NLM

Since $\sum F_{x} = 0$
$H - T \cos (3 0 ^ {\circ}) = 0$
$\Rightarrow \quad H = (392)\cos(30^{\circ}) = 339 \text{ N to the right}$
Hence, the correct answer is (C).

Since $\sum F_y = 0$
$V + T \sin (3 0 ^ {\circ}) - 2 0 0 = 0$
$\Rightarrow V = 1 9 6 - (3 9 2) \sin (3 0 ^ {\circ}) = 0$
Hence, the correct answer is (A).

FBD of particle in ground frame of reference is shown in figure. We observe that

$\tan \theta = \frac {r}{R - h}$
Also, $N\cos \theta = mg$ and
$N \sin \theta = m r \omega^ {2}\tag{... (1}$
$\ldots(2)$
Dividing equation (2) by (1), we get
$\tan \theta = \frac {r \omega^ {2}}{g}$
$\Rightarrow \frac {r}{R - h} = \frac {r \omega^ {2}}{g}$
$\Rightarrow \omega^ {2} = \frac {g}{R - h}\tag{... (3}$
$\Rightarrow \quad \omega = \sqrt {\frac {g}{R - h}}$
Hence, the correct answer is (D).

From equation (3),
$h = R - \frac {g}{\omega^ {2}}$
For non-zero value of h, we must have
$R > \frac {g}{\omega^ {2}}$
$\Rightarrow \omega > \sqrt {g / R}$
So, minimum value of $\omega$ should be
$\omega_ {\mathrm{min}} = \sqrt {\frac {g}{R}} = \sqrt {\frac {9 . 8}{0 . 1}} \mathrm{rads} ^ {- 1}$
$\Rightarrow \omega_ {\mathrm{min}} = 9. 9 \mathrm{rads} ^ {- 1}$
Hence, the correct answer is (C).

Since, $h = R - \frac{g}{\omega^2}$
If R and $\omega$ are known precisely, then
$\begin{array}{r l} & {\Delta h = - \frac {\Delta g}{\omega^ {2}}} \\ {\Rightarrow} & {\Delta g = \omega^ {2} \Delta h \qquad \mathrm{(neglectingthenegativesign)}} \\ {\Rightarrow} & {\left(\Delta g\right) _ {\min} = \left(\omega_ {\min}\right) ^ {2} \Delta h = (9. 8 9) ^ {2} \times 1 0 ^ {- 4} \mathrm{ms} ^ {- 2}} \\ {\Rightarrow} & {\left(\Delta g\right) _ {\min} = 9. 7 8 \times 1 0 ^ {- 3} \mathrm{ms} ^ {- 2}} \end{array}$
Hence, the correct answer is (B).

$CP = CO = \text{Radius of circle}(R)$
$\therefore \angle C P O = \angle P O C = 6 0 ^ {\circ}$
$\therefore \angle O C P \text { is also } 6 0 ^ {\circ}.$

Therefore, $\Delta OCP$ is an equilateral triangle.
Hence, OP = R
Natural length of spring is 3R/4.
So, extension in the spring is
$x = R - \frac {3 R}{4} = \frac {R}{4}$
$\Rightarrow$ Spring force, $F = kx = \left(\frac{mg}{R}\right)\left(\frac{R}{4}\right) = \frac{mg}{4}$
The free body diagram of the ring will be as shown.
Here, $F = k \Delta x = \frac{mg}{4}$
$\left\{\because \Delta x = \frac {R}{4} \right\}$
And N = normal reaction.
Hence, the correct answer is (C).

Tangential acceleration $a_{T}$ :

The ring will move towards the x-axis just after the release. So, net force along x-axis i.e. the tangential force is
$F _ {T} = F _ {x} = F \sin 6 0 ^ {\circ} + m g \sin 6 0 ^ {\circ}$
$\Rightarrow F _ {T} = \left(\frac {m g}{4}\right) \frac {\sqrt {3}}{2} + m g \left(\frac {\sqrt {3}}{2}\right)$
$\Rightarrow F _ {T} = \frac {5 \sqrt {3}}{8} m g$
$\Rightarrow a _ {T} = a _ {x} = \frac {F _ {x}}{m} = \frac {5 \sqrt {3}}{8} g$
Hence, the correct answer is (D).

