NLM

The water jet striking the block at the rate of $2kgs^{-1}$ at a speed of $10ms^{-1}$ will exert a force on the block,
$F = v \frac {d m}{d t} = 1 0 \times 2 = 2 0 N$

Under the action of this force of 20 N, the block of mass 4 kg will move with an acceleration given by
$a = \frac {F}{m} = \frac {2 0 N}{4 k g} = 5 m s ^ {- 2}.$

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

$m\frac{dv}{dt} = F_0e^{-bt}\Rightarrow \frac{dv}{dt} = \frac{F_0}{m} e^{-bt}\Rightarrow \int_0^v dv = \frac{F_0}{m}\int_0^t e^{-bt}dt$
$\Rightarrow v = \frac {F _ {0}}{m} \left[ \frac {e ^ {- b t}}{- b} \right] _ {0} ^ {t} \Rightarrow v = \frac {F _ {0}}{m b} (1 - e ^ {- b t}).$

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

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

The free-body diagram of the block is as shown in adjacent figure. $\uparrow N$
$\because (f _ {s}) _ {\max} = 12.0 N (\text { Given })$
$\therefore (T _ {1}) _ {\max} = 12.0 N$

The free-body diagram of knot is as shown in adjacent figure.
$T _ {2} \sin 3 0 ^ {\circ} = W$
$T _ {2} \cos 3 0 ^ {\circ} = T _ {1}$
Divide (i) by (ii), we get, $\tan 30^{\circ} = \frac{W}{T_1}$

or $W = T_{1} \tan 30^{\circ} = 0.577T_{1}$
$\therefore W _ {\max} = 0.5 7 7 \left(T _ {1}\right) _ {\max} = 0.5 7 7 \times 1 2.0 N = 6.9 2 N.$

E

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

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

Let $T$ be the tension in the string.
Let a be acceleration of the system.
The equations of motion are
$M a = M g \sin 5 3 ^ {\circ} - T$
and $Ma = T - Mg\sin 37^{\circ}$
Adding (i) and (ii), we get
$\begin{array}{r l} a & = \frac {M g (\sin 5 3 ^ {\circ} - \sin 3 7 ^ {\circ})}{2 M} \\ & = g \cos 4 5 ^ {\circ} \sin 8 ^ {\circ} \end{array}$
$\left[ \because \sin A - \sin B = 2 \cos \left(\frac {A + B}{2}\right) \sin \left(\frac {A - B}{2}\right) \right]$
$= 1 0 \times \frac {1}{\sqrt {2}} \times 0.139$
$= 0.98 m s ^ {- 2} \approx 1 m s ^ {- 2}.$

Special theory of relativity is based on two postulates
(i) All laws of physics are the same in all inertial reference frames.
(ii) The speed of light in vacuum has the same value in all inertial frames, regardless of the velocity of the observer or the velocity of source emitting the light.

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

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
$\frac {t _ {1} ^ {2} \cos 6 0 ^ {\circ}}{t _ {2} ^ {2}} = \frac {1}{\cos 6 0 ^ {\circ}} \Rightarrow \frac {t _ {1} ^ {2}}{t _ {2} ^ {2}} = \frac {1}{\cos^ {2} 6 0 ^ {\circ}} = \frac {4}{1} \Rightarrow \frac {t _ {1}}{t _ {2}} = \frac {2}{1}.$

B

Force, $F = (M \text{ kg s}^{-1}) (\nu \text{ m s}^{-1})$
$= M\nu \mathrm{kg} \mathrm{ms}^{-2} = M\nu \mathrm{N}$ .

Since acceleration of lift is zero.

D

Due to Newton's third law.

A

Force equation for 'M'
M a = T
... (i)
Force equation for, m
m g - T = m a

... (ii)
on solving (i) and (ii)
$T = \left(\frac {M m}{M + m}\right) g.$

Relative vertical acceleration of $A$ with respect to $B$
$= g (\sin^ {2} 6 0 ^ {\circ} - \sin^ {2} 3 0 ^ {\circ})$
$= 9. 8 \left(\frac {3}{4} - \frac {1}{4}\right) = 4.9 m / s ^ {2}.$

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

E

$m = 5kg, a = 5m / s^2$
$\because F \leq fr_{\max} \leq \mu N$
$F = ma = 5 \times 5 = 25N$
∴ $fr = F \Rightarrow fr = 25N$ .

Here friction force provides centripetal force so $f = m\omega^{2}r$ but $f \leq \mu mg$ so $m\omega^{2}r \leq \mu mg \Rightarrow r \leq \frac{\mu g}{\omega^{2}}$ .

A