All practice questions

Given the universal set

\mathcal{E} = \{x \in \mathbb{Z} : 1 \le x \le 20\}

, and subsets:

A = \{x \in \mathcal{E} : x \text{ is a multiple of } 3\}

B = \{x \in \mathcal{E} : x \text{ is a multiple of } 4\}

C = \{x \in \mathcal{E} : x \text{ is a prime number}\}

What is the number of elements in the set $(A \cup B)' \cap C$ ?

7

8

5

10

Show answer & explanation

Answer: 7

Answer

7

The universal set contains integers from 1 to 20. Subset

A

has multiples of 3, subset

B

has multiples of 4, and subset

C

contains the prime numbers

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

. The set

A \cup B

contains all multiples of 3 or 4 within the range, namely

\{3, 4, 6, 8, 9, 12, 15, 16, 18, 20\}

. The complement

(A \cup B)'

consists of elements in the universal set not in

A \cup B

:

\{1, 2, 5, 7, 10, 11, 13, 14, 17, 19\}

. Taking the intersection of

(A \cup B)'

with

C

filters this list to only the prime numbers:

\{2, 5, 7, 11, 13, 17, 19\}

, giving a cardinality of 7.

Step-by-Step Solution

1

List the elements of the universal set and the given subsets.

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

A = \{3, 6, 9, 12, 15, 18\}

B = \{4, 8, 12, 16, 20\}

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

Explicit listing allows accurate evaluation of set operations.

2

Find the union

A \cup B

.

A \cup B = \{3, 4, 6, 8, 9, 12, 15, 16, 18, 20\}

The union combines all elements present in either

A

or

B

.

3

Determine the complement

(A \cup B)'

relative to

\mathcal{E}

.

(A \cup B)' = \{1, 2, 5, 7, 10, 11, 13, 14, 17, 19\}

The complement contains all elements of the universal set

\mathcal{E}

that are not in

A \cup B

.

4

Find the intersection

(A \cup B)' \cap C

and count its cardinality.

(A \cup B)' \cap C = \{2, 5, 7, 11, 13, 17, 19\}

, which contains 7 elements.

The intersection yields elements common to both

(A \cup B)'

and prime set

C

.

Key Concept

Set Complement and Intersections

Question 4302Question →

A binary operation $\Delta$ defined on the set of rational numbers $\mathbb{Q}$ is given by $a \Delta b = \frac{ab}{4}$ . What is the inverse element of $6$ under this operation?

\frac{2}{3}

\frac{8}{3}

\frac{1}{6}

\frac{3}{8}

Show answer & explanation

Answer: $\frac{8}{3}$

Answer

The inverse element of

6

under the given binary operation is

\frac{8}{3}

.

To find the inverse of an element under a binary operation, the identity element

e

must first be found using

a \Delta e = a

, which yields

\frac{ae}{4} = a \implies e = 4

. Then, setting

6 \Delta x = 4

gives

\frac{6x}{4} = 4

, which simplifies to

3x = 8

, giving the inverse

x = \frac{8}{3}

.

Step-by-Step Solution

1

Find the identity element

e

of the binary operation.

e = 4

By definition of identity element,

a \Delta e = a

. Substituting into the definition gives

\frac{ae}{4} = a \implies ae = 4a \implies e = 4

.

2

Set up the inverse equation for the element

6

.

6 \Delta x = 4

, where

x

is the inverse of

6

.

By definition of inverse element,

a \Delta a^{-1} = e

.

3

Solve for the inverse

x

.

x = \frac{8}{3}

Applying the operation rule:

\frac{6x}{4} = 4 \implies \frac{3x}{2} = 4 \implies 3x = 8 \implies x = \frac{8}{3}

.

Key Concept

Identity and Inverse Elements of a Binary Operation

Question 4303Question →

An observer standing at a position between two tall parallel vertical cliffs fires a signal pistol. The observer hears the first echo reflected from the nearer cliff after $1.2\text{ s}$ and the second echo reflected from the farther cliff after $1.8\text{ s}$ . If the speed of sound in air is $340\text{ m s}^{-1}$ , what is the total distance between the two cliffs?

102\text{ m}

510\text{ m}

816\text{ m}

1020\text{ m}

Show answer & explanation

Answer: $510\text{ m}$

Answer

The total distance between the two cliffs is

510\text{ m}

.

Because an echo involves two-way travel of sound, the distance

d

from an observer to a reflecting barrier is given by

d = \frac{v t}{2}

. For the nearer cliff, the distance is

d_1 = \frac{340 \times 1.2}{2} = 204\text{ m}

. For the farther cliff, the distance is

d_2 = \frac{340 \times 1.8}{2} = 306\text{ m}

. The total distance between the parallel cliffs is

d_1 + d_2 = 204\text{ m} + 306\text{ m} = 510\text{ m}

.