Since, $N + F \cos 60^{\circ} = mg \cos 60^{\circ}$
$\Rightarrow N = m g \cos 6 0 ^ {\circ} - F \cos 6 0 ^ {\circ}$
$\Rightarrow N = \frac {m g}{2} - \frac {m g}{4} \left(\frac {1}{2}\right) = \frac {m g}{2} - \frac {m g}{8}$
$\Rightarrow N = \frac {3}{8} m g$
Hence, the correct answer is (C).

Given $m_{1}=10$ kg, $m_{2}=5$ kg, $\omega=10$ rads $^{-1}$
$r = 0. 3 \mathrm{m}, r _ {1} = 0.1 2 4 \mathrm{m}$
$\Rightarrow r _ {2} = r - r _ {1} = 0.1 7 6 \mathrm{m}$
Masses $m_{1}$ and $m_{2}$ are at rest with respect to rotating table. Let f be the friction between mass $m_{1}$ and table.

Free body diagram of $m_{1}$ and $m_{2}$ with respect to table (non-inertial frame of reference are shown in figures).
$T \xleftarrow {m _ {2}} F _ {2} = m ^ {2} r ^ {2} \omega^ {2} (\text { P s e u d o f o r c e })$
$\begin{array}{c} F _ {1} = m _ {1} r _ {1} \omega^ {2} \\ (\text { P s e u d o f o r c e }) \end{array} \xleftarrow {\quad m _ {1} \quad f \quad T}$
Equilibrium of $m_{2}$ gives
$T = m _ {2} r _ {2} \omega^ {2}$...(1)
Since, $m_{2}r_{2}\omega^{2}< m_{1}r_{1}\omega^{2}$
Therefore, $m_{1}r_{1}\omega^{2} > T$
and friction of $M_{1}$ will be inward (toward centre)
Equilibrium of $m_{1}$ gives
$f + T = m _ {1} r _ {1} \omega^ {2}$....(2)
From equations (1) and (2), we get
$ f = m _ {1} r _ {1} \omega^ {2} - m _ {2} r _ {2} \omega^ {2}$
$f = (m _ {1} r _ {1} - m _ {2} r _ {2}) \omega^ {2}$
$\Rightarrow f = (1 0 \times 0. 1 2 4 - 5 \times 0. 1 7 6) (1 0) ^ {2} \mathrm{N} = 3 6 \mathrm{N}$...(3)
Therefore, frictional force on $m_{1}$ is 36 N (inwards)
Hence, the correct answer is (D).

From equation (3),
$f = \left(m _ {1} r _ {1} - m _ {2} r _ {2}\right) \omega^ {2}$
Masses will start slipping when this force is greater than $f_{\max}$ or $(m_{1}r_{1}-m_{2}r_{2})\omega^{2}>f_{\max}>\mu m_{1}g$ .
Minimum values of $\omega$ is
$\omega_ {\mathrm{min}} = \sqrt {\frac {\mu m _ {1} g}{m _ {1} r _ {1} - m _ {2} r _ {2}}} = \sqrt {\frac {0.5 \times 1 0 \times 9.8}{1 0 \times 0.1 2 4 - 5 \times 0.1 7 6}}$
$\Rightarrow \omega_ {\mathrm{min}} = 1 1. 6 7 \mathrm{rads} ^ {- 1}$
Hence, the correct answer is (B).

From equation (3), frictional force $f = 0$ where $m_{1}r_{1} = m_{2}r_{2}$
$\Rightarrow \quad \frac {r _ {1}}{r _ {2}} = \frac {m _ {2}}{m _ {1}} = \frac {5}{1 0} = \frac {1}{2}$
and $r = r_{1} + r_{2} = 0.3 \, m$
$\Rightarrow r_{1}=0.1\ m\ and\ r_{2}=0.2\ m$
i.e. mass $m_{2}$ should placed at 0.2 m and $m_{1}$ at 0.1 m from the centre O.
Hence, the correct answer is (C).

