos \alpha = v _ {2} \cos \beta \\ \Rightarrow & \frac {v _ {2}}{v _ {1}} = \frac {\cos \alpha}{\cos \beta} \end{array}$
Hence, the correct answer is (D).

Since one end of pulley $P_{1}$ is fixed and the other end moves with an acceleration of $6 \, ms^{-2}$ . So, the acceleration of pulley is
$a_{P_{1}}=\frac{6+0}{2}=3\ ms^{-2}$ , towards left
Also, if $P_{1}$ moves, say left by $x$ , then $B$ moves up by $\frac{x}{3}$ . Hence, acceleration of $B$ must be one third
acceleration of $P_{1}$ . Hence
$a _ {B} = \frac {(a) _ {P _ {1}}}{3} = 1 \mathrm{ms} ^ {- 2}$
Hence, the correct answer is (D).

$\begin{array}{l} \frac {1}{K _ {s}} = \frac {1}{K} + \frac {1}{3 K} + \frac {1}{9 K} + \frac {1}{2 7 K} + \dots \\ \Rightarrow \quad \frac {1}{K _ {s}} = \frac {1}{K} \left(1 + \frac {1}{3} + \frac {1}{9} + \frac {1}{2 7} + \dots\right) \\ \Rightarrow \quad \frac {1}{K _ {s}} = \frac {1}{K} \left(\frac {1}{1 - \frac {1}{3}}\right) \\ \Rightarrow \quad K _ {s} = \frac {2 K}{3} \end{array}$
Hence, the correct answer is (C).