JEE Physics Laws Of Motion MCQs Set 06

Answer: (c) 19 N

Explanation:
1. Limiting friction \(f_{max}=\mu mg=0.8\times4\times10=32\) N.
2. The applied force 19 N is less than 32 N, so the block does not move.
3. For a body at rest the static friction equals the applied force, so \(f=19\) N.

Teacher's Note:
1. Compare the applied force with the limiting friction first.
2. Static friction adjusts itself up to its maximum value; it is not always equal to \(\mu N\).

Answer: (c) 5 mg

Explanation:
1. The floor of the rocket pushes the man up with a normal reaction \(R\).
2. \(R-mg=m(4g)\), so \(R=5mg\).
3. The apparent weight is the normal reaction, which is 5 mg.

Teacher's Note:
1. Apparent weight equals the normal reaction on the body.
2. Upward acceleration increases it to \(m(g+a)\).

Answer: (a) mg

Explanation:
1. Each mass hangs at rest, so the tension in the string is \(T=mg\).
2. The two tensions pulling on P are equal and the angle between them is \(60^\circ+60^\circ=120^\circ\).
3. Resultant \(=\sqrt{T^2+T^2+2T^2\cos120^\circ}=T=mg\).

Teacher's Note:
1. Add the two equal tensions as vectors, not as numbers.
2. When two equal vectors make an angle of 120°, their resultant has the same magnitude as each of them.

Answer: (c) rolling friction is greater than sliding friction

Explanation:
1. Rougher surfaces have a larger coefficient of friction, so statement (a) is true.
2. Friction always opposes the relative or impending relative motion, so (b) is true.
3. Rolling friction is much smaller than sliding friction, which is why wheels are used, so (c) is false.
4. The coefficient for wood on wood is below 1, so (d) is true.

Teacher's Note:
1. For a "not true" question, test every option and pick the false one.
2. Remember the order: static friction is greater than sliding friction, which is greater than rolling friction.

Answer: (b) 417 N

Explanation:
1. The retardation is \(a=\frac{u^2}{2s}=\frac{(100)^2}{2\times0.06}=\frac{10^4}{0.12}\approx8.33\times10^4\ \text{m/s}^2\).
2. Force \(=ma=5\times10^{-3}\times8.33\times10^4\approx417\) N.

Teacher's Note:
1. Convert grams to kilograms and centimetres to metres before substituting.
2. The force on the block equals the force on the bullet in magnitude by Newton's third law.

Answer: (b) the resultant force on it is zero

Explanation:
1. By Newton's first law, a body keeps its state of rest or of uniform motion in a straight line unless a net external force acts on it.
2. So the straight-line uniform motion lasts only while the resultant force is zero.
3. Hence the correct statement is the one that says the resultant force is zero.

Teacher's Note:
1. Uniform straight-line motion means zero acceleration, so the net force must be zero.
2. A force perpendicular to the motion would change the direction of the velocity.

Answer: (b) \(\left(\frac{5}{2\mu}\right)mg\)

Explanation:
1. The whole system (total mass \(5m\)) accelerates with \(a=\frac{F}{5m}\).
2. The normal force between B and C is the force that gives C its acceleration: \(N=2m\times a=\frac{2F}{5}\).
3. B does not slip when friction balances its weight: \(\mu N=mg\), so \(\mu\frac{2F}{5}=mg\).
4. Hence the minimum force is \(F=\frac{5mg}{2\mu}\).

Teacher's Note:
1. Find the common acceleration first, then the contact force on the block that is pressed against the wall-like block.
2. Friction on B must at least equal its weight mg.

Answer: (b) \(\mu=0\)

Explanation:
1. In this position the diagonal through the lower edge is vertical, so the centre of the cube lies directly above the floor contact.
2. The floor normal force N balances the weight Mg and passes through the same vertical line.
3. No horizontal force is needed, so the wall exerts no force and friction is not required.
4. Hence the minimum coefficient of friction is \(\mu=0\).

Teacher's Note:
1. Draw the free-body diagram and check whether any horizontal force is actually needed.
2. If the weight and the normal reaction are collinear, friction plays no part.

Answer: (d) \(g(\sin\theta-\mu_k\cos\theta)\)

Explanation:
1. Along the incline the driving force is \(mg\sin\theta\) (down the plane).
2. Perpendicular to the plane \(N=mg\cos\theta\), so kinetic friction is \(\mu_kmg\cos\theta\) (up the plane).
3. \(ma=mg\sin\theta-\mu_kmg\cos\theta\), so \(a=g(\sin\theta-\mu_k\cos\theta)\).

