1. (Physics) Which statement best defines force?
ⓐ. Force is the rate at which energy is transferred through a surface.
ⓑ. Force is the amount of matter contained in a moving object at a given instant.
ⓒ. Force is the property by which a body keeps moving without any interaction.
ⓓ. Force is an interaction that can change the state of motion of a body.
Correct Answer: Force is an interaction that can change the state of motion of a body.
Explanation: Force is an external interaction that can change the speed, direction, or state of rest of an object. It may also change the shape of a body when the interaction is strong enough. This is why force is not treated as the same thing as energy, mass, or motion itself. In mechanics, force is understood from the change it produces in motion or deformation.
2. A body starts with an initial velocity of \(5\,\text{m s}^{-1}\) and moves with uniform acceleration \(2\,\text{m s}^{-2}\) for \(3\,\text{s}\). What is its final velocity?
ⓐ. \(9\,\text{m s}^{-1}\)
ⓑ. \(11\,\text{m s}^{-1}\)
ⓒ. \(13\,\text{m s}^{-1}\)
ⓓ. \(15\,\text{m s}^{-1}\)
Correct Answer: \(11\,\text{m s}^{-1}\)
Explanation: \(\textbf{Given:}\)
Initial velocity, \(u = 5\,\text{m s}^{-1}\)
Acceleration, \(a = 2\,\text{m s}^{-2}\)
Time, \(t = 3\,\text{s}\)
\(\textbf{Required:}\)
Final velocity, \(v\)
\(\textbf{Relevant Formula:}\)
\[
v = u + at
\]
\(\textbf{Why this formula applies:}\)
When motion has uniform acceleration and we know initial velocity and time, this equation directly gives the final velocity.
\(\textbf{Identify known values:}\)
\[
u = 5,\quad a = 2,\quad t = 3
\]
\(\textbf{Substitution:}\)
\[
v = 5 + (2)(3)
\]
\(\textbf{Intermediate Simplification:}\)
\[
v = 5 + 6
\]
\(\textbf{Final Simplification:}\)
\[
v = 11\,\text{m s}^{-1}
\]
\(\textbf{Unit Check:}\)
Velocity is measured in \(\text{m s}^{-1}\), so the final unit is correct.
\(\textbf{Final Answer:}\)
\[11\,\text{m s}^{-1}\]
3. A particle moves along a straight line. Its velocity changes uniformly from \(2\,\text{m s}^{-1}\) to \(8\,\text{m s}^{-1}\) in \(3\,\text{s}\). What is its acceleration?
ⓐ. \(4\,\text{m s}^{-2}\)
ⓑ. \(3\,\text{m s}^{-2}\)
ⓒ. \(2\,\text{m s}^{-2}\)
ⓓ. \(1\,\text{m s}^{-2}\)
Correct Answer: \(2\,\text{m s}^{-2}\)
Explanation: \(\textbf{Given:}\)
Initial velocity, \(u = 2\,\text{m s}^{-1}\)
Final velocity, \(v = 8\,\text{m s}^{-1}\)
Time, \(t = 3\,\text{s}\)
\(\textbf{Required:}\)
Acceleration, \(a\)
\(\textbf{Relevant Formula:}\)
\[
a = \frac{v-u}{t}
\]
\(\textbf{Why this formula applies:}\)
Acceleration is defined as rate of change of velocity with time.
\(\textbf{Substitution:}\)
\[
a = \frac{8-2}{3}
\]
\(\textbf{Intermediate Simplification:}\)
\[
a = \frac{6}{3}
\]
\(\textbf{Final Simplification:}\)
\[
a = 2\,\text{m s}^{-2}
\]
\(\textbf{Unit Check:}\)
Acceleration is measured in \(\text{m s}^{-2}\).
\(\textbf{Final Answer:}\)
\[2\,\text{m s}^{-2}\]
4. The following observations are made for a body moving in a circle:
Case 1: speed is doubled, radius unchanged
Case 2: radius is doubled, speed unchanged
If the original centripetal acceleration is \(a_c\), which statement is correct?
