Tension Force and Pulley Block System Questions in English

(A) Let $T$ be the tension in the upper half of the rope. The lower half of the rope is vertical and supports the mass of $10 \; kg$,so the tension in the lower half is $T_{lower} = mg = 10 \times 10 = 100 \; N$.
At the mid-point where force $F$ is applied,we consider the equilibrium of forces.
The horizontal component of the tension $T$ in the upper rope must balance the applied force $F$: $T \sin 45^{\circ} = F$.
The vertical component of the tension $T$ in the upper rope must balance the downward tension from the lower rope: $T \cos 45^{\circ} = T_{lower} = 100 \; N$.
Dividing the two equations: $\frac{T \sin 45^{\circ}}{T \cos 45^{\circ}} = \frac{F}{100}$.
$\tan 45^{\circ} = \frac{F}{100}$.
Since $\tan 45^{\circ} = 1$,we get $1 = \frac{F}{100}$,which implies $F = 100 \; N$.

(D) Let the masses be $m_1 = 8\; kg$ and $m_2 = 12\; kg$. The system is connected by a light inextensible string over a frictionless pulley.
Since $m_2 > m_1$,the mass $m_2$ moves downward with acceleration $a$,and $m_1$ moves upward with the same acceleration $a$.
Applying Newton's second law of motion:
For mass $m_1$: $T - m_1 g = m_1 a$ --- $(i)$
For mass $m_2$: $m_2 g - T = m_2 a$ --- $(ii)$
Adding equations $(i)$ and $(ii)$:
$(m_2 - m_1) g = (m_1 + m_2) a$
$a = \left( \frac{m_2 - m_1}{m_1 + m_2} \right) g = \left( \frac{12 - 8}{12 + 8} \right) \times 10 = \frac{4}{20} \times 10 = 2\; m/s^2$
Now,substitute $a$ into equation $(ii)$ to find tension $T$:
$T = m_2(g - a) = 12(10 - 2) = 12 \times 8 = 96\; N$
Thus,the acceleration is $2\; m/s^2$ and the tension is $96\; N$.

(B) Mass of the block,$m = 25 \; kg$
Mass of the man,$M = 50 \; kg$
Acceleration due to gravity,$g = 10 \; m/s^2$
Force required to lift the block,$F = mg = 25 \times 10 = 250 \; N$
Weight of the man,$W = Mg = 50 \times 10 = 500 \; N$
Case $(a)$: When the man lifts the block directly,he pulls the rope upwards. By Newton's third law,the rope pulls the man downwards with a force $F = 250 \; N$. The total downward force on the floor is the sum of the man's weight and the reaction force from the rope.
Action on the floor $= W + F = 500 + 250 = 750 \; N$.
Case $(b)$: When the man lifts the block using a pulley,he pulls the rope downwards. By Newton's third law,the rope pulls the man upwards with a force $F = 250 \; N$. The net downward force on the floor is the man's weight minus the upward reaction force from the rope.
Action on the floor $= W - F = 500 - 250 = 250 \; N$.
Conclusion: Since the floor yields at $700 \; N$,the man should adopt method $(b)$ to lift the block,as the force exerted on the floor $(250 \; N)$ is less than the limit of $700 \; N$.

(D) For a system of two masses $m_1$ and $m_2$ connected by a string over a frictionless pulley,the acceleration $a$ is given by the formula:
$a = \frac{(m_2 - m_1)g}{m_1 + m_2}$
Here,$m_1 = 4 \, kg$ and $m_2 = 6 \, kg$.
Substituting the values into the formula:
$a = \frac{(6 - 4)g}{6 + 4}$
$a = \frac{2g}{10}$
$a = \frac{g}{5}$
Thus,the acceleration of the system is $\frac{g}{5}$.

(A) Let $a_{1}$ and $a_{2}$ be the accelerations of the ball (upward) and the rod (downward),respectively.
From the constraint relation,$2 a_{1} = a_{2} \dots (i)$
For the ball,the equation of motion is: $2T - \frac{9}{5}mg = \frac{9}{5}ma_{1} \dots (ii)$
For the rod,the equation of motion is: $mg - T = ma_{2} \dots (iii)$
Substituting $T = mg - ma_{2}$ from $(iii)$ into $(ii)$:
$2(mg - ma_{2}) - \frac{9}{5}mg = \frac{9}{5}ma_{1}$
$2mg - 2ma_{2} - 1.8mg = 1.8ma_{1}$
$0.2g - 2a_{2} = 1.8a_{1}$
Using $a_{2} = 2a_{1}$ from $(i)$:
$0.2g - 2(2a_{1}) = 1.8a_{1}$
$0.2g = 5.8a_{1} \implies a_{1} = \frac{0.2g}{5.8} = \frac{g}{29} \, m/s^2$ (upward)
Then $a_{2} = 2a_{1} = \frac{2g}{29} \, m/s^2$ (downward)
The relative acceleration of the ball with respect to the rod is $a_{rel} = a_{1} + a_{2} = \frac{g}{29} + \frac{2g}{29} = \frac{3g}{29}$.
Using the equation of motion $s = ut + \frac{1}{2}a_{rel}t^2$ with $u=0$ and $s=1 \, m$:
$1 = 0 + \frac{1}{2} \left(\frac{3 \times 10}{29}\right) t^2$
$1 = \frac{15}{29} t^2 \implies t^2 = \frac{29}{15} \approx 1.933$
$t = \sqrt{1.933} \approx 1.39 \, s \approx 1.4 \, s$.

