Apparent weight and Pseudo Force Questions in English

(C) The elevator is moving upwards with an acceleration $a = 1.2 \, m/s^2$. When the bolt detaches from the ceiling,it is in a frame of reference accelerating upwards.
The effective acceleration of the bolt relative to the elevator is $g_{eff} = g + a$,where $g = 9.8 \, m/s^2$ is the acceleration due to gravity.
Given the height $h = 2.7 \, m$,the time $t$ taken for the bolt to reach the floor is given by the kinematic equation $h = \frac{1}{2} g_{eff} t^2$.
Substituting the values:
$2.7 = \frac{1}{2} (9.8 + 1.2) t^2$
$2.7 = \frac{1}{2} (11) t^2$
$2.7 = 5.5 t^2$
$t^2 = \frac{2.7}{5.5} \approx 0.49$
$t = \sqrt{0.49} = 0.7 \, s$.

(C) When milk is churned,it undergoes circular motion.
Due to the rotation,a pseudo force known as centrifugal force acts on the particles in the rotating frame of reference.
This force acts radially outward from the center of rotation.
Since cream particles are lighter than the other components of milk,they are pushed towards the center of the vessel during the churning process,allowing them to be separated.
Therefore,the separation of cream from milk is due to centrifugal force.

(A) When observing the system from a frame of reference that is stationary with respect to the mass $m$,we are in a non-inertial frame.
In this frame,the mass appears to be at rest,so the net force acting on it must be zero.
The forces acting on the mass are:
$1$. The tension $T$ in the thread,acting along the thread towards the pivot $O$.
$2$. The weight $W = mg$ acting vertically downwards.
$3$. The centrifugal force $F = m\omega^2r$ acting radially outwards from the center of the circular path.
Since the mass is in equilibrium in this rotating frame,these three forces must balance each other.
Looking at the provided options,the correct representation of these three forces is shown in image $18-b87$ (Option $A$). Therefore,the correct option is $A$.

(C) When the car moves in a circular path,the plumb bob experiences a pseudo-force (centrifugal force) in the frame of the car,directed radially outward,given by $F_c = \frac{mv^2}{r}$.
The gravitational force acting on the bob is $mg$,directed vertically downward.
Let $\theta$ be the angle the rod makes with the vertical. In the equilibrium position,the net force on the bob in the car's frame is zero.
Taking the ratio of the horizontal force to the vertical force,we get:
$\tan \theta = \frac{F_c}{mg} = \frac{mv^2/r}{mg} = \frac{v^2}{rg}$.
Given $v = 10 \, m/s$,$r = 10 \, m$,and taking $g = 10 \, m/s^2$:
$\tan \theta = \frac{10^2}{10 \times 10} = \frac{100}{100} = 1$.
Therefore,$\theta = \tan^{-1}(1) = 45^\circ$.

(D) The reading on a spring balance represents the normal force or the apparent weight of the body.
When a lift descends with an acceleration $a$,the apparent weight $W'$ is given by the formula $W' = m(g - a)$.
In this problem,the mass $m = 2 \, kg$ and the acceleration of the lift $a = g$.
Substituting these values into the formula:
$W' = m(g - g) = m(0) = 0$.
Therefore,the reading on the spring balance will be $0 \, kg$.

(A) The problem asks for the reading on the spring balance when the lift moves up with a constant velocity of $2 \, m/s$.
When the lift moves with a constant velocity,its acceleration $a = 0$.
The apparent weight $R$ is given by the formula $R = m(g + a)$.
Substituting $a = 0$ and $m = 2 \, kg$,we get $R = 2(g + 0) = 2g \, N$.
Since the spring balance is calibrated to read mass in $kg$,the reading is $R/g = 2 \, kg$.
Note: The condition regarding the lift descending with acceleration $g$ is a separate scenario where the reading would be $m(g - g) = 0 \, kg$,but the question specifically asks for the reading when moving up with constant velocity.