Step-by-Step Solution

1

Calculate the distance from the observer to the nearer cliff

d_1 = \frac{v \times t_1}{2} = \frac{340\text{ m s}^{-1} \times 1.2\text{ s}}{2} = 204\text{ m}

An echo involves sound traveling from the observer to the cliff and back, so the one-way distance is half the total path length.

2

Calculate the distance from the observer to the farther cliff

d_2 = \frac{v \times t_2}{2} = \frac{340\text{ m s}^{-1} \times 1.8\text{ s}}{2} = 306\text{ m}

Similarly, the sound travels to the farther cliff and back in

1.8\text{ s}

.

3

Sum the two one-way distances to find the total distance between the two cliffs

D = d_1 + d_2 = 204\text{ m} + 306\text{ m} = 510\text{ m}

Since the observer is positioned between the two parallel cliffs, the total separation distance is the sum of the individual distances to each cliff.

Key Concept

Echo distance calculation involving two-way sound propagation

Question 4304Question →

Find the value of $k$ if the point $P(k, 3)$ is equidistant from the points $A(1, 5)$ and $B(7, 1)$ .

Show answer & explanation

Answer: 4

Answer

The value of

k

is 4.

Using the distance formula, the squared distance

PA^2 = (k-1)^2 + (3-5)^2 = (k-1)^2 + 4

, and

PB^2 = (k-7)^2 + (3-1)^2 = (k-7)^2 + 4

. Equating

PA^2 = PB^2

gives

(k-1)^2 = (k-7)^2

. Expanding both sides yields

k^2 - 2k + 1 = k^2 - 14k + 49

. Subtracting

k^2

from both sides gives

12k = 48

, which leads to

k = 4

.

Step-by-Step Solution

1

Write the expressions for the squared distances

PA^2

and

PB^2

using the distance formula.

PA^2 = (k - 1)^2 + 4

and

PB^2 = (k - 7)^2 + 4

The distance formula between two points

(x_1, y_1)

and

(x_2, y_2)

is

d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2

.

2

Equate

PA^2

and

PB^2

since point

P

is equidistant from points

A

and

B

.

(k - 1)^2 + 4 = (k - 7)^2 + 4 \implies (k - 1)^2 = (k - 7)^2

Subtracting 4 from both sides simplifies the equality of squared distances.

3

Expand both sides and isolate

k

to find its numerical value.

k^2 - 2k + 1 = k^2 - 14k + 49 \implies 12k = 48 \implies k = 4

Canceling

k^2

terms yields a simple linear equation.

Key Concept

Equidistant points and the distance formula in coordinate geometry

Estimated Time:1m 30s

Question 4305Question →

The maximum acceleration of a body oscillating in simple harmonic motion is $8\text{ m/s}^2$ . If the period of oscillation is $\pi\text{ s}$ , calculate the amplitude of the oscillation in meters.

Show answer & explanation

Answer: 2

Answer

The amplitude of the oscillation is

2.0\text{ m}

.

The correct answer of

2.0\text{ m}

is obtained by first deriving the angular frequency

\omega = \frac{2\pi}{T} = 2\text{ rad/s}

, and then using the relation

a_{\text{max}} = \omega^2 A

to solve for amplitude:

A = \frac{8}{2^2} = 2.0\text{ m}

.

Step-by-Step Solution

1

Calculate angular frequency (

\omega

) from the given period (

T

).

\omega = \frac{2\pi}{T} = \frac{2\pi}{\pi} = 2\text{ rad/s}

Angular frequency specifies the rate of phase change in oscillations.

2

Apply the maximum acceleration formula for simple harmonic motion to determine amplitude (

A

).

a_{\text{max}} = \omega^2 A \implies 8 = 2^2 \times A \implies A = 2.0\text{ m}

In simple harmonic motion, maximum acceleration occurs at the extreme position and equals

\omega^2 A

.

Key Concept

Simple Harmonic Motion Acceleration and Period Relationship

Question 4306Question →

Match each advanced vocabulary word in the left column with its precise antonym (word opposite in meaning) in the right column.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct matches are: Obstreperous aligns with Submissive and quiet; Munificent aligns with Miserly and stingy; Perennial aligns with Short-lived and temporary; Equivocal aligns with Explicit and unambiguous.

Each vocabulary item in the left column is matched strictly to the phrase in the right column that expresses its direct opposite semantic meaning.

Step-by-Step Solution

1

Analyze the literal definition and denotation of each target word in the left column.