(C) Let the acceleration of the system be $a$ towards the left.
For the mass $4M$ on the ice slab,the forces acting are the pulling force $2Mg$ to the left and the tension $T$ to the right. Applying Newton's second law: $2Mg - T = 4Ma$ (Equation $1$).
For the hanging mass $M$,the forces acting are the tension $T$ upwards and its weight $Mg$ downwards. Since the system accelerates towards the left,the mass $M$ moves upwards. Applying Newton's second law: $T - Mg = Ma$ (Equation $2$).
Adding Equation $1$ and Equation $2$: $(2Mg - T) + (T - Mg) = 4Ma + Ma$.
$Mg = 5Ma$,which gives $a = \frac{g}{5}$.
Substituting the value of $a$ in Equation $2$: $T = Mg + M(\frac{g}{5}) = Mg + \frac{Mg}{5} = \frac{6Mg}{5}$.
Comparing this with the given expression $\frac{x}{5}Mg$,we get $x = 6$.

(B) The acceleration of a system of two masses $M_{1}$ and $M_{2}$ connected by a string over a frictionless pulley is given by the formula:
$a = \frac{|M_{2} - M_{1}|}{M_{1} + M_{2}} g$
Case $1$: Given $M_{2} = 2M_{1}$.
$a_{1} = \frac{|2M_{1} - M_{1}|}{M_{1} + 2M_{1}} g = \frac{M_{1}}{3M_{1}} g = \frac{g}{3}$
Case $2$: Given $M_{2} = 3M_{1}$.
$a_{2} = \frac{|3M_{1} - M_{1}|}{M_{1} + 3M_{1}} g = \frac{2M_{1}}{4M_{1}} g = \frac{g}{2}$
Now,calculating the ratio $\frac{a_{1}}{a_{2}}$:
$\frac{a_{1}}{a_{2}} = \frac{g/3}{g/2} = \frac{2}{3}$
Therefore,the correct option is $B$.

(C) Free Body Diagram ($F$.$B$.$D$) of the monkey while moving downward:
Using Newton's second law: $mg - T = ma_1$
$500 - T = 50 \times 4 \Rightarrow T = 300\,N$.
Free Body Diagram ($F$.$B$.$D$) of the monkey while moving upward:
Using Newton's second law of motion: $T - mg = ma_2$
$T - 500 = 50 \times 5 \Rightarrow T = 750\,N$.
Breaking strength of the string is $350\,N$.
Since $750\,N > 350\,N$,the string will break while the monkey is moving upward.

(D) Let $\lambda$ be the linear mass density of the chain,so $\lambda = \frac{m}{L}$.
The mass of the chain on one side is $m_1 = \lambda l$ and on the other side is $m_2 = \lambda (L - l)$.
The net force on the chain is $F_{net} = (m_2 - m_1)g = \lambda(L - l - l)g = \lambda(L - 2l)g$.
The total mass of the chain is $m = \lambda L$.
The acceleration $a$ of the chain is given by $a = \frac{F_{net}}{m} = \frac{\lambda(L - 2l)g}{\lambda L} = \frac{(L - 2l)g}{L}$.
Given that $a = \frac{g}{2}$ when $l = \frac{L}{x}$,we substitute these values:
$\frac{g}{2} = \frac{(L - 2(\frac{L}{x}))g}{L}$
$\frac{1}{2} = 1 - \frac{2}{x}$
$\frac{2}{x} = 1 - \frac{1}{2} = \frac{1}{2}$
$x = 4$.

(B) Initially,the system is in equilibrium. For mass $M$,the spring force $F_s = Mg$. Since the pulley is massless and the string is continuous,the tension $T$ in the string is equal to the spring force,so $T = Mg$.
For mass $m$,the forces are tension $T$ upwards and gravity $mg$ downwards. Since it is in equilibrium,$T = mg$,which implies $Mg = mg$. However,the problem states $M > m$,which implies the initial setup must have been held in equilibrium by the string connected to the ground.
Immediately after the string connected to $m$ is detached,the tension $T$ in the string becomes zero. The spring force $F_s$ cannot change instantaneously,so $F_s = Mg$ remains acting on $M$ upwards.
For mass $M$,the forces are $Mg$ downwards and $F_s = Mg$ upwards. Thus,the net force on $M$ is zero,and its acceleration is $0$.
For mass $m$,the only force acting is gravity $mg$ downwards. Thus,its acceleration is $g$ downwards. Wait,let's re-evaluate: The tension $T$ was supporting $m$. If the string to the ground is cut,the tension $T$ in the string connected to the pulley becomes $Mg$. The forces on $m$ are $T = Mg$ upwards and $mg$ downwards. The net force is $Mg - mg$ upwards. Therefore,the acceleration of $m$ is $a_m = (Mg - mg)/m = (M-m)g/m$ upwards. The acceleration of $M$ is zero because the spring force $Mg$ balances its weight $Mg$.