ⓐ. In Case 1 it becomes \(2a_c\), and in Case 2 it becomes \(2a_c\)
ⓑ. In Case 1 it becomes \(4a_c\), and in Case 2 it becomes \(\frac{a_c}{2}\)
ⓒ. In Case 1 it becomes \(\frac{a_c}{2}\), and in Case 2 it becomes \(4a_c\)
ⓓ. In both cases it remains \(a_c\)
Correct Answer: In Case 1 it becomes \(4a_c\), and in Case 2 it becomes \(\frac{a_c}{2}\)
Explanation: \(\textbf{Given:}\)
Centripetal acceleration depends on speed and radius.
\(\textbf{Required:}\)
Effect of changing speed or radius on \(a_c\)
\(\textbf{Relevant Formula:}\)
\[
a_c = \frac{v^2}{r}
\]
\(\textbf{Step 1 — Case 1: speed doubled}\)
If \(v\) becomes \(2v\), then
\[
a_c' = \frac{(2v)^2}{r} = \frac{4v^2}{r} = 4a_c
\]
\(\textbf{Step 2 — Case 2: radius doubled}\)
If \(r\) becomes \(2r\), then
\[
a_c' = \frac{v^2}{2r} = \frac{1}{2}\cdot\frac{v^2}{r} = \frac{a_c}{2}
\]
\(\textbf{Final Answer:}\)
In Case 1 it becomes \(4a_c\), and in Case 2 it becomes \(\frac{a_c}{2}\).
5. Which statement correctly explains why satellites remain in orbit around Earth?
ⓐ. They remain in orbit because no force acts on them once they enter space.
ⓑ. They remain in orbit because gravity provides the required centripetal force.
ⓒ. They remain in orbit because their mass becomes zero above the atmosphere.
ⓓ. They remain in orbit because air resistance keeps bending their path continuously.
Correct Answer: They remain in orbit because gravity provides the required centripetal force.
Explanation: A satellite has a tangential velocity that tends to carry it forward, but Earth’s gravitational attraction continuously pulls it inward. That inward pull acts as the centripetal force required for circular or nearly circular motion. If gravity were absent, the satellite would move away along a straight line instead of orbiting Earth. Orbital motion is therefore the result of the balance between forward motion and inward gravitational pull.
6. A body of mass \(4\,\text{kg}\) is acted upon by a net force of \(20\,\text{N}\). What acceleration is produced?
ⓐ. \(2\,\text{m s}^{-2}\)
ⓑ. \(3\,\text{m s}^{-2}\)
ⓒ. \(4\,\text{m s}^{-2}\)
ⓓ. \(5\,\text{m s}^{-2}\)
Correct Answer: \(5\,\text{m s}^{-2}\)
Explanation: \(\textbf{Given:}\)
Force, \(F = 20\,\text{N}\)
Mass, \(m = 4\,\text{kg}\)
\(\textbf{Required:}\)
Acceleration, \(a\)
\(\textbf{Relevant Formula:}\)
\[
F = ma
\]
\(\textbf{Why this formula applies:}\)
Newton's second law relates the net force on a body to its mass and acceleration.
\(\textbf{Rearrange:}\)
\[
a = \frac{F}{m}
\]
\(\textbf{Substitution:}\)
\[
a = \frac{20}{4}
\]
\(\textbf{Final Simplification:}\)
\[
a = 5\,\text{m s}^{-2}
\]
\(\textbf{Final Answer:}\)
\[5\,\text{m s}^{-2}\]
7. Two resistors are connected in series:
Resistor 1: \(2\,\Omega\)
Resistor 2: \(4\,\Omega\)
What is the equivalent resistance of the combination?
ⓐ. \(1.33\,\Omega\)
ⓑ. \(2\,\Omega\)
ⓒ. \(6\,\Omega\)
ⓓ. \(8\,\Omega\)
Correct Answer: \(6\,\Omega\)
Explanation: \(\textbf{Given:}\)
\[
R_1 = 2\,\Omega,\quad R_2 = 4\,\Omega
\]
\(\textbf{Required:}\)
Equivalent resistance in series
\(\textbf{Relevant Formula:}\)
\[
R_{\text{eq}} = R_1 + R_2
\]
\(\textbf{Why this formula applies:}\)
In a series combination, the same current flows through each resistor and the resistances add directly.
