All practice questions

Question 7841Question →

The mean mass of $6$ packages in a delivery van is $20\text{ kg}$ . If one package weighing $35\text{ kg}$ is unloaded from the van, what is the mean mass of the remaining $5$ packages?

17\text{ kg}

23\text{ kg}

31\text{ kg}

14\text{ kg}

Show answer & explanation

Answer: $17\text{ kg}$

Answer

The mean mass of the remaining packages is

17\text{ kg}

.

The total weight of the initial

6

packages is

6 \times 20\text{ kg} = 120\text{ kg}

. Unloading one package of

35\text{ kg}

leaves

120 - 35 = 85\text{ kg}

. Dividing this remaining mass by the remaining

5

packages yields

\frac{85}{5} = 17\text{ kg}

.

Step-by-Step Solution

1

Calculate the initial total mass of the 6 packages

\text{Total mass} = 6 \times 20 = 120\text{ kg}

The sum of a set of numbers is equal to the product of their count and their mean.

2

Calculate the total mass after unloading the 35 kg package

\text{New total mass} = 120 - 35 = 85\text{ kg}

Removing a package decreases the total combined mass by its individual weight.

3

Calculate the new mean mass for the remaining 5 packages

\text{New mean} = \frac{85}{5} = 17\text{ kg}

The mean of ungrouped data is found by dividing the new sum by the updated count of items (

6 - 1 = 5

).

Key Concept

Calculation of the mean of ungrouped data after removing an item

Question 7842Question →

A gas cylinder fitted with a pressure relief valve contains a fixed mass of an ideal gas at an initial absolute pressure of $2.50 \times 10^5\text{ Pa}$ and a temperature of $27^\circ\text{C}$ . The valve is designed to open when the internal gauge pressure exceeds $3.20 \times 10^5\text{ Pa}$ . Assuming the atmospheric pressure is $1.00 \times 10^5\text{ Pa}$ , at what temperature will the relief valve open?

231^\circ\text{C}

45.4^\circ\text{C}

111^\circ\text{C}

34.6^\circ\text{C}

Show answer & explanation

Answer: $231^\circ\text{C}$

Answer

231^\circ\text{C}

The correct answer is

231^\circ\text{C}

. To solve this, first convert the initial temperature to Kelvin:

T_1 = 27 + 273 = 300\text{ K}

. Next, calculate the final absolute pressure by adding the atmospheric pressure to the given gauge pressure:

P_2 = 3.20 \times 10^5\text{ Pa} + 1.00 \times 10^5\text{ Pa} = 4.20 \times 10^5\text{ Pa}

. Applying Gay-Lussac's Law (

\frac{P_1}{T_1} = \frac{P_2}{T_2}

) gives

T_2 = 300 \times \frac{4.20 \times 10^5}{2.50 \times 10^5} = 504\text{ K}

. Converting back to Celsius gives

504 - 273 = 231^\circ\text{C}

.

Step-by-Step Solution

1

Convert the initial temperature from Celsius to Kelvin and determine the initial absolute pressure.

T_1 = 27 + 273 = 300\text{ K}

,

P_1 = 2.50 \times 10^5\text{ Pa}

.

Gas laws require thermodynamic (absolute) temperature in Kelvin.

2

Calculate the final total absolute pressure (

P_2

) at which the valve opens.

P_2 = P_{\text{gauge}} + P_{\text{atm}} = 3.20 \times 10^5\text{ Pa} + 1.00 \times 10^5\text{ Pa} = 4.20 \times 10^5\text{ Pa}

.

Gauge pressure measures the pressure difference relative to atmospheric pressure; gas laws use absolute pressure.

3

Apply Gay-Lussac's (Pressure) Law at constant volume to find the final Kelvin temperature (

T_2

).

\frac{P_1}{T_1} = \frac{P_2}{T_2} \implies T_2 = 300 \times \frac{4.20 \times 10^5}{2.50 \times 10^5} = 504\text{ K}

.

For a fixed volume of ideal gas, absolute pressure is directly proportional to absolute temperature.

4

Convert the final temperature back to degrees Celsius.

t_2 = 504 - 273 = 231^\circ\text{C}

.

The question asks for the temperature in degrees Celsius.

Key Concept

Pressure Law (Gay-Lussac's Law) and Absolute vs Gauge Pressure

Estimated Time:2m 0s

Question 7843Question →

An autonomous drone navigating an obstacle course experiences three mutually perpendicular velocity vectors simultaneously: a horizontal forward velocity of $12\text{ m s}^{-1}$ due East, a horizontal crosswind drift velocity of $9\text{ m s}^{-1}$ due North, and a vertical downdraft velocity of $8\text{ m s}^{-1}$ directed straight downward. What is the magnitude of the resultant velocity vector of the drone in $\text{m s}^{-1}$ ?