(D) The reading of a spring balance in a lift is given by the apparent weight $R = m(g + a)$,where $m$ is the mass,$g$ is the acceleration due to gravity,and $a$ is the acceleration of the lift.
Given: $m = 2\, kg$ and $a = g$ (upward acceleration).
Substituting these values into the formula:
$R = 2(g + g) = 2(2g) = 4g\, N$.
Since the spring balance is calibrated to read mass in $kg$,the reading corresponds to $4\, kg$.

(A) When the lift is stationary,the acceleration of the coin relative to the floor is $g$. The time taken to cover distance $h$ is $t_1 = \sqrt{\frac{2h}{g}}$.
When the lift moves upward with constant acceleration $a$,the effective acceleration of the coin relative to the lift floor is $g_{eff} = g + a$.
The time taken to cover the same distance $h$ is $t_2 = \sqrt{\frac{2h}{g + a}}$.
Since $g + a > g$,it follows that $\frac{2h}{g + a} < \frac{2h}{g}$,which implies $t_2 < t_1$ or $t_1 > t_2$.

(B) The apparent weight $W'$ of a person in an elevator is given by $W' = m(g \pm a)$,where $m$ is the mass of the person,$g$ is the acceleration due to gravity,and $a$ is the acceleration of the elevator.
When the elevator moves downward with an acceleration $a$,the apparent weight is $W' = m(g - a)$.
Since $(g - a) < g$,the apparent weight $W'$ is less than the actual weight $W = mg$.
Therefore,the person feels lighter when the elevator moves downward with constant acceleration.

(D) When the lift is stationary,the normal force $N$ exerted by the weighing machine is equal to the actual weight $W = mg = 40 \times 10 = 400 \, N$.
When the lift accelerates upwards with an acceleration $a = 2 \, m/s^2$,the apparent weight $W'$ is given by the equation $W' = m(g + a)$.
Substituting the values: $W' = 40 \times (10 + 2) = 40 \times 12 = 480 \, N$.
Since the weighing machine is calibrated to show weight in $kg$ (where $1 \, kg$ corresponds to $10 \, N$ in this problem),the reading will be $480 / 10 = 48 \, kg$.

(D) The apparent weight of the body is $W' = 4.8 \, kgf = 4.8 \, g \, N$.
The actual mass of the body is $m = 4 \, kg$,so its actual weight is $W = 4 \, g \, N$.
Since the apparent weight is greater than the actual weight $(4.8 \, g > 4 \, g)$,the lift must be accelerating upwards.
The formula for apparent weight in an accelerating lift is $R = m(g + a)$.
Substituting the values: $4.8 \, g = 4(g + a)$.
Dividing both sides by $4$: $1.2 \, g = g + a$.
Therefore,$a = 1.2 \, g - g = 0.2 \, g$.
Using $g = 9.8 \, m/s^2$,we get $a = 0.2 \times 9.8 = 1.96 \, m/s^2$ in the upward direction.

(D) The apparent weight $N$ of a person in an elevator is given by $N = m(g + a)$,where $a$ is the upward acceleration.
For case $A$ (Stands still): $a = 0$,so $N = mg = 40 \times 9.8 = 392 \, N$.
For case $B$ (Constant velocity): $a = 0$,so $N = mg = 392 \, N$.
For case $C$ (Accelerates downward): $a = -4 \, m/s^2$,so $N = m(g - 4) = 40(9.8 - 4) = 40(5.8) = 232 \, N$.
For case $D$ (Accelerates upward): $a = 4 \, m/s^2$,so $N = m(g + 4) = 40(9.8 + 4) = 40(13.8) = 552 \, N$.
Comparing all cases,the force is greatest when the elevator accelerates upward.

