NLM

Let $\vec{F}_1$ and $\vec{F}_2$ be two forces and $\vec{F}$ be their resultant. Let $\vec{F}_1$ be the smaller force.

In $\Delta ABC$ , $F_{2}^{2} = F^{2} + F_{1}^{2}$ or $F^{2} = F_{2}^{2} - F_{1}^{2}$
$(8 \sqrt {3}) ^ {2} = F _ {2} ^ {2} - F _ {1} ^ {2}$
$1 9 2 = (F _ {2} + F _ {1}) (F _ {2} - F _ {1})$...(i)
$\because F _ {1} + F _ {2} = 1 6 N \text {(Given)}$
Put this value in (i), we get, $F_{2} - F_{1} = \frac{192}{16} = 12N \ldots (\mathrm{iii})$ Solving (ii) and (iii), we get, $F_{1} = 2N$ , $F_{2} = 14N$ ∴ Forces are $2N$ and $14N$ .

Common acceleration of the system,
$a = \frac {1 4 N}{4 k g + 2 k g + 1 k g} = \frac {1 4}{7} m s ^ {- 2} = 2 m s ^ {- 2}$
Let R be the contact force between 4 kg and 2 kg blocks. The free body diagram of 4 kg block is as shown in the figure.
$14 N \longrightarrow \boxed {4 k g} \xleftarrow {} R$
The equation of motion is $14 - R = 4a$
$R = 1 4 - 4 \times 2 = 6 N.$

According to conservation of linear momentum
$0 = m _ {1} v _ {1} + m _ {2} v _ {2}$
$0 = 1 \times 2 + 2 \times v _ {2}$

$v _ {2} = - 1 m / s$
-ve sign shows that 2 kg is pulled in a opposite direction to that of mass 1 kg.

Here, $m = 10kg$ , $F = 10N$
The situation is as shown in the figure.


Motion along vertical direction
Acceleration along vertical direction
$a _ {y} = \frac {F}{m} = \frac {1 0 N}{1 0 k g} = 1 m s ^ {- 2}$
Distance travelled by the block in 4s in vertical direction is
$S _ {y} = \frac {1}{2} a _ {y} t ^ {2} = \frac {1}{2} \times (1 m s ^ {- 2}) (4 s) ^ {2} = 8 m$
Motion along horizontal direction
Distance travelled by the block in 4s in horizontal direction is, $S_{x} = (1.5ms^{-1})(4s) = 6m$
The displacement of the block at the end of 4 s is
$= \sqrt {S _ {x} ^ {2} + S _ {y} ^ {2}} = \sqrt {(8 m) ^ {2} + (6 m) ^ {2}} = 1 0 m.$

Here, $M = 15\mathrm{kg}$ , $\theta = 30^{\circ}$ , $g = 10m / s^{2}$
As the block is held by the string,
$\therefore T = M g \sin 3 0 ^ {\circ} = 1 5 \times 1 0 \times \frac {1}{2} = 7 5 N.$

$m\frac{dv}{dt} = F_0e^{-bt}\Rightarrow \frac{dv}{dt} = \frac{F_0}{m} e^{-bt}$ $\Rightarrow \int_0^v dv = \frac{F_0}{m}\int_0^t e^{-bt}dt$ $\Rightarrow v = \frac{F_0}{m}\left[\frac{e^{-bt}}{-b}\right]_0^t\Rightarrow v = \frac{F_0}{mb} (1 - e^{-bt}).$

For the box remains stationary, $ma_{max} = \mu_{s}mg$
$a _ {\max} = \mu_ {s} g = 0.2 \times 1 0 m s ^ {- 2} = 2 m s ^ {- 2}.$

In the given condition the required centripetal force is provided by frictional force between the road and tyre.
$\frac {m v ^ {2}}{R} = \mu m g \quad \therefore \quad v = \sqrt {\mu R g}.$

$F - mg = 2$ $F = 2 + mg = 3N$ $a = \frac{\text{unbalanced force}}{\text{mass}} = \frac{2}{0.1} = 20m/s^2$ $\therefore S = \frac{1}{2}at^2 = \frac{1}{2} \times 20 \times 1 = 10m$ $\therefore W \text{ by tension} = F \times 10 = 3 \times 10 = 30J$ $W \text{ against gravity} = mg \times s = 1 \times 10 = 10J.$

Centripetal acceleration, $a_{c} = \omega^{2} r$ , where $\omega = \frac{2\pi}{T}$ Since, $T_{1} = T_{2}$
$\begin{array}{r l} \Rightarrow & \omega_ {1} = \omega_ {2} \\ \Rightarrow & \frac {a _ {c _ {1}}}{a _ {c _ {2}}} = \frac {r _ {1}}{r _ {2}} \end{array}$
Hence, the correct answer is (B).

$w = mg = 0.1 \times 10 = 1 \mathrm{~N}$

$F _ {1} = \text { component of weight } = 1 \sin \theta = \sin \theta$
$F_{2}=$ component of applied force = 1cosθ = cosθ
Now, at $\theta = 45^{\circ}: F_{1} = F_{2}$ and block remains stationary without the help of friction.
For $\theta > 45^{\circ}$ , $F_{1} > F_{2}$ , so friction will act towards $Q$ .
For $\theta < 45^{\circ}$ , $F_{2} > F_{1}$ and friction will act towards $P$ .
Hence, (A) and (C) are correct.

