NLM

Acceleration of block till the slipping continues is $a = \frac{f_{\max}}{m}$
$\Rightarrow a = \frac {\mu m g}{m} = \mu g = (0. 5) (1 0) = 5 \mathrm{ms} ^ {- 2}$
Slipping will continue till the velocity of the block also becomes $3 \, ms^{-1}$ . So, using $v = u + at$ , we get
3 = 0 + 5 t
$\left\{\because u = 0 \right\}$
$\Rightarrow t = 0. 6 \mathrm{s}$
During this time, the displacement of the block is
$s _ {1} = \frac {1}{2} a t ^ {2} = \frac {1}{2} (5) (0. 6) ^ {2} = 0. 9 \mathrm{m}$
and the displacement of the belt is
$s _ {2} = v t = (3) (0. 6) = 1. 8 \mathrm{m}$
Hence displacement of the block w.r.t., the belt is
$\left| s _ {1} - s _ {2} \right| = 0. 9 \mathrm{m}$
Hence, the correct answer is (A).

$mg \sin \alpha - T - f = m \ddot{x}$
$\Rightarrow m g \sin \alpha - \frac {\lambda (l - l _ {0})}{l _ {0}} - k \dot {x} = m \ddot {x}$
where, k is any known constant.
At Equilibrium Position, $\dot{x} = \ddot{x} = 0$
$\Rightarrow l = l _ {0} \left(1 + \frac {m g}{\lambda} \sin \alpha\right)$
Hence, the correct answer is (B).

The maximum acceleration and maximum retardation for the car can be $\mu g = \left(\frac{1}{2}\right)(10) = 5 \, \text{ms}^{-2}$ . The corresponding velocity-time graph is as follows

Let $t_0$ be the time taken by the car to accelerate. The time taken by it to decelerate is also $t_0$ . So,
$v _ {m} = (\mu g) t _ {0} = 5 t _ {0}$
Since, displacement = area under v-t graph. So, we get
$5 0 0 = \frac {1}{2} (2 t _ {0}) (5 t _ {0})$
$\Rightarrow t _ {0} = 1 0 \mathrm{s}$
Hence total time of journey is $t = 2t_{0} = 20$ s.
Hence, the correct answer is (C).

Impulse is the area under F-t graph, as well as the change in momentum. So,
$m u = \text { Area } = \frac {1}{2} F _ {0} T$
$\Rightarrow F _ {0} = \frac {2 m u}{T}$
Hence, the correct answer is (C).

For equilibrium of the system i.e., M and m, we have
$T + \mu_ {1} n = m g$
$n = m a\tag{... (1}$
$\ldots(2)$

Since the mass m is at rest, so n = 0. This is a justified and an expected answer $(f_{2} = 0)$ , because when M does not move, then no force is exerted by M on m and hence n = 0. Let us now draw free body diagram for M, without taking into account the frictional force on M due to m (as it is zero). So, for M to be at rest we have

$N = M g + T$.... (3)
$3 T = \mu_ {2} N$.......(4)
Also, from (1), we get
T = m g
$\left\{\because n = 0 \right\}$
From (3), we get
N = (M + m) g
From (4), we get
$ 3 m g = \mu_ {2} (M + m) g $
$\Rightarrow 3 m = \mu_ {2} M + \mu_ {2} m $
$\Rightarrow m = \frac {\mu_ {2} M}{3 - \mu_ {2}}$
$\Rightarrow \quad m = \frac {\left(\frac {1}{3}\right) (8)}{3 - \left(\frac {1}{3}\right)} = 1 \mathrm{kg}$
Hence, the correct answer is (C).

For $a \leq \mu g$ , the blocks will remain stationary with respect to the surface of the plank, so $x = 0$ .
For $a \geq \mu g$ , both the blocks will move with respect to the surface of the plank with an acceleration ( $a - \mu g$ ). However at any instant no relative motion between the blocks exists, so still they don't stretch the spring.
Hence, the correct answer is (D).

Since the block is about to slide, so we have
$\begin{array}{r l} & m g \sin \theta = \mu N \\ \Rightarrow & \not {m} g \sin \theta = \mu (\not {m} g \cos \theta) \\ \Rightarrow & \mu = \tan \theta \end{array}$
This condition is independent of the mass, so it will still be on the verge of motion down the incline. Hence, the correct answer is (D).
$\begin{array}{r l} & \text { In CASE - 1, sliding starts when } \\ & \quad m g \sin \theta \geq \mu m g \cos \theta \\ & \Rightarrow \tan \theta \geq \mu \\ & \text { While in CASE - 2, sliding starts when } \\ & \quad m (g + a _ {0}) \sin \alpha \geq \mu m (g + a _ {0}) \cos \alpha \\ & \Rightarrow \tan \alpha \geq \mu \\ & \text { From (1) and (2), we get } \\ & \quad \theta = \alpha \end{array}\tag{... (1}$
$\ldots(2)$
Hence, the correct answer is (D).