Show answer & explanation

Answer: 17

Answer

17 m s⁻¹

Because the three velocity components are mutually perpendicular, the magnitude of the overall resultant velocity is calculated using the 3D Pythagorean theorem:

\sqrt{12^2 + 9^2 + 8^2} = \sqrt{144 + 81 + 64} = \sqrt{289} = 17\text{ m s}^{-1}

.

Step-by-Step Solution

1

Identify the perpendicular vector components

v_x = 12\text{ m s}^{-1}

,

v_y = 9\text{ m s}^{-1}

,

v_z = 8\text{ m s}^{-1}

The three given velocity vectors act along mutually orthogonal spatial axes (East, North, and Downward).

2

Apply the 3D vector resultant magnitude formula

v_r = \sqrt{v_x^2 + v_y^2 + v_z^2}

Since the vector components are perpendicular to one another, the magnitude of their resultant is given by the extension of the Pythagorean theorem to three dimensions.

3

Substitute values and evaluate

v_r = \sqrt{144 + 81 + 64} = \sqrt{289} = 17\text{ m s}^{-1}

Squaring each component, adding them, and taking the principal square root yields the total magnitude of the velocity vector.

Key Concept

Magnitude of three mutually perpendicular vector components using the 3D Pythagorean theorem

Question 7844Question →

In discharge tube experiments, electrical conduction in gases transitions through distinct physical regimes as the internal gas pressure is progressively reduced. Match each discharge phenomenon with its corresponding physical cause or operational pressure condition.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings are: (1) Formation of the luminous positive column matches continuous de-excitation and radiative recombination of gas atoms at pressures around

1.0\text{ mmHg}

; (2) Appearance and expansion of the Crookes dark space matches increase in the electron mean free path at pressures around

0.01\text{ mmHg}

; (3) Emission of high-velocity cathode rays matches bombardment of the cathode by energetic positive ions releasing secondary electrons below

0.01\text{ mmHg}

; and (4) Complete cessation of electric current flow matches extensive evacuation below

10^{-4}\text{ mmHg}

leaving insufficient gas molecules for ionization.

Conduction through gases relies heavily on pressure. At moderate low pressure (

1.0\text{ mmHg}

), excited gas atoms emit light forming the positive column. Decreasing pressure to

0.01\text{ mmHg}

increases the electron mean free path to produce the Crookes dark space and generates energetic cathode rays through ion bombardment. Extreme evacuation (

< 10^{-4}\text{ mmHg}

) removes all gas charge carriers, halting electric conduction.

Step-by-Step Solution

1

Analyze the pressure regime of

1.0\text{ mmHg}

in a discharge tube.

Identified the positive column as the main luminous region filling most of the tube due to atom excitation and light emission.

At this pressure, gas density is sufficient to undergo repeated inelastic collisions that excite atoms and produce visible glow.

2

Examine the physical origin of the Crookes dark space at

0.01\text{ mmHg}

.

Understood that lower gas density increases electron mean free path.

Electrons near the cathode travel a longer distance before hitting gas particles, creating a dark gap where collisions do not occur.

3

Determine how cathode rays are emitted at very low pressures.

Linked cathode ray emission to positive ion impact on the cathode surface.

High electric fields accelerate residual positive ions to strike the cathode, causing secondary electron emission.

4

Evaluate the extreme vacuum limit below

10^{-4}\text{ mmHg}

.

Concluded that current stops when gas particles are virtually absent.

Gases conduct electricity via ion and electron production from collision ionization; eliminating gas molecules prevents charge transport.

Key Concept

Pressure-dependent regimes of gas conduction and cathode ray generation

Estimated Time:2m 0s

Question 7845Question →

What is the simplified form of the expression $\frac{3\sqrt{2} + 2\sqrt{3}}{\sqrt{3} - \sqrt{2}}$ ?

12 + 5\sqrt{6}

5\sqrt{15} + 5\sqrt{10}

12 - 5\sqrt{6}

\sqrt{6}

Show answer & explanation

Answer: $12 + 5\sqrt{6}$

Answer

The simplified form is

12 + 5\sqrt{6}

.

Multiplying both numerator and denominator by the conjugate of the denominator,

(\sqrt{3} + \sqrt{2})

, clears the square roots in the denominator (resulting in

3 - 2 = 1

). Expanding the numerator gives

3\sqrt{6} + 6 + 6 + 2\sqrt{6}

, which simplifies cleanly to

12 + 5\sqrt{6}

.

Step-by-Step Solution

1

Identify the conjugate of the denominator

The conjugate of

(\sqrt{3} - \sqrt{2})

is

(\sqrt{3} + \sqrt{2})

.

To rationalize a binomial denominator of the form

(\sqrt{a} - \sqrt{b})

, multiply numerator and denominator by

(\sqrt{a} + \sqrt{b})

.