(B) Let the mass of the man be $m$.
When the lift is stationary,the apparent weight is $W_1 = mg$.
When the lift is moving downward with uniform acceleration $a$,the apparent weight is $W_2 = m(g - a)$.
According to the problem,the ratio of the weights is $W_1 : W_2 = 3 : 2$.
Therefore,$\frac{mg}{m(g - a)} = \frac{3}{2}$.
Canceling $m$ from the numerator and denominator,we get $\frac{g}{g - a} = \frac{3}{2}$.
Cross-multiplying gives $2g = 3(g - a)$.
$2g = 3g - 3a$.
$3a = 3g - 2g$.
$3a = g$.
Thus,$a = g/3$.

(D) The apparent weight $R$ of a person in an elevator moving with an upward acceleration $a$ is given by the formula $R = M(g + a)$.
Given that the elevator is moving vertically up with an acceleration $a = g$.
Substituting the value of $a$ in the formula,we get $R = M(g + g)$.
Therefore,$R = M(2g) = 2Mg$.
Thus,the force exerted on the floor by the passenger is $2Mg$.

(B) The apparent weight $W'$ of a body in a lift moving downwards with an acceleration $a$ is given by the formula: $W' = m(g - a)$.
Given:
Mass $m = 50 \, kg$
Acceleration $a = 9.8 \, m/s^2$
Acceleration due to gravity $g = 9.8 \, m/s^2$
Substituting the values:
$W' = 50 \times (9.8 - 9.8)$
$W' = 50 \times 0 = 0 \, N$.
Therefore,the apparent weight of the boy is $0 \, N$.

(D) The apparent weight $W'$ of a body of mass $m$ in a lift moving with an acceleration $a$ is given by $W' = m(g + a)$ if the lift is accelerating upwards and $W' = m(g - a)$ if it is accelerating downwards.
In this question,it is stated that the lift ascends,but it is not specified whether the lift is moving with constant velocity or with acceleration.
If the lift moves with a constant velocity,the acceleration $a = 0$,so the reading remains $mg$ (no change).
If the lift moves with an upward acceleration,the reading increases.
If the lift moves with a downward acceleration (deceleration while ascending),the reading decreases.
Therefore,the reading depends on the state of motion (acceleration) of the lift,which is related to its velocity change over time.

(D) When a body of mass $m$ moves upwards with an acceleration $a$,the apparent weight $R$ is given by the formula: $R = m(g + a)$.
Given: Mass $m = 10 \ kg$,acceleration $a = 2 \ m/s^2$,and acceleration due to gravity $g = 9.8 \ m/s^2$.
Substituting the values into the formula:
$R = 10 \times (9.8 + 2)$
$R = 10 \times 11.8$
$R = 118 \ N$.
Therefore,the apparent weight is $118 \ N$.

(D) The apparent weight $R$ of a body of mass $m$ moving with an acceleration $a$ in a downward direction is given by the formula: $R = m(g - a)$.
Given:
Mass $m = 1.0 \, kg$
Acceleration $a = 10 \, m/s^2$
Acceleration due to gravity $g = 10 \, m/s^2$
Substituting the values into the formula:
$R = 1.0 \times (10 - 10)$
$R = 1.0 \times 0$
$R = 0 \, kg \, wt$.
Therefore,the apparent weight of the body is $0$.

(D) When the rope of a lift breaks,the lift falls freely under gravity.
Therefore,the acceleration of the lift becomes equal to the acceleration due to gravity,i.e.,$a = g$.
The apparent weight (normal force or tension exerted by the floor) $N$ on a person of mass $m$ inside the lift is given by $N = m(g - a)$.
Substituting $a = g$ into the equation,we get $N = m(g - g) = 0$.
Thus,the person experiences weightlessness,and the tension exerted by the surface is $0$.

(D) The apparent weight $R$ of a person in an accelerating lift is given by the formula $R = m(g + a)$,where $m$ is the mass,$g$ is the acceleration due to gravity,and $a$ is the acceleration of the lift.
Given: $m = 50\,kg$,$g = 10\,m/s^2$,and $a = 2\,m/s^2$.
Substituting the values: $R = 50 \times (10 + 2) = 50 \times 12 = 600\,N$.
The reading of the balance is in $kg$ weight,so $R = 600/10 = 60\,kg$.