Teacher's Note:
1. Gravity's component along the plane is mg sin θ and friction opposes it.
2. If the body is pushed up the plane, the friction term changes its sign.

Answer: (c) 20 m/s²

Explanation:
1. The cart pushes the block with a normal force \(N=ma\), which supplies the block's horizontal acceleration.
2. The block does not fall when friction balances the weight: \(\mu N=mg\), so \(\mu ma=mg\).
3. \(a=\frac{g}{\mu}=\frac{10}{0.5}=20\ \text{m/s}^2\).

Teacher's Note:
1. The normal force on a vertical surface comes from the horizontal acceleration.
2. Friction must balance weight, so a smaller coefficient of friction needs a larger acceleration.

Answer: (b) 4 N

Explanation:
1. The retardation is \(a=\frac{0-4}{2}=-2\ \text{m/s}^2\).
2. The friction force is \(f=ma=2\times2=4\) N, opposing the motion.
3. To move with constant velocity the applied force must balance friction, so \(F=4\) N.

Teacher's Note:
1. First find friction from the stopping motion.
2. Constant velocity needs zero net force, so the push equals the friction.

Answer: (a) mg

Explanation:
1. Wall (2) makes \(74^\circ\) with the floor, so the normal from wall (2) makes \(53^\circ\) with the line of wall (1).
2. Resolve all forces along wall (1), where the normal from wall (1) has no component: \(N_2\cos53^\circ=mg\sin37^\circ\).
3. Since \(\cos53^\circ=\sin37^\circ\), we get \(N_2=mg\).

Teacher's Note:
1. Resolve along the surface of the wall whose normal force you want to eliminate.
2. Smooth walls exert forces only perpendicular to themselves.

Answer: (a) 5 N

Explanation:
1. The maximum friction is \(\mu mg=0.6\times1\times10=6\) N.
2. The friction needed to accelerate the block with the truck is \(ma=1\times5=5\) N, which is less than 6 N.
3. So the block does not slip and the frictional force is 5 N.

Teacher's Note:
1. Compare the required friction with the limiting friction.
2. Do not write \(\mu N\) as the answer unless the block is about to slip.

Answer: (c) \(\frac{3g}{2\sqrt2}\)

Explanation:
1. While moving up, both \(mg\sin\theta\) and friction act down the plane.
2. \(a=g(\sin45^\circ+\mu\cos45^\circ)=g\left(\frac{1}{\sqrt2}+\frac12\times\frac{1}{\sqrt2}\right)\).
3. \(a=\frac{3g}{2\sqrt2}\).

Teacher's Note:
1. On the way up the plane, friction and the weight component point in the same direction.
2. The retardation is the magnitude of this acceleration.

Answer: (b) 1/3 m/s

Explanation:
1. Only the component of the impulse along the string makes the string taut: \(10\cos60^\circ=5\) kg m/s.
2. Along the string the two masses then move together with a common speed v: \(5=(10+5)v\).
3. \(v=\frac{1}{3}\) m/s, which is the speed of the 10 kg mass.

Teacher's Note:
1. The string transmits impulse only along its own length.
2. After the jerk the two masses have the same velocity component along the string.

Answer: (d) The force of interaction between A and B is F/3

Explanation:
1. The acceleration of every block is \(a=\frac{F}{3M}\), so (c) is true.
2. The net force on A is \(Ma=\frac{F}{3}\), so (b) is true, and the vertical forces on A balance, so (a) is true.
3. The force of A on B must accelerate B and C: \(2M\times\frac{F}{3M}=\frac{2F}{3}\), so statement (d) is false.

Teacher's Note:
1. The contact force between two blocks must accelerate everything beyond that contact.
2. Net force and applied force are different quantities.

Answer: (b) \(30\sqrt3,\ 60\)

Explanation:
1. The knot is in equilibrium. Horizontally \(T\sin30^\circ=30\), so \(T=60\) N.
2. Vertically \(T\cos30^\circ=W\), so \(W=60\times\frac{\sqrt3}{2}=30\sqrt3\) N.
3. So W = \(30\sqrt3\) N and T = 60 N.

Teacher's Note:
1. Resolve the tension of the slanted cord into horizontal and vertical parts.
2. Horizontal forces balance and vertical forces balance separately.