2

Multiply the numerator and denominator by the conjugate

\frac{(3\sqrt{2} + 2\sqrt{3})(\sqrt{3} + \sqrt{2})}{(\sqrt{3} - \sqrt{2})(\sqrt{3} + \sqrt{2})}

This removes the radical terms from the denominator using the difference of squares identity.

3

Expand the numerator and simplify the denominator

Denominator:

(\sqrt{3})^2 - (\sqrt{2})^2 = 3 - 2 = 1

.
Numerator:

3\sqrt{2}\cdot\sqrt{3} + 3\sqrt{2}\cdot\sqrt{2} + 2\sqrt{3}\cdot\sqrt{3} + 2\sqrt{3}\cdot\sqrt{2} = 3\sqrt{6} + 6 + 6 + 2\sqrt{6}

.

Apply the distributive property and basic radical simplification rules.

4

Combine like terms in the numerator

(6 + 6) + (3\sqrt{6} + 2\sqrt{6}) = 12 + 5\sqrt{6}

.

Collect rational numbers together and like surd terms together.

Key Concept

Rationalization of Denominators with Binomial Surds

Question 7846Question →

If $x$ and $y$ are real numbers satisfying the simultaneous equations $x - y = 4$ and $x^2 + y^2 = 26$ , what is the value of the product $xy$ ?

Show answer & explanation

Answer: 5

Answer

The value of the product

xy

is 5.

Expressing

x

from the linear equation yields

x = y + 4

. Substituting this into

x^2 + y^2 = 26

gives

(y + 4)^2 + y^2 = 26

, which expands and simplifies to

y^2 + 4y - 5 = 0

. Factoring gives solutions

y = 1

(with

x = 5

) and

y = -5

(with

x = -1

). Both solution pairs

(5, 1)

and

(-1, -5)

result in the product

xy = 5

.

Step-by-Step Solution

1

Express

x

in terms of

y

using the linear equation

x = y + 4

Isolate

x

to substitute into the second equation.

2

Substitute

x = y + 4

into the quadratic equation

x^2 + y^2 = 26

(y + 4)^2 + y^2 = 26

Reduce the system to a single quadratic equation in

y

.

3

Expand and simplify into standard quadratic form

y^2 + 4y - 5 = 0

Expanding gives

2y^2 + 8y + 16 = 26

, which simplifies by subtracting 26 and dividing by 2.

4

Solve for

y

by factoring

y = 1

or

y = -5

The factors of

y^2 + 4y - 5

are

(y + 5)(y - 1) = 0

.

5

Compute corresponding

x

values and the product

xy

For

y = 1

,

x = 5 \implies xy = 5

; for

y = -5

,

x = -1 \implies xy = 5

Substitute each

y

back into

x = y + 4

and evaluate

xy

.

Key Concept

Simultaneous Linear and Quadratic Equations

Question 7847Question →

A cannonball is fired from level ground with an initial speed of $20\text{ m/s}$ at an angle of $30^\circ$ above the horizontal. Taking the acceleration due to gravity $g = 10\text{ m/s}^2$ , what is the total time of flight of the cannonball before it returns to ground level?

2.0\text{ s}

1.0\text{ s}

4.0\text{ s}

0.5\text{ s}

Show answer & explanation

Answer: $2.0\text{ s}$

Answer

2.0\text{ s}

The correct answer of

2.0\text{ s}

is determined by resolving the initial speed into its vertical component

u_y = 20 \sin 30^\circ = 10\text{ m/s}

and using the total time of flight formula

T = \frac{2 u \sin \theta}{g} = \frac{2(10)}{10} = 2.0\text{ s}

.

Step-by-Step Solution

1

Calculate the vertical component of the initial velocity (

u_y

).

u_y = u \sin \theta = 20 \times \sin 30^\circ = 20 \times 0.5 = 10\text{ m/s}

Only the vertical component of velocity determines the time the projectile remains in the air.

2

Calculate the total time of flight (

T

) using the kinematic formula for full trajectory.

T = \frac{2 u_y}{g} = \frac{2 \times 10}{10} = 2.0\text{ s}

The total time of flight includes both the time to ascend to peak height and descend back to the launch height under gravity

g

.

Key Concept

Total Time of Flight in Projectile Motion

Question 7848Question →

The table below shows the distribution of examination scores of $50$ candidates in a selection test:

Score Class	Frequency ( $f$ )
$10 - 19$	$5$
$20 - 29$	$9$
$30 - 39$	$16$
$40 - 49$	$14$
$50 - 59$	$6$

Using the cumulative frequency distribution (ogive), what is the estimated $70^{\text{th}}$ percentile score?

33.1

43.1

43.6

38.1

Show answer & explanation

Answer: $43.1$

Answer

The estimated

70^{\text{th}}

percentile score is

43.1

.