Acceleration of block A
Maximum friction force that can be obtained at A is
$\left(f _ {\max}\right) _ {A} = \mu_ {A} (m g \cos 4 5 ^ {\circ}) = \frac {2}{3} \left(\frac {m g}{\sqrt {2}}\right) = \frac {\sqrt {2} m g}{3}$

Similarly,
$\begin{array}{l} \left(f _ {\max}\right) _ {B} = \mu_ {B} \left(2 m g \cos 4 5 ^ {\circ}\right) = \frac {1}{3} \left(\frac {2 m g}{\sqrt {2}}\right) \\ \Rightarrow \left(f _ {\max}\right) _ {B} = \frac {\sqrt {2} m g}{3} \end{array}$
Therefore, maximum value of friction that can be obtained on the system is
$\left(f _ {\max}\right) = \left(f _ {\max}\right) _ {A} + \left(f _ {\max}\right) _ {B} = \frac {2 \sqrt {2} m g}{3}\tag{... (1}$
Net pulling force on the system is
$F = F _ {1} - F _ {2} = \frac {2 m g}{\sqrt {2}} - \frac {m g}{\sqrt {2}} = \frac {m g}{\sqrt {2}}\tag{... (2}$
From (1) and (2), we can see that
Net pulling force $< f_{\mathrm{max}}$
Therefore, the system will not move or the acceleration of block A will be zero.
Hence, the correct answer is (D).

The correct answer is (B).

The correct answer is (C).
Combined solution to 17, 18
Tension in the string and friction at A
Net external force on the system (block A and B)
$F = F _ {1} - F _ {2} = \frac {m g}{\sqrt {2}}$
Therefore, total friction force on the blocks should also be equal to $\frac{mg}{\sqrt{2}}$
$\Rightarrow f _ {A} + f _ {B} = F = \frac {m g}{\sqrt {2}}$
Now since the blocks will start moving from block B first (if they move), therefore, $f_{B}$ will reach its limiting value first and if still some force is needed, it will be provided by $f_{A}$ .
Here $\left(f_{\max}\right)_B < F$ .
Therefore, $f_{B}$ will be in its limiting value and rest will be provided by $f_{A}$ .
Hence $f_{B}=\left(f_{\max}\right)_{B}=\frac{\sqrt{2}mg}{3}$
and $f_{A}=F-f_{B}=\frac{mg}{\sqrt{2}}-\frac{\sqrt{2}mg}{3}=\frac{mg}{3\sqrt{2}}$
The FBD of the whole system will be as shown in the figure.

Therefore, friction on $A$ is
$f _ {A} = \frac {m g}{3 \sqrt {2}}$
(down the plane)
Now for tension $T$ in the string, we may consider either equilibrium of $A$ or $B$ .
Equilibrium of A gives
$T = F _ {2} + f _ {A} = \frac {m g}{\sqrt {2}} + \frac {m g}{3 \sqrt {2}} = \frac {4 m g}{3 \sqrt {2}} \mathrm{or} \frac {2 \sqrt {2} m g}{3}$
Similarly, equilibrium of $B$ gives $T + f_{B} = F_{1}$ .
$T = F _ {1} - f _ {B} = \frac {2 m g}{\sqrt {2}} - \frac {\sqrt {2} m g}{3} = \frac {4 m g}{3 \sqrt {2}} = \frac {2 \sqrt {2}}{3} \mathrm{mg}$
Therefore, tension in the string is $\frac{2\sqrt{2}}{3}$ mg.

Free body diagram of mass M is

Hence, the correct answer is (B).

The maximum value of $f_{1}$ is
$\left(f _ {1}\right) _ {\max} = (0. 3) (2 0) (1 0) = 6 0 \mathrm{N}$
The maximum value of $f_{2}$ is
$\left(f _ {2}\right) _ {\mathrm{max}} = (0. 3) (5) (1 0) = 1 5 \mathrm{N}$
Forces on $m_{1}$ and $m_{2}$ in horizontal direction are as follows

