All practice questions

Question 12761Question →

A composite solid consists of a metal block $A$ of mass $2.0\text{ kg}$ and a metal block $B$ of mass $1.0\text{ kg}$ joined in thermal contact. Block $A$ has a specific heat capacity of $300\text{ J kg}^{-1}\text{K}^{-1}$ . When the composite solid absorbs $18\text{ kJ}$ of thermal energy, its overall temperature increases by $15\text{ K}$ . Assuming no heat loss to the surroundings, what is the specific heat capacity of metal block $B$ ?

600\text{ J kg}^{-1}\text{K}^{-1}

1200\text{ J kg}^{-1}\text{K}^{-1}

400\text{ J kg}^{-1}\text{K}^{-1}

200\text{ J kg}^{-1}\text{K}^{-1}

Show answer & explanation

Answer: $600\text{ J kg}^{-1}\text{K}^{-1}$

Answer

600\text{ J kg}^{-1}\text{K}^{-1}

The total heat energy supplied (

18\text{ kJ} = 18000\text{ J}

) causes a temperature rise of

15\text{ K}

, giving a total heat capacity

C_{\text{total}} = 1200\text{ J K}^{-1}

for the composite solid. Subtracting block

A

's heat capacity (

C_A = 2.0\text{ kg} \times 300\text{ J kg}^{-1}\text{K}^{-1} = 600\text{ J K}^{-1}

) leaves a heat capacity of

600\text{ J K}^{-1}

for block

B

. Dividing block

B

's heat capacity by its mass (

1.0\text{ kg}

) yields its specific heat capacity of

600\text{ J kg}^{-1}\text{K}^{-1}

.

Step-by-Step Solution

1

Calculate the total heat capacity (

C_{\text{total}}

) of the composite system.

C_{\text{total}} = \frac{Q}{\Delta T} = \frac{18000\text{ J}}{15\text{ K}} = 1200\text{ J K}^{-1}

Heat capacity is defined as the total heat energy absorbed per unit change in temperature.

2

Determine the heat capacity (

C_A

) of metal block

A

.

C_A = m_A \times c_A = 2.0\text{ kg} \times 300\text{ J kg}^{-1}\text{K}^{-1} = 600\text{ J K}^{-1}

Heat capacity is the product of mass and specific heat capacity.

3

Determine the heat capacity (

C_B

) of metal block

B

.

C_B = C_{\text{total}} - C_A = 1200\text{ J K}^{-1} - 600\text{ J K}^{-1} = 600\text{ J K}^{-1}

Total heat capacity of a composite body is the sum of the individual heat capacities of its components.

4

Calculate the specific heat capacity (

c_B

) of metal block

B

.

c_B = \frac{C_B}{m_B} = \frac{600\text{ J K}^{-1}}{1.0\text{ kg}} = 600\text{ J kg}^{-1}\text{K}^{-1}

Specific heat capacity is the heat capacity per unit mass of that specific substance.

Key Concept

Additivity of Heat Capacity in Composite Systems (

C_{\text{total}} = m_A c_A + m_B c_B

)

Question 12762Question →

A screw jack with a pitch of $4\text{ mm}$ and a tommy bar of length $56\text{ cm}$ is used to raise a heavy load of mass $1100\text{ kg}$ . If an effort force of $50\text{ N}$ is applied at the outer end of the tommy bar, what is the efficiency of the screw jack? (Take $\pi = \frac{22}{7}$ and $g = 10\text{ m/s}^2$ )

25%

400%

157%

2.5%

Show answer & explanation

Answer: 25%

Answer

The efficiency of the screw jack is 25%

The mechanical advantage is calculated as

MA = \frac{11,000\text{ N}}{50\text{ N}} = 220

. The velocity ratio for a screw jack is

VR = \frac{2 \pi R}{p} = \frac{2 \times \frac{22}{7} \times 0.56\text{ m}}{0.004\text{ m}} = 880

. Dividing

MA

by

VR

and multiplying by

100\%

yields an efficiency of

\frac{220}{880} \times 100\% = 25\%

.

Step-by-Step Solution

1

Calculate the load force (weight) from the given mass.

W = m \times g = 1100\text{ kg} \times 10\text{ m/s}^2 = 11,000\text{ N}

Mass must be converted to weight force in newtons to compute mechanical advantage.

2

Calculate the Mechanical Advantage (MA).

MA = \frac{\text{Load}}{\text{Effort}} = \frac{11,000\text{ N}}{50\text{ N}} = 220

Mechanical advantage is the ratio of output load force to input effort force.

3

Calculate the Velocity Ratio (VR) of the screw jack.