(A) The system consists of two sides connected by a pulley. On one side, we have mass $m_3 = 4 \, kg$. On the other side, we have masses $m_1$ and $m_2$ connected in series, so the total mass on that side is $M_{right} = m_1 + m_2 = 4 \, kg + 2 \, kg = 6 \, kg$.
The net force $F_{net}$ acting on the system is the difference between the weights on both sides:
$F_{net} = (M_{right} \cdot g) - (m_3 \cdot g) = (6 \, kg \cdot 10 \, m/s^2) - (4 \, kg \cdot 10 \, m/s^2) = 60 \, N - 40 \, N = 20 \, N$.
The total mass of the system is $M_{total} = m_1 + m_2 + m_3 = 4 \, kg + 2 \, kg + 4 \, kg = 10 \, kg$.
Using Newton's second law, $a = \frac{F_{net}}{M_{total}} = \frac{20 \, N}{10 \, kg} = 2 \, m/s^2$.
Therefore, the acceleration of the blocks is $2 \, m/s^2$.

(B) Given:
$F_1 = 20 \, N$ (force on $4 \, kg$ block to the right)
$F_2 = 10 \, N$ (force on $2 \, kg$ block to the left)
$m_1 = 2 \, kg$,$m_2 = 4 \, kg$
Let $T$ be the tension in the string.
Since $F_1 > F_2$,the system accelerates towards the right.
The net force on the system is $F_{net} = F_1 - F_2 = 20 - 10 = 10 \, N$.
The total mass of the system is $M = m_1 + m_2 = 2 + 4 = 6 \, kg$.
The acceleration of the system is $a = \frac{F_{net}}{M} = \frac{10}{6} = \frac{5}{3} \, m/s^2$.
Now,consider the free body diagram of the $2 \, kg$ block:
The forces acting on it are $T$ (to the right) and $F_2$ (to the left).
$T - F_2 = m_1 a$
$T - 10 = 2 \times \frac{5}{3}$
$T = 10 + \frac{10}{3} = \frac{30 + 10}{3} = \frac{40}{3} \, N$.

(B) Let the total mass of the block be $m$. The force applied is $F = 40 \, N$.
According to Newton's second law,the acceleration $a$ of the block is given by $a = \frac{F}{m} = \frac{40}{m}$.
Now,consider the block to be divided into two equal halves,each of mass $\frac{m}{2}$.
Let $T$ be the tension at the middle of the block. This tension $T$ acts as the force pulling the rear half of the block.
Applying Newton's second law to the rear half of the block:
$T = (\text{mass of rear half}) \times a$
$T = \left(\frac{m}{2}\right) \times \left(\frac{40}{m}\right)$
$T = \frac{40}{2} = 20 \, N$.
Therefore,the tension at the middle of the block is $20 \, N$.

(C) The system is in equilibrium,so the acceleration $a = 0$.
For block $C$ (mass $m = 5 \ kg$): The tension $T_1$ supports the weight of block $C$.
$T_1 = mg = 5 \times 10 = 50 \ N$.
For block $B$ (mass $m = 5 \ kg$): The tension $T_2$ supports the weight of block $B$ and block $C$.
$T_2 = (m + m)g = 10 \times 10 = 100 \ N$.
For block $A$ (mass $m = 5 \ kg$): The tension $T_3$ supports the weight of blocks $A$,$B$,and $C$.
$T_3 = (m + m + m)g = 15 \times 10 = 150 \ N$.
Therefore,the ratio is:
$\frac{T_3}{T_1} = \frac{150}{50} = 3$.

(C) The total tension $T$ at the rigid support is the sum of the tensions exerted by each man on the rope.
For man $A$ $(m_A = 60 \, kg)$ ascending with acceleration $a_A = 2 \, m/s^2$:
$T_A - m_A g = m_A a_A$
$T_A = m_A(g + a_A) = 60(10 + 2) = 60 \times 12 = 720 \, N$
For man $B$ $(m_B = 50 \, kg)$ descending with constant velocity $(a_B = 0)$:
$T_B = m_B g = 50 \times 10 = 500 \, N$
For man $C$ $(m_C = 40 \, kg)$ descending with acceleration $a_C = 1 \, m/s^2$:
$m_C g - T_C = m_C a_C$
$T_C = m_C(g - a_C) = 40(10 - 1) = 40 \times 9 = 360 \, N$
The total tension at the rigid support is $T = T_A + T_B + T_C = 720 + 500 + 360 = 1580 \, N$.

(C) The system consists of a pulley with a mass $m$ on one side and a combined mass of $2m + 3m = 5m$ on the other side.
For an Atwood machine with masses $M_1$ and $M_2$,the acceleration $a$ is given by $a = \left( \frac{M_1 - M_2}{M_1 + M_2} \right) g$.
Here,$M_1 = 5m$ and $M_2 = m$.
Substituting these values into the formula:
$a = \left( \frac{5m - m}{5m + m} \right) g$
$a = \left( \frac{4m}{6m} \right) g$
$a = \frac{2}{3} g$
Therefore,the block of mass $m$ will move upward with an acceleration of $\frac{2g}{3}$.