(D) Let the downward direction be positive.
The acceleration of the lift relative to the Earth is $a_{lift} = a$ (downward).
The acceleration of the ball relative to the lift is $a_{ball/lift} = -a_0$ (upward,so negative).
Using the relative acceleration formula: $a_{ball/earth} = a_{ball/lift} + a_{lift/earth}$.
Substituting the values: $a_{ball/earth} = -a_0 + a = (a - a_0)$.
Since the result is positive,the acceleration of the ball as observed by an observer on the Earth is $(a - a_0)$ in the downward direction.

(C) $1$. For the man standing stationary on the ground (inertial frame),the ball is under the influence of gravity alone. Therefore,its acceleration is $g$ downwards.
$2$. For the man inside the lift (non-inertial frame),we must apply a pseudo force. The lift is accelerating downwards with $a$,so a pseudo force $ma$ acts upwards on the ball.
$3$. The net force on the ball in the lift frame is $F_{net} = mg - ma$ (downwards).
$4$. The acceleration observed by the man in the lift is $a_{lift} = F_{net} / m = (mg - ma) / m = g - a$ downwards.
$5$. Thus,the accelerations are $g - a$ and $g$ respectively.

(C) When a lift moves upwards with a uniform acceleration $a$,the apparent weight $R$ (reading on the weighing scale) is given by the formula:
$R = m(g + a)$
Given:
Mass of the man,$m = 80\,kg$
Acceleration of the lift,$a = 5\,m/s^2$
Acceleration due to gravity,$g = 10\,m/s^2$
Substituting the values into the formula:
$R = 80 \times (10 + 5)$
$R = 80 \times 15$
$R = 1200\,N$
Therefore,the reading on the scale is $1200\,N$.

(B) When the lift is stationary,the reading of the spring balance is equal to the actual weight of the bag: $W = mg = 49 \, N$.
Taking $g = 9.8 \, m/s^2$,we find the mass of the bag: $m = \frac{49}{9.8} = 5 \, kg$.
When the lift moves downward with an acceleration $a = 5 \, m/s^2$,the apparent weight $R$ is given by the formula: $R = m(g - a)$.
Substituting the values: $R = 5 \times (9.8 - 5) = 5 \times 4.8 = 24 \, N$.
Therefore,the reading of the spring balance will be $24 \, N$.

(A) The apparent weight $R$ of a person in a lift moving downwards with an acceleration $a$ is given by the formula $R = m(g - a)$.
Given:
Mass $m = 50 \, kg$
Acceleration of lift $a = 10 \, m/s^2$
Acceleration due to gravity $g = 10 \, m/s^2$
Substituting the values into the formula:
$R = 50 \times (10 - 10)$
$R = 50 \times 0$
$R = 0 \, N$.
Therefore,the reading of the spring balance will be $0 \, N$.

(D) To keep the block stationary relative to the wedge,the net force acting on the block along the inclined plane must be zero.
$1$. Resolve the forces acting on the block in the frame of the wedge:
- Gravitational force $mg$ acting downwards.
- Pseudo force $ma$ acting horizontally in the direction opposite to the acceleration of the wedge.
- Normal reaction $R$ exerted by the wedge perpendicular to the inclined surface.
$2$. For the block not to slip,the component of $mg$ down the plane must be balanced by the component of the pseudo force $ma$ up the plane:
$mg \sin \theta = ma \cos \theta$
$a = g \tan \theta$
$3$. The normal force $R$ balances the components of $mg$ and $ma$ perpendicular to the inclined plane:
$R = mg \cos \theta + ma \sin \theta$
Substitute $a = g \tan \theta = g \frac{\sin \theta}{\cos \theta}$:
$R = mg \cos \theta + m(g \frac{\sin \theta}{\cos \theta}) \sin \theta$
$R = mg \frac{\cos^2 \theta + \sin^2 \theta}{\cos \theta}$
$R = \frac{mg}{\cos \theta}$