In CASE-1, sliding starts when
$mg\sin\theta \geq \mu mg\cos\theta$
$\Rightarrow \tan\theta \geq \mu$ … (1)
While in CASE-2, sliding starts when
$m(g+a_0)\sin\alpha \geq \mu m(g+a_0)\cos\alpha$
$\Rightarrow \tan\alpha \geq \mu$ … (2)
From (1) and (2), we get
$\theta=\alpha$
Hence, the correct answer is (D).

$T_1=m(2)\omega^2$ … (1)
($\because$ centrifugal force acting on $Q$ is $2m\omega^2$ radially outward)
$T_1+m(1)\omega^2=T_2$
$\Rightarrow T_2=3m\omega^2$
$\Rightarrow \frac{T_1}{T_2}=\frac{2}{3}$
Hence, the correct answer is (C).

$T \cos \theta = mg \cos \alpha$ and
... (1)
$T \sin \theta = m a _ {0} + m g \sin \alpha$...(2)
$\Rightarrow \tan \theta = \frac {a _ {0} + g \sin \alpha}{g \cos \alpha}$
Hence, the correct answer is (D).

$\tan \theta = \frac{v^2}{rg} = 1$
$\Rightarrow \quad \theta = 4 5 ^ {\circ}$
Hence, the correct answer is (C).

As long as $\tan\theta\leq\mu$ , the block will not slide down, so we have the contact force $F_{C}=\sqrt{N^{2}+f^{2}}$

where $N = mg \cos \theta$ and $f = mg \sin \theta$
So, for $\tan\theta\leq\mu$ , we have
$F _ {C} = \sqrt {\left(m g \cos \theta\right) ^ {2} + \left(m g \sin \theta\right) ^ {2}} = m g$
However, when $\tan\theta > \mu$ , then the block starts sliding down and now $mg\sin\theta > f$ . So,
$\begin{array}{r l} & F _ {C} = \sqrt {N ^ {2} + f ^ {2}} = \sqrt {N ^ {2} + (\mu N) ^ {2}} \\ \Rightarrow & F _ {C} = N \sqrt {1 + \mu^ {2}} = m g \cos \theta \sqrt {1 + \mu^ {2}} \end{array}$
Since, $\cos\theta$ decreases as $\theta$ increases hence, the contact force starts decreasing and finally it becomes zero at $\theta=90^{\circ}$ . Therefore, graph between $F_{C}$ and $\theta$ will be as follows

Hence, the correct answer is (D).

Impulse is the area under $F - t$ graph, as well as the change in momentum. So,
$m u = \pi \left(\frac {F _ {0}}{2}\right) \left(\frac {T}{2}\right)$
$\Rightarrow u = \frac {\pi F _ {0} T}{4 m}$
Hence, the correct answer is (C).

At position B, speed of ball is $v^{2} = 2gh$
where $h = R \cos(60^{\circ}) = (0.5) \left( \frac{1}{2} \right) = 0.25 \, \text{m}$
$\Rightarrow v ^ {2} = (2) (1 0) (0. 2 5) = 5 \mathrm{m} ^ {2} s ^ {- 2}\tag{... (1}$
Let N be the normal reaction between the ball and the wedge in this position. Then

Since, we have
$N - m g \cos (6 0 ^ {\circ}) = \frac {m v ^ {2}}{R}$
$\Rightarrow N = \frac {m v ^ {2}}{R} + m g \cos (6 0 ^ {\circ})$
Since $v^{2}=5$
$\{\because \text { of } (1) \}$
$\Rightarrow \quad N = \frac {(1) (5)}{(0 . 5)} + (1) (1 0) \left(\frac {1}{2}\right) = 1 5 \mathrm{N}$
Now for horizontal equilibrium of wedge Force exerted by vertical wall on wedge is
$F = N \sin (6 0 ^ {\circ}) = \frac {1 5 \sqrt {3}}{2} \mathrm{N}$
Hence, the correct answer is (C).

