NLM

B

According to Newton's Third Law of motion action and reactions are equal and opposite. Hence, the correct answer is (D).

Horizontal range also depends upon the velocity of projection.
Hence, the correct answer is (D).

In the direction of normal reaction net acceleration is zero. Hence forces in this direction will be balanced. Hence $N = mg \cos \theta$ .
Hence, the correct answer is (A).

By the definition of inertial and non-inertial frame. Hence, the correct answer is (B).

Coefficient of friction $\mu = \tan (\theta)$ . The value of $\tan \theta$ way exceed unity. Hence, the correct answer is (C).

$F = \frac {\Delta P}{\Delta t}$
If $\Delta t$ is more, then F will be less.
Hence, the correct answer is (A).

$a _ {1} = g - \frac {F}{M}$
$a _ {2} = g - \frac {F}{m}$
$\Rightarrow a _ {1} > a _ {2}$
Hence, the correct answer is (A).

While running boy pushes the ground in backward direction and available friction pushed him in forward direction.
Hence, the correct answer is (D).

In equilibrium, net force on the body = 0, therefore, it acceleration a = 0. If the body is at rest it will remain at rest. If the body is moving with constant speed along a straight line, it will continue to do so.
Hence, the correct answer is (D).

Inertia is the property by virtue of which the body is unable to change by itself not only the state of rest but also the state of motion.
Hence, the correct answer is (A).

Due to change in normal reaction pulling is easier. Hence, the correct answer is (D).

Contact force is sum of friction and normal reaction. Hence, the correct answer is (D).

Static frictional force is self adjusting. Hence, the correct answer is (A).

Hence, the correct answer is (B).

For pulling condition:
$N + F \sin \theta = m g$
$N = M g - F \sin \theta\tag{.(1}$

For pushing condition:
$N = F \sin \theta + M g$......(2)

Hence, the correct answer is (D).

The correct answer is (B).

The correct answer is (A).

The correct answer is (B).
Combined solution to 1, 2, 3
Let length of the rod be 2l. Using the three conditions of equilibrium. Anticlockwise moment is taken as positive.

For translational equilibrium, we have
$\begin{array}{c c c c} & \Sigma F _ {x} = 0 \\ \Rightarrow & N _ {B} - f _ {A} = 0 \\ \Rightarrow & N _ {B} = f _ {A} & \ldots (1) \end{array} \quad \begin{array}{c c c c} & \Sigma F _ {y} = 0 \\ \Rightarrow & N _ {A} - W = 0 \\ \Rightarrow & N _ {A} = W \end{array}\tag{... (2}$
For rotational equilibrium, we have
$\begin{array}{r l} & \Sigma \tau_ {0} = 0 \\ \Rightarrow & N _ {A} (2 l \cos 3 0 ^ {\circ}) - N _ {B} (2 l \sin 3 0 ^ {\circ}) - W (l \cos 3 0 ^ {\circ}) = 0 \end{array}$
$\Rightarrow \sqrt {3} N _ {A} - N _ {B} - \frac {\sqrt {3}}{2} W = 0\tag{... (3}$
Solving these three equations, we get
$f _ {A} = \frac {\sqrt {3}}{2} W, N _ {A} = W, N _ {B} = \frac {\sqrt {3}}{2} W$

Consider the torques about an axis perpendicular to the page and through the left end of the horizontal beam.

$\begin{array}{l l} & \sum \tau = + (T \sin (3 0 ^ {\circ})) d - (1 9 6 \mathrm{N}) d = 0 \\ \Rightarrow & T = 3 9 2 \mathrm{N} \end{array}$
Hence, the correct answer is (D).

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).