Now there are only two possibilities
(1) Either both $m_1$ and $m_2$ will remain stationary (with respect to ground) or
(2) Both $m_{1}$ and $m_{2}$ will move (with respect to ground)
First case is possible when
$T \leq \left(f _ {1}\right) _ {\max} \mathrm{i.e.} T \leq 6 0 \mathrm{N}$
and $T \leq \left( f_{2} \right)_{\max}$ i.e. $T \leq 15 N$
These conditions will be satisfied when $T \leq 15$ N
say $T = 14\mathrm{N}$ then $f_{1} = f_{2} = 14\mathrm{N}$
Therefore, the condition $f_{1} = 2f_{2}$ will not be satisfied. So, $m_{1}$ and $m_{2}$ both can't remain stationary.
In the second case, when $m_{1}$ and $m_{2}$ both move
$f _ {2} = \left(f _ {2}\right) _ {\mathrm{max}} = 1 5 \mathrm{N}$
$\Rightarrow f _ {1} = 2 f _ {2} = 3 0 \mathrm{N}$
Now since $f_{1} < \left(f_{2}\right)_{\max}$ , there is no relative motion between $m_{1}$ and M. i.e., all the masses move with same acceleration, say a.
$\Rightarrow f _ {2} = 1 5 \mathrm{N}$
and $f_{1}=30\ N$
Hence, the correct answer is (D).

The correct answer is (A).

The correct answer is (D).

The correct answer is (C).
Combined solution to 21, 22, 23
Free body diagrams and equations of motion are as shown


For $m_{1}$ , $30 - T = 20a$
For $m_{2}$ , $T - 15 = 5a$
... (1)
For $M$ , $F - 30 = 50a$
$\ldots(2)$
... (3)
Solving these three equations, we get
$F = 6 0 \mathrm{N}$
$T = 1 8 \mathrm{N} \text { and } a = \frac {3}{5} \mathrm{ms} ^ {- 2}$

$F_{x} = \frac{d}{dt}\big(p_{x}\big) = -2\sin t$
$F _ {y} = \frac {d}{d t} \big (p _ {y} \big) = 2 \cos t$
$\Rightarrow F = \sqrt {F _ {x} ^ {2} + F _ {y} ^ {2}}$
$\begin{array}{l l} \Rightarrow & F = \sqrt {4 \sin^ {2} t + 4 \cos^ {2} t} \\ \Rightarrow & F = \sqrt {4 (\sin^ {2} t + \cos^ {2} t)} = 2 \text { unit } \\ & \text { Hence,the correct answer is(B). } \end{array}$

$\begin{array}{l} \vec {F} = - 2 \sin t \hat {i} + 2 \cos t \hat {j} \\ \vec {p} = 2 \cos t \hat {i} + 2 \sin t \hat {j} \\ \vec {F} \cdot \vec {p} = (- 2 \sin t \hat {i} + 2 \cos t \hat {j}) \cdot (2 \cos t \hat {i} + 2 \sin t \hat {j}) \\ \Rightarrow \quad \vec {F} \cdot \vec {p} = - 4 \sin t \cos t + 4 \sin t \cos t = 0 \\ \text {So, the angle between} \vec {F} \text {and} \vec {p} \text {is} 9 0 ^ {\circ}. \\ \text {Hence, the correct answer is (D).} \end{array}$

$p = \sqrt {p _ {x} ^ {2} + p _ {y} ^ {2}}$
$p = \sqrt {4 \cos^ {2} t + 4 \sin^ {2} t} = 2 \text {units}$
Hence, the correct answer is (B).

$N = m (g + a _ {0}) \cos \alpha$
and
$ m a = m (g + a _ {0}) \sin \alpha$
$\Rightarrow a = (g + a _ {0}) \sin \alpha$

$\begin{array}{l} {F _ {1} = m (g + a _ {0}) \mathrm{cos} \alpha} \\ {F _ {2} = m (g + a _ {0}) \mathrm{sin} \alpha} \\ {F _ {3} = m a _ {0}} \\ {F _ {4} = m g} \end{array}$
Hence, the correct answer is (B).

C

Since $OB = \frac{l}{\cos\alpha} = \frac{1}{2} at^2$
$\begin{array}{r l} \Rightarrow & \frac {l}{\cos \alpha} = \frac {1}{2} (g + a _ {0}) \sin \alpha t ^ {2} \\ \Rightarrow & T = \sqrt {\frac {2 l}{(g + a _ {0}) \sin \alpha \cos \alpha}} \\ \Rightarrow & T = \sqrt {\frac {4 l}{(g + a _ {0}) \sin 2 \alpha}} \end{array}$
Hence, the correct answer is (B).