All practice questions

If $y = \ln(2 + \sin(3x)) + e^{4x}$ , find the value of $\frac{dy}{dx}$ at $x = 0$ .

Show answer & explanation

Answer: 5.5

Answer

The value of

\frac{dy}{dx}

at

x = 0

is

5.5

.

Differentiating

\ln(2 + \sin(3x))

by the chain rule gives

\frac{3\cos(3x)}{2 + \sin(3x)}

, and differentiating

e^{4x}

gives

4e^{4x}

. Evaluating

\frac{3\cos(3x)}{2 + \sin(3x)} + 4e^{4x}

at

x = 0

yields

\frac{3(1)}{2 + 0} + 4(1) = 1.5 + 4 = 5.5

.

Step-by-Step Solution

1

Differentiate the logarithmic component

\ln(2 + \sin(3x))

using the chain rule

\frac{d}{dx}[\ln(2 + \sin(3x))] = \frac{3\cos(3x)}{2 + \sin(3x)}

By the chain rule,

\frac{d}{dx}[\ln(u)] = \frac{1}{u}\frac{du}{dx}

, where

u = 2 + \sin(3x)

and

\frac{du}{dx} = 3\cos(3x)

.

2

Differentiate the exponential component

e^{4x}

\frac{d}{dx}[e^{4x}] = 4e^{4x}

The standard rule for exponential differentiation states that

\frac{d}{dx}[e^{kx}] = k e^{kx}

.

3

Combine the results to state the derivative function

\frac{dy}{dx}

\frac{dy}{dx} = \frac{3\cos(3x)}{2 + \sin(3x)} + 4e^{4x}

The derivative of a sum of functions is the sum of their individual derivatives.

4

Evaluate

\frac{dy}{dx}

at

x = 0

\frac{dy}{dx}\Big|_{x=0} = \frac{3\cos(0)}{2 + \sin(0)} + 4e^0 = \frac{3(1)}{2 + 0} + 4(1) = 1.5 + 4 = 5.5

Substituting

x = 0

uses the values

\cos(0) = 1

,

\sin(0) = 0

, and

e^0 = 1

.

Key Concept

Differentiation of Logarithmic, Trigonometric, and Exponential Functions using the Chain Rule

Question 862Question →

In a cathode-ray experiment, electrons are accelerated from rest through a potential difference of $100\text{ V}$ . Taking the specific charge ( $\frac{e}{m}$ ) of an electron to be $1.80 \times 10^{11}\text{ C/kg}$ , calculate the final speed of the electrons as they pass through the anode aperture, expressed in units of $10^6\text{ m/s}$ .

Show answer & explanation

Answer: 6

Answer

The final speed of the electrons is

6.0 \times 10^6\text{ m/s}

, which corresponds to a coefficient value of 6.0.

By applying conservation of energy, the kinetic energy acquired by electrons in a cathode-ray tube equals the electric work done on them (

eV = \frac{1}{2}mv^2

). Solving for velocity gives

v = \sqrt{2V(e/m)}

. Substituting

V = 100\text{ V}

and

e/m = 1.80 \times 10^{11}\text{ C/kg}

gives

v = 6.0 \times 10^6\text{ m/s}

.

Step-by-Step Solution

1

Set up the energy conservation relation for cathode ray electrons.

Electrical work done

W = eV

equals kinetic energy

K = \frac{1}{2}mv^2

.

Electrons starting from rest gain kinetic energy equal to the electrical potential energy lost across the potential difference.

2

Rearrange the formula to isolate the electron velocity

v

.

v^2 = 2V\left(\frac{e}{m}\right) \implies v = \sqrt{2V\left(\frac{e}{m}\right)}

.

Isolating velocity allows direct calculation using the given specific charge

(\frac{e}{m})

and accelerating voltage

V

.

3

Substitute given values into the equation and compute

v

.

v = \sqrt{2 \times 100 \times 1.80 \times 10^{11}} = \sqrt{36 \times 10^{12}} = 6.0 \times 10^6\text{ m/s}

.

Evaluating the square root yields the speed in meters per second.

Key Concept

Electron acceleration in cathode ray tubes and specific charge relation

Question 863Question →

A metallic conductor wire of cross-sectional area $2.5 \times 10^{-6}\,\text{m}^2$ has a resistance of $10.0\,\Omega$ at $0\,^\circ\text{C}$ . The temperature coefficient of resistance of the material is $5.0 \times 10^{-3}\,^\circ\text{C}^{-1}$ . The conductor contains a free-electron density of $5.0 \times 10^{28}\,\text{m}^{-3}$ . When the wire is heated to $100\,^\circ\text{C}$ and connected across a potential difference of $60\,\text{V}$ , what is the drift velocity of the conduction electrons in the wire in millimeters per second ( $\text{mm/s}$ )? (Take elementary charge $e = 1.6 \times 10^{-19}\,\text{C}$ .)

Show answer & explanation

Answer: 0.2

Answer

The drift velocity of the conduction electrons is

0.2\,\text{mm/s}

.

The resistance increases from

10.0\,\Omega

to

15.0\,\Omega

when heated from

0\,^\circ\text{C}

to

100\,^\circ\text{C}

. Applying

60\,\text{V}

results in a current of

4.0\,\text{A}

. Combining this with the cross-sectional area and electron density gives a drift velocity of

2.0 \times 10^{-4}\,\text{m/s}

, which equals

0.2\,\text{mm/s}

.

Step-by-Step Solution

1

Calculate the resistance at the operating temperature (

100\,^\circ\text{C}

)

R_{100} = 15.0\,\Omega

Resistance varies with temperature according to

R_T = R_0(1 + \alpha \Delta T)

.