The

70^{\text{th}}

percentile corresponds to the score at the

35^{\text{th}}

candidate (

70\%

of

50

). This falls within the

40 - 49

score class, which has a lower boundary of

39.5

, a frequency of

14

, and a class width of

10

. Interpolating linearly gives

39.5 + \left(\frac{35 - 30}{14}\right) \times 10 = 43.07 \approx 43.1

.

Step-by-Step Solution

1

Construct the cumulative frequency table to find class boundaries and cumulative frequencies.

Cumulative frequencies:

10-19

(

CF = 5

),

20-29

(

CF = 14

),

30-39

(

CF = 30

),

40-49

(

CF = 44

),

50-59

(

CF = 50

).

Cumulative frequencies are required to locate the position of the desired percentile.

2

Determine the rank position of the

70^{\text{th}}

percentile (

P_{70}

).

Rank position =

\frac{70}{100} \times 50 = 35^{\text{th}}

position.

The

70^{\text{th}}

percentile corresponds to the score below which

70\%

of the total candidates fall.

3

Identify the class interval containing the

35^{\text{th}}

cumulative frequency and state its parameters.

The target class is

40 - 49

. Parameters: Lower class boundary

L = 39.5

, preceding cumulative frequency

CF_{\text{prev}} = 30

, class frequency

f = 14

, class width

c = 10

.

Since

30 < 35 \le 44

, the

35^{\text{th}}

entry falls within the

40 - 49

class.

4

Apply the linear interpolation formula for percentiles from an ogive.

P_{70} = L + \left(\frac{\frac{70N}{100} - CF_{\text{prev}}}{f}\right) \times c = 39.5 + \left(\frac{35 - 30}{14}\right) \times 10 = 39.5 + \frac{50}{14} \approx 43.07 \approx 43.1

.

This calculates the exact score estimate on the cumulative frequency curve.

Key Concept

Estimating Percentiles from Grouped Data and Ogives

Question 7849Question →

An electron inside an excited atom drops from an energy state of $-2.10\text{ eV}$ to a lower energy state of $-4.575\text{ eV}$ . What is the wavelength of the emitted photon in nanometers ( $\text{nm}$ )? (Take Planck's constant $h = 6.6 \times 10^{-34}\text{ J s}$ , speed of light $c = 3.0 \times 10^8\text{ m/s}$ , and $1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$ ).

Show answer & explanation

Answer: 500

Answer

The wavelength of the emitted photon is

500\text{ nm}

.

The energy of the emitted photon is calculated from the energy change of the electron,

\Delta E = -2.10\text{ eV} - (-4.575\text{ eV}) = 2.475\text{ eV}

. Converting this to Joules gives

2.475 \times 1.6 \times 10^{-19}\text{ J} = 3.96 \times 10^{-19}\text{ J}

. Substituting into the wavelength formula

\lambda = \frac{hc}{\Delta E} = \frac{6.6 \times 10^{-34} \times 3.0 \times 10^8}{3.96 \times 10^{-19}} = 5.0 \times 10^{-7}\text{ m} = 500\text{ nm}

.

Step-by-Step Solution

1

Calculate the energy difference between the initial excited state and final state

\Delta E = -2.10\text{ eV} - (-4.575\text{ eV}) = 2.475\text{ eV}

The energy of the emitted photon equals the difference between the two energy levels.

2

Convert the photon energy from electron-volts (eV) to Joules (J)

\Delta E = 2.475 \times 1.6 \times 10^{-19}\text{ J} = 3.96 \times 10^{-19}\text{ J}

Standard SI units must be used to calculate wavelength in meters.

3

Apply the Planck-Einstein relation

\lambda = \frac{hc}{\Delta E}

to calculate the wavelength in meters and convert to nanometers

\lambda = \frac{(6.6 \times 10^{-34}\text{ J s})(3.0 \times 10^8\text{ m/s})}{3.96 \times 10^{-19}\text{ J}} = 5.0 \times 10^{-7}\text{ m} = 500\text{ nm}

Converting meters to nanometers requires multiplying by

10^9

.

Key Concept

Photon emission wavelength during atomic energy level transitions

Estimated Time:1m 30s

Question 7850Question →

Given the universal set $U = \{x \in \mathbb{Z} : 1 \le x \le 20\}$ , with subsets $P = \{x \in U : x \text{ is a prime number}\}$ and $Q = \{x \in U : x \text{ is an odd integer}\}$ , what is the cardinal number of $(P \cup Q)'$ ?

9

11

10

8

Show answer & explanation

Answer: $9$

Answer

9

The universal set contains

20

elements. The union

P \cup Q

consists of all odd numbers and prime numbers between

1

and

20

, giving

11

unique elements:

\{1, 2, 3, 5, 7, 9, 11, 13, 15, 17, 19\}

. Subtracting these

11

elements from the total

20

elements in

U

gives

n((P \cup Q)') = 20 - 11 = 9

.

Step-by-Step Solution

1

Identify the elements of the universal set

U

and subsets

P

and

Q

.

U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20\}

, so

n(U) = 20

.