(B) The tension $T$ in the rope is equal to the applied force $F$,so $T = F = \frac{3}{2} mg$.
Applying Newton's second law to the mass $m$:
$F_{\text{net}} = ma$
Here,the upward force is tension $T$ and the downward force is the weight $mg$:
$T - mg = ma$
Substituting the value of $T$:
$\frac{3}{2} mg - mg = ma$
$\frac{1}{2} mg = ma$
$a = \frac{g}{2}$

(B) The mass per unit length of the rope is $\mu = \frac{M}{L}$.
Consider a point at a distance $l$ from the rigid support. The length of the rope below this point is $(L-l)$.
The mass of this lower portion of the rope is $m' = \mu \times (L-l) = \frac{M}{L}(L-l)$.
Since the rope is in equilibrium,the tension $T$ at distance $l$ must support the weight of the lower portion of the rope.
Applying Newton's second law for the lower portion,$\sum F_y = m' a_y$. Since the system is at rest,$a_y = 0$.
Therefore,$T - m'g = 0$,which gives $T = m'g$.
Substituting the value of $m'$,we get $T = \frac{M}{L}(L-l)g$.

(C) Let $T$ be the tension in the string.
According to Newton's second law of motion for the downward motion of the body:
$mg - T = ma$
Given that the acceleration $a = \frac{g}{6}$,we substitute this into the equation:
$mg - T = m(\frac{g}{6})$
$T = mg - \frac{mg}{6} = \frac{5mg}{6}$
The tension $T$ acts upwards while the displacement $x$ is downwards. Therefore,the angle between the force (tension) and the displacement is $180^{\circ}$.
The work done by the string is given by:
$W = T \cdot x \cdot \cos(180^{\circ})$
$W = (\frac{5mg}{6}) \cdot x \cdot (-1)$
$W = -\frac{5mgx}{6}$

(D) The system consists of a trolley of mass $M_0$ and two blocks of mass $M$ and $m$ connected by a string over a pulley. All surfaces are smooth.
If $F=0$,the system is not accelerating,but the block $m$ will experience a gravitational force pulling it downwards,causing it to move. Thus,the blocks cannot remain stationary.
If the trolley accelerates with acceleration $a = F / (M + m + M_0)$,a pseudo force acts on the blocks in the frame of the trolley. For the blocks to be stationary relative to the trolley,the forces acting on them must balance. Specifically,for block $m$,the pseudo force $ma$ must balance the tension $T$,and for block $M$,the tension $T$ must balance the pseudo force $Ma$. This requires $ma = T$ and $T = Ma$,which implies $m=M$. If $M=m$,there exists a unique value of $F$ that makes the system stationary relative to the trolley. Since the question implies a general case where such a condition can be met,both statements $(a)$ and $(b)$ are correct.

(A) The maximum tension the rope can withstand is $T_{max} = 600 \,N$.
The weight of the monkey is $W = mg = 40 \times 10 = 400 \,N$.
Applying Newton's second law for the monkey climbing up with maximum acceleration $a$:
$T_{max} - mg = ma$
$600 - 400 = 40a$
$200 = 40a$
$a = 5 \,m/s^2$
Using the equation of motion $S = ut + \frac{1}{2}at^2$,where initial velocity $u = 0$,distance $S = 10 \,m$,and $a = 5 \,m/s^2$:
$10 = 0 + \frac{1}{2} \times 5 \times t^2$
$10 = 2.5 t^2$
$t^2 = 4$
$t = 2 \,s$
Thus,the time taken is $2 \,s$.

(C) To find the tension in the string,we analyze the forces acting on the bob of mass $m$ in the non-inertial frame of the car.
In the car's frame,the bob experiences a gravitational force $mg$ downwards and a pseudo force $ma$ horizontally backwards.
The effective acceleration due to gravity in this frame is $g_{eff} = \sqrt{g^2+a^2}$.
The component of the effective weight acting along the string is $m g_{eff} = m \sqrt{g^2+a^2}$.
Since the string is being pulled with an acceleration '$a$' vertically,the bob also experiences an upward acceleration '$a$' relative to the car.
Applying Newton's second law along the direction of the string:
$T - m \sqrt{g^2+a^2} = ma$
Therefore,the tension in the string is $T = m \sqrt{g^2+a^2} + ma$.

(C) The total mass of the system is $M_{total} = M + M/2 = 3M/2$.
Given the applied force $F = 2Mg$,the acceleration $a$ of the entire system is given by $a = F / M_{total} = (2Mg) / (3M/2) = 4g/3$.
The force exerted by the rope on the block is the tension $T$ at the point of contact between the rope and the block. This force is responsible for accelerating the block of mass $M$.
Using Newton's second law for the block: $T = M \times a$.
Substituting the value of $a$: $T = M \times (4g/3) = 4Mg/3$.

(A) For a system of two masses $m_1$ and $m_2$ connected by a light string over a smooth fixed pulley,the magnitude of acceleration $a$ is given by the formula:
$a = \frac{|m_1 - m_2| g}{m_1 + m_2}$
Given that the acceleration $a = g / 8$,we have:
$\frac{|m_1 - m_2| g}{m_1 + m_2} = \frac{g}{8}$
Assuming $m_1 > m_2$,we get:
$\frac{m_1 - m_2}{m_1 + m_2} = \frac{1}{8}$
Cross-multiplying gives:
$8(m_1 - m_2) = m_1 + m_2$
$8m_1 - 8m_2 = m_1 + m_2$
$7m_1 = 9m_2$
Therefore,the ratio of the masses is:
$\frac{m_1}{m_2} = \frac{9}{7}$

(D) Let the total mass of the system be $M = M_1 + M_2 + M_3 = 4 \ kg + 6 \ kg + 10 \ kg = 20 \ kg$.
The total downward force due to gravity is $W = Mg = 20 \ kg \times 10 \ m/s^2 = 200 \ N$.
The system is moving upward with an acceleration $a = 2 \ m/s^2$.
Applying Newton's second law for the entire system: $T_1 - Mg = Ma$.
Substituting the values: $T_1 - 200 = 20 \times 2$.
$T_1 - 200 = 40$.
$T_1 = 240 \ N$.