VR = \frac{2 \pi R}{p} = \frac{2 \times \frac{22}{7} \times 0.56\text{ m}}{0.004\text{ m}} = \frac{3.52\text{ m}}{0.004\text{ m}} = 880

Velocity ratio is the distance moved by the effort in one full revolution (

2\pi R

) divided by the distance moved by the load in one revolution (pitch

p

).

4

Calculate the efficiency of the machine.

\text{Efficiency} = \frac{MA}{VR} \times 100\% = \frac{220}{880} \times 100\% = 25\%

Efficiency is defined as the ratio of Mechanical Advantage to Velocity Ratio multiplied by 100%.

Key Concept

Mechanical Advantage, Velocity Ratio, and Efficiency of a Screw Jack

Question 12763Question →

When white anhydrous copper(II) tetraoxosulfate(VI) powder is exposed to moist air, it turns blue by absorbing water vapor without forming a solution. Under the same conditions, solid sodium hydroxide pellets absorb moisture from the atmosphere until they completely dissolve to form a liquid solution. Which terms correctly classify the behavior of anhydrous copper(II) tetraoxosulfate(VI) and solid sodium hydroxide, respectively?

Hygroscopic and deliquescent

Deliquescent and hygroscopic

Efflorescent and deliquescent

Hygroscopic and efflorescent

Show answer & explanation

Answer: Hygroscopic and deliquescent

Answer

Anhydrous copper(II) tetraoxosulfate(VI) is hygroscopic, while solid sodium hydroxide is deliquescent.

Hygroscopic substances absorb moisture from the atmosphere without dissolving or forming a liquid solution, as seen with anhydrous copper(II) tetraoxosulfate(VI) turning blue (

\text{CuSO}_4 + 5\text{H}_2\text{O} \rightarrow \text{CuSO}_4\cdot 5\text{H}_2\text{O}

). Deliquescent substances, such as solid sodium hydroxide (

\text{NaOH}

), absorb moisture from the atmosphere and dissolve in that absorbed water to form a saturated solution.

Step-by-Step Solution

1

Analyze the behavior of anhydrous copper(II) tetraoxosulfate(VI) in moist air.

It absorbs atmospheric water vapor to form hydrated copper(II) tetraoxosulfate(VI) (blue) without forming a liquid solution. Substances that absorb atmospheric moisture without dissolving are classified as hygroscopic.

Hygroscopy involves water absorption without phase change into a liquid solution.

2

Analyze the behavior of solid sodium hydroxide in moist air.

It absorbs sufficient atmospheric water to dissolve entirely and form a saturated solution. Substances that absorb water from the air and dissolve in it are classified as deliquescent.

Deliquescence occurs when a substance absorbs water until its vapor pressure matches atmospheric humidity, dissolving the solid.

3

Match the respective terms in order.

Hygroscopic and deliquescent.

The question asks for the behavior of anhydrous copper(II) tetraoxosulfate(VI) followed by sodium hydroxide.

Key Concept

Distinction between Hygroscopy, Deliquescence, and Efflorescence

Question 12764Question →

At the Earth's magnetic equator, a freely suspended magnetic dip needle comes to rest horizontally, resulting in an angle of dip of $0^\circ$ .

True

False

Show answer & explanation

Answer: True

Answer

True

The statement is true because the Earth's magnetic field lines at the magnetic equator are parallel to the Earth's surface. Consequently, the vertical component of the magnetic field is zero, causing a freely suspended dip needle to settle completely horizontally at an angle of

0^\circ

.

Step-by-Step Solution

1

Define the angle of dip (magnetic inclination).

The angle of dip

\theta

is the angle between the Earth's magnetic field direction and the horizontal plane at a given location.

This definition helps determine how a dip needle orientates itself relative to the surface of the Earth.

2

Examine the Earth's magnetic field components at the magnetic equator.

At the magnetic equator, the vertical component of Earth's magnetic field is zero (

B_v = 0

), so the total field is entirely horizontal (

B = B_h

).

The magnetic flux lines are parallel to the geographical surface along the magnetic equator.

3

Evaluate the statement.

Since the field vector is horizontal, the angle of dip

\theta = \tan^{-1}\left(\frac{B_v}{B_h}\right) = \tan^{-1}(0) = 0^\circ

, making the statement True.

The dip needle aligns with the total magnetic field vector, resting horizontally at

0^\circ

.

Key Concept

Angle of Dip at the Magnetic Equator

Question 12765Question →

The length and width of a rectangular metal sheet are measured as $(8.0 \pm 0.2)\text{ cm}$ and $(5.0 \pm 0.1)\text{ cm}$ , respectively. What is the percentage error in the calculated area of the metal sheet?

Show answer & explanation

Answer: 4.5

Answer

The percentage error in the calculated area of the metal sheet is

4.5\%

.

