NLM

Let $\mu$ be the friction coefficient between sphere and each wall. Free body diagram of sphere is

Net force on the sphere in horizontal direction is zero. So,
$\begin{array}{r l} & N _ {1} \cos 6 0 ^ {\circ} + \mu N _ {2} \cos 6 0 ^ {\circ} = N _ {2} \cos 3 0 ^ {\circ} + \mu N _ {1} \cos 3 0 ^ {\circ} \\ \Rightarrow & N _ {1} + \mu N _ {2} = \sqrt {3} (N _ {2} + \mu N _ {1}) \end{array}$
$\Rightarrow N _ {1} (1 - \sqrt {3} \mu) = N _ {2} (\sqrt {3} - \mu)$
$\Rightarrow \frac {N _ {1}}{N _ {2}} = \frac {\sqrt {3} - \mu}{1 - \sqrt {3} \mu}$
Substituting $\mu=\frac{1}{3}$ we get,
$\frac {N _ {1}}{N _ {2}} = \frac {\sqrt {3} - \frac {1}{3}}{1 - \frac {\sqrt {3}}{3}} = \frac {3 \sqrt {3} - 1}{3 - \sqrt {3}} = 1 + \frac {4}{\sqrt {3}}$
$\Rightarrow \quad \frac {f _ {1}}{f _ {2}} = \frac {\mu N _ {1}}{\mu N _ {2}} = 1 + \frac {4}{\sqrt {3}}$
Hence, the correct answer is (A).

Blocks A and C both move due to friction. But less friction is available to A as compared to C because normal reaction between A and B is less. Maximum friction between A and B can be
$f _ {\mathrm{max}} = \mu m _ {A} g = \left(\frac {1}{2}\right) m g$
So, maximum acceleration of A can be
$a _ {\mathrm{max}} = \frac {f _ {\mathrm{max}}}{m} = \frac {g}{2}$
Further $a_{\mathrm{max}} = \frac{m_Dg}{3m + m_D}$
$\Rightarrow \frac {g}{2} = \frac {m _ {D} g}{3 m + m _ {D}}$
$\Rightarrow m _ {D} = 3 m$
Hence, the correct answer is (C).

Let $f$ be the friction force, then
$f = \sqrt {F _ {C} ^ {2} + F _ {T} ^ {2}}$
After 2 seconds, $\omega = \alpha t = \frac{2}{3}$ rads $^{-1}$
$\Rightarrow F _ {T} = m a _ {T} = m (\alpha R) = (3 6 \times 1 0 ^ {- 5}) \left(\frac {1}{3}\right) \left(\frac {1}{4}\right) N$
$\begin{array}{r l} \Rightarrow & F _ {T} = 3 \times 1 0 ^ {- 5} \mathrm{N} \\ & F _ {C} = m \omega^ {2} R = (3 6 \times 1 0 ^ {- 5}) \bigg (\frac {4}{9} \bigg) \bigg (\frac {1}{4} \bigg) = 4 \times 1 0 ^ {- 5} \mathrm{N} \\ \Rightarrow & f = \sqrt {F _ {C} ^ {2} + F _ {T} ^ {2}} = 5 \times 1 0 ^ {- 5} \mathrm{N} = 5 0 \mu \mathrm{N} \end{array}$
Hence, the correct answer is (C).

$m\omega^{2}r\cos\theta$

$m\omega^{2}r\cos \theta = mg\sin \theta$
$\Rightarrow m \omega^ {2} l \cos^ {2} \theta = m g \sin \theta$
$\left\{\because r = \ell \cos \theta \right\}$
$\Rightarrow \omega = \sqrt {\frac {g \sin \theta}{l \cos^ {2} \theta}} = \sqrt {\frac {g \sin \theta}{l}} \sec \theta$
Hence, the correct answer is (A).

$a = \frac {a _ {A} + a _ {B}}{2} = \frac {3 + 4}{2} = \frac {7}{2} = 3. 5 \mathrm{ms} ^ {- 2}$
Hence, the correct answer is (C).

Reading of spring balance is
\begin{array}{l} \text {Reading} = \frac {\text {Thrust}}{g} = \frac {4 m _ {1} m _ {2}}{m _ {1} + m _ {2}} = \frac {1 0}{3} \mathrm{kg} \\ \Rightarrow \text {Reading} = 3. 3 3 \mathrm{kg(< 6kg)} \\ \text {Hence, the correct answer is (B).} \end{array}

Net horizontal force on wedge
$F _ {H} = m g \cos \theta \sin \theta$
Net normal reaction from the ground is
$N = 2 m g + m g \cos^ {2} \theta$
Since, $\mu N = F_{H}$
$\Rightarrow \mu = 0. 2 0$
Hence, the correct answer is (B).