(A) For an Atwood machine with masses $m_1$ and $m_2$ $(m_2 > m_1)$,the acceleration $a$ is given by:
$a = \left( \frac{m_2 - m_1}{m_1 + m_2} \right) g$
Given $a = \frac{g}{\sqrt{2}}$,we substitute this into the equation:
$\frac{g}{\sqrt{2}} = \left( \frac{m_2 - m_1}{m_1 + m_2} \right) g$
$\frac{1}{\sqrt{2}} = \frac{m_2 - m_1}{m_1 + m_2}$
$m_1 + m_2 = \sqrt{2} m_2 - \sqrt{2} m_1$
$m_1 + \sqrt{2} m_1 = \sqrt{2} m_2 - m_2$
$m_1(1 + \sqrt{2}) = m_2(\sqrt{2} - 1)$
$\frac{m_1}{m_2} = \frac{\sqrt{2} - 1}{\sqrt{2} + 1}$

(B) The system consisting of the body and the chain is in equilibrium.
The total downward force acting on the branch is the sum of the weight of the body and the weight of the chain.
Weight of the body,$W_b = 200 \,N$.
Weight of the chain,$W_c = m \times g = 10 \,kg \times 10 \,m/s^2 = 100 \,N$.
Since the system is in equilibrium,the tension $T$ in the chain at the point where it is attached to the branch must balance the total weight.
$T = W_b + W_c = 200 \,N + 100 \,N = 300 \,N$.
Therefore,the branch pulls the chain with a force of $300 \,N$.

(A) For an Atwood machine with masses $m_1$ and $m_2$ (where $m_2 > m_1$),the acceleration $a$ of the system is given by the formula: $a = \left( \frac{m_2 - m_1}{m_1 + m_2} \right) g$.
Given that the acceleration $a = \frac{g}{8}$,we equate the two expressions: $\frac{g}{8} = \left( \frac{m_2 - m_1}{m_1 + m_2} \right) g$.
Canceling $g$ from both sides,we get: $\frac{1}{8} = \frac{m_2 - m_1}{m_1 + m_2}$.
Cross-multiplying gives: $m_1 + m_2 = 8(m_2 - m_1)$.
Expanding the equation: $m_1 + m_2 = 8m_2 - 8m_1$.
Rearranging the terms to group $m_1$ and $m_2$: $m_1 + 8m_1 = 8m_2 - m_2$.
This simplifies to: $9m_1 = 7m_2$.
Therefore,the ratio $\frac{m_2}{m_1} = \frac{9}{7}$.

(B) Let the angle made by the string with the horizontal be $\theta$. The length of each half of the string is $a$. When the separation between the particles is $2x$,the horizontal distance of each particle from the center $P$ is $x$.
From the geometry,$\cos \theta = \frac{x}{a}$ and $\sin \theta = \frac{\sqrt{a^2-x^2}}{a}$.
Considering the vertical equilibrium of the midpoint of the string: $2T \sin \theta = F$,so $T = \frac{F}{2 \sin \theta}$.
The horizontal force on each particle is $T \cos \theta = ma$,where $a$ is the acceleration of the particle.
Substituting $T$: $ma = \left( \frac{F}{2 \sin \theta} \right) \cos \theta = \frac{F}{2} \cot \theta$.
Since $\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{x/a}{\sqrt{a^2-x^2}/a} = \frac{x}{\sqrt{a^2-x^2}}$,
we get $ma = \frac{F}{2} \left( \frac{x}{\sqrt{a^2-x^2}} \right)$,
which gives the acceleration $a = \frac{F}{2m} \left( \frac{x}{\sqrt{a^2-x^2}} \right)$.

(C) Let the acceleration of the system be $a$ and the tension in the string be $T$.
For the $6 \ kg$ mass hanging vertically,the equation of motion is: $6g - T = 6a \quad \dots(i)$
For the $4 \ kg$ mass on the frictionless table,the equation of motion is: $T = 4a \quad \dots(ii)$
Adding equations $(i)$ and $(ii)$,we get:
$6g - T + T = 6a + 4a$
$6g = 10a$
Substituting $g = 10 \ ms^{-2}$:
$6 \times 10 = 10a$
$60 = 10a$
$a = 6 \ ms^{-2}$
Thus,the acceleration of the $6 \ kg$ mass is $6 \ ms^{-2}$.

(D) The system consists of two masses $m$ and $2m$ connected by a string over a pulley. The acceleration $a$ of the system is given by $a = \frac{(m_2 - m_1)g}{m_1 + m_2}$.
Substituting the values,$a = \frac{(2m - m)g}{2m + m} = \frac{mg}{3m} = \frac{g}{3}$.
The tension $T$ in the wire can be found using the equation of motion for mass $m$: $T - mg = ma$,so $T = m(g + a)$.
Substituting $a = \frac{g}{3}$,we get $T = m(g + \frac{g}{3}) = m(\frac{4g}{3}) = \frac{4mg}{3}$.
Stress is defined as force per unit area,so $\text{Stress} = \frac{T}{A} = \frac{4mg}{3A}$.