P = \{2, 3, 5, 7, 11, 13, 17, 19\}

Q = \{1, 3, 5, 7, 9, 11, 13, 15, 17, 19\}

Explicitly listing the set elements helps correctly calculate the union.

2

Determine the union

P \cup Q

.

P \cup Q = \{1, 2, 3, 5, 7, 9, 11, 13, 15, 17, 19\}

, which contains

11

elements.

The union combines all unique elements present in set

P

, set

Q

, or both.

3

Calculate the complement set

(P \cup Q)'

and its cardinality.

(P \cup Q)' = U \setminus (P \cup Q) = \{4, 6, 8, 10, 12, 14, 16, 18, 20\}

, so

n((P \cup Q)') = 20 - 11 = 9

.

The complement of a set contains all elements in the universal set

U

that are not in the given set.

Key Concept

Set Operations and Complement of Sets

Estimated Time:1m 0s

Question 7851Question →

A micrometer screw gauge with a pitch of $0.5\text{ mm}$ and $50$ divisions on its circular thimble scale has a negative zero error such that when the jaws are fully closed, the $46\text{th}$ division on the thimble scale aligns with the main scale datum line. When this instrument is used to measure the total thickness of a stack of $40$ identical sheets of paper, the main scale reads $10.0\text{ mm}$ and the thimble scale reads $16$ divisions. What is the actual thickness of a single sheet of paper?

0.253\text{ mm}

0.255\text{ mm}

0.254\text{ mm}

0.266\text{ mm}

Show answer & explanation

Answer: $0.255\text{ mm}$

Answer

The actual thickness of a single sheet of paper is

0.255\text{ mm}

.

The least count of the micrometer is

\frac{0.5\text{ mm}}{50} = 0.01\text{ mm}

. Because the 46th division coincides with the datum line upon closure, the zero error is negative:

-(50 - 46) \times 0.01\text{ mm} = -0.04\text{ mm}

. The observed reading is

10.0\text{ mm} + 0.16\text{ mm} = 10.16\text{ mm}

. Correcting for zero error gives a true total thickness of

10.16\text{ mm} - (-0.04\text{ mm}) = 10.20\text{ mm}

. Dividing this total thickness by 40 sheets gives an individual sheet thickness of

0.255\text{ mm}

.

Step-by-Step Solution

1

Determine the least count (LC) of the micrometer screw gauge.

\text{LC} = \frac{\text{Pitch}}{\text{Number of thimble divisions}} = \frac{0.5\text{ mm}}{50} = 0.01\text{ mm}

.

The least count is the minimum distance measurable by one circular scale division.

2

Calculate the zero error of the instrument.

\text{Zero error} = -(50 - 46) \times 0.01\text{ mm} = -0.04\text{ mm}

.

Since the 46th division aligns with the datum line when jaws are closed, the zero mark lies 4 divisions below the line, indicating a negative zero error.

3

Find the observed reading for the stack of 40 sheets.

\text{Observed reading} = \text{Main scale} + (\text{Thimble reading} \times \text{LC}) = 10.0\text{ mm} + (16 \times 0.01\text{ mm}) = 10.16\text{ mm}

.

The total observed value combines the main scale reading and the fractional thimble scale reading.

4

Calculate the true (corrected) total thickness of the stack.

\text{True reading} = \text{Observed reading} - \text{Zero error} = 10.16\text{ mm} - (-0.04\text{ mm}) = 10.20\text{ mm}

.

Zero error correction requires subtracting the zero error (with sign) from the observed reading.

5

Calculate the thickness of a single sheet of paper.

\text{Thickness of one sheet} = \frac{10.20\text{ mm}}{40} = 0.255\text{ mm}

.

Dividing the total corrected thickness by the total number of sheets gives the thickness per sheet.

Key Concept

Measurement of length using micrometer screw gauge with negative zero error correction

Estimated Time:2m 0s

Question 7852Question →

A periodic water wave of frequency $5.0\text{ Hz}$ has a wavelength of $1.2\text{ m}$ in deep water. When the wave enters a shallow section of the ripple tank, its speed reduces to $4.0\text{ m s}^{-1}$ . What is the wavelength of the wave in the shallow section?

0.67\text{ m}

0.80\text{ m}

1.5\text{ m}

1.8\text{ m}

Show answer & explanation

Answer: $0.80\text{ m}$

Answer

0.80\text{ m}

When a wave travels from one medium to another (e.g., deep to shallow water), its frequency

f

remains constant because frequency is dependent only on the wave source. Using the wave equation

v = f\lambda

, the wavelength in the shallow water is calculated as

\lambda = \frac{v}{f} = \frac{4.0\text{ m s}^{-1}}{5.0\text{ Hz}} = 0.80\text{ m}

.

Step-by-Step Solution

1

Determine the invariant property across media boundaries

The frequency

f

remains constant at

5.0\text{ Hz}

when a wave passes from deep water to shallow water.