Maximum frictional force between A and B is
$f _ {1} = \mu_ {1} m _ {A} g = (0. 2) (2) (1 0) \mathrm{N}$
$\Rightarrow f _ {1} = 4 \mathrm{N}$
Limiting friction between $B$ and ground is
$f _ {2} = (0. 4) (6) (1 0) = 2 4 \mathrm{N}$

For $A$ , $4 = 2a_{A} = 2a$
$\Rightarrow a = 2 \mathrm{ms} ^ {- 2}$
$\mathrm{For} B, F - 2 8 = 4 a$
$\Rightarrow F - 2 8 = 8$
$\Rightarrow F = 3 6 \mathrm{N}$
Hence, the correct answer is (B).

Since $S_{1} + S_{2} = 2\pi r$
$\Rightarrow 4 t + 6 t = 2 \pi r$
$\Rightarrow t = \frac {2 \pi r}{1 0} = \frac {2 \times 3 . 1 4 \times 4}{1 0} = 2. 5 \mathrm{s}$
Hence, the correct answer is (B).

Since $a = g\tan \theta$
$\Rightarrow M g - T = M a\tag{... (1}$
$T = (m + M ^ {\prime}) a\tag{... (2}$
Put (2) in (1), we get
$\begin{array}{r l} & M g - (m + M ^ {\prime}) a = M a \\ \Rightarrow & M g = (m + M + M ^ {\prime}) g \tan \theta \\ \Rightarrow & M g (1 - \tan \theta) = (m + M ^ {\prime}) (g \tan \theta) \\ \Rightarrow & M = \frac {(m + M ^ {\prime}) \tan \theta}{1 - \tan \theta} \\ \Rightarrow & M = \frac {m + M ^ {\prime}}{\cot \theta - 1} \end{array}$
Hence, the correct answer is (D).

C

Free body diagram

For equilibrium, we have
$T \sin (3 0 ^ {\circ}) = 2 0$
$T \cos (3 0 ^ {\circ}) = W$
$\Rightarrow \tan (3 0 ^ {\circ}) = \frac {2 0}{W}\tag{182.}$
$a = \frac {2 m g - \frac {m g}{3}}{m} = \frac {5 g}{3}$
$\Rightarrow W = 2 0 \cot (3 0 ^ {\circ})$
Hence, the correct answer is (D).

C

Since $\mu = 0.8 = \frac{4}{5}$
$F \cos \alpha \tan \theta = W - F \sin \alpha$
$\Rightarrow$ Angle of repose $\alpha = 39^{\circ}$
Since angle which incline makes with the horizontal is $37^{\circ}(<39^{\circ})$ , so the block will not slide down the incline and friction between the block and the incline will be static in nature. So,
$\Rightarrow F \left(\cos \alpha \frac {\sin \theta}{\cos \theta} + \sin \alpha\right) = W$
$\Rightarrow F \sin (\alpha + \theta) = W \cos \theta$
$f _ {s} = F _ {\mathrm{applied}} = m g \sin (3 7 ^ {\circ})$
$\Rightarrow F = \frac {W \cos \theta}{\sin (\theta + \alpha)}$

Now $F = F_{\mathrm{min}}$ when $\theta +\alpha = 90^{\circ}$
$\Rightarrow F _ {\mathrm{min}} = W \cos \theta$
Hence, the correct answer is (A).
Now, if $T$ be the tension in the string, then
$T + f _ {s} = m g \sin (3 7 ^ {\circ})$
$\Rightarrow T = 0$
$\Rightarrow \mu m g = m a$
$\left\{\because f _ {s} = m g \sin (3 7 ^ {\circ}) \right\}$
$\Rightarrow a = \mu g\tag{... (1}$
Hence, the correct answer is (D).

Mass of hanging part of chain is
$m'=\frac{m}{\frac{3\pi R}{2}}\left(\frac{3\pi R}{2}-\pi R\right)=\frac{m}{3}$
So, acceleration of chain is given by
$ a=\frac{2mg-\frac{mg}{3}}{m}=\frac{5g}{3}$
Hence, the correct answer is (D).