Obstreperous = unruly/noisy; Munificent = highly generous; Perennial = long-lasting/enduring; Equivocal = ambiguous/unclear.

Establishing the baseline meaning of each term is necessary to identify its exact opposite.

2

Evaluate the choices in the right column to locate the corresponding antonym for each term.

Submissive and quiet is the opposite of noisy and unruly (Obstreperous). Miserly and stingy is the opposite of generous (Munificent). Short-lived and temporary is the opposite of long-lasting (Perennial). Explicit and unambiguous is the opposite of ambiguous (Equivocal).

Connecting each word to its polar opposite demonstrates vocabulary mastery.

Key Concept

Advanced Lexical Antonyms and Contrastive Meaning

Question 4307Question →

Given the universal set $U = \{x \in \mathbb{Z} : 1 \le x \le 15\}$ , let $A = \{x \in U : x \text{ is a multiple of } 3\}$ and $B = \{x \in U : x \text{ is an even number}\}$ . What is the cardinality of $(A \cup B)'$ ?

5

10

3

8

Show answer & explanation

Answer: $5$

Answer

The cardinality of

(A \cup B)'

is

5

.

The universal set contains

15

elements. The set of multiples of

3

within

U

contains

5

elements, and the set of even numbers contains

7

elements. Two numbers (

6

and

12

) belong to both sets. Subtracting the overlapping count gives

10

unique elements in the union. Subtracting

10

from the universal set total of

15

yields

5

elements in the complement.

Step-by-Step Solution

1

List elements of sets

U

,

A

, and

B

, and determine their cardinalities

U = \{1, 2, 3, \dots, 15\}

, so

n(U) = 15

.

A = \{3, 6, 9, 12, 15\}

(

n(A) = 5

).

B = \{2, 4, 6, 8, 10, 12, 14\}

(

n(B) = 7

).

Establishing explicit set memberships allows accurate counting of set operations.

2

Find the intersection

A \cap B

and compute the union cardinality

n(A \cup B)

A \cap B = \{6, 12\}

, so

n(A \cap B) = 2

. Using inclusion-exclusion:

n(A \cup B) = n(A) + n(B) - n(A \cap B) = 5 + 7 - 2 = 10

.

The union includes all elements that are either multiples of 3, even, or both, avoiding double counting.

3

Calculate the complement cardinality

n((A \cup B)')

relative to

U

n((A \cup B)') = n(U) - n(A \cup B) = 15 - 10 = 5

. Explicitly,

(A \cup B)' = \{1, 5, 7, 11, 13\}

.

The complement set

(A \cup B)'

consists of all elements in the universal set

U

that are not in

A \cup B

.

Key Concept

Complement of Set Union and Inclusion-Exclusion Principle

Estimated Time:1m 0s

Question 4308Question →

A stone is projected vertically upwards from the top edge of a cliff $80\text{ m}$ high with an initial speed of $30\text{ m/s}$ . Taking acceleration due to gravity $g = 10\text{ m/s}^2$ , calculate the total time, in seconds, taken by the stone to reach the ground at the base of the cliff.

Show answer & explanation

Answer: 8

Answer

The total time taken by the stone to reach the ground at the base of the cliff is

8\text{ s}

.

Using the equation of motion

s = ut - \frac{1}{2}gt^2

with

s = -80\text{ m}

,

u = 30\text{ m/s}

, and

g = 10\text{ m/s}^2

yields the quadratic equation

t^2 - 6t - 16 = 0

. Solving gives

t = 8\text{ s}

(ignoring the unphysical negative root

t = -2\text{ s}

). Alternatively, breaking the motion into two parts: time to reach maximum height (

30\text{ m/s} / 10\text{ m/s}^2 = 3\text{ s}

, covering

45\text{ m}

) plus time to fall from maximum height of

125\text{ m}

to the ground (

t = \sqrt{2(125)/10} = 5\text{ s}

), giving a total time of

3 + 5 = 8\text{ s}

.

Step-by-Step Solution

1

Set up the kinematic equation with appropriate vector signs

Displacement

s = -80\text{ m}

, initial velocity

u = +30\text{ m/s}

, acceleration

a = -g = -10\text{ m/s}^2

Since the ground is below the release point, displacement is negative when taking the upward direction as positive.

2

Substitute values into

s = ut + \frac{1}{2}at^2

-80 = 30t - 5t^2

Relates displacement, initial speed, time, and constant gravitational acceleration.

3

Form and solve the quadratic equation

5t^2 - 30t - 80 = 0 \implies t^2 - 6t - 16 = 0 \implies (t - 8)(t + 2) = 0

Simplifies the algebraic expression to find the time roots.