For a calculated quantity involving multiplication (

A = L \times W

), the total relative error is the sum of the relative errors of the individual measurements. The fractional error in length is

\frac{0.2}{8.0} = 0.025

(

2.5\%

) and in width is

\frac{0.1}{5.0} = 0.020

(

2.0\%

). Summing these gives

0.045

, which is equivalent to

4.5\%

.

Step-by-Step Solution

1

Determine the fractional error in the length measurement

\frac{0.2\text{ cm}}{8.0\text{ cm}} = 0.025

Fractional error is given by the ratio of absolute error to the measured value.

2

Determine the fractional error in the width measurement

\frac{0.1\text{ cm}}{5.0\text{ cm}} = 0.020

Fractional error is calculated as the absolute uncertainty divided by the measured dimension.

3

Combine the fractional errors for the calculated area

\frac{\Delta A}{A} = 0.025 + 0.020 = 0.045

For quantities multiplied together (

A = L \times W

), individual fractional errors sum to give the total fractional error.

4

Convert the fractional error to percentage error

0.045 \times 100\% = 4.5\%

Multiplying the relative error by 100 yields the percentage error.

Key Concept

Error Propagation in Products

Question 12766Question →

A swimming pool has an apparent depth of $1.8\text{ m}$ when viewed vertically from directly above. If the refractive index of water relative to air is $\frac{4}{3}$ , what is the real depth of the pool in meters?

Show answer & explanation

Answer: 2.4

Answer

The real depth of the pool is

2.4\text{ m}

.

The refractive index

n

of a medium is defined as the ratio of the real depth to the apparent depth (

n = \frac{\text{Real Depth}}{\text{Apparent Depth}}

). Multiplying the observed apparent depth of

1.8\text{ m}

by the refractive index

\frac{4}{3}

gives the true real depth of

2.4\text{ m}

.

Step-by-Step Solution

1

Identify the relationship between refractive index, real depth, and apparent depth.

Refractive index

n = \frac{\text{Real Depth}}{\text{Apparent Depth}}

.

Light bending at the boundary causes an object submerged in a denser medium to appear closer to the surface.

2

Rearrange the equation to make Real Depth the subject.

\text{Real Depth} = n \times \text{Apparent Depth}.

To calculate the true depth from the observed apparent depth and the optical density of water.

3

Substitute the given values into the formula.

\text{Real Depth} = \frac{4}{3} \times 1.8\text{ m} = 2.4\text{ m}.

Multiplying the apparent depth by the refractive index yields the actual physical depth.

Key Concept

Refraction of Light and Real/Apparent Depth

Question 12767Question →

Consider the nuclear fission reaction represented by the equation below:

{^{235}_{92}\text{U}} + {^{1}_{0}\text{n}} \rightarrow {^{144}_{56}\text{Ba}} + {^{89}_{36}\text{Kr}} + x\,^{1}_{0}\text{n}

What is the value of

x

, representing the number of neutrons emitted in this reaction?

1

2

3

4

Show answer & explanation

Answer: 3

Answer

The number of emitted neutrons,

x

, is equal to 3.

By the law of conservation of mass number, the sum of nucleon numbers on the left side (

235 + 1 = 236

) must equal the sum on the right side (

144 + 89 + x

). Solving the equation

236 = 233 + x

gives

x = 3

.

Step-by-Step Solution

1

Calculate the total mass number on the reactant side.

Total reactant mass number =

235 + 1 = 236

.

According to the law of conservation of nucleon number, the total mass number before the reaction must equal the total mass number after the reaction.

2

Sum the known mass numbers on the product side.

Known product mass number =

144 + 89 = 233

.

Adding the mass numbers of Barium-144 and Krypton-89.

3

Set up and solve the mass conservation equation for

x

.

236 = 233 + x(1) \implies x = 3

.

Each neutron has a mass number of 1, so solving

236 - 233

gives

x = 3

.

Key Concept

Conservation of Mass Number in Nuclear Fission

Question 12768Question →

A ticker-tape timer connected to an alternating current supply operates at a frequency of $50\text{ Hz}$ . If a tape pulled through the timer displays $6$ consecutive dots, what is the total time interval represented by this section of tape?

0.10\text{ s}

0.12\text{ s}

0.02\text{ s}

0.25\text{ s}

Show answer & explanation

Answer: $0.10\text{ s}$

Answer

The total time interval represented by the section of tape is

0.10\text{ s}

.

The frequency of

50\text{ Hz}

means each tick interval takes

T = \frac{1}{50} = 0.02\text{ s}

. For

6

consecutive dots, there are

5

intervals between them. The total time elapsed is

5 \times 0.02\text{ s} = 0.10\text{ s}

.

Step-by-Step Solution

1

Calculate the period

T

of a single tick interval

T = \frac{1}{f} = \frac{1}{50\text{ Hz}} = 0.02\text{ s}

The period of oscillation of the timer represents the time elapsed between two successive dots.