(D) Let the acceleration of the system be $a$. The total mass of the system is $M = 5 \text{ kg} + 3 \text{ kg} + 1 \text{ kg} = 9 \text{ kg}$.
The net driving force is the difference in weights: $F_{\text{net}} = (5g) - (3g + 1g) = 5g - 4g = g$.
Using Newton's second law,$F_{\text{net}} = Ma$,so $g = 9a$,which gives $a = \frac{g}{9}$.
Now,consider the free body diagram of block $B$ ($1 \text{ kg}$ mass). The forces acting on it are its weight $(1g)$ downwards and the tension $T_2$ upwards. Since the system accelerates such that block $B$ moves upwards,we have:
$T_2 - 1g = 1a$
$T_2 = g + a = g + \frac{g}{9} = \frac{10g}{9}$.
Thus,the tension in the string connecting $A$ and $B$ is $\frac{10g}{9}$.

(B) In an Atwood machine,the tension $T$ in the string is given by $T = \frac{2 m_1 m_2 g}{m_1 + m_2}$. The extension $x$ in the spring is proportional to the tension $T$ $(x = T/k)$.
Case $1$: $m_1 = 2 \ kg, m_2 = 2 \ kg$. Tension $T_1 = \frac{2 \times 2 \times 2 \times g}{2 + 2} = 2g$.
Case $2$: $m_1 = 3 \ kg, m_2 = 2 \ kg$. Tension $T_2 = \frac{2 \times 3 \times 2 \times g}{3 + 2} = 2.4g$.
Case $3$: $m_1 = 1 \ kg, m_2 = 2 \ kg$. Tension $T_3 = \frac{2 \times 1 \times 2 \times g}{1 + 2} = 1.33g$.
Comparing the tensions,we get $T_2 > T_1 > T_3$.
Since $x \propto T$,the extensions follow the same order: $x_2 > x_1 > x_3$.

(D) Let the tension in the strings be $T$. Since the pulleys are smooth and massless,the tension $T$ in the string is equal to the weight of the hanging masses $M$,so $T = Mg$.
For the central mass $\sqrt{2}M$ to be in equilibrium,the vertical components of the tension forces must balance its weight.
The vertical component of the tension in each of the two strings is $T \cos \theta$.
Therefore,the total upward force is $2T \cos \theta$.
Equating this to the weight of the central mass:
$2T \cos \theta = (\sqrt{2}M)g$
Since $T = Mg$,we substitute this into the equation:
$2(Mg) \cos \theta = \sqrt{2}Mg$
$2 \cos \theta = \sqrt{2}$
$\cos \theta = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$
$\theta = \cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)$

(B) The system consists of two blocks $A$ and $B$ connected by a string. Since the table is frictionless and the system is pulled with an acceleration $a = 2 \,ms^{-2}$, we can analyze the motion of block $A$ separately.
Block $A$ is being pulled by the tension $T$ in the string. According to Newton's second law of motion, the force acting on block $A$ is given by:
$T = M_A \times a$
Given:
$M_A = 20 \,kg$
$a = 2 \,ms^{-2}$
Substituting the values:
$T = 20 \,kg \times 2 \,ms^{-2} = 40 \,N$
Therefore, the tension in the string is $40 \,N$.

(B) Let the tension in the string be $T$ and the acceleration of the system be $a$.
For the mass $m$ (moving upwards): $T - mg = ma$ --- $(1)$
For the mass $\frac{3}{2}m$ (moving downwards): $\frac{3}{2}mg - T = \frac{3}{2}ma$ --- $(2)$
Adding equation $(1)$ and $(2)$:
$(T - mg) + (\frac{3}{2}mg - T) = ma + \frac{3}{2}ma$
$\frac{1}{2}mg = \frac{5}{2}ma$
$a = \frac{g}{5}$
Thus,the mass $m$ ascends with an acceleration of $\frac{g}{5}$.

(A) Let the acceleration of the system be $a$ and the tension in the string be $T$.
For the block with mass $(M+m)$ (moving downwards):
$(M+m)g - T = (M+m)a$ --- $(1)$
For the block with mass $M$ (moving upwards):
$T - Mg = Ma$ --- $(2)$
Adding equations $(1)$ and $(2)$:
$(M+m)g - Mg = (M+m+M)a$
$mg = (2M+m)a$
$a = \frac{mg}{2M+m}$

(B) The tension $T$ in the rope when the lift accelerates upwards with acceleration $a$ is given by $T = M(g + a)$.
The stress $\sigma$ is defined as the force per unit area,so $\sigma = \frac{T}{A}$,where $A = \pi r^2 = \pi (\frac{d}{2})^2 = \frac{\pi d^2}{4}$.
Given the maximum safe stress is $S$,we have $S = \frac{M(g + a)}{\frac{\pi d^2}{4}}$.
Rearranging for the diameter $d$,we get $d^2 = \frac{4 M(g + a)}{\pi S}$.
Therefore,the minimum diameter is $d = [\frac{4 M(g + a)}{\pi S}]^{\frac{1}{2}}$.