4

Select the physical root

t = 8\text{ s}

Time elapsed must be a positive quantity.

Key Concept

Kinematics of Vertical Motion under Gravity with Displacement from Elevation

Estimated Time:2m 0s

Question 4309Question →

An ultrasonic rangefinder mounted on a drone sends a sound pulse vertically downward to measure its altitude above flat ground. If the echo is detected by the sensor $0.08\text{ s}$ after emission and the speed of sound in air is $340\text{ m s}^{-1}$ , what is the altitude of the drone in meters?

Show answer & explanation

Answer: 13.6

Answer

13.6 meters

The sound pulse emitted by the drone travels down to the ground and reflects back to the sensor. The relationship between speed

v

, total round-trip time

t

, and altitude

d

is given by

2d = v \times t

. Substituting

v = 340\text{ m s}^{-1}

and

t = 0.08\text{ s}

yields

d = \frac{340 \times 0.08}{2} = 13.6\text{ m}

.

Step-by-Step Solution

1

Identify the total time taken by the sound pulse for the round trip.

Total round-trip time

t = 0.08\text{ s}

and speed of sound

v = 340\text{ m s}^{-1}

.

Echo detection measures the time for sound to travel to a barrier and return.

2

Apply the echo distance relationship

2d = v \times t

to solve for altitude

d

.

d = \frac{340 \times 0.08}{2} = 13.6\text{ m}

.

Dividing the total path distance by 2 yields the one-way distance to the ground.

Key Concept

Calculation of distance using echoes and two-way sound wave propagation

Alternative Method

Determine the one-way travel time first:

t_{\text{one-way}} = \frac{0.08}{2} = 0.04\text{ s}

. Then calculate altitude directly using distance = speed × one-way time:

d = 340 \times 0.04 = 13.6\text{ m}

.

Estimated Time:1m 0s

Question 4310Question →

In a group of 100 candidates preparing for an entrance examination, 48 registered for Mathematics, 45 for Physics, and 40 for Chemistry. If 18 candidates registered for both Mathematics and Physics, 15 for both Physics and Chemistry, 20 for both Mathematics and Chemistry, and 8 registered for none of these three subjects, how many candidates registered for all three subjects?

Show answer & explanation

Answer: 12; 12 candidates; 12 students

Answer

12 candidates registered for all three subjects.

By the Principle of Inclusion-Exclusion, the total number of candidates taking at least one subject is

100 - 8 = 92

. Summing the individual totals gives

48 + 45 + 40 = 133

. Subtracting the pairwise intersections gives

133 - (18 + 15 + 20) = 80

. Adding the intersection of all three sets must equal 92, yielding

92 - 80 = 12

.

Step-by-Step Solution

1

Determine the total number of candidates who registered for at least one of the three subjects.

n(M \cup P \cup C) = 100 - 8 = 92

Subtracting the number of candidates who registered for none of the subjects from the universal set gives the cardinality of the union.

2

Apply the Principle of Inclusion-Exclusion for three sets.

n(M \cup P \cup C) = n(M) + n(P) + n(C) - n(M \cap P) - n(P \cap C) - n(M \cap C) + n(M \cap P \cap C)

This formula relates the individual set sizes, pair intersections, and triple intersection to the union.

3

Substitute the known values into the inclusion-exclusion equation.

92 = 48 + 45 + 40 - 18 - 15 - 20 + n(M \cap P \cap C)

Insert the given cardinalities into the formula.

4

Simplify and solve for

n(M \cap P \cap C)

.

92 = 80 + n(M \cap P \cap C) \implies n(M \cap P \cap C) = 92 - 80 = 12

Isolate the unknown triple intersection term.

Key Concept

Principle of Inclusion-Exclusion for Three Sets

Question 4311Question →

A projectile is launched from ground level with a horizontal velocity component of $15\text{ m/s}$ and a vertical velocity component of $20\text{ m/s}$ . Neglecting air resistance, what is the magnitude of the velocity of the projectile at its maximum height?

15\text{ m/s}

0\text{ m/s}

20\text{ m/s}

25\text{ m/s}

Show answer & explanation

Answer: $15\text{ m/s}$

Answer

The magnitude of the velocity of the projectile at its maximum height is

15\text{ m/s}

.

In projectile motion under gravity without air resistance, the horizontal component of velocity remains constant throughout flight (

v_x = 15\text{ m/s}

). At the highest point (apex), the vertical component of velocity momentarily becomes zero (

v_y = 0\text{ m/s}

). Therefore, the magnitude of the velocity at the maximum height is equal to the horizontal component, which is

15\text{ m/s}

.