2

Determine the number of time intervals (spaces) between 6 consecutive dots

Number of intervals =

6 - 1 = 5\text{ intervals}

Time measurement on a ticker tape is based on the number of spaces between dots, not the count of dots themselves.

3

Calculate the total time interval

t

t = 5 \times 0.02\text{ s} = 0.10\text{ s}

Multiply the number of spaces by the time period per space.

Key Concept

Measurement of time using a ticker-tape timer

Question 12769Question →

A mass of air at $30^\circ\text{C}$ has a relative humidity of $50\%$ . The saturated vapour pressure of water is $30\text{ mmHg}$ at $30^\circ\text{C}$ and $9\text{ mmHg}$ at $10^\circ\text{C}$ . If the temperature of the air is lowered to $10^\circ\text{C}$ , what percentage of the initial water vapour condenses out?

$40\%

$60\%

$70\%

$30\%

Show answer & explanation

Answer: $40\%

Answer

The percentage of initial water vapour that condenses out when cooled to

10^\circ\text{C}

is

40\%

.

The correct answer is

40\%

. Initially, the air contains water vapour exerting a partial pressure of

0.50 \times 30\text{ mmHg} = 15\text{ mmHg}

. Upon cooling to

10^\circ\text{C}

, the air becomes saturated at

9\text{ mmHg}

, causing

15\text{ mmHg} - 9\text{ mmHg} = 6\text{ mmHg}

worth of vapour to condense into liquid. The condensed amount as a fraction of the initial vapour is

6 / 15 = 0.40

, or

40\%

.

Step-by-Step Solution

1

Calculate the initial partial vapour pressure of water at

30^\circ\text{C}

.

\text{Partial Vapour Pressure} = \text{Relative Humidity} \times \text{SVP at } 30^\circ\text{C} = 0.50 \times 30\text{ mmHg} = 15\text{ mmHg}

.

Relative humidity is the ratio of actual partial vapour pressure to the saturated vapour pressure at that temperature.

2

Determine the amount of vapour pressure that must condense when cooled to

10^\circ\text{C}

.

\text{Vapour pressure condensed} = 15\text{ mmHg} - 9\text{ mmHg} = 6\text{ mmHg}

.

At

10^\circ\text{C}

, the air can hold at most its saturated vapour pressure of

9\text{ mmHg}

, so any excess vapour above

9\text{ mmHg}

condenses into liquid water.

3

Calculate the percentage of the initial water vapour that condenses out.

\text{Percentage condensed} = \left(\frac{6\text{ mmHg}}{15\text{ mmHg}}\right) \times 100\% = 40\%

.

The question asks for the fraction of the initial vapour originally present that leaves the gaseous state.

Key Concept

Relative Humidity and Dew Point Condensation

Estimated Time:1m 30s

Question 12770Question →

Match each physical phenomenon or quantity involving electromagnetic forces listed on the left with its corresponding governing mathematical equation on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Magnetic force on a current-carrying conductor pairs with

F = B I L \sin \theta

; Force per unit length between parallel conductors pairs with

\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}

; Radius of circular trajectory of a charged particle pairs with

r = \frac{m v}{q B}

; Torque on a current-carrying coil pairs with

\tau = B I A N \sin \theta

.

Each electromagnetic phenomenon matches directly with its derived expression from the magnetic force laws: magnetic force on a wire is

F = B I L \sin \theta

, force between parallel wires is

\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}

, orbit radius of charge is

r = \frac{m v}{q B}

, and coil torque is

\tau = B I A N \sin \theta

.

Step-by-Step Solution

1

Analyze the magnetic force on a straight conductor

The Lorentz force law applied to current elements yields

F = B I L \sin \theta

.

Free charges moving inside the wire experience magnetic force perpendicular to both current and magnetic field vector.

2

Determine the mutual force formula for parallel conductors

The magnetic field from wire 1 is

B_1 = \frac{\mu_0 I_1}{2\pi d}

, giving force per length

\frac{F}{L} = B_1 I_2 = \frac{\mu_0 I_1 I_2}{2\pi d}

.

Each conductor sits within the circular magnetic field lines generated by the other conductor.

3

Derive the motion equation for a charged particle in a magnetic field

Setting

q v B = \frac{m v^2}{r}

yields

r = \frac{m v}{q B}

.

The magnetic force provides the required inward centripetal acceleration for circular motion.

4

Identify the expression for torque on a magnetic dipole / coil

The couple produced by forces on opposite sides of a rectangular loop gives

\tau = B I A N \sin \theta

.

Opposite sides experience forces in opposing directions separated by a moment arm.