\(\textbf{Substitution:}\)
\[
R_{\text{eq}} = 2 + 4
\]
\(\textbf{Final Simplification:}\)
\[
R_{\text{eq}} = 6\,\Omega
\]
\(\textbf{Final Answer:}\)
\[6\,\Omega\]
8. Which statement about inertia is correct?
ⓐ. Inertia is the force that keeps every moving body moving forever.
ⓑ. Inertia is the tendency of a body to resist any change in its state of motion.
ⓒ. Inertia is the acceleration produced in a body due to the absence of force.
ⓓ. Inertia is the energy stored in a body because of its motion only.
Correct Answer: Inertia is the tendency of a body to resist any change in its state of motion.
Explanation: Inertia is a property of matter by virtue of which a body resists changes in its state of rest or uniform motion. It is not itself a force, and it does not act separately from the body. Bodies with greater mass usually show greater inertia, which is why heavier objects are harder to start, stop, or turn. This idea forms the physical basis of Newton’s first law of motion.
9. A current of \(2\,\text{A}\) flows through a resistor of \(6\,\Omega\). What is the potential difference across it?
ⓐ. \(3\,\text{V}\)
ⓑ. \(12\,\text{V}\)
ⓒ. \(9\,\text{V}\)
ⓓ. \(18\,\text{V}\)
Correct Answer: \(12\,\text{V}\)
Explanation: \(\textbf{Given:}\)
Current, \(I = 2\,\text{A}\)
Resistance, \(R = 6\,\Omega\)
\(\textbf{Required:}\)
Potential difference, \(V\)
\(\textbf{Relevant Formula:}\)
\[
V = IR
\]
\(\textbf{Why this formula applies:}\)
Ohm's law directly relates potential difference, current, and resistance for an ohmic conductor.
\(\textbf{Substitution:}\)
\[
V = (2)(6)
\]
\(\textbf{Final Simplification:}\)
\[
V = 12\,\text{V}
\]
\(\textbf{Final Answer:}\)
\[12\,\text{V}\]
10. A block is pulled with a constant horizontal force \(F\) on a smooth surface. Which conclusion is correct if the mass of the block is doubled while the applied force remains unchanged?
ⓐ. The acceleration becomes four times the original value.
ⓑ. The acceleration remains unchanged because force is constant.
ⓒ. The acceleration becomes double because the body is heavier.
ⓓ. The acceleration becomes half of the original value.
Correct Answer: The acceleration becomes half of the original value.
Explanation: From Newton’s second law, acceleration is inversely proportional to mass when the net force remains fixed. If the same force acts on a body of twice the mass, the acceleration must reduce to half its earlier value. This does not mean the force becomes weaker; it means the same force now produces a smaller change in motion because there is more inertia. The relation is captured directly by \(a = \frac{F}{m}\).
11. A convex lens has focal length \(20\,\text{cm}\). What is its power?
ⓐ. \(+2\,\text{D}\)
ⓑ. \(+5\,\text{D}\)
ⓒ. \(+10\,\text{D}\)
ⓓ. \(+20\,\text{D}\)
Correct Answer: \(+5\,\text{D}\)
Explanation: \(\textbf{Given:}\)
Focal length, \(f = 20\,\text{cm} = 0.20\,\text{m}\)
\(\textbf{Required:}\)
Power, \(P\)
\(\textbf{Relevant Formula:}\)
\[
P = \frac{1}{f}
\]
\(\textbf{Why this formula applies:}\)
Power of a lens in dioptre is the reciprocal of focal length in metre. For a convex lens, power is positive.
\(\textbf{Substitution:}\)
\[
P = \frac{1}{0.20}
\]
\(\textbf{Final Simplification:}\)
\[
P = +5\,\text{D}
\]
\(\textbf{Final Answer:}\)
\[+5\,\text{D}\]
12. An object is projected vertically upward with speed \(20\,\text{m s}^{-1}\). Taking \(g = 10\,\text{m s}^{-2}\), what is the maximum height reached?
ⓐ. \(20\,\text{m}\)
ⓑ. \(30\,\text{m}\)
ⓒ. \(40\,\text{m}\)
ⓓ. \(50\,\text{m}\)
Correct Answer: \(20\,\text{m}\)
Explanation: \(\textbf{Given:}\)
Initial velocity, \(u = 20\,\text{m s}^{-1}\)
Acceleration due to gravity, \(g = 10\,\text{m s}^{-2}\)
At maximum height, final velocity \(v = 0\)
\(\textbf{Required:}\)
Maximum height, \(h\)
\(\textbf{Relevant Formula:}\)
\[
v^2 = u^2 - 2gh
\]
\(\textbf{Why this formula applies:}\)
For upward vertical motion under uniform gravitational retardation, this equation relates velocity and displacement.