Frequency is determined solely by the source of the wave vibration, not the medium of propagation.

2

Calculate the wavelength in the shallow section using the wave equation

\lambda_2 = \frac{v_2}{f} = \frac{4.0\text{ m s}^{-1}}{5.0\text{ Hz}} = 0.80\text{ m}

Applying the relationship

v = f\lambda

for the second medium.

Key Concept

Frequency invariance of waves across boundaries and application of the wave equation

v = f\lambda

Question 7853Question →

X-rays are high-frequency electromagnetic waves that undergo deflection when passing through strong electric or magnetic fields.

True

False

Show answer & explanation

Answer: False

Answer

The statement is False because X-rays carry no electrical charge and therefore pass through electric and magnetic fields without deflection.

X-rays are electromagnetic radiation made of photons with zero electric charge. Because they carry no charge, they travel in straight lines and remain entirely unaffected by surrounding electric or magnetic fields.

Step-by-Step Solution

1

Identify the physical nature of X-rays.

X-rays are electromagnetic radiation (photons) of high frequency and short wavelength.

Electromagnetic waves travel at the speed of light and consist of transverse oscillating electric and magnetic fields, carrying energy but no electric charge.

2

Analyze the effect of electric and magnetic fields on uncharged radiation.

The electric force

F_E = qE

and magnetic force

F_B = qvB sin\theta

are both zero when

q = 0

.

Since photons have zero net electrical charge (

q = 0

), they experience no net deflecting force in external electric or magnetic fields.

Key Concept

Neutrality of X-rays in electromagnetic fields

Question 7854Question →

If $\theta$ is an acute angle such that $\sin \theta = \frac{5}{13}$ , what is the exact numerical value of $169(\sin^2 \theta - \cos^2 \theta)$ ?

Show answer & explanation

Answer: -119

Answer

The exact numerical value of the expression is

-119

.

Using the Pythagorean identity

\sin^2 \theta + \cos^2 \theta = 1

, we find

\cos^2 \theta = 1 - \frac{25}{169} = \frac{144}{169}

. Then

\sin^2 \theta - \cos^2 \theta = \frac{25 - 144}{169} = -\frac{119}{169}

. Multiplying by

169

gives

-119

.

Step-by-Step Solution

1

Calculate

\cos^2 \theta

using the Pythagorean trigonometric identity.

\cos^2 \theta = \frac{144}{169}

Since

\sin^2 \theta + \cos^2 \theta = 1

, subtracting

\sin^2 \theta = \frac{25}{169}

from

1

yields

\frac{144}{169}

.

2

Compute the difference

\sin^2 \theta - \cos^2 \theta

.

\sin^2 \theta - \cos^2 \theta = -\frac{119}{169}

Subtracting

\frac{144}{169}

from

\frac{25}{169}

gives

-\frac{119}{169}

.

3

Scale the difference by

169

.

169 \times \left(-\frac{119}{169}\right) = -119

Multiplying

-\frac{119}{169}

by

169

cancels the denominator, leaving

-119

.

Key Concept

Pythagorean Trigonometric Identity

Question 7855Question →

A girl stands at a specific distance from a flat vertical wall and claps her hands once. If she hears the echo $0.4\text{ s}$ later, what is her distance from the wall in meters? (Take the speed of sound in air as $340\text{ m/s}$ )

Show answer & explanation

Answer: 68

Answer

The distance of the girl from the wall is

68\text{ m}

.

An echo involves the sound traveling to the reflecting surface and returning to the source, covering a total distance of

2d

. Using

2d = v \times t

, we obtain

d = \frac{340 \times 0.4}{2} = 68\text{ m}

.

Step-by-Step Solution

1

Identify the given values from the problem statement.

Time for echo

t = 0.4\text{ s}

, speed of sound

v = 340\text{ m/s}

.

An echo is a reflected sound wave that travels to the wall and back, completing a round trip.

2

Apply the echo calculation formula.

d = \frac{v \times t}{2}

The total distance traveled by the sound is

2d

. Thus, the one-way distance

d

to the reflecting surface is half of the total distance.

3

Substitute the values and calculate the distance.

d = \frac{340 \times 0.4}{2} = 68\text{ m}

Multiplying the speed by half the elapsed time gives the distance to the wall.

Key Concept

Echo distance calculation

Question 7856Question →

Given the simultaneous equations $x - y = 1$ and $x^2 + y^2 = 25$ , where $x$ and $y$ are both positive real numbers, what is the value of $x + y$ ?

7

9

1

3

Show answer & explanation

Answer: $7$

Answer

The value of

x + y

is

7

.

Expressing

x

as

y + 1

and substituting into

x^2 + y^2 = 25

yields

2y^2 + 2y - 24 = 0

. Factoring gives

y = 3

(rejecting

y = -4

as

y > 0

). Substituting

y = 3

back gives

x = 4

. Adding these values together yields

x + y = 7

.