Key Concept

Formulas for magnetic forces on current-carrying conductors, moving charges, parallel wires, and coils

Question 12771Question →

A sample of a pure iron oxide synthesized in a laboratory contains $5.60\text{ g}$ of iron and $2.40\text{ g}$ of oxygen. According to the Law of Definite Proportions, what is the mass of iron, in grams, present in a $20.0\text{ g}$ sample of the same iron oxide collected from a natural deposit?

Show answer & explanation

Answer: 14

Answer

The mass of iron present in the 20.0 g sample of iron oxide is 14.0 g.

According to the Law of Definite Proportions, a pure chemical compound always contains the same elements combined together in the exact same proportion by mass, regardless of its source or method of preparation. In the laboratory sample,

8.00\text{ g}

of iron oxide contains

5.60\text{ g}

of iron, giving an iron mass composition of

70\%

. Therefore, a

20.0\text{ g}

natural sample of the same compound must also contain

70\%

iron by mass, which equals

14.0\text{ g}

.

Step-by-Step Solution

1

Calculate total mass of the laboratory sample

5.60\text{ g} + 2.40\text{ g} = 8.00\text{ g}

The total mass of the compound is the sum of the constituent element masses (Law of Conservation of Mass).

2

Find the mass percentage/fraction of iron

\frac{5.60\text{ g}}{8.00\text{ g}} = 0.70

(or

70\%

)

By the Law of Definite Proportions, the mass ratio of elements in a pure compound is constant.

3

Calculate mass of iron in the 20.0 g natural sample

0.70 \times 20.0\text{ g} = 14.0\text{ g}

Applying the constant mass composition percentage to the new sample mass.

Key Concept

Law of Definite Proportions (Constant Composition)

Estimated Time:1m 30s

Question 12772Question →

When solid ammonium chloride ( $\text{NH}_4\text{Cl}$ ) is dissolved in water inside a beaker, the temperature of the reaction mixture drops noticeably from $27^\circ\text{C}$ to $18^\circ\text{C}$ . Which of the following statements correctly accounts for the thermochemical behavior of this process?

The process is endothermic because thermal energy is absorbed from the surroundings, resulting in a positive enthalpy change (

\Delta H > 0

).

The process is exothermic because thermal energy is absorbed from the surroundings, resulting in a negative enthalpy change (

\Delta H < 0

).

The process is endothermic because thermal energy is released to the surroundings, resulting in a negative enthalpy change (

\Delta H < 0

).

The process is exothermic because thermal energy is released to the surroundings, resulting in a positive enthalpy change (

\Delta H > 0

).

Show answer & explanation

Answer: The process is endothermic because thermal energy is absorbed from the surroundings, resulting in a positive enthalpy change ( $\Delta H > 0$ ).

Answer

The dissolution process is endothermic because thermal energy is absorbed from the surroundings, resulting in a positive enthalpy change (

\Delta H > 0

).

The correct option identifies that a observed temperature drop in the surrounding solution signifies heat absorption by the chemical system. By definition, processes absorbing heat from their surroundings are endothermic and have a positive enthalpy change (

\Delta H > 0

).

Step-by-Step Solution

1

Analyze the temperature change observed during the process.

The temperature of the solution decreased from

27^\circ\text{C}

to

18^\circ\text{C}

, indicating heat was absorbed by the system from the surrounding solution.

When a reaction system absorbs heat from its immediate surroundings, the temperature of the surroundings falls.

2

Classify the type of thermochemical process.

A process that absorbs thermal energy from the surroundings is classified as endothermic.

Endothermic processes require heat input to break solute-solute and solvent-solvent interactions.

3

Determine the sign of the enthalpy change (

\Delta H

).

Since

H_{\text{products}} > H_{\text{reactants}}

,

\Delta H = H_{\text{products}} - H_{\text{reactants}} > 0

(positive).

In endothermic reactions, the heat content (enthalpy) of the products is greater than that of the reactants.

Key Concept

Endothermic Dissolution and Enthalpy Sign Convention

Estimated Time:1m 0s

Question 12773Question →

An object is placed $15.0\text{ cm}$ in front of a thin diverging lens with a focal length of $10.0\text{ cm}$ . What is the image distance formed by the lens?

-6.0\text{ cm}

(

6.0\text{ cm}

on the same side as the object)

+30.0\text{ cm}

(

30.0\text{ cm}

on the opposite side of the object)

+6.0\text{ cm}

(

6.0\text{ cm}

on the opposite side of the object)

-30.0\text{ cm}

(

30.0\text{ cm}

on the same side as the object)

Show answer & explanation

Answer: $-6.0\text{ cm}$ ( $6.0\text{ cm}$ on the same side as the object)

Answer

The image distance is

-6.0\text{ cm}

, indicating a virtual image located

6.0\text{ cm}

in front of the lens on the same side as the object.