\(\textbf{Substitution:}\)
\[
0 = (20)^2 - 2(10)h
\]
\(\textbf{Intermediate Simplification:}\)
\[
0 = 400 - 20h
\]
\[
20h = 400
\]
\(\textbf{Final Simplification:}\)
\[
h = 20\,\text{m}
\]
\(\textbf{Final Answer:}\)
\[20\,\text{m}\]
13. Which statement about electric power is correct?
ⓐ. Electric power is the charge flowing per unit time in a conductor.
ⓑ. Electric power is the force acting on each unit positive charge only.
ⓒ. Electric power is the opposition offered by a conductor to the flow of charge.
ⓓ. Electric power is the energy consumed or transferred per unit time.
Correct Answer: Electric power is the energy consumed or transferred per unit time.
Explanation: Electric power describes the rate at which electrical energy is converted, used, or transferred in a circuit. It can be written as \(P = VI\), \(P = I^2R\), or \(P = \frac{V^2}{R}\) depending on the known quantities. Power is not the same thing as current, voltage, or resistance, though it depends on them. This is why power is best understood as energy per unit time in an electrical context.
14. A particle moves with speed \(10\,\text{m s}^{-1}\) in a circle of radius \(5\,\text{m}\). What is its centripetal acceleration?
ⓐ. \(5\,\text{m s}^{-2}\)
ⓑ. \(10\,\text{m s}^{-2}\)
ⓒ. \(20\,\text{m s}^{-2}\)
ⓓ. \(50\,\text{m s}^{-2}\)
Correct Answer: \(20\,\text{m s}^{-2}\)
Explanation: \(\textbf{Given:}\)
Speed, \(v = 10\,\text{m s}^{-1}\)
Radius, \(r = 5\,\text{m}\)
\(\textbf{Required:}\)
Centripetal acceleration, \(a_c\)
\(\textbf{Relevant Formula:}\)
\[
a_c = \frac{v^2}{r}
\]
\(\textbf{Substitution:}\)
\[
a_c = \frac{10^2}{5}
\]
\(\textbf{Intermediate Simplification:}\)
\[
a_c = \frac{100}{5}
\]
\(\textbf{Final Simplification:}\)
\[
a_c = 20\,\text{m s}^{-2}
\]
\(\textbf{Final Answer:}\)
\[20\,\text{m s}^{-2}\]
15. Two observations are made for the same conductor:
Case 1: length is doubled, cross-sectional area unchanged
Case 2: cross-sectional area is doubled, length unchanged
What happens to resistance in the two cases, if material and temperature remain unchanged?
ⓐ. In Case 1 it doubles, and in Case 2 it becomes half
ⓑ. In Case 1 it becomes half, and in Case 2 it doubles
ⓒ. In both cases it doubles
ⓓ. In both cases it becomes half
Correct Answer: In Case 1 it doubles, and in Case 2 it becomes half
Explanation: \(\textbf{Given:}\)
Material and temperature remain unchanged.
\(\textbf{Required:}\)
Effect on resistance when length or area is changed
\(\textbf{Relevant Formula:}\)
\[
R = \rho \frac{L}{A}
\]
\(\textbf{Step 1 — Case 1: length doubled}\)
If \(L\) becomes \(2L\), then
\[
R' = \rho \frac{2L}{A} = 2R
\]
So resistance doubles.
\(\textbf{Step 2 — Case 2: area doubled}\)
If \(A\) becomes \(2A\), then
\[
R' = \rho \frac{L}{2A} = \frac{R}{2}
\]
So resistance becomes half.
\(\textbf{Final Answer:}\)
In Case 1 it doubles, and in Case 2 it becomes half.
16. (Chemistry) For a reactant \(R\), its concentration changes from \(0.80\,\text{mol L}^{-1}\) to \(0.50\,\text{mol L}^{-1}\) in \(10\,\text{s}\). What is the average rate of disappearance of \(R\)?
