NLM

Maximum amount of heat is liberated when the friction between the block and belt is kinetic, i.e., when the block moves with a velocity relative to the belt.
Since, $a = \mu g$
Also, $v^{2} = u^{2} + 2as$
$\Rightarrow 0 = v _ {0} ^ {2} + 2 (- \mu g) \ell$
$\Rightarrow v _ {0} = \sqrt {2 \mu g \ell}$
Hence, the correct answer is (B).

Since, $a_{A} = 6a_{B}$
$\Rightarrow a _ {B} = \frac {a _ {A}}{6} = \frac {1 2}{6} = 2 \mathrm{ms} ^ {- 2}$
Hence, the correct answer is (B).

Let acceleration of block A and B be $a_{1}$ and $a_{2}$ respectively and $a_{p}$ be acceleration of pulley P and let tension in string be T. Then, according to principle of virtual work, we have
$\begin{array}{l} (3 T) a _ {p} = T a _ {1} + (8 T) a _ {2} \\ \Rightarrow a _ {p} = \frac {9 g}{3} = 3 g (\text { upwards }) \end{array}$
Hence, the correct answer is (D).

For first $\frac{H}{3}$ , we have
$\begin{array}{r l} & a _ {1} = \frac {2 m g}{3 m} = \frac {2 g}{3} \\ \Rightarrow & v _ {1} ^ {2} = 2 \bigg (\frac {2 g}{3} \bigg) \bigg (\frac {H}{3} \bigg) \\ \Rightarrow & v _ {1} ^ {2} = \frac {4 g H}{9} \end{array}$
For next $H$ , we have
$\begin{array}{r l} & a _ {2} = \frac {m g}{2 m} = \frac {g}{2} \\ \Rightarrow & v ^ {2} = v _ {1} ^ {2} + 2 a _ {2} H \\ \Rightarrow & v ^ {2} = \frac {4 g H}{9} + 2 \left(\frac {g}{2}\right) H = \frac {1 3}{9} g H \\ \Rightarrow & v = \frac {\sqrt {1 3 g H}}{3} \end{array}$
Hence, the correct answer is (C).

Acceleration of $A$ w.r.t. ground $(G)$ is
$\vec {a} _ {A G} = a _ {0} \cos \theta \hat {i} + a _ {0} \sin \theta \hat {j}$
Acceleration of B w.r.t. ground (G) is
$\vec {a} _ {B G} = a _ {0} \sin \theta \hat {j}$
So, acceleration of A w.r.t. B is
$\vec {a} _ {A B} = \vec {a} _ {A G} - \vec {a} _ {B G} = a _ {0} \cos \theta \hat {i}$
Hence, the correct answer is (C).

$2 V _ {A} \cos \theta = V _ {0}$
Differentiating, we get
$2 a _ {A} \cos \theta - 2 V _ {A} \sin \theta \frac {d \theta}{d t} = 0\tag{... (1}$
Since $Y_{A} = b\cot \theta$
$\Rightarrow V _ {A} = \frac {d Y _ {A}}{d t} = (b \sec^ {2} \theta) \frac {d \theta}{d t}$
Put $V_{A}$ in (1), we get
$a _ {A} = \frac {V _ {0} ^ {2}}{4 b}$
Hence, the correct answer is (C).

(A) is correct when the bicycle is being pedalled because during pedalling, the force is acting in the backward direction on the rear wheel and thus frictional force acts in the forward direction.
(D) is correct when bicycle is not pedalled.
Hence, (A) and (D) are correct.

All accelerated frames are Non-inertial frames. Since earth rotates about its own axis and revolves around the sun, so it is a Non-Inertial frame (STRICTLY SPEAKING), whereas for a good number of cases we assume the earth to be an Inertial frame of reference as the value of acceleration is extremely small.
Hence, (B) and (D) are correct.