(D) The apparent weight $R$ of a person in an elevator moving upwards with acceleration $a$ is given by the formula:
$R = m(g + a)$
Given:
Mass $m = 75\, kg$
Acceleration $a = 5\, m/s^2$
Acceleration due to gravity $g = 10\, m/s^2$
Substituting the values into the formula:
$R = 75(10 + 5)$
$R = 75 \times 15$
$R = 1125\, N$
Therefore,the apparent weight of the man is $1125\, N$.

(A) When a lift accelerates upwards with an acceleration $a$,the normal force $N$ (or tension $T$ in the spring balance) acting on the object is given by Newton's second law: $N - mg = ma$.
Therefore,the apparent weight $N = m(g + a)$.
Since the acceleration $a > 0$ during the start of the ascent,the reading $N$ becomes greater than the actual weight $mg$.
Thus,the reading of the spring balance increases.

(B) The apparent weight $R$ of a person in an elevator moving upwards with acceleration $a$ is given by the formula:
$R = m(g + a)$
Given:
Mass of the man,$m = 80\, kg$
Acceleration of the elevator,$a = 6\, m/s^2$
Acceleration due to gravity,$g = 10\, m/s^2$
Substituting the values into the formula:
$R = 80(10 + 6)$
$R = 80 \times 16$
$R = 1280\, N$
Therefore,the apparent weight of the man is $1280\, N$.

(D) When the thief jumps from the wall,both the thief and the box are in a state of free fall under the influence of gravity alone.
In a state of free fall,the acceleration of the system is equal to the acceleration due to gravity $(a = g)$.
The apparent weight $W'$ of an object is given by the formula $W' = m(g - a)$.
Substituting $a = g$ into the equation,we get $W' = m(g - g) = 0$.
Therefore,the thief experiences a load of zero while in the air.

(B) When a body of mass $m$ is in a lift moving upwards with an acceleration $a$,the forces acting on the body are the normal reaction $R$ (upwards) and the weight $mg$ (downwards).
According to Newton's second law of motion,the net force is equal to the product of mass and acceleration:
$F_{net} = ma$
Here,the net force is $R - mg$ because the lift is accelerating upwards.
Therefore,the equation of motion is:
$R - mg = ma$
Rearranging the terms to solve for $R$:
$R = mg + ma$

(B) Let the mass of the block be $m$. To keep the block stationary relative to the inclined surface,we analyze the forces in the frame of the incline.
$1$. The forces acting on the block are its weight $(mg)$ acting downwards,the normal force $(N)$ perpendicular to the surface,and the pseudo force $(ma)$ acting horizontally in the direction opposite to the acceleration of the incline.
$2$. Resolving the forces along the inclined surface:
- The component of the gravitational force down the incline is $mg \sin \alpha$.
- The component of the pseudo force up the incline is $ma \cos \alpha$.
$3$. For the block to remain stationary relative to the incline,these two forces must balance each other:
$ma \cos \alpha = mg \sin \alpha$
$4$. Dividing both sides by $m \cos \alpha$:
$a = g \frac{\sin \alpha}{\cos \alpha}$
$a = g \tan \alpha$

(C) When a vessel containing liquid is accelerated horizontally with an acceleration $a$,the effective acceleration experienced by the liquid in the frame of the vessel is the vector sum of the gravitational acceleration $g$ (downwards) and the pseudo-acceleration $-a$ (backwards).
The free surface of the liquid must be perpendicular to the net effective acceleration vector. The angle $\theta$ that the surface makes with the horizontal is given by $\tan \theta = \frac{a}{g}$.
Since the acceleration $a$ is towards the right,the pseudo-force acts towards the left. Consequently,the liquid level rises on the left side and falls on the right side of the vessel. This corresponds to the surface profile shown in diagram $C$.