(A) The particle is moving in a uniform circular motion on a smooth horizontal table.
In this motion,the only horizontal force acting on the particle towards the center of the circle is the tension $T$ in the string.
According to Newton's second law of motion,the net force acting on a body in circular motion must be equal to the centripetal force required for that motion.
The centripetal force $F_c$ is given by the formula $F_c = \frac{m v^{2}}{l}$.
Since the tension $T$ is the only force providing this centripetal force,we have:
Net force (directed towards the centre) $= T = \frac{m v^{2}}{l}$.
Therefore,the net force on the particle directed towards the centre is $T$.

(B) The acceleration $a$ of the system is given by $a = \left(\frac{m_1 - m_2}{m_1 + m_2}\right) g$.
Using the equation of motion $S = ut + \frac{1}{2}at^2$,where $S = 3 \ m$,$u = 0$,and $t = 3 \ s$:
$3 = 0 + \frac{1}{2} \times a \times (3)^2$
$3 = \frac{1}{2} \times a \times 9$
$a = \frac{6}{9} = \frac{2}{3} \ ms^{-2}$.
Now,substitute $a$ into the acceleration formula:
$\frac{2}{3} = \left(\frac{m_1 - m_2}{m_1 + m_2}\right) \times 10$
$\frac{2}{30} = \frac{m_1 - m_2}{m_1 + m_2}$
$\frac{1}{15} = \frac{m_1 - m_2}{m_1 + m_2}$
$m_1 + m_2 = 15m_1 - 15m_2$
$16m_2 = 14m_1$
$\frac{m_1}{m_2} = \frac{16}{14} = \frac{8}{7}$.

(A) Given:
Mass of the body,$m = 200 \, kg$
Breaking force of the rope,$T = 50 \, kg\text{-wt} = 50 \times 10 \, N = 500 \, N$
Acceleration due to gravity,$g = 10 \, ms^{-2}$
When the body is lowered with an acceleration $a$,the equation of motion is:
$mg - T = ma$
Substituting the values:
$(200 \times 10) - 500 = 200a$
$2000 - 500 = 200a$
$1500 = 200a$
$a = \frac{1500}{200} = 7.5 \, ms^{-2}$
Thus,the maximum acceleration is $7.5 \, ms^{-2}$.

(C) Let $T$ be the tension in the string.
The maximum vertical force on the clamp is given by $T \sin 30^{\circ} = 40 \,N$.
Substituting the value of $\sin 30^{\circ} = 1/2$:
$T \cdot (1/2) = 40$
$T = 80 \,N$
Now,consider the free-body diagram of the monkey of mass $m = 5 \,kg$ climbing with acceleration $a$.
The forces acting on the monkey are tension $T$ upwards and weight $mg$ downwards.
According to Newton's second law: $T - mg = ma$
$a = (T - mg) / m$
Substituting the values $T = 80 \,N$,$m = 5 \,kg$,and $g = 10 \,ms^{-2}$:
$a = (80 - 5 \times 10) / 5$
$a = (80 - 50) / 5$
$a = 30 / 5 = 6 \,ms^{-2}$
Thus,the maximum acceleration is $6 \,ms^{-2}$.

(C) Let the weights be $W_1 = 2 \,N$ and $W_2 = 3 \,N$. The masses are $m_1 = W_1/g$ and $m_2 = W_2/g$.
The pulley is accelerated upwards with $a = g$.
In the frame of the pulley, each mass experiences a pseudo-force $F_p = ma$ acting downwards.
Thus, the effective acceleration due to gravity for each mass becomes $g_{eff} = g + a = g + g = 2g$.
The effective weights are $W_1' = m_1(2g) = 2W_1 = 4 \,N$ and $W_2' = m_2(2g) = 2W_2 = 6 \,N$.
The tension $T$ in a string over a pulley with masses $m_1$ and $m_2$ is given by $T = \frac{2m_1m_2}{m_1+m_2} g_{eff}$.
Substituting the effective weights: $T = \frac{2 W_1' W_2'}{W_1' + W_2'} = \frac{2 \times 4 \times 6}{4 + 6} = \frac{48}{10} = 4.8 \,N$.

(D) Consider a wire of negligible mass suspended from a ceiling. $A$ force $F$ is applied at the lower end of the wire.
Since the wire is massless,the net force on any segment of the wire must be zero for it to be in equilibrium.
If we take any cross-section of the wire and consider the lower part of the wire below that section,the only forces acting on this part are the tension $T$ acting upwards at the cross-section and the weight $F$ acting downwards at the end.
Applying Newton's second law,$T - F = ma$. Since the mass $m$ of the wire is zero,$T - F = 0$,which gives $T = F$.
Therefore,the tension at any cross-section of the wire is $F$.