Step-by-Step Solution

1

Identify the velocity components at the maximum height of a projectile

Vertical component

v_y = 0\text{ m/s}

and horizontal component

v_x = u_x = 15\text{ m/s}

.

Gravity acts vertically, reducing vertical velocity to zero at the peak, while horizontal velocity remains constant in the absence of air resistance.

2

Calculate the total magnitude of velocity at maximum height

v = \sqrt{v_x^2 + v_y^2} = \sqrt{15^2 + 0^2} = 15\text{ m/s}

.

The magnitude of the resultant velocity vector is derived using the Pythagorean theorem.

Key Concept

Velocity at Maximum Height in Projectile Motion

Question 4312Question →

A solid trophy consists of a right circular cone mounted on top of a right circular cylinder with the same base radius of $6\text{ cm}$ . The height of the cylinder is $10\text{ cm}$ and the slant height of the cone is $10\text{ cm}$ . What is the total volume of the trophy?

216\pi\text{ cm}^3

456\pi\text{ cm}^3

480\pi\text{ cm}^3

648\pi\text{ cm}^3

Show answer & explanation

Answer: $456\pi\text{ cm}^3$

Answer

The total volume of the trophy is

456\pi\text{ cm}^3

.

First, use the Pythagorean theorem on the cone to find its vertical height

h_{\text{cone}} = \sqrt{10^2 - 6^2} = 8\text{ cm}

. The volume of the cone is

\frac{1}{3}\pi (6)^2 (8) = 96\pi\text{ cm}^3

. The volume of the cylinder is

\pi (6)^2 (10) = 360\pi\text{ cm}^3

. Adding both gives a total volume of

456\pi\text{ cm}^3

.

Step-by-Step Solution

1

Calculate the vertical height of the cone

h_{\text{cone}} = \sqrt{10^2 - 6^2} = \sqrt{100 - 36} = \sqrt{64} = 8\text{ cm}

The volume formula for a cone requires the vertical height (

h

), which forms a right-angled triangle with the radius (

r=6\text{ cm}

) and slant height (

l=10\text{ cm}

).

2

Calculate the volume of the conical part

V_{\text{cone}} = \frac{1}{3} \pi r^2 h_{\text{cone}} = \frac{1}{3} \times \pi \times 6^2 \times 8 = 96\pi\text{ cm}^3

The formula for the volume of a right circular cone is

V = \frac{1}{3}\pi r^2 h

.

3

Calculate the volume of the cylindrical part

V_{\text{cylinder}} = \pi r^2 h_{\text{cylinder}} = \pi \times 6^2 \times 10 = 360\pi\text{ cm}^3

The formula for the volume of a right circular cylinder is

V = \pi r^2 h

.

4

Sum the volumes of both 3D shapes to get the total volume

V_{\text{total}} = 96\pi + 360\pi = 456\pi\text{ cm}^3

The total volume of a composite solid is the sum of the volumes of its constituent parts.

Key Concept

Volume of Composite 3D Solids

Estimated Time:2m 0s

Question 4313Question →

A spherical particle of radius $3.0 \times 10^{-3}\text{ m}$ and density $8.0 \times 10^3\text{ kg/m}^3$ is released from rest and falls vertically through a tall column of a viscous liquid of density $2.0 \times 10^3\text{ kg/m}^3$ . If the coefficient of viscosity of the fluid is $0.40\text{ Pa}\cdot\text{s}$ and the acceleration due to gravity $g = 10\text{ m/s}^2$ , calculate the magnitude of its terminal velocity in $\text{m/s}$ .

Show answer & explanation

Answer: 0.3

Answer

The magnitude of the terminal velocity of the falling sphere is

0.3\text{ m/s}

.

When a body falls at terminal velocity through a viscous medium, its weight is balanced by the sum of buoyancy upthrust and Stokes' viscous drag force. Applying

v_t = \frac{2 r^2 (\rho_s - \rho_f) g}{9 \eta}

with sphere radius

r = 0.003\text{ m}

, sphere density

\rho_s = 8000\text{ kg/m}^3

, fluid density

\rho_f = 2000\text{ kg/m}^3

, viscosity

\eta = 0.40\text{ Pa}\cdot\text{s}

, and

g = 10\text{ m/s}^2

yields

v_t = 0.3\text{ m/s}

.

Step-by-Step Solution

1

Formulate the dynamic equilibrium condition at terminal velocity.

At terminal velocity, the net acceleration is zero, leading to the force balance equation

W = U + F_v

, where

W

is the gravitational weight of the sphere,

U

is the buoyant upthrust, and

F_v

is the retarding viscous force.

Terminal velocity occurs when the downward force of gravity is precisely balanced by the sum of upward resistive and buoyancy forces.