Using the thin lens equation

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

with

u = +15.0\text{ cm}

and

f = -10.0\text{ cm}

for the diverging lens gives

\frac{1}{v} = -\frac{1}{10} - \frac{1}{15} = -\frac{1}{6}

, resulting in

v = -6.0\text{ cm}

. The negative sign confirms the image is virtual and formed on the same side as the object.

Step-by-Step Solution

1

Identify the given optical parameters and apply the proper sign convention.

Object distance

u = +15.0\text{ cm}

; Focal length for a diverging lens

f = -10.0\text{ cm}

.

By the real-is-positive sign convention, real object distance

u

is positive, while the focal length

f

of a concave/diverging lens is strictly negative.

2

Set up the thin lens formula

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

and solve for

\frac{1}{v}

.

\frac{1}{-10.0} = \frac{1}{15.0} + \frac{1}{v} \implies \frac{1}{v} = -\frac{1}{10.0} - \frac{1}{15.0}

Isolating the reciprocal image distance term requires subtracting

\frac{1}{u}

from both sides.

3

Calculate the common denominator and evaluate

v

.

\frac{1}{v} = \frac{-3 - 2}{30.0} = -\frac{5.0}{30.0} = -\frac{1}{6.0} \implies v = -6.0\text{ cm}

Inverting the reciprocal yields the final signed image distance.

Key Concept

Thin Lens Formula and Sign Conventions for Diverging Lenses

Estimated Time:1m 15s

Question 12774Question →

A stopwatch with a zero error of $-0.25\text{ s}$ is used to record the time taken for an object to travel down an inclined path. If the stopwatch displays a time reading of $14.65\text{ s}$ at the end of the motion, what is the true time elapsed in seconds?

Show answer & explanation

Answer: 14.9

Answer

The true time elapsed is 14.90 seconds.

The correct answer is obtained by subtracting the zero error from the observed reading. Since the stopwatch has a negative zero error of

-0.25\text{ s}

, the true time is

14.65\text{ s} - (-0.25\text{ s}) = 14.90\text{ s}

.

Step-by-Step Solution

1

Apply the zero error correction formula for measurement instruments

Actual Time = Displayed Time - Zero Error

Zero error represents a constant bias on the instrument that must be subtracted from the uncorrected reading.

2

Substitute the given values into the formula

Actual Time = 14.65 s - (-0.25 s)

The instrument starts behind zero by 0.25 s, so the zero error value is negative.

3

Perform the calculation

Actual Time = 14.90 s

Subtracting a negative quantity is mathematically equivalent to adding its positive magnitude.

Key Concept

Zero Error Correction in Stopwatch Time Measurement

Question 12775Question →

An object undergoing simple harmonic motion completes $20$ complete oscillations in a time duration of $10.0\text{ s}$ . What is the period of oscillation of the object in seconds?

Show answer & explanation

Answer: 0.5

Answer

The period of oscillation is

0.5\text{ s}

.

The period of simple harmonic motion is the time taken to complete one single oscillation. Dividing the total time (

10.0\text{ s}

) by the number of oscillations (

20

) yields

0.5\text{ s}

.

Step-by-Step Solution

1

Apply the definition of oscillation period

T = \frac{t}{N}

Period

T

measures the time required for a single complete cycle.

2

Calculate the numerical value for period

T = \frac{10.0\text{ s}}{20} = 0.5\text{ s}

Dividing total elapsed time by total completed oscillations gives time per oscillation.

Key Concept

Period of Simple Harmonic Motion

Question 12776Question →

How much energy, in $\text{MeV}$ , is released when a nuclear fission process results in a mass defect of $0.05 \text{ u}$ ? (Take $1 \text{ u} = 931.5 \text{ MeV}$ )

Show answer & explanation

Answer: 46.575

Answer

46.575 MeV

The total energy released in a nuclear fission reaction is found by multiplying the mass defect by the energy equivalent of 1 atomic mass unit. Multiplying

0.05 \text{ u}

by

931.5 \text{ MeV/u}

gives

46.575 \text{ MeV}

.

Step-by-Step Solution

1

Identify the mass defect and the mass-to-energy conversion factor.

Mass defect

\Delta m = 0.05 \text{ u}

, and

1 \text{ u} = 931.5 \text{ MeV}

.

In nuclear fission, mass lost during the reaction is converted directly into energy according to Einstein's mass-energy equivalence principle.

2

Calculate the total energy released.

E = 0.05 \text{ u} \times 931.5 \text{ MeV/u} = 46.575 \text{ MeV}

.

Multiplying the mass defect in atomic mass units by

931.5 \text{ MeV/u}

gives the total released energy in

\text{MeV}

.