Let F be the force exerted on the mirror by the photon. Since a photon will be reflected from the mirror with the same value of momentum so, change in momentum $\Delta p = p - (-p) = 2p = \frac{2E}{c}$ .
(Because for a photon E = pc)

$\Rightarrow F = \frac {\Delta p}{\Delta t}$
$\Rightarrow F = \frac {2 E}{c \Delta T}\tag{... (1}$
Also
$T\cos \theta = mg$ and
... (2)
$T \sin \theta = F = \frac {2 E}{c \Delta T}\tag{... (3}$
$\Rightarrow \tan \theta = \frac {2 E}{m g c \Delta T}\tag{... (4}$
Further by definition intensity of a photon beam is defined as the energy incident per unit area of a surface per unit time.
$\Rightarrow I = \frac {E}{A \Delta t}$
$\Rightarrow \frac {E}{\Delta t} = I A\tag{... (5}$
Substituting (5) in (4), we get
$\tan \theta = \frac {2 I A}{m g c}$
Also from (1), we get
$F = \frac {2}{c} (I A)$
$\Rightarrow \text { Radiation Pressure } = \frac {F}{A} = \frac {2 I}{c}$
Hence, (A) and (C) are correct.

When P is gradually increased, so till the moment P becomes equal to $f_{0}$ , mass A will not move and hence the string will not develop any tension due to pull P. So, T = 0 for $P < f_{0}$ . Further for A and B to move collectively with acceleration a (say), we must have
$\begin{array}{l l} & a = \frac {P - (f _ {1}) _ {\lim} - (f _ {2}) _ {\lim}}{2 m} \\ \Rightarrow & a = \frac {P - 2 f _ {0}}{2 m} \end{array}\tag{... (1}$
So, the system will not move till $P > 2f_{0}$ , but T will be developed as soon as P becomes greater than $f_{0}$ . Hence
\begin{array}{r l} & P = T + f _ {0} \\ \Rightarrow & T = P - f _ {0} \text {for} f _ {0} < P < 2 f _ {0} \end{array}
So, (B) is also correct.
Once $P > 2f_{0}$ , then for
BLOCK A
$ P - T - f _ {0} = m a $
$\Rightarrow P - T - f _ {0} = m \left(\frac {P - 2 f _ {0}}{2 m}\right)$
$\Rightarrow P - T - f _ {0} = \frac {P}{2} - f _ {0}$
$ \Rightarrow P - T = \frac {P}{2}$
$\Rightarrow T = \frac {P}{2}$....(2)
because of.....(1)
Hence, (A), (B) and (C) are correct.

For no motion $F_{\mathrm{applied}} \leq f(= \mu N)$
$\begin{array}{r l} \Rightarrow & F \cos 6 0 ^ {\circ} \leq \mu (\mathrm{mg} + F \sin 6 0 ^ {\circ}) \\ \Rightarrow & \frac {F}{2} \leq \frac {1}{2 \sqrt {3}} \left(\sqrt {3} g + \frac {F \sqrt {3}}{2}\right) \\ \Rightarrow & \frac {F}{2} \leq g \\ \Rightarrow & F _ {\max} = 2 0 \mathrm{N} \end{array}$
Hence, (A) is correct.

If $T$ is the tension in the string and $N$ is the normal reaction exerted by $B$ on $A$ , then from FBD of $A$ , we get
$T \sin (3 0 ^ {\circ}) = m _ {A} g + 0. 2 \mathrm{N}\tag{... (1}$
$T \cos (3 0 ^ {\circ}) = N\tag{\( \ldots(2) \}$
$\Rightarrow T \sin (3 0 ^ {\circ}) = m _ {A} g + 0. 2 T \cos (3 0 ^ {\circ})$
$\Rightarrow T \bigl [ \sin (3 0 ^ {\circ}) - 0. 2 \cos (3 0 ^ {\circ}) \bigr ] = 1 0 0$
$\Rightarrow T\cong 306N\text{ and } N\cong 265N$

Free body diagram for B is shown here. If a be the acceleration of B, then

$\begin{array}{r l} & m _ {B} g - 2 (0. 2 \mathrm{N}) = m _ {B} a \\ \Rightarrow & a = \frac {2 0 0 - (0 . 4) (2 6 5)}{2 0} \\ \Rightarrow & a = 4. 7 \mathrm{ms} ^ {- 2} \end{array}$
Hence, (A), (C) and (D) are correct.