2

Substitute algebraic expressions for weight, upthrust, and Stokes' viscous drag.

W = \frac{4}{3}\pi r^3 \rho_s g

,

U = \frac{4}{3}\pi r^3 \rho_f g

, and

F_v = 6\pi \eta r v_t

.

Archimedes' principle defines the upthrust force equal to the weight of displaced liquid, while Stokes' law governs viscous resistance on spherical bodies.

3

Solve the equilibrium equation for terminal velocity

v_t

.

v_t = \frac{2 r^2 (\rho_s - \rho_f) g}{9 \eta}

.

Equating

\frac{4}{3}\pi r^3 (\rho_s - \rho_f) g = 6\pi \eta r v_t

and simplifying cancels common factors of

\pi

and

r

.

4

Substitute the specified numerical parameters into the derived expression.

v_t = \frac{2 \times (3.0 \times 10^{-3})^2 \times (8000 - 2000) \times 10}{9 \times 0.40} = \frac{2 \times 9.0 \times 10^{-6} \times 6000 \times 10}{3.6} = \frac{1.08}{3.6} = 0.3\text{ m/s}

.

Direct calculation yields the exact value of terminal velocity.

Key Concept

Terminal Velocity, Stokes' Law, and Archimedes' Principle

Question 4314Question →

A uniform spherical planet has twice the mass and twice the radius of the Earth. If the acceleration due to gravity at the Earth's surface is $g$ , what is the acceleration due to gravity at the surface of this planet?

0.25g

0.5g

1.0g

2.0g

Show answer & explanation

Answer: $0.5g$

Answer

The acceleration due to gravity at the surface of the planet is

0.5g

.

The acceleration due to gravity at a planet's surface is given by

g = \frac{GM}{R^2}

. Doubling the mass doubles the field strength, but doubling the radius reduces the field strength by a factor of

2^2 = 4

due to the inverse-square law. Combining these changes results in

\frac{2}{4}g = 0.5g

.

Step-by-Step Solution

1

Write the general expression for surface acceleration due to gravity.

g = \frac{GM}{R^2}

, where

G

is the gravitational constant,

M

is planetary mass, and

R

is planetary radius.

Establishes the fundamental formula relating gravity to mass and radius.

2

Substitute the given parameters for the new planet (

M_p = 2M

and

R_p = 2R

).

g_p = \frac{G(2M)}{(2R)^2} = \frac{2GM}{4R^2}

.

Applies the proportional changes to mass and radius into the formula.

3

Simplify the expression to find

g_p

in terms of

g

.

g_p = \frac{2}{4} \left(\frac{GM}{R^2}\right) = 0.5g

.

Evaluates the fractional change relative to Earth's surface gravity.

Key Concept

Dependence of surface gravitational field strength on planetary mass and radius via the inverse-square law.

Estimated Time:1m 15s

Question 4315Question →

A glass vessel has a linear expansivity of $1.0 \times 10^{-5} \text{ K}^{-1}$ and is filled with a liquid. If the apparent cubic expansivity of the liquid in this vessel is $1.5 \times 10^{-4} \text{ K}^{-1}$ , calculate the real cubic expansivity of the liquid in units of $10^{-4} \text{ K}^{-1}$ .

Show answer & explanation

Answer: 1.8

Answer

The real cubic expansivity of the liquid is

1.8 \times 10^{-4} \text{ K}^{-1}

(giving

1.8

in units of

10^{-4} \text{ K}^{-1}

).

The real cubic expansivity of a liquid equals the sum of its apparent cubic expansivity and the volume expansivity of the vessel (\gamma_r = \gamma_a + \gamma_v). First, convert the linear expansivity of the glass vessel to volume expansivity:

\gamma_v = 3\alpha = 3 \times 1.0 \times 10^{-5} \text{ K}^{-1} = 0.3 \times 10^{-4} \text{ K}^{-1}

. Adding this to the apparent cubic expansivity (

1.5 \times 10^{-4} \text{ K}^{-1}

) yields a real cubic expansivity of

1.8 \times 10^{-4} \text{ K}^{-1}

.

Step-by-Step Solution

1

Determine the cubic expansivity of the vessel (\gamma_v)

\gamma_v = 3.0 \times 10^{-5} \text{ K}^{-1} = 0.3 \times 10^{-4} \text{ K}^{-1}

The volume (cubic) expansivity of a solid vessel is three times its linear expansivity (\gamma_v = 3\alpha).

2

Calculate the real cubic expansivity of the liquid (\gamma_r)

\gamma_r = 1.8 \times 10^{-4} \text{ K}^{-1}

Real cubic expansivity is the sum of apparent cubic expansivity and vessel cubic expansivity (\gamma_r = \gamma_a + \gamma_v).