A

Since, $F\cos\alpha-N\cos\theta=0$
$\Rightarrow F\cos\alpha=N\cos\theta$ … (1)
Also, $N\sin\theta+F\sin\alpha-W=0$
$\Rightarrow N\sin\theta=W-F\sin\alpha$ … (2)
From (1), $N=\frac{F\cos\alpha}{\cos\theta}$, put in (2), we get
$F\cos\alpha\tan\theta=W-F\sin\alpha$
$\Rightarrow F\left(\frac{\cos\alpha\sin\theta}{\cos\theta}+\sin\alpha\right)=W$
$\Rightarrow F\sin(\alpha+\theta)=W\cos\theta$
$\Rightarrow F=\frac{W\cos\theta}{\sin(\theta+\alpha)}$
Now $F=F_{\min}$ when $\alpha+\theta=90^\circ$
$\Rightarrow F_{\min}=W\cos\theta$
Hence, the correct answer is (A).

METHOD I
For A, $f=ma$
$\Rightarrow \mu mg=ma$
$\Rightarrow a=\mu g$
For B, $T-\mu mg=2ma$
For C, $mg-T=ma$
From (1), (2) and (3), we get
$mg-\mu mg=3m(\mu g)$
$\Rightarrow 4\mu mg=mg$
$\Rightarrow \mu=\frac{1}{4}$
METHOD II
Acceleration of the system is
$a_{\text{system}}=\frac{mg}{4m}=\frac{g}{4}$
For Block A not to slip on $B$, we have
$F_{\text{pseudo}}=f$
$\Rightarrow m\left(\frac{g}{4}\right)=\mu mg$
$\Rightarrow \mu=\frac{1}{4}$
Hence, the correct answer is (C).

If common maximum acceleration is 'a' then,
$\begin{array}{r l} & {\left(m _ {A} + m _ {B}\right) a = f} \\ {\Rightarrow} & {(2 + 4) a = 2 t} \\ {\Rightarrow} & {3 a = t} \end{array}\tag{... (1}$
For block B, we have
$\begin{array}{r l} & f _ {L} = m _ {B} \times a \\ \Rightarrow & 0. 5 \times 2 \times 1 0 = 4 a \\ \Rightarrow & a = 2. 5 \mathrm{ms} ^ {- 2} \end{array}$
From equation (1), we get
$t = 7. 5 \mathrm{sec}$
Hence, the correct answer is (B).

Since $M$ is at rest, so the tension in the string is
$T = \frac {M g}{2}$
Let acceleration of m and $m'$ be a, so one will move downward and other will move upward.
For $m$ , we have
$m g - \frac {M g}{2} = m a\tag{... (1}$
For $m'$ , we have
$\frac {M g}{2} - m ^ {\prime} g = m ^ {\prime} a$.......(2)
Solving equations (1) and (2), we get
$\frac {4}{M} = \frac {1}{m} + \frac {1}{m ^ {\prime}}$
Hence, the correct answer is (A).

$T = 4g + 5(10 + 2)$
$\Rightarrow T = 4 0 + 6 0 = 1 0 0 \mathrm{N}$
Hence, the correct answer is (B).

A

For the sliding not to occur, we have $\tan\theta\leq\mu$ .
Since $\tan\theta=\frac{dy}{dx}=\frac{2x}{a}=\frac{2\sqrt{ya}}{a}=2\sqrt{\frac{y}{a}}$
$\Rightarrow 2 \sqrt {\frac {y}{a}} \leq \mu$
$\Rightarrow y \leq \frac {a \mu^ {2}}{4}$
Hence, the correct answer is (C).

Acceleration of block relative to wedge, is
$a^{\prime} = a\cos \theta -g\sin \theta$ (up the plane)
$\Rightarrow \quad a ^ {\prime} = 1 0 (\sqrt {3}) \left(\frac {\sqrt {3}}{2}\right) - 1 0 \left(\frac {1}{2}\right) = 1 0 \mathrm{ms} ^ {- 2}$
Since $l=\frac{1}{2}a't^{2}$
$\Rightarrow t = \sqrt {\frac {2 l}{a ^ {\prime}}} = \sqrt {\frac {2 . 5}{1 0}} = \frac {1}{2} \sec$
Hence, the correct answer is (D).

Since, acceleration of body w.r.t. lift is $a_{net} = g + a$ .
So, time of flight of body, $t = \frac{2u}{g + a}$
$\Rightarrow a = \frac {2 u - g t}{t} = \frac {(2 \times 5) - (1 0 \times 0 . 8)}{0 . 8} = 2. 5 \mathrm{ms} ^ {- 2}$
Hence, the correct answer is (C).