(B) The common acceleration of the system is given by:
$a = \left( \frac{m_A - m_B}{m_A + m_B} \right) g = \left( \frac{1.5 - 0.5}{1.5 + 0.5} \right) g = \frac{1}{2} g = 5 \ m/s^2$.
When block $A$ is released from a height of $80 \ cm$ $(0.8 \ m)$,it moves downwards with acceleration $a = 5 \ m/s^2$. The velocity $v$ of the blocks when block $A$ hits the ground is given by $v^2 = u^2 + 2as$:
$v^2 = 0 + 2(5)(0.8) = 8 \ (m/s)^2$.
At this instant,block $B$ is at a height of $80 \ cm$ from the ground and has an upward velocity $v = \sqrt{8} \ m/s$.
After block $A$ hits the ground,the string becomes slack,and block $B$ moves under gravity with deceleration $g = 10 \ m/s^2$.
Using the third equation of motion for block $B$ to find the additional height $h$ it rises:
$v_f^2 = v^2 - 2gh$
$0 = 8 - 2(10)h$
$h = \frac{8}{20} = 0.4 \ m = 40 \ cm$.
The maximum height reached by block $B$ from the ground is the initial height plus the additional height:
$H_{max} = 80 \ cm + 40 \ cm = 120 \ cm$.

(B) Let $M = 10 \ kg$ be the mass on the horizontal surface and $m$ be the mass of the hanging block.
Since the system is released from rest,the initial velocity $u = 0$.
The distance covered by the hanging block is $s = 2 \ m$ in time $t = 2 \ s$.
Using the equation of motion $s = ut + \frac{1}{2}at^2$:
$2 = 0 + \frac{1}{2} \times a \times (2)^2$
$2 = 2a \Rightarrow a = 1 \ ms^{-2}$.
Now,applying Newton's second law to the system:
For the hanging block: $mg - T = ma$
For the block on the surface: $T = Ma$
Adding these two equations: $mg = (M + m)a$
$m(g - a) = Ma$
$m(10 - 1) = 10 \times 1$
$9m = 10 \Rightarrow m = \frac{10}{9} \ kg \approx 1.11 \ kg$.
The weight of the hanging block is $W = mg = \frac{10}{9} \times 10 = \frac{100}{9} \ N \approx 11.11 \ N$.

(B) Let the string form a triangle with the horizontal surface when pulled. The length of each half of the string is $a$. Let the horizontal distance of each particle from the midpoint be $x$. The vertical height of the midpoint $P$ is $y = \sqrt{a^2 - x^2}$.
Let $\theta$ be the angle the string makes with the horizontal surface. Then $\cos \theta = \frac{x}{a}$ and $\sin \theta = \frac{y}{a} = \frac{\sqrt{a^2 - x^2}}{a}$.
For the midpoint $P$,the vertical force balance is $2T \sin \theta = F$,so $T = \frac{F}{2 \sin \theta}$.
For each particle of mass $m$,the horizontal force is $T \cos \theta = ma_{p}$,where $a_{p}$ is the acceleration of the particle.
Substituting $T$,we get $a_{p} = \frac{T \cos \theta}{m} = \frac{F \cos \theta}{2m \sin \theta} = \frac{F}{2m} \cot \theta$.
Since $\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{x/a}{\sqrt{a^2 - x^2}/a} = \frac{x}{\sqrt{a^2 - x^2}}$,the acceleration is $a_{p} = \frac{F}{2m} \frac{x}{\sqrt{a^2 - x^2}}$.

(B) Given: Mass $m = 4 \text{ kg}$,Spring constant $K = 8 \times 10^3 \text{ Nm}^{-1}$,$g = 10 \text{ ms}^{-2}$.
From the free body diagram of the lower pulley,the tension $T$ in the string supporting the $4 \text{ kg}$ mass is $T = mg = 4 \times 10 = 40 \text{ N}$.
The lower pulley is supported by two segments of the string,each with tension $T$. Thus,the force exerted by the lower pulley on the upper pulley is $2T = 2 \times 40 = 80 \text{ N}$.
The upper pulley is supported by the spring and the string fixed to the ground. The total downward force on the upper pulley is the sum of the tension from the string fixed to the ground $(2T)$ and the force from the lower pulley $(2T)$.
Therefore,the total force $F$ in the spring is $F = 2T + 2T = 4T = 4 \times 40 = 160 \text{ N}$.
Using Hooke's Law,the extension $x$ in the spring is $x = \frac{F}{K} = \frac{160}{8 \times 10^3} = 20 \times 10^{-3} \text{ m} = 2 \times 10^{-2} \text{ m} = 2 \text{ cm}$.

(A) Given:
Mass of block $1$, $m_1 = 0.36 \,kg$
Mass of block $2$, $m_2 = 0.72 \,kg$
Acceleration due to gravity, $g = 10 \,m/s^2$
Time, $t = 1 \,s$
The acceleration $a$ of the system is given by:
$a = \frac{(m_2 - m_1)g}{m_1 + m_2} = \frac{(0.72 - 0.36) \times 10}{0.72 + 0.36} = \frac{0.36 \times 10}{1.08} = \frac{3.6}{1.08} = \frac{10}{3} \,m/s^2$
The tension $T$ in the string is:
$T = m_1(g + a) = 0.36 \times (10 + \frac{10}{3}) = 0.36 \times \frac{40}{3} = 0.12 \times 40 = 4.8 \,N$
The displacement $s$ of the block $m_1$ in $t = 1 \,s$ is:
$s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2} \times \frac{10}{3} \times (1)^2 = \frac{5}{3} \,m$
The work done by the string on block $m_1$ is:
$W = T \times s \times \cos(0^\circ) = 4.8 \times \frac{5}{3} = 1.6 \times 5 = 8 \,J$