Key Concept

Mass-Energy Conversion in Nuclear Reactions

Question 12777Question →

A solid cylindrical metal rod of length $0.50\text{ m}$ and cross-sectional area $4.0 \times 10^{-4}\text{ m}^2$ is perfectly insulated along its lateral surface. One end of the rod is maintained at $100^\circ\text{C}$ in boiling water, while the other end is in contact with an ice block at $0^\circ\text{C}$ . If $0.072\text{ kg}$ of ice melts in $10\text{ minutes}$ due to thermal energy conducted through the rod, what is the thermal conductivity of the metal? (Take the specific latent heat of fusion of ice as $3.36 \times 10^5\text{ J}\cdot\text{kg}^{-1}$ )

630\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}

504\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}

7000\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}

3.23\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}

Show answer & explanation

Answer: $504\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}$

Answer

The thermal conductivity of the metal rod is

504\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}

.

The correct answer is derived by first finding the total heat absorbed during the phase change of ice using

Q = m L = 0.072 \times 3.36 \times 10^5 = 24,192\text{ J}

. Dividing by the time in seconds (

600\text{ s}

) gives a heat flow rate of

40.32\text{ W}

. Substituting this into the conduction formula

\frac{Q}{t} = \frac{k A \Delta T}{d}

gives

40.32 = \frac{k (4.0 \times 10^{-4})(100)}{0.50} = 0.08 k

, yielding

k = 504\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}

.

Step-by-Step Solution

1

Calculate the total heat energy

Q

required to melt

0.072\text{ kg}

of ice at

0^\circ\text{C}

.

Q = m L = 0.072\text{ kg} \times 3.36 \times 10^5\text{ J}\cdot\text{kg}^{-1} = 24,192\text{ J}

During melting at constant temperature, heat transfer is governed by the latent heat of fusion formula.

2

Convert the elapsed time into seconds and calculate the rate of heat transfer

\frac{Q}{t}

.

t = 10\text{ min} = 600\text{ s}

;

\frac{Q}{t} = \frac{24,192\text{ J}}{600\text{ s}} = 40.32\text{ W}

Thermal conductivity formulas require rate of heat transfer in Joules per second (Watts).

3

Apply Fourier's law of thermal conduction

\frac{Q}{t} = \frac{k A (T_1 - T_2)}{d}

to solve for thermal conductivity

k

.

40.32 = \frac{k \times (4.0 \times 10^{-4}) \times (100 - 0)}{0.50} \implies 40.32 = 0.08 k \implies k = 504\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}

Rearranging the steady-state thermal conduction equation yields

k = \frac{(Q/t) \cdot d}{A \cdot \Delta T}

.

Key Concept

Thermal Conduction Rate and Latent Heat of Fusion

Question 12778Question →

A dip needle placed within the magnetic meridian at a location on Earth's surface records an angle of dip of $60^\circ$ . If the vertical component of Earth's magnetic field at this location is $3.46 \times 10^{-5}\text{ T}$ , what is the horizontal component of Earth's magnetic field? (Take $\tan 60^\circ = 1.73$ , $\sin 60^\circ = 0.87$ , $\cos 60^\circ = 0.50$ )

2.0 \times 10^{-5}\text{ T}

6.0 \times 10^{-5}\text{ T}

1.73 \times 10^{-5}\text{ T}

4.0 \times 10^{-5}\text{ T}

Show answer & explanation

Answer: $2.0 \times 10^{-5}\text{ T}$

Answer

The horizontal component of Earth's magnetic field is

2.0 \times 10^{-5}\text{ T}

.

The horizontal component

B_h

and vertical component

B_v

of Earth's magnetic field are related by

\tan \theta = \frac{B_v}{B_h}

, where

\theta

is the inclination or dip angle. Substituting

B_v = 3.46 \times 10^{-5}\text{ T}

and

\tan 60^\circ = 1.73

yields

B_h = \frac{3.46 \times 10^{-5}}{1.73} = 2.0 \times 10^{-5}\text{ T}

.

Step-by-Step Solution

1

Identify the relationship between the vertical component (

B_v

), horizontal component (

B_h

), and dip angle (

\theta

).

\tan \theta = \frac{B_v}{B_h}

The angle of dip

\theta

is defined by the direction of Earth's total magnetic field relative to the horizontal plane.

2

Rearrange the equation to express

B_h

in terms of

B_v

and

\tan \theta

.

B_h = \frac{B_v}{\tan \theta}

We need to solve for the horizontal component

B_h

.

3

Substitute the given values

B_v = 3.46 \times 10^{-5}\text{ T}

and

\tan 60^\circ = 1.73

into the equation.

B_h = \frac{3.46 \times 10^{-5}}{1.73} = 2.0 \times 10^{-5}\text{ T}

Performing the numerical division yields the correct horizontal field intensity.