Step-by-Step Solution

1

Express

x

in terms of

y

using the linear equation.

x = y + 1

Rearranging

x - y = 1

allows substitution into the quadratic equation.

2

Substitute

x = y + 1

into the quadratic equation

x^2 + y^2 = 25

.

(y + 1)^2 + y^2 = 25 \implies y^2 + 2y + 1 + y^2 = 25 \implies 2y^2 + 2y - 24 = 0

This reduces the system to a single quadratic equation in terms of

y

.

3

Solve the quadratic equation for

y

.

y^2 + y - 12 = 0 \implies (y + 4)(y - 3) = 0 \implies y = 3

or

y = -4

Dividing by

2

simplifies the equation, and factoring gives the potential roots for

y

.

4

Select the positive value of

y

and calculate

x

and

x + y

.

Since

y > 0

,

y = 3

. Then

x = 3 + 1 = 4

, so

x + y = 4 + 3 = 7

.

The question specifies that both

x

and

y

are positive real numbers.

Key Concept

Solving Simultaneous Linear and Quadratic Equations by Substitution

Question 7857Question →

An electron beam traveling horizontally enters a region containing mutually perpendicular uniform electric and magnetic fields. The electric field intensity is $4.4 \times 10^3\text{ V m}^{-1}$ and the magnetic flux density is $5.0 \times 10^{-4}\text{ T}$ . If the beam passes through undeflected and subsequently enters a region containing only the magnetic field $B$ , what is the radius of the circular path executed by the electrons? (Take the specific charge of an electron $\frac{e}{m} = 1.76 \times 10^{11}\text{ C kg}^{-1}$ )

0.10\text{ m}

0.88\text{ m}

10.0\text{ m}

0.20\text{ m}

Show answer & explanation

Answer: $0.10\text{ m}$

Answer

The radius of the circular path executed by the electrons in the magnetic field is

0.10\text{ m}

.

Under velocity selection conditions, equal electric and magnetic forces (

eE = evB

) establish electron speed

v = \frac{E}{B} = 8.8 \times 10^6\text{ m s}^{-1}

. When moving solely through magnetic field

B

, magnetic force provides centripetal force (

evB = \frac{mv^2}{r}

), yielding orbital radius

r = \frac{v}{(e/m)B} = \frac{8.8 \times 10^6}{1.76 \times 10^{11} \times 5.0 \times 10^{-4}} = 0.10\text{ m}

.

Step-by-Step Solution

1

Calculate the electron velocity using the undeflected crossed fields condition (velocity selector)

v = \frac{E}{B} = \frac{4.4 \times 10^3\text{ V m}^{-1}}{5.0 \times 10^{-4}\text{ T}} = 8.8 \times 10^6\text{ m s}^{-1}

When electric and magnetic forces are equal and opposite (

eE = evB

), the electrons travel in a straight line undeflected.

2

Equate magnetic force to centripetal force in the region with magnetic field only

evB = \frac{mv^2}{r} \implies r = \frac{mv}{eB} = \frac{v}{\left(\frac{e}{m}\right)B}

The magnetic Lorentz force provides the necessary centripetal force for circular motion.

3

Substitute numerical values to find the radius

r

r = \frac{8.8 \times 10^6}{1.76 \times 10^{11} \times 5.0 \times 10^{-4}} = \frac{8.8 \times 10^6}{8.8 \times 10^7} = 0.10\text{ m}

Direct algebraic simplification yields the orbital radius in meters.

Key Concept

Deflection of cathode rays (electrons) in crossed electric and magnetic fields (velocity selector) and circular orbital dynamics in uniform magnetic fields.

Estimated Time:2m 0s

Question 7858Question →

The table below shows the distribution of weekly overtime hours worked by $10$ technicians in a manufacturing plant:

Overtime Hours ( $x$ )	2	4	6	8
Number of Technicians ( $f$ )	4	3	2	1

What is the variance of the overtime hours?

4

2

1.6

4.4

Show answer & explanation

Answer: 4

Answer

The variance of the overtime hours is

4

.

The correct answer is

4

. First, compute the mean

\bar{x} = \frac{40}{10} = 4

. Then compute the sum of weighted squared deviations

\sum f(x - \bar{x})^2 = 4(2-4)^2 + 3(4-4)^2 + 2(6-4)^2 + 1(8-4)^2 = 16 + 0 + 8 + 16 = 40

. Dividing this by total frequency

\sum f = 10

yields

\sigma^2 = \frac{40}{10} = 4

.

Step-by-Step Solution

1

Calculate the total frequency (

\sum f

) and the sum of the products of values and frequencies (

\sum fx

).

\sum f = 4 + 3 + 2 + 1 = 10

and

\sum fx = (2 \times 4) + (4 \times 3) + (6 \times 2) + (8 \times 1) = 8 + 12 + 12 + 8 = 40

.