(C) physical balance works on the principle of comparing the gravitational force (weight) acting on the body with the gravitational force acting on the standard weights placed on the other pan.
Since the weight of the body $W = mg$ and the weight of the standard masses $W' = m'g$ are both affected by the same acceleration $a$ of the lift,the effective weight becomes $W_{eff} = m(g + a)$ and $W'_{eff} = m'(g + a)$.
When the balance is in equilibrium,$W_{eff} = W'_{eff}$,which implies $m(g + a) = m'(g + a)$.
Therefore,$m = m'$.
Thus,the mass measured by a physical balance remains $m$ and is independent of the acceleration of the lift.

(C) When a system falls freely under gravity,it experiences an acceleration equal to the acceleration due to gravity,$g$,directed downwards.
In the frame of reference of the falling rod,the effective acceleration of the weights is zero because the pseudo-force acting on each mass $m$ is $F_p = -mg$,which exactly cancels the gravitational force $F_g = mg$.
Since the net force on each spring is zero,there is no extension in any of the springs.
Therefore,all weights will remain at their natural (unstretched) length from the rod,meaning they will all be at the same distance.

(C) Given velocity of the cube: $\vec v = 5t\,\hat i + 2t\,\hat j$.
Acceleration of the cube: $\vec a = \frac{d\vec v}{dt} = 5\,\hat i + 2\,\hat j\,m/s^2$.
Here,$a_x = 5\,m/s^2$ and $a_y = 2\,m/s^2$.
In the frame of the cube,the sphere experiences a pseudo force $\vec F_p = -m\vec a = -m(5\,\hat i + 2\,\hat j)$.
The forces acting on the sphere in the cube's frame are:
$1$. Weight: $\vec W = -mg\,\hat j = -2(10)\,\hat j = -20\,\hat j\,N$.
$2$. Pseudo force: $\vec F_p = -2(5\,\hat i + 2\,\hat j) = -10\,\hat i - 4\,\hat j\,N$.
$3$. Normal forces from the walls of the cube: $\vec N_x$ (from wall $AD$) and $\vec N_y$ (from wall $DC$).
Since the sphere is at rest relative to the cube,the net force is zero:
$\vec N_x + \vec N_y + \vec W + \vec F_p = 0$.
$\vec N_x = -(-10\,\hat i) = 10\,\hat i\,N$.
$\vec N_y = -(-20\,\hat j - 4\,\hat j) = 24\,\hat j\,N$.
The total force exerted by the sphere on the cube is the vector sum of the normal forces: $\vec N = \vec N_x + \vec N_y = 10\,\hat i + 24\,\hat j\,N$.
The magnitude is $N = \sqrt{10^2 + 24^2} = \sqrt{100 + 576} = \sqrt{676} = 26\,N$.

(D) The normal force $N$ acting on the body of mass $m$ inside the lift can be calculated using Newton's second law.
For a lift moving downwards with acceleration $a$,the equation of motion is $mg - N = ma$.
Given that the lift is moving downwards with an acceleration equal to the acceleration due to gravity,we have $a = g$.
Substituting $a = g$ into the equation: $mg - N = mg$.
This simplifies to $N = mg - mg = 0$.
The frictional force $f$ is given by $f = \mu N$.
Since the normal force $N = 0$,the frictional resistance $f = \mu \times 0 = 0$.
Therefore,the correct option is $D$.

(B) The cork is held submerged by the spring force $F_s$ and the buoyant force $F_B$, balancing its weight $W = mg$.
$F_s + W = F_B$ or $F_s = F_B - W = V \rho_w g - V \rho_c g = V(\rho_w - \rho_c)g$.
When the elevator accelerates downwards with acceleration $a$, the effective gravity becomes $g' = (g - a)$.
The new spring force is $F_s' = V(\rho_w - \rho_c)(g - a)$.
Since $(g - a) < g$, the spring force $F_s'$ decreases.
As $F_s = kx$, where $k$ is the spring constant and $x$ is the extension, a decrease in $F_s$ leads to a decrease in the extension $x$.
Therefore, the length of the spring decreases.