Key Concept

Resolution of Earth's Magnetic Field Components

Question 12779Question →

Match each kinetic theory concept or microscopic property of an ideal gas on the left with its corresponding mathematical expression or derivation result on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct matches pair the momentum change per collision with

2 m v_x

, the single-molecule average wall force with

\frac{m v_x^2}{L}

, the kinetic energy density with

\frac{3}{2} P

, and the root-mean-square speed with

\sqrt{\frac{3 R T}{M}}

.

Each kinetic theory quantity is derived directly from fundamental principles of mechanics applied to gas particles. Elastic collision with a wall yields a momentum reversal of magnitude

2 m v_x

. Taking the round-trip collision frequency over length

L

yields an average force of

\frac{m v_x^2}{L}

. Linking microscopic kinetic energy density to pressure gives

\frac{E_k}{V} = \frac{3}{2} P

, and linking pressure to the ideal gas law for one mole yields

v_{\text{rms}} = \sqrt{\frac{3 R T}{M}}

.

Step-by-Step Solution

1

Analyze momentum transfer during elastic collision of a molecule with a wall.

Initial momentum along the x-axis is

p_i = m v_x

and final momentum after elastic reflection is

p_f = -m v_x

. The change in momentum is

\Delta p_x = p_f - p_i = -2 m v_x

, which has a magnitude of

2 m v_x

.

Elastic collision conserves kinetic energy and reverses velocity direction perpendicular to the wall.

2

Calculate the time rate of momentum transfer to determine average force.

The round-trip distance between opposite walls separated by length

L

is

2L

, so the time between collisions with the same wall is

\Delta t = \frac{2L}{v_x}

. The average force is

F_x = \frac{\Delta p}{\Delta t} = \frac{2 m v_x}{2L / v_x} = \frac{m v_x^2}{L}

.

Newton's second law expresses force as the average rate of change of momentum.

3

Relate total translational kinetic energy density to gas pressure.

From kinetic theory, gas pressure is given by

P = \frac{1}{3} \frac{N m}{V} v_{\text{rms}}^2

. Since total kinetic energy

E_k = \frac{1}{2} N m v_{\text{rms}}^2

, we can express pressure as

P = \frac{2}{3} \left(\frac{E_k}{V}\right)

. Rearranging gives energy density

\frac{E_k}{V} = \frac{3}{2} P

.

Translational kinetic energy density is directly proportional to pressure with a factor of 3/2.

4

Derive the formula for root-mean-square velocity from macroscopic and microscopic gas equations.

Substitute density

\rho = \frac{M}{V}

(where

M

is molar mass) into

P = \frac{1}{3} \rho v_{\text{rms}}^2

, obtaining

P = \frac{M v_{\text{rms}}^2}{3 V}

. Since

P V = R T

for one mole of ideal gas,

R T = \frac{1}{3} M v_{\text{rms}}^2 \implies v_{\text{rms}} = \sqrt{\frac{3 R T}{M}}

.

Connects microscopic speed distribution parameter with thermodynamic temperature and molar mass.

Key Concept

Kinetic Theory of Matter and Pressure of Gases

Question 12780Question →

Match each physical modification of a vibrating string or air pipe system on the left with its corresponding effect on the system's frequency on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Quadrupling tension corresponds to increasing frequency by a factor of 2; quadrupling linear mass density corresponds to reducing frequency to half; doubling tension corresponds to increasing frequency by a factor of

\sqrt{2}

; transitioning a closed pipe from fundamental mode to first overtone corresponds to increasing frequency by a factor of 3.

Each physical modification correctly maps to its quantitative outcome based on wave mechanics: string frequency scales with

\sqrt{T}

and

1/\sqrt{\mu}

, while closed pipe overtones follow odd harmonic multipliers (

1, 3, 5, \dots

).

Step-by-Step Solution

1

Examine the fundamental frequency formula for a stretched string under tension:

f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

.

Frequency is directly proportional to

\sqrt{T}

and inversely proportional to

\sqrt{\mu}

.

This establishes how changes in tension and mass per unit length scale the fundamental frequency.

2

Calculate scaling factors for the string modifications.

Quadrupling

T

multiplies frequency by

\sqrt{4} = 2

. Quadrupling

\mu

multiplies frequency by

1/\sqrt{4} = 0.5

. Doubling

T

multiplies frequency by

\sqrt{2}

.

Applying square roots to the parameter change factors gives the resultant frequency change.

3

Analyze harmonic ratios for air columns in pipes closed at one end.

The fundamental mode frequency is

f_1 = \frac{v}{4L}

. The first overtone is the third harmonic (

f_3 = \frac{3v}{4L} = 3f_1

).

Closed air columns produce only odd harmonics (

n = 1, 3, 5, \dots

).

Key Concept

Parameter scaling of transverse waves on stretched strings and harmonic modes in closed air columns

Answer

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