Let T be the tension in the string.
Let a be acceleration of the system.
The equations of motion are
$M a = M g \sin 5 3 ^ {\circ} - T$...(i)
and $Ma = T - Mg\sin 37^{\circ}$
... (ii)
Adding (i) and (ii), we get
$\begin{array}{r l} a & = \frac {M g (\sin 5 3 ^ {\circ} - \sin 3 7 ^ {\circ})}{2 M} \\ & = g \cos 4 5 ^ {\circ} \sin 8 ^ {\circ} \end{array}$
$\left[ \because \sin A - \sin B = 2 \cos \left(\frac {A + B}{2}\right) \sin \left(\frac {A - B}{2}\right) \right]$
$= 10 \times \frac {1}{\sqrt {2}} \times 0.139$
$= 0.9 8 m s ^ {- 2} \approx 1 m s ^ {- 2}.$

Here, $m_{1} = 1kg$ , $m_{2} = 2kg$
The acceleration of the system is
$a = \frac {(m _ {2} - m _ {1}) g}{m _ {1} + m _ {2}} = \frac {(2 - 1) g}{1 + 2} = \frac {g}{3} = \frac {1 0}{3}$
Acceleration of the centre of mass is 1kg

$a _ {c m} = \frac {m _ {1} a _ {1} + m _ {2} a _ {2}}{m _ {1} + m _ {2}} = \frac {1 (- a) + 2 (a)}{1 + 2}$
$= \frac {1 \left(\frac {- g}{3}\right) + 2 \left(\frac {g}{3}\right)}{3} = \frac {g}{9} = \frac {1 0}{9}$
The distance travelled by the centre of mass in two seconds is
$S = \frac {1}{2} a _ {c m} t ^ {2} = \frac {1}{2} \times \frac {1 0}{9} \times (2) ^ {2} = \frac {2 0}{9} m.$

The component of applied force F in the direction of motion is $F \cos 30^{\circ}$ . The work done by the applied force is
$W = (F \cos 3 0 ^ {\circ}) S = \sqrt {3} \times 1 0 ^ {3} \times \frac {\sqrt {3}}{2} \times 1 0 J$
$= 1 5 \times 1 0 ^ {3} J = 1 5 k J.$

Let $L$ be the length and $H$ be height of the inclined plane respectively
Acceleration of the block slide down the smooth incline plane is
$a = g \cos 6 0 ^ {\circ}$
$\therefore L = \frac {1}{2} g \cos 6 0 ^ {\circ} t _ {1} ^ {2}$
$[ \because u = 0 ]$...(i)
Acceleration of another block dropped vertically down from the same inclined plane is
$a = g$
$\therefore H = \frac {1}{2} a t _ {2} ^ {2} = \frac {1}{2} g t _ {2} ^ {2}$
$[ \because u = 0 ]$
From figure,
$\cos 6 0 ^ {\circ} = \frac {H}{L} \Rightarrow H = L \cos 6 0 ^ {\circ}$
$\therefore L \cos 6 0 ^ {\circ} = \frac {1}{2} g t _ {2} ^ {2}$...(ii)
Divide (i) by (ii), we get
$\frac {t _ {1} ^ {2} \cos 6 0 ^ {\circ}}{t _ {2} ^ {2}} = \frac {1}{\cos 6 0 ^ {\circ}} \Rightarrow \frac {t _ {1} ^ {2}}{t _ {2} ^ {2}} = \frac {1}{\cos^ {2} 6 0 ^ {\circ}} = \frac {4}{1} \Rightarrow \frac {t _ {1}}{t _ {2}} = \frac {2}{1}.$

$T \cos \theta$ component will cancel mg.

$T \sin \theta$ component will provide necessary centripetal force to the ball towards centre C.
$\Rightarrow T \sin \theta = m r \omega^ {2} = m (l \sin \theta) \omega^ {2}$
$\Rightarrow T = m l \omega^ {2}$
$\Rightarrow \omega = \sqrt {\frac {T}{m l}}$
$\Rightarrow \omega_ {\mathrm{max}} = \sqrt {\frac {T _ {\mathrm{max}}}{m l}}$
$\Rightarrow \omega_ {\mathrm{max}} = \sqrt {\frac {3 2 4}{0.5 \times 0.5}}$
$\Rightarrow \omega_ {\mathrm{max}} = 3 6 \mathrm{rads} ^ {- 1}$
Hence, the correct answer is (D).

Since acceleration of lift is zero.

Due to Newton's third law.

Force equation for 'M'
$M a = T$... (i)
Force equation for, $m$
$m g - T = m a$... (ii)
on solving (i) and (ii)
$T = \left(\frac {M m}{M + m}\right) g.$

Impulse $= \Delta P = m(V_{f} - V_{i}) = 0.4[1 - (-1)] = 0.8Ns$ .

$m = 5kg, a = 5m / s^2$
$\because F \leq f r _ {\max} \leq \mu N$
$F = m a = 5 \times 5 = 2 5 N$
$\therefore f r = F \Rightarrow f r = 2 5 N.$