(B) In the accelerated frame of reference (the trolley),a fictitious force (pseudo force) $F_p = ma$ acts on the bob of the pendulum in the direction opposite to the acceleration of the trolley.
The forces acting on the bob in the frame of the trolley are:
$1$. Gravitational force $mg$ acting vertically downwards.
$2$. Pseudo force $ma$ acting horizontally backwards.
$3$. Tension $T$ in the string.
In the equilibrium position (mean position),the net force on the bob is zero. Resolving the forces,we have:
$\tan \theta = \frac{\text{Pseudo force}}{\text{Gravitational force}} = \frac{ma}{mg} = \frac{a}{g}$
Therefore,$\theta = \tan^{-1}(\frac{a}{g})$ in the backward direction.

(D) The sphere moves with acceleration $a = g$ towards the left,so the particle experiences a pseudo force $F_p = ma = mg$ towards the right.
Let the particle be at the top of the sphere at $t = 0$. As it slides to an angle $\theta$ from the vertical,the vertical displacement is $h = R(1 - \cos \theta)$ and the horizontal displacement is $d = R \sin \theta$.
Applying the work-energy theorem relative to the sphere:
$W_g + W_{pseudo} = \Delta K.E.$
Work done by gravity: $W_g = mg \cdot h = mgR(1 - \cos \theta)$.
Work done by pseudo force: $W_{pseudo} = F_p \cdot d = (mg)(R \sin \theta)$.
Since the initial kinetic energy is zero,the final kinetic energy is $\frac{1}{2}mv^2$:
$mgR(1 - \cos \theta) + mgR \sin \theta = \frac{1}{2}mv^2$.
Dividing by $m$ and multiplying by $2$:
$2gR(1 - \cos \theta + \sin \theta) = v^2$.
Therefore,$v = \sqrt{2gR(1 + \sin \theta - \cos \theta)}$.

(B) When the bead is in equilibrium relative to the wire,the forces acting on it in the frame of the wire are gravity ($mg$ downwards),pseudo force ($ma$ in the negative $X$-direction),and the normal force $(N)$.
At equilibrium,the net force along the wire is zero.
Resolving forces: $N \cos \theta = mg$ and $N \sin \theta = ma$,where $\theta$ is the angle the tangent makes with the horizontal.
Dividing the two equations,we get $\tan \theta = \frac{a}{g}$.
The slope of the parabola $Y = Kx^2$ is given by $\frac{dy}{dx} = 2Kx$.
Since the slope is equal to $\tan \theta$,we have $2Kx = \frac{a}{g}$.
Solving for $x$,we get $x = \frac{a}{2Kg}$.

(B) The weight of a man in a stationary lift is $W_1 = mg$.
When the lift moves downwards with an acceleration '$a$',the apparent weight of the man is $W_2 = m(g - a)$.
Given the ratio $\frac{W_1}{W_2} = \frac{3}{2}$.
Substituting the expressions: $\frac{mg}{m(g - a)} = \frac{3}{2}$.
$\frac{g}{g - a} = \frac{3}{2}$.
Cross-multiplying gives: $2g = 3(g - a)$.
$2g = 3g - 3a$.
$3a = 3g - 2g$.
$3a = g$.
Therefore,$a = \frac{g}{3}$.

(C) To prevent the block from falling,the upward frictional force $f$ must balance the downward gravitational force $mg$.
$1$. The cart accelerates with $\alpha$. In the frame of the cart,a pseudo force $F_{fic} = m\alpha$ acts on the block,pushing it against the cart. This force provides the normal reaction $N = m\alpha$.
$2$. The maximum static frictional force available is $f_{max} = \mu N = \mu m\alpha$.
$3$. For the block not to fall,the frictional force $f$ must be at least equal to the weight of the block: $f \ge mg$.
$4$. Substituting $f = \mu m\alpha$,we get $\mu m\alpha \ge mg$.
$5$. Solving for $\alpha$,we obtain $\alpha \ge \frac{g}{\mu}$.

