Physics 2010 Final Practice: Step-by-Step Guidance for Fluids, Oscillations & Waves

Q1. A solid object floats in water with exactly half its volume submerged. Its density compared to water is:

Background

Topic: Buoyancy and Density

This question tests your understanding of Archimedes' principle and the relationship between the density of an object and the fluid it floats in.

Key Terms and Formulas

• Buoyant Force: $F_b = \rho_{fluid} V_{sub} g$

• Floating Condition: $F_b = mg$

• Density: $\rho = \frac{m}{V}$

Step-by-Step Guidance

1. Recall that for a floating object, the weight of the displaced fluid equals the weight of the object.

2. Write the equation: $\rho_{fluid} V_{sub} g = \rho_{object} V_{object} g$

3. Since half the volume is submerged, $V_{sub} = \frac{1}{2} V_{object}$.

4. Substitute this into the equation and simplify to relate $\rho_{object}$ and $\rho_{fluid}$.

Try solving on your own before revealing the answer!

Final Answer: (C) Half that of water

Since $V_{sub} = \frac{1}{2} V_{object}$, $\rho_{object} = \frac{1}{2} \rho_{fluid}$. The object's density is half that of water.

Q2. Water flows from a wide pipe into a narrow pipe. The pressure in the narrow pipe (same height) is:

Background

Topic: Fluid Dynamics (Bernoulli's Principle)

This question tests your understanding of how fluid speed and pressure are related in a pipe of varying diameter.

Key Terms and Formulas

• Continuity Equation: $A_1 v_1 = A_2 v_2$

• Bernoulli's Equation (constant height): $P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2$

Step-by-Step Guidance

1. Recognize that the fluid speeds up in the narrow pipe due to conservation of mass (continuity equation).

2. Apply Bernoulli's equation at the same height to relate pressure and speed.

3. Since $v_2 > v_1$, compare $P_2$ and $P_1$ using the equation.

Try solving on your own before revealing the answer!

Final Answer: (B) Lower, because water is faster (Bernoulli)

As speed increases in the narrow pipe, pressure decreases according to Bernoulli's principle.

Q3. A spring-mass system on a frictionless surface: as the mass passes through equilibrium moving right, its acceleration is:

Background

Topic: Simple Harmonic Motion (SHM)

This question tests your understanding of the relationship between position, velocity, and acceleration in SHM.

Key Terms and Formulas

• Acceleration in SHM: $a = -\omega^2 x$

• At equilibrium: $x = 0$

Step-by-Step Guidance

1. Recall that acceleration in SHM is proportional to displacement and directed toward equilibrium.

2. At equilibrium, substitute $x = 0$ into the acceleration formula.

3. Consider the direction and magnitude of acceleration at this point.

Try solving on your own before revealing the answer!

Final Answer: (B) Zero

At equilibrium ($x = 0$), acceleration is zero in SHM.

Q4. Two waves of equal amplitude and frequency traveling in opposite directions on a string produce:

Background

Topic: Wave Interference and Standing Waves

This question tests your understanding of how waves interact to form standing waves.

Key Terms and Formulas

• Standing Wave: Formed by superposition of two waves traveling in opposite directions.

• Nodes and Antinodes: Points of zero and maximum amplitude, respectively.

Step-by-Step Guidance

1. Recall what happens when two identical waves travel in opposite directions.

2. Consider the resulting pattern on the string (look for nodes and antinodes).

3. Compare this to the options given.

Try solving on your own before revealing the answer!

Final Answer: (B) A standing wave with nodes and antinodes

Superposition of two identical waves in opposite directions creates a standing wave pattern.

Q5. Sound intensity decreases as $1/r^2$. Moving from 2 m to 6 m from a point source, intensity:

Background

Topic: Sound Intensity and the Inverse Square Law

This question tests your understanding of how sound intensity changes with distance from a point source.

Key Terms and Formulas

• Intensity: $I = \frac{P}{4\pi r^2}$

• Inverse Square Law: $I \propto \frac{1}{r^2}$

Step-by-Step Guidance

1. Write the ratio of intensities at two distances: $\frac{I_2}{I_1} = \left(\frac{r_1}{r_2}\right)^2$

2. Plug in $r_1 = 2$ m and $r_2 = 6$ m.

3. Simplify the ratio to find the factor by which intensity decreases.

Try solving on your own before revealing the answer!

Final Answer: (B) Decreases by a factor of 9

Intensity decreases by $(6/2)^2 = 9$ times when moving from 2 m to 6 m.

Q6. An open organ pipe and a closed organ pipe of the same length L: ratio of their fundamental frequencies is:

Background

Topic: Standing Waves in Pipes

This question tests your understanding of the fundamental frequencies for open and closed pipes.

Key Terms and Formulas

• Open Pipe: $f_{open} = \frac{v}{2L}$

• Closed Pipe: $f_{closed} = \frac{v}{4L}$

Step-by-Step Guidance

1. Write the expressions for the fundamental frequencies for both pipes.

2. Take the ratio $\frac{f_{open}}{f_{closed}}$.

3. Simplify the ratio to match one of the answer choices.

Try solving on your own before revealing the answer!

Final Answer: (B) 2:1

The open pipe's fundamental frequency is twice that of the closed pipe of the same length.

Q7. A mass on a spring has amplitude A. Amplitude is doubled and mass is quadrupled (spring constant unchanged). New period is:

Background

Topic: Period of Simple Harmonic Motion

This question tests your understanding of how period depends on mass and amplitude in SHM.

Key Terms and Formulas

• Period: $T = 2\pi \sqrt{\frac{m}{k}}$

• Amplitude does not affect period for a mass-spring system.