$\Rightarrow F \cos \theta = m a$
$\Rightarrow \frac {m g}{3} \cos \theta = m \frac {d v}{d t}$
$\Rightarrow \frac {m g}{3} \cos (k s) = m \frac {d v}{d s} \frac {d s}{d t}$
$\Rightarrow \frac {m g}{3} \cos (k s) = m v \frac {d v}{d s}$
$\Rightarrow v d v = \frac {g}{3} \cos (k s) d s$
$\Rightarrow \frac {v ^ {2}}{2} = \frac {g}{3 k} \sin (k s)$
$\Rightarrow v = \sqrt {\frac {2 g}{3 k} \sin \theta}$
Hence, the correct answer is (C).

Since, the radii of both the balls is same, so the force of buoyancy is also the same.
When the balls fall with a uniform velocity, then condition of equilibrium is attained, so we get
For $A$ , $F + T = m_{1}g$
... (1)
For $B$ , $m_2g + T = F$
$\ldots(2)$
From (1), equating the value of T, we get
$2 F = m _ {1} g + m _ {2} g$
$\Rightarrow F = \left(\frac {m _ {1} + m _ {2}}{2}\right) g\tag{... (3}$

Substitute (3) either in (1) or in (2), we get
$T = m _ {1} g - \left(\frac {m _ {1} + m _ {2}}{2}\right) g$
$\Rightarrow T = \left(\frac {m _ {1} - m _ {2}}{2}\right) g$
Hence, the correct answer is (B).

Let the mass of the block be $m$ . Force required to drag it up along the plane is
$F _ {1} = \left(m g \sin \theta + \mu m g \cos \theta\right)$
and force required to lift it is

$F _ {2} = m g$
Since it is given that $F_{1} < F_{2}$
$\Rightarrow m g \sin \theta + \mu m g \cos \theta < m g$
$\Rightarrow \mu < \frac {1 - \sin \theta}{\cos \theta}$
$\Rightarrow \mu < \frac {1 - \sin 3 0 ^ {\circ}}{\cos 3 0 ^ {\circ}}$
$\Rightarrow \mu < \frac {1}{\sqrt {3}}$
$\mathrm{So}, \mu_ {M A X} = \frac {1}{\sqrt {3}}$
Hence, the correct answer is (C).

If $m_{1}=m_{2}$ , block will slide down when
$ \mu g \cos \theta < g \sin \theta $
$\Rightarrow \frac {1}{2} \left(\frac {\sqrt {3}}{2}\right) < \frac {1}{2}$
which is true
If $m_{1} < m_{2}$ , block will slide down when
$\mu (g - a) \cos \theta < (g - a) \sin \theta$
$\Rightarrow \left(\frac {1}{2}\right) \left(\frac {\sqrt {3}}{2}\right) < \frac {1}{2}$
$\text { Here } a = \left(\frac {m _ {2} - m _ {1}}{m _ {1} + m _ {2}}\right) g$
If $m_{1} > m_{2}$ , block will slide down when
$\mu (g + a) \cos \theta < (g + a) \sin \theta$
$\Rightarrow \frac {1}{2} \left(\frac {\sqrt {3}}{2}\right) < \frac {1}{2}$
Here $a=\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right)g$
So, block can't remain stationary when $m_1 > m_2$ .
Therefore, block will slide down under all conditions.
Hence, the correct answer is (D).

The free body diagram of the block is shown here.

So, we get
$N = m g \cos \theta + F \sin \theta\tag{92.}$
and $F_{\mathrm{net}} = F\cos \theta -mg\sin \theta$
{up the plane}
Hence, the correct answer is (D).

Let x be the maximum displacement of block downwards. Then from conservation of mechanical energy, we have the decrease in potential energy of 2 kg block must be equal to the increase in elastic potential energy of both the springs. So, we have
$\begin{array}{l} m g x = \frac {1}{2} \left(k _ {1} + k _ {2}\right) x ^ {2} \\ \Rightarrow x = \frac {2 m g}{k _ {1} + k _ {2}} = \frac {(2) (4) (1 0)}{1 0 0 + 3 0 0} = 0. 2 \mathrm{m} \end{array}$
Acceleration of block in this position is
$\begin{array}{l l} & a = \frac {(k _ {1} + k _ {2}) x - m g}{m} \\ \Rightarrow & a = \frac {(4 0 0) (0 . 2) - (4) (1 0)}{4} = \frac {8 0 - 4 0}{4} \\ \Rightarrow & a = 1 0 \mathrm{ms} ^ {- 2} \end{array}\tag{{upwards}}$
{upwards}
Hence, the correct answer is (D).

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).