These sums are required to compute the mean of the distribution.

2

Determine the mean (

\bar{x}

) of the distribution.

\bar{x} = \frac{\sum fx}{\sum f} = \frac{40}{10} = 4

.

The mean is used as the central reference point to find deviations.

3

Find the squared deviation

(x - \bar{x})^2

for each score and multiply by its respective frequency

f

.

For

x = 2

:

4(2 - 4)^2 = 16

.
For

x = 4

:

3(4 - 4)^2 = 0

.
For

x = 6

:

2(6 - 4)^2 = 8

.
For

x = 8

:

1(8 - 4)^2 = 16

.
Total sum

\sum f(x - \bar{x})^2 = 16 + 0 + 8 + 16 = 40

.

Variance evaluates the average of these weighted squared deviations.

4

Compute the variance (

\sigma^2

) by dividing the sum of weighted squared deviations by the total frequency.

\sigma^2 = \frac{\sum f(x - \bar{x})^2}{\sum f} = \frac{40}{10} = 4

.

This gives the population variance for the given frequency distribution.

Key Concept

Variance for Frequency Distributions

Estimated Time:1m 30s

Question 7859Question →

A trapezium has an area of $180\text{ cm}^2$ and a perpendicular height of $12\text{ cm}$ . If the lengths of its two parallel sides are in the ratio $2:3$ , calculate the length, in $\text{cm}$ , of the longer parallel side.

Show answer & explanation

Answer: 18

Answer

The length of the longer parallel side is

18\text{ cm}

.

By representing the parallel sides as

2x

and

3x

, the trapezium area formula

A = \frac{1}{2}(a + b)h

gives

180 = \frac{1}{2}(2x + 3x)(12) = 30x

. Solving for

x

yields

x = 6

. Therefore, the longer parallel side is

3(6) = 18\text{ cm}

.

Step-by-Step Solution

1

Express the parallel sides algebraically from the ratio

2:3

.

Let the shorter parallel side

a = 2x

and the longer parallel side

b = 3x

.

Using a common multiplier

x

preserves the given side ratio.

2

Substitute the expressions and known values into the trapezium area formula.

180 = \frac{1}{2}(2x + 3x) \times 12

The area

A

of a trapezium is given by

A = \frac{1}{2}(a + b)h

.

3

Solve for the variable

x

.

180 = 6 \times 5x \implies 30x = 180 \implies x = 6

Simplifying

\frac{1}{2} \times 12 = 6

and multiplying by

(2x + 3x) = 5x

.

4

Calculate the length of the longer parallel side.

Longer side

= 3x = 3 \times 6 = 18\text{ cm}

The longer side corresponds to the

3x

term in the ratio.

Key Concept

Area of a trapezium involving algebraic ratio problem solving

Question 7860Question →

A Vernier caliper with a least count of $0.01\text{ cm}$ is used to determine the outer diameter of a cylindrical tube. When the measuring jaws are brought together without any object between them, the zero mark of the vernier scale lies to the right of the main scale zero mark, with the $3\text{rd}$ vernier division coinciding with a main scale line. When measuring the tube, the main scale reads $4.20\text{ cm}$ and the $6\text{th}$ vernier division coincides with a main scale line. What is the actual outer diameter of the tube?

4.23\text{ cm}

4.29\text{ cm}

4.26\text{ cm}

4.17\text{ cm}

Show answer & explanation

Answer: $4.23\text{ cm}$

Answer

The actual outer diameter of the tube is

4.23\text{ cm}

.

The observed reading of the Vernier caliper is calculated as Main Scale Reading + (Coinciding Division × Least Count) =

4.20\text{ cm} + (6 \times 0.01\text{ cm}) = 4.26\text{ cm}

. Because the vernier zero lies to the right when closed, it has a positive zero error of

+0.03\text{ cm}

. Subtracting this zero error from the observed reading gives the true measurement of

4.23\text{ cm}

.

Step-by-Step Solution

1

Determine the zero error of the Vernier caliper

Zero Error

= + (3 \times 0.01\text{ cm}) = +0.03\text{ cm}

Since the zero of the vernier scale lies to the right of the main scale zero, the instrument has a positive zero error.

2

Calculate the observed reading

Observed Reading

= 4.20\text{ cm} + (6 \times 0.01\text{ cm}) = 4.26\text{ cm}

The observed measurement is the main scale reading plus the product of the coinciding vernier division and the least count.

3

Calculate the actual (corrected) reading

Actual Reading

= \text{Observed Reading} - \text{Zero Error} = 4.26\text{ cm} - (+0.03\text{ cm}) = 4.23\text{ cm}

Zero error must always be subtracted algebraically from the observed reading to obtain the accurate measurement.

Key Concept

Zero Error Correction in Vernier Calipers

Estimated Time:1m 30s

Answer

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