For $m_{1}$
... (1)

For $m_{2}$
$\begin{array}{r l} & T + m _ {2} g \sin \beta = m _ {2} a _ {0} \cos \beta \\ \Rightarrow & T = m _ {2} a _ {0} \cos \beta - m _ {2} g \sin \beta \end{array}\tag{... (2}$

Equating (1) and (2), we get
$\begin{array}{l l} & m _ {1} g \sin \alpha - m _ {1} a _ {0} \cos \alpha = m _ {2} a _ {0} \cos \beta - m _ {2} g \sin \beta \\ \Rightarrow & \big (m _ {1} \sin \alpha + m _ {2} \sin \beta \big) g = \big (m _ {1} \cos \alpha + m _ {2} \cos \beta \big) a _ {0} \\ \Rightarrow & a _ {0} = \left(\frac {m _ {1} \sin \alpha + m _ {2} \sin \beta}{m _ {1} \cos \alpha + m _ {2} \cos \beta}\right) g \end{array} ..\tag{... (3}$
Substituting (3) either in (2) or in (1), we get
$T = \left[ \frac {m _ {1} m _ {2}}{m _ {1} \cos \alpha + m _ {2} \cos \beta} \right] g \sin (\alpha - \beta)$......(4)
Hence, (B) and (D) are correct.

Free body diagram for the block is shown for convenience

According to Newton's Second Law, we have
$\begin{array}{l} f - m g \sin (3 0 ^ {\circ}) = m a \\ \Rightarrow \quad f = (1) (1 0) \left(\frac {1}{2}\right) + (1) (1) = 6 \mathrm{N} \end{array}$
Also, we have $N = mg\cos (30^{\circ}) = 5\sqrt{3} \mathrm{~N}$ .
So, net contact force between the block and the belt is
$\begin{array}{r l} & F _ {c} = \sqrt {N ^ {2} + f ^ {2}} = \sqrt {7 5 + 3 6} = \sqrt {1 1 1} \mathrm{N} \\ \Rightarrow & F _ {c} \cong 1 0.5 \mathrm{N} \end{array}$
Hence, (A), (B) and (D) are correct.

$a = g\tan \alpha$
$\mathrm{Since,} \sin \alpha = \frac {1}{x}$
(m x) g - T = (m x) a

From our knowledge of constraint equations, we observe that $a_{A}=a_{B}=a$ (say). Also, since the string is light and inextensible, so tension throughout the string is also the same, say T.
$\Rightarrow v \frac {d v}{d x} = g \frac {x}{L}$
For B,
Solving (1) and (2), we get
$a = \frac {3 g}{4} \mathrm{and} T = \frac {m g}{4}$
For A,
$\Rightarrow T + \frac {m g}{2} = m a$
Hence, (B) and (D) are correct.

The readings of the spring balance (under various situations discussed) and the reading of the weighing machine are equal to the forces exerted by them on the block.
T = m (L - x) a
$\Rightarrow \tan \alpha = \frac {1}{\sqrt {x ^ {2} - 1}}$
$\Rightarrow x g = L a$
$\Rightarrow (m x) g - m (L - x) a = (m x) a$
$\Rightarrow a = \frac {g}{\sqrt {x ^ {2} - 1}}\tag{{Using (1)}}$
$\text { and } N = \frac {m g}{\cos \alpha} = \frac {m g x}{\sqrt {x ^ {2} - 1}}$
$\Rightarrow a = g \frac {x}{L}$

Hence, (B), (C) and (D) are correct.