Answer: (a) \(\frac{P+Q\sin\theta}{mg+Q\cos\theta}\)

Explanation:
1. Vertical balance: \(N=mg+Q\cos\theta\).
2. Horizontal balance: friction \(=P+Q\sin\theta\).
3. The block stays at rest if \(\mu N\ge P+Q\sin\theta\), so \(\mu\ge\frac{P+Q\sin\theta}{mg+Q\cos\theta}\).

Teacher's Note:
1. When Q makes an angle with the vertical, its vertical part is \(Q\cos\theta\) and its horizontal part is \(Q\sin\theta\).
2. The vertical part increases the normal reaction and hence the limiting friction.

Answer: (b) \(\frac{9g}{11}\)

Explanation:
1. Ascending: \(N=m(g+a)\). Descending: \(N'=m(g-a)\).
2. Given \(N'=\frac{10}{100}N\), so \(g-a=\frac{g+a}{10}\).
3. \(10g-10a=g+a\), so \(9g=11a\) and \(a=\frac{9g}{11}\).

Teacher's Note:
1. The reading of a weighing machine is the normal reaction, not the weight.
2. The reading increases when the lift accelerates upward and decreases when it accelerates downward.

Answer: (a) 200 N

Explanation:
1. The second boy moves with constant velocity, so the rope pulls him up with a force \(f_2=8\times10=80\) N.
2. For the first boy: \(f_1-10g=10\times2\), so \(f_1=120\) N.
3. The massless rope is in equilibrium, so the tension at the support is \(f_1+f_2=200\) N.

Teacher's Note:
1. A boy moving with constant velocity has zero acceleration, so the rope force on him equals his weight.
2. The tension at the support is the sum of the downward pulls of the two boys.

Answer: (a) 4.3 m/s²

Explanation:
1. The string pulling on the 5 kg block stays far below its limiting friction \(0.8\times50=40\) N, so the 5 kg block stays at rest.
2. Because that end of the string is fixed, the 1 kg block moves twice as fast as the 3 kg block: \(a_1=2a\).
3. For the 1 kg block: \(T=1\times2a\). For the 3 kg block: \(30-2T=3a\).
4. Substituting, \(30=7a\), so \(a=\frac{30}{7}\approx4.3\ \text{m/s}^2\).

Teacher's Note:
1. Use the constraint on string length before writing the force equations.
2. Check afterwards that the tension (about 8.6 N) is less than the limiting friction of 40 N on the 5 kg block.

Answer: (c) 1000 m

Explanation:
1. The retardation is \(a=\mu_kg=0.5\times10=5\ \text{m/s}^2\).
2. From \(v^2-u^2=-2as\): \(0-100^2=-2\times5\times s\).
3. \(s=\frac{10000}{10}=1000\) m.

Teacher's Note:
1. The stopping distance is \(\frac{u^2}{2\mu g}\), which depends on the square of the speed.
2. Doubling the speed makes the stopping distance four times larger.

Answer: (c) 3 m/s

Explanation:
1. The force is 5 N up to t = 2 s, then falls linearly, passes through zero at t = 3 s and reaches -5 N at t = 4 s.
2. The impulse from t = 2 s to 4 s is the area under the graph: \(\frac12(1)(5)+\frac12(1)(-5)=0\).
3. The change in momentum is zero, so \(v_4=v_2=-3\) m/s and the speed is 3 m/s.

Teacher's Note:
1. The area under a force-time graph is the impulse, and areas below the axis are negative.
2. Equal positive and negative areas mean no net change in velocity.

Answer: (b) \(\frac{-\alpha v^2}{M}\)

Explanation:
1. The dust is at rest, so the ship must give it speed v. This acts as a retarding thrust \(F=-v\frac{dM}{dt}\).
2. \(Ma=-v(\alpha v)\).
3. \(a=-\frac{\alpha v^2}{M}\).

Teacher's Note:
1. For variable mass with the picked-up matter initially at rest, the thrust equals the ship's speed multiplied by the rate of mass gain.
2. The negative sign shows that the ship slows down.

Answer: (d) 0

Explanation:
1. For a lift moving down with acceleration a, the reading is \(\frac{m(g-a)}{g}\) in kilogram-weight.
2. In free fall \(a=g\).
3. The reading becomes zero, because the body is weightless.

Teacher's Note:
1. A spring balance shows the tension in the spring, which is m(g - a) for a lift accelerating down.
2. Free fall means apparent weightlessness.