Step-by-Step Guidance

1. Recall the formula for the period of a mass-spring system.

2. Note that amplitude does not appear in the formula.

3. Substitute the new mass ($4m$) into the formula and compare to the original period.

Try solving on your own before revealing the answer!

Final Answer: (C) Doubled, because mass quadrupled

Period increases by a factor of $\sqrt{4} = 2$; amplitude does not affect period.

Q8. A pendulum clock keeps perfect time on Earth. On a planet where g is 4 times larger, the clock runs:

Background

Topic: Pendulum Period and Gravity

This question tests your understanding of how the period of a pendulum depends on gravitational acceleration.

Key Terms and Formulas

• Pendulum Period: $T = 2\pi \sqrt{\frac{L}{g}}$

Step-by-Step Guidance

1. Write the formula for the period of a simple pendulum.

2. Substitute $g_{planet} = 4g_{Earth}$ into the formula.

3. Compare the new period to the original period.

Try solving on your own before revealing the answer!

Final Answer: (B) Twice as fast

Period decreases by a factor of $\sqrt{1/4} = 1/2$, so the clock runs twice as fast.

Q9. A tuning fork held near a guitar string causes the string to resonate. This demonstrates:

Background

Topic: Resonance and Natural Frequency

This question tests your understanding of forced resonance and natural frequencies in oscillating systems.

Key Terms and Formulas

• Resonance: When a system is driven at its natural frequency, it oscillates with large amplitude.

Step-by-Step Guidance

1. Recall what happens when an external frequency matches a system's natural frequency.

2. Identify which answer choice describes this phenomenon.

Try solving on your own before revealing the answer!

Final Answer: (C) Forced resonance at the string's natural frequency

The string resonates because the tuning fork's frequency matches its natural frequency.

Q10. Which factor does NOT affect wave speed on a stretched string?

Background

Topic: Wave Speed on a String

This question tests your understanding of the factors that determine the speed of a wave on a string.

Key Terms and Formulas

• Wave Speed: $v = \sqrt{\frac{T}{\mu}}$

• $T$ = tension, $\mu$ = linear density

Step-by-Step Guidance

1. List the variables in the wave speed formula.

2. Check which of the answer choices are not present in the formula.

3. Consider whether amplitude affects wave speed.

Try solving on your own before revealing the answer!

Final Answer: (C) Amplitude of the wave

Wave speed depends on tension and linear density, not amplitude.

Q11. At a point where two identical waves interfere constructively, the combined amplitude is:

Background

Topic: Wave Interference

This question tests your understanding of constructive interference and how amplitudes combine.

Key Terms and Formulas

• Constructive Interference: $A_{total} = A_1 + A_2$

Step-by-Step Guidance

1. Recall the rule for adding amplitudes during constructive interference.

2. For two identical waves, add their amplitudes together.

Try solving on your own before revealing the answer!

Final Answer: (A) Doubled

Constructive interference doubles the amplitude when the waves are identical.

Q12. A wave pulse moves right on a string fixed at the right end. When it reaches the end, the reflected pulse:

Background

Topic: Reflection of Waves

This question tests your understanding of how a wave pulse behaves when it reflects from a fixed boundary.

Key Terms and Formulas

• Fixed End Reflection: The pulse is inverted (flipped upside down).

Step-by-Step Guidance

1. Recall the rule for wave reflection at a fixed boundary.

2. Determine the orientation of the reflected pulse compared to the incident pulse.

Try solving on your own before revealing the answer!

Final Answer: (B) Is inverted (flipped upside down)

Reflection at a fixed end inverts the wave pulse.

Q13. An object floats in a liquid. When the liquid is heated and expands (lower density), the object:

Background

Topic: Buoyancy and Density Changes

This question tests your understanding of how changes in fluid density affect floating objects.

Key Terms and Formulas

• Buoyant Force: $F_b = \rho_{fluid} V_{sub} g$

• Floating Condition: $F_b = mg$

Step-by-Step Guidance

1. Consider what happens to the fluid's density when it is heated and expands.

2. Analyze how a decrease in fluid density affects the volume of the object that must be submerged to balance its weight.

3. Predict whether the object will sink lower, rise higher, or stay the same.

Try solving on your own before revealing the answer!

Final Answer: (A) Sinks lower (more submerged)

Lower fluid density means more of the object must be submerged to displace the same weight of fluid.

Q14. Increasing the damping coefficient of an oscillator causes the oscillation frequency to:

Background

Topic: Damped Oscillations

This question tests your understanding of how damping affects the frequency of oscillation.

Key Terms and Formulas

• Damped Frequency: $\omega' = \sqrt{\omega_0^2 - (\frac{b}{2m})^2}$

• Undamped Frequency: $\omega_0 = \sqrt{\frac{k}{m}}$

Step-by-Step Guidance

1. Recall the formula for the frequency of a damped oscillator.

2. Consider what happens to $\omega'$ as the damping coefficient $b$ increases.

3. Compare the damped frequency to the undamped frequency.

Try solving on your own before revealing the answer!

Final Answer: (C) Decrease slightly compared to undamped

Increased damping lowers the oscillation frequency slightly.

Q15. The Doppler shift for light is used in astronomy to determine whether a star is:

Background

Topic: Doppler Effect for Light

This question tests your understanding of how the Doppler effect is used to study motion in astronomy.

Key Terms and Formulas

• Doppler Effect: Change in observed frequency due to relative motion.

• Redshift/Blueshift: Used to determine if an object is moving away or toward us.

Step-by-Step Guidance

1. Recall what information the Doppler effect provides about a light source.

2. Identify which property of a star can be determined using this effect.

Try solving on your own before revealing the answer!

Final Answer: (B) Moving toward or away from Earth

The Doppler shift reveals the radial motion of stars relative to Earth.