2

Find the current in the wire using Ohm's Law

I = 4.0\,\text{A}

Current is given by

I = V / R_{100}

.

3

Determine current density

J

J = 1.6 \times 10^6\,\text{A/m}^2

Current density is total current per unit cross-sectional area,

J = I / A

.

4

Calculate the electron drift velocity

v_d

v_d = 2.0 \times 10^{-4}\,\text{m/s} = 0.2\,\text{mm/s}

Drift velocity relates to current density by

v_d = J / (n e)

.

Key Concept

Temperature Dependence of Resistance and Microscopic Model of Electric Current

Estimated Time:2m 0s

Question 864Question →

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 865Question →

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 866Question →

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 867Question →

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 868Question →

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 869Question →

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 870Question →

Find the real value of $x$ that satisfies the logarithmic equation $\log_5(x^2 - 4) - \log_5(x - 2) = 2$ .

Show answer & explanation

Answer: 23

Answer

The value of

x

that satisfies the equation is 23.

Applying the logarithmic quotient rule gives

\log_5 \left(\frac{x^2 - 4}{x - 2}\right) = 2

. Factoring the numerator gives

\log_5 \left(\frac{(x - 2)(x + 2)}{x - 2}\right) = \log_5(x + 2) = 2

. Converting to exponential form yields

x + 2 = 5^2 = 25

, which simplifies to

x = 23

.

Step-by-Step Solution

1

Apply the logarithmic quotient rule

\log_5\left(\frac{x^2 - 4}{x - 2}\right) = 2

The difference of two logarithms of the same base equals the logarithm of their quotient.

2

Factor the numerator and simplify the expression

\log_5(x + 2) = 2

Factoring

x^2 - 4

into

(x-2)(x+2)

allows canceling the common factor

(x-2)

in the denominator.

3

Convert the logarithmic equation into exponential form

x + 2 = 5^2 = 25

By definition,

\log_b(y) = c

is equivalent to

b^c = y

.

4

Solve the linear equation for

x

x = 23

Subtracting 2 from both sides gives

x = 23

.

Key Concept

Logarithmic Quotient Rule and Logarithm-to-Exponent Conversion

Question 871Question →

The speed of sound $v$ in a gas depends on the gas pressure $P$ and density $\rho$ according to the relation $v = k P^a \rho^b$ , where $k$ is a dimensionless constant. What is the numerical value of the exponent $a$ ?

Show answer & explanation

Answer: 0.5

Answer

The numerical value of the exponent

a

is

0.5

.

Applying dimensional homogeneity to the relation

v = k P^a \rho^b

yields

[M^0 L^1 T^{-1}] = [M L^{-1} T^{-2}]^a [M L^{-3}]^b

. Equating the powers of time

T

gives

-2a = -1

, which simplifies to

a = 0.5

.

Step-by-Step Solution

1

Determine the base dimensions of all physical quantities in the relationship

[v] = M^0 L T^{-1}

,

[P] = M L^{-1} T^{-2}

, and

[\rho] = M L^{-3}

Physical quantities must be expressed in fundamental dimensions (

M, L, T

) to apply dimensional analysis.

2

Formulate the dimensional balance equation

M^0 L^1 T^{-1} = M^{a+b} L^{-a-3b} T^{-2a}

By the principle of dimensional homogeneity, the total dimensions on the left side must equal those on the right side.

3

Equate exponents of

T

to solve for

a

-2a = -1 \implies a = 0.5

Comparing powers of time

T

directly isolates the variable

a

.

Key Concept

Dimensional Homogeneity and Derivation of Exponents

Question 872Question →

In a sports academy of 120 athletes, 70 play football, 60 play basketball, and 50 play tennis. If 10 athletes play none of these three sports and 15 athletes play all three sports, how many athletes play exactly two of these sports?

Show answer & explanation

Answer: 40

Answer

40 athletes play exactly two of the sports.

The correct answer is 40. Subtracting the 10 athletes who play no sports from the total of 120 leaves 110 athletes playing at least one sport. Using inclusion-exclusion, the sum of pairwise intersections is

S_2 = 70 + 60 + 50 + 15 - 110 = 85

. Since

S_2

contains the region of all three sports counted three times, subtracting

3 \times 15 = 45

gives 40 athletes who play exactly two sports.

Step-by-Step Solution

1

Determine the cardinality of the union of all three sets

n(F \cup B \cup T) = 120 - 10 = 110

Athletes who play none of the three sports are excluded from the total universal set.

2

Apply the Principle of Inclusion-Exclusion for three sets to find the sum of 2-set intersections

S_2 = n(F) + n(B) + n(T) + n(F \cap B \cap T) - n(F \cup B \cup T) = 70 + 60 + 50 + 15 - 110 = 85

The formula relates the total union, individual set cardinalities, pairwise intersections, and triple intersection.

3

Subtract three times the triple intersection from

S_2

to isolate regions corresponding to exactly two sports

Exactly two sports =

S_2 - 3 \times n(F \cap B \cap T) = 85 - 3(15) = 40

Each pairwise intersection sum

S_2

includes the triple intersection region three times.

Key Concept

Three-set inclusion-exclusion principle and region cardinality decomposition

Estimated Time:1m 30s

Question 873Question →

The rate of change of a function $f(x)$ with respect to $x$ is defined by $f'(x) = 3x^2 - 4x + 6\sin(3x)$ . If $f(0) = 7$ , determine the value of the constant of integration, $C$ .

Show answer