Let $m$ be the mass per unit length of the chain. Then
$(mx)g-T=(mx)a$ … (1)
$T=m(L-x)a$ … (2)
$\Rightarrow (mx)g-m(L-x)a=(mx)a$
$\Rightarrow xg=La$
$\Rightarrow a=g\frac{x}{L}$
$\Rightarrow v\frac{dv}{dx}=g\frac{x}{L}$
$\Rightarrow v,dv=\frac{g}{L}x,dx$
$\Rightarrow \int_0^v v,dv=\frac{g}{L}\int_0^x x,dx$
$\Rightarrow \frac{v^2}{2}=\frac{g}{L}\frac{x^2}{2}$
$\Rightarrow v=x\sqrt{\frac{g}{L}}$
Hence, (A) and (C) are correct.

mg = $\mu N$ and
$\begin{array}{l} {N = m r \omega_ {\mathrm{min}} ^ {2}} \\ {\Rightarrow \quad m g = \mu m r \omega_ {\mathrm{min}} ^ {2}} \end{array}$

$\Rightarrow \quad \omega_ {\min} = \sqrt {\frac {g}{\mu R}} \text { and } v _ {\min} = R \omega_ {\min} = \sqrt {\frac {g R}{\mu}}$
Hence, (A) and (D) are correct.

For $W_{2}$ :
$N _ {2} + T \sin 4 5 = W _ {2} = 1 0 0\tag{... (1}$
$T \cos 4 5 = 0. 2 5 N _ {2}\tag{.(2}$

For $W_{1}$ :
$P = 0. 2 5 \left(N _ {1} + N _ {2}\right)\tag{.(3}$
$N _ {2} + W _ {1} = N _ {1}\tag{... (4}$

Solving, we get $P = 90 \mathrm{~N}$ , $T = 20\sqrt{2} \mathrm{~N}$
Hence, (B) and (D) are correct.

$W\sin \theta = P + \mu N_{1}$ and $N_{1} = W\cos \theta$
$\Rightarrow \quad W\sin\theta = P + \mu W\cos\theta$
$\Rightarrow \quad P = W (\sin \theta - \mu \cos \theta)$
$\Rightarrow \quad P = W (\sin \theta - \tan \phi \cos \theta)$
$\left\{\because \mu = \tan \phi \right\}$
$\Rightarrow \quad \frac {P}{W} = \frac {\sin (\theta - \phi)}{\cos \phi}\tag{... (1}$

$P \cos \theta + \mu N _ {2} = W \sin \theta$
$W \cos \theta + P \sin \theta = N _ {2}$
$\Rightarrow P \cos \theta + \mu (W \cos \theta + P \sin \theta) = W \sin \theta$
$\Rightarrow P (\cos \theta + \mu \sin \theta) = W (\sin \theta - \mu \cos \theta)$
$\begin{array}{l l} \Rightarrow & P \cos (\theta - \phi) = W \sin (\theta - \phi) \\ \Rightarrow & \frac {P}{W} = \tan (\theta - \phi) \end{array}\tag{... (2}$
Equating (1) and (2), we get
$\cos \phi = \cos (\theta - \phi)$
$\Rightarrow \quad \theta - \phi = \phi$
$\Rightarrow \quad \theta = 2 \phi$
$\Rightarrow \frac {P}{W} = \tan (2 \phi - \phi)$
$\Rightarrow \frac {P}{W} = \tan \phi$
Hence, (A) and (D) are correct.

$mg\cos \alpha = N$
$m g \sin \alpha = \mu N\tag{... (1}$
$\Rightarrow \cot \alpha = \frac {1}{\mu} = 3\tag{... (2}$
Hence, (A) is correct.

$T = Mg$

Also the weight of pulley $mg$ is acting vertically downward. So total downward force is $(m + M)g$ and horizontal force is $T = Mg$ . The resultant of these two is
$\sqrt {M ^ {2} + (m + M) ^ {2}} g$
Hence, (D) is correct.

Reading of the spring balance is equal to the tension in the rope which equals the force of friction between the rope and the boy as well as the force exerted by the rope and the boy on each other.
Hence, (A), (C) and (D) are correct.

Due to Newton's Third Law, the block will exert an equal and opposite force on the table. Since $F > W$ , so the block will have an upward acceleration.

Hence, (B) and (C) are correct.