(C) To keep the block stationary on the wedge,we analyze the forces in the non-inertial frame of the wedge.
$1$. The forces acting on the block are: gravitational force $mg$ (downwards),normal reaction $N$ (perpendicular to the surface),and pseudo force $ma$ (towards the left).
$2$. Resolving the forces into components parallel and perpendicular to the inclined surface:
- Perpendicular to the surface: $N = mg \cos \theta + ma \sin \theta$. However,for the block to be stationary relative to the wedge,the net force along the incline must be zero.
- Parallel to the incline: $ma \cos \theta = mg \sin \theta$.
$3$. From the parallel component equation:
$ma \cos \theta = mg \sin \theta$
$a \cos \theta = g \sin \theta$
$a = g \frac{\sin \theta}{\cos \theta}$
$a = g \tan \theta$

(D) When the car takes a left turn,it undergoes centripetal acceleration directed towards the center of the turn (left).
Due to inertia,the air inside the car tends to move in a straight line,but because the car is enclosed,the air experiences a pseudo-force directed towards the right.
Since the air is denser than the helium-filled balloon,the air is pushed towards the right side of the car due to this pseudo-force.
This displacement of the denser air towards the right creates a pressure gradient that pushes the lighter helium balloon towards the left.
Therefore,the balloon will be thrown to the left side.

(A) In a stationary lift,the acceleration of a block sliding down a smooth inclined plane is $a_{rel} = g \sin \theta$.
When the lift is moving with an acceleration $A$ (upward being positive),the effective acceleration due to gravity becomes $g_{eff} = g + A$.
In this problem,the lift is descending with a retardation $a$. Retardation while descending means the acceleration vector is directed upwards. Thus,$A = +a$.
The effective acceleration becomes $g_{eff} = g + a$.
Therefore,the acceleration of the block relative to the incline is $a_{rel} = g_{eff} \sin \theta = (g + a) \sin \theta$.

(C) The apparent weight $W'$ of a person in a lift is given by $W' = m(g + a)$,where $a$ is the acceleration of the lift.
If the lift is moving upwards with a constant velocity,the acceleration $a = 0$,so the reading is $m \times g = 60\, kg$.
When the lift suddenly stops while moving upwards,it experiences a downward acceleration (retardation),so $a$ becomes negative. The apparent weight becomes $W' = m(g - a)$.
Since the reading decreases from $60\, kg$ to $50\, kg$,it indicates that the lift experienced a downward acceleration (deceleration) while moving upwards.
Therefore,we conclude that the lift was moving upwards with constant speed and was suddenly stopped.

(D) When the lift accelerates upwards with acceleration $a$,a pseudo force $F_p = ma$ acts downwards on any mass $m$ inside the lift.
For a spring balance,the reading is the normal force or tension,which becomes $T = m(g + a)$. Since $g + a > g$,the reading of the spring balance increases.
For a physical balance,the pseudo force $F_p = ma$ acts on both pans equally. Since the balance compares the weights on both sides,the effect of the pseudo force cancels out,and the equilibrium position remains undisturbed.

(A) For the block not to fall,the upward frictional force $f$ must balance the downward gravitational force $mg$.
$f = mg$
The normal force $R$ exerted by the board on the block is provided by the pseudo-force acting on the block due to the acceleration $a$ of the board:
$R = ma$
The frictional force is given by $f \leq \mu R$. To prevent falling,we need the maximum possible friction to be at least equal to the weight:
$f_{max} = \mu R = \mu (ma)$
Setting $f_{max} \geq mg$:
$\mu (ma) \geq mg$
Dividing both sides by $ma$:
$\mu \geq g/a$