Key Concept

Real and Apparent Cubic Expansivity of Liquids

Question 4316Question →

A binary operation $\circ$ on the set of real numbers $\mathbb{R}$ is defined by $a \circ b = a + b + 2ab$ . If the identity element of the operation is $e$ , what is the value of $x$ such that the inverse of $x$ under $\circ$ is equal to $2$ ?

Show answer & explanation

Answer: -0.4

Answer

The value of

x

is

-0.4

.

First find the identity element

e

by solving

a \circ e = a

, which yields

a + e + 2ae = a \implies e(1 + 2a) = 0 \implies e = 0

. Next, by definition of an inverse,

x \circ x^{-1} = e

. Substituting

x^{-1} = 2

and

e = 0

gives

x \circ 2 = 0

. Expanding this using the binary operation rule yields

x + 2 + 2(x)(2) = 0 \implies 5x + 2 = 0 \implies x = -0.4

.

Step-by-Step Solution

1

Find the identity element

e

of the operation

\circ

e = 0

By definition of identity element,

a \circ e = a \implies a + e + 2ae = a

, which simplifies to

e(1 + 2a) = 0

, giving

e = 0

.

2

Set up the inverse equation using

x^{-1} = 2

x \circ 2 = 0

The inverse of

x

satisfies

x \circ x^{-1} = e

. Since

x^{-1} = 2

and

e = 0

,

x \circ 2 = 0

.

3

Solve for

x

x = -0.4

Expanding

x \circ 2 = 0

gives

x + 2 + 4x = 0 \implies 5x = -2 \implies x = -0.4

.

Key Concept

Identity and Inverse Elements in Binary Operations

Question 4317Question →

The area of the region bounded by the parabola $y = kx - x^2$ (where $k > 0$ ) and the $x$ -axis between its $x$ -intercepts at $x = 0$ and $x = k$ is equal to $36$ square units. What is the value of the positive constant $k$ ?

Show answer & explanation

Answer: 6

Answer

The value of the positive constant

k

is 6.

The area bounded by

y = kx - x^2

and the x-axis from

x = 0

to

x = k

is obtained by integrating

kx - x^2

, which yields

\frac{k^3}{6}

. Setting

\frac{k^3}{6} = 36

gives

k^3 = 216

, whose cube root is

k = 6

.

Step-by-Step Solution

1

Set up the definite integral representing the area bounded by the curve and the x-axis between the intercepts

x = 0

and

x = k

.

\int_{0}^{k} (kx - x^2) \, dx = 36

The area under a curve

y = f(x)

above the x-axis from

x = a

to

x = b

is given by

\int_{a}^{b} f(x) \, dx

.

2

Find the antiderivative and evaluate it at the limits

x = k

and

x = 0

.

\left[ \frac{kx^2}{2} - \frac{x^3}{3} \right]_{0}^{k} = \left(\frac{k(k)^2}{2} - \frac{k^3}{3}\right) - 0 = \frac{k^3}{6}

Applying the integration power rule

\int x^n \, dx = \frac{x^{n+1}}{n+1}

and simplifying

\frac{k^3}{2} - \frac{k^3}{3} = \frac{k^3}{6}

.

3

Set the evaluated expression equal to 36 and solve for

k

.

\frac{k^3}{6} = 36 \implies k^3 = 216 \implies k = 6

Multiplying both sides by 6 yields

k^3 = 216

, and taking the cube root gives

k = 6

.

Key Concept

Definite Integral and Area Under Curve

Question 4318Question →

The speed $v$ of a transverse wave traveling along a stretched string under tension $T$ with mass per unit length $\mu$ is given by $v = k T^x \mu^y$ , where $k$ is a dimensionless constant. What is the numerical value of the exponent $x$ ?

Show answer & explanation

Answer: 0.5

Answer

The numerical value of the exponent

x

is 0.5.

Applying the principle of dimensional homogeneity, the dimensions on both sides must match. Speed

[v] = L T^{-1}

, tension force

[T] = M L T^{-2}

, and mass per unit length

[\mu] = M L^{-1}

. Substituting these into

v = k T^x \mu^y

yields

M^0 L^1 T^{-1} = M^{x+y} L^{x-y} T^{-2x}

. Comparing the exponents of

T

gives

-2x = -1

, leading to

x = 0.5

.

Step-by-Step Solution

1

Determine the dimensions of speed

v

, tension force

T

, and linear density

\mu

.

[v] = L T^{-1}

,

[T] = M L T^{-2}

,

[\mu] = M L^{-1}

Tension is a force (

F=ma

) with dimensions

[M L T^{-2}]

, and

\mu

is mass per unit length (

m/l

) with dimensions

[M L^{-1}]

.