ⓐ. \(0.02\,\text{mol L}^{-1}\text{s}^{-1}\)
ⓑ. \(-0.03\,\text{mol L}^{-1}\text{s}^{-1}\)
ⓒ. \(0.03\,\text{mol L}^{-1}\text{s}^{-1}\)
ⓓ. \(0.05\,\text{mol L}^{-1}\text{s}^{-1}\)
Correct Answer: \(0.03\,\text{mol L}^{-1}\text{s}^{-1}\)
Explanation: \(\textbf{Given:}\)
Initial concentration of \(R\), \([R]_1 = 0.80\,\text{mol L}^{-1}\)
Final concentration of \(R\), \([R]_2 = 0.50\,\text{mol L}^{-1}\)
Time interval, \(\Delta t = 10\,\text{s}\)
\(\textbf{Required:}\)
Average rate of disappearance of \(R\)
\(\textbf{Relevant Formula:}\)
\[\text{Average rate} = -\frac{\Delta [R]}{\Delta t}\]
\(\textbf{Why this formula applies:}\)
For a reactant, concentration decreases with time, so a negative sign is used to keep the rate positive.
\(\textbf{Identify the concentration change:}\)
\[\Delta [R] = [R]_2 - [R]_1 = 0.50 - 0.80 = -0.30\,\text{mol L}^{-1}\]
\(\textbf{Substitution:}\)
\[\text{Average rate} = -\frac{-0.30}{10}\]
\(\textbf{Intermediate Simplification:}\)
\[\text{Average rate} = \frac{0.30}{10}\]
\(\textbf{Final Simplification:}\)
\[\text{Average rate} = 0.03\,\text{mol L}^{-1}\text{s}^{-1}\]
\(\textbf{Unit Check:}\)
Concentration divided by time gives \(\text{mol L}^{-1}\text{s}^{-1}\)
\(\textbf{Final Answer:}\)
\[0.03\,\text{mol L}^{-1}\text{s}^{-1}\]
17. For a reaction, the following data are obtained:
Experiment 1: \([A] = 0.10\,\text{mol L}^{-1}\), \([B] = 0.10\,\text{mol L}^{-1}\), rate \(= 2.0 \times 10^{-3}\,\text{mol L}^{-1}\text{s}^{-1}\)
Experiment 2: \([A] = 0.20\,\text{mol L}^{-1}\), \([B] = 0.10\,\text{mol L}^{-1}\), rate \(= 4.0 \times 10^{-3}\,\text{mol L}^{-1}\text{s}^{-1}\)
Experiment 3: \([A] = 0.20\,\text{mol L}^{-1}\), \([B] = 0.20\,\text{mol L}^{-1}\), rate \(= 1.6 \times 10^{-2}\,\text{mol L}^{-1}\text{s}^{-1}\)
What is the overall order of the reaction?
ⓐ. \(1\)
ⓑ. \(2\)
ⓒ. \(4\)
ⓓ. \(3\)
Correct Answer: \(3\)
Explanation: \(\textbf{Given:}\)
Rate law, \(r = k[A]^m[B]^n\)
\(\textbf{Experimental Data:}\)
From Experiment 1 to Experiment 2:
\([A]\) doubles, \([B]\) constant
Rate doubles from \(2.0 \times 10^{-3}\) to \(4.0 \times 10^{-3}\)
From Experiment 2 to Experiment 3:
\([A]\) constant
\([B]\) doubles from \(0.10\) to \(0.20\,\text{mol L}^{-1}\)
Rate increases from \(4.0 \times 10^{-3}\) to \(1.6 \times 10^{-2}\)
\(\textbf{Required:}\)
Overall order \(= m+n\)
\(\textbf{Step 1 — Find order with respect to }A\textbf{:}\)
Since doubling \([A]\) doubles the rate:
\[
2^m = 2
\]
So,
\[
m = 1
\]
\(\textbf{Step 2 — Find order with respect to }B\textbf{:}\)
From Experiment 2 to Experiment 3, rate increases by a factor of
\[
\frac{1.6 \times 10^{-2}}{4.0 \times 10^{-3}} = 4
\]
Since \([B]\) doubles:
\[
2^n = 4
\]
So,
\[
n = 2
\]
\(\textbf{Step 3 — Find overall order:}\)
\[
m+n = 1+2 = 3
\]
\(\textbf{Final Answer:}\)
Overall order \(= 3\)
18. Two reactions are studied at the same temperature.
Reaction X:
At \(0.20\,\text{mol L}^{-1}\), \(t_{1/2} = 10\,\text{s}\)
At \(0.40\,\text{mol L}^{-1}\), \(t_{1/2} = 20\,\text{s}\)
Reaction Y:
At \(0.20\,\text{mol L}^{-1}\), \(t_{1/2} = 15\,\text{s}\)
At \(0.40\,\text{mol L}^{-1}\), \(t_{1/2} = 15\,\text{s}\)
Which conclusion is correct?
ⓐ. Both X and Y are zero order.
ⓑ. X is first order and Y is zero order.
ⓒ. X is zero order and Y is first order.
ⓓ. Both X and Y are first order.
Correct Answer: X is zero order and Y is first order.
Explanation: \(\textbf{Given:}\)
Reaction X has a half-life that is directly proportional to the initial concentration.
Reaction Y has a half-life that is independent of the initial concentration.
\(\textbf{Required:}\)
Identify the likely orders of Reaction X and Reaction Y.
\(\textbf{Relevant Principles:}\)
For zero-order kinetics:
\[
t_{1/2} = \frac{[R]_0}{2k}
\]
For first-order kinetics:
\[
t_{1/2} = \frac{0.693}{k}
\]
\(\textbf{Step 1 — Apply to Reaction X:}\)
When \([R]_0\) doubles from \(0.20\) to \(0.40\,\text{mol L}^{-1}\), \(t_{1/2}\) also doubles from \(10\,\text{s}\) to \(20\,\text{s}\).
This matches the direct proportionality expected for zero-order kinetics.
\(\textbf{Step 2 — Apply to Reaction Y:}\)
\(t_{1/2}\) remains constant at \(15\,\text{s}\) at both concentrations.
This matches first-order behavior, where half-life does not depend on initial concentration.
\(\textbf{Final Answer:}\)
Reaction X is zero order and Reaction Y is first order.
19. For a first-order reaction, the rate constant is \(2.0 \times 10^{-3}\,\text{s}^{-1}\). What is the half-life of the reaction?
ⓐ. \(173\,\text{s}\)
ⓑ. \(231\,\text{s}\)
ⓒ. \(346.5\,\text{s}\)
ⓓ. \(693\,\text{s}\)
Correct Answer: \(346.5\,\text{s}\)
Explanation: \(\textbf{Given:}\)
Rate constant, \(k = 2.0 \times 10^{-3}\,\text{s}^{-1}\)
\(\textbf{Required:}\)
Half-life, \(t_{1/2}\)
\(\textbf{Relevant Formula:}\)
\[
t_{1/2} = \frac{0.693}{k}
\]
\(\textbf{Why this formula applies:}\)
For a first-order reaction, half-life depends only on the rate constant.
\(\textbf{Substitution:}\)
\[
t_{1/2} = \frac{0.693}{2.0 \times 10^{-3}}
\]
\(\textbf{Intermediate Simplification:}\)
\[
t_{1/2} = \frac{0.693}{0.002}
\]
\(\textbf{Final Simplification:}\)
\[
t_{1/2} = 346.5\,\text{s}
\]
\(\textbf{Unit Check:}\)
Since \(k\) has unit \(\text{s}^{-1}\), the half-life has unit \(\text{s}\).
\(\textbf{Final Answer:}\)
\[346.5\,\text{s}\]
20. Which statement correctly explains why the rate constant of a reaction increases with temperature?
ⓐ. Temperature increases the fraction of molecules with energy above activation energy.
ⓑ. Temperature makes every collision productive without any energy barrier at all.
ⓒ. Temperature changes all reactions into equilibrium-controlled processes immediately.
ⓓ. Temperature always reduces the activation energy value of the reaction itself.
Correct Answer: Temperature increases the fraction of molecules with energy above activation energy.
Explanation: As temperature rises, the kinetic energy distribution shifts so that a larger fraction of molecules has energy equal to or greater than the activation energy. That increases the number of effective collisions per unit time, so the rate constant increases. The activation energy itself does not automatically become smaller just because temperature rises. This effect is described through the Arrhenius relation and is one of the main reasons reaction rates are temperature-sensitive.