2

Substitute the dimensional formulas into the equation

v = k T^x \mu^y

and collect powers of base dimensions.

M^0 L^1 T^{-1} = M^{x+y} L^{x-y} T^{-2x}

Combining exponents for base dimensions

M

,

L

, and

T

allows applying the principle of dimensional homogeneity.

3

Equate the exponent of

T

on both sides to solve for

x

.

-2x = -1 \implies x = 0.5

The exponent of

T

on the left side is

-1

and on the right side is

-2x

.

Key Concept

Dimensional Analysis and Determination of Exponents

Estimated Time:1m 30s

Question 4319Question →

If $2x + y = 7$ and $x^2 + xy = 6$ , what are the possible values of $x$ ?

1

or

6

-1

or

-6

2

or

3

-2

or

-3

Show answer & explanation

Answer: $1$ or $6$

Answer

The possible values of

x

are

1

or

6

.

From the linear equation

2x + y = 7

, we get

y = 7 - 2x

. Substituting this into

x^2 + xy = 6

gives

x^2 + x(7 - 2x) = 6

, which simplifies to

-x^2 + 7x = 6

, or

x^2 - 7x + 6 = 0

. Factoring gives

(x - 1)(x - 6) = 0

, leading to

x = 1

or

x = 6

.

Step-by-Step Solution

1

Express

y

in terms of

x

using the linear equation.

y = 7 - 2x

Isolation of one variable allows substitution into the non-linear equation.

2

Substitute

y = 7 - 2x

into the second equation

x^2 + xy = 6

.

x^2 + x(7 - 2x) = 6 \implies x^2 + 7x - 2x^2 = 6 \implies -x^2 + 7x - 6 = 0

This reduces the system to a single quadratic equation in

x

.

3

Rearrange into standard quadratic form

ax^2 + bx + c = 0

.

x^2 - 7x + 6 = 0

Standard quadratic form enables easy factorization.

4

Factor the quadratic equation and solve for

x

.

(x - 1)(x - 6) = 0 \implies x = 1 \text{ or } x = 6

Setting each factor to zero yields the values of

x

.

Key Concept

Solving simultaneous linear and quadratic equations by substitution

Question 4320Question →

A motorist travelling at a constant speed of $20\text{ m s}^{-1}$ directly towards a tall vertical cliff sounds a horn. If the motorist hears the echo of the horn $2.0\text{ s}$ later and the speed of sound in air is $340\text{ m s}^{-1}$ , what was the distance of the car from the cliff at the moment the horn was sounded?

Show answer & explanation

Answer: 360

Answer

The distance of the car from the cliff at the instant the horn was sounded was

360\text{ m}

.

When the motorist sounds the horn at an initial distance

D

from the cliff, the sound wave travels toward the cliff. In the

2.0\text{ s}

it takes for the echo to return, the car advances

40\text{ m}

toward the cliff (

20\text{ m s}^{-1} \times 2.0\text{ s}

). The returning echo meets the motorist at a distance of

(D - 40)\text{ m}

from the cliff. Consequently, the sound covers a total distance of

D + (D - 40) = 2D - 40\text{ m}

. Because the sound wave travels at

340\text{ m s}^{-1}

for

2.0\text{ s}

, the actual distance covered by sound is

340 \times 2.0 = 680\text{ m}

. Setting

2D - 40 = 680

gives

2D = 720\text{ m}

, which yields

D = 360\text{ m}

.

Step-by-Step Solution

1

Calculate the distance covered by the car while moving toward the cliff during the echo time interval.

d_{\text{car}} = 20\text{ m s}^{-1} \times 2.0\text{ s} = 40\text{ m}

.

The car continues to move closer to the cliff for the entire

2.0\text{ s}

period.

2

Set up an expression for the total distance covered by the sound wave.

d_{\text{sound}} = D + (D - 40) = 2D - 40\text{ m}

.

The sound travels forward a distance

D

to the cliff and reflects back to the car's updated location, which is

(D - 40)\text{ m}

from the cliff.

3

Calculate the distance travelled by sound using the given speed of sound.

d_{\text{sound}} = 340\text{ m s}^{-1} \times 2.0\text{ s} = 680\text{ m}

.

Sound propagates through air at

340\text{ m s}^{-1}

.

4

Equate the geometric path expression to the physical sound distance and solve for

D

.

2D - 40 = 680 \implies 2D = 720 \implies D = 360\text{ m}

.

Solving the equation yields the initial position of the car relative to the cliff.

Key Concept

Echo distance calculation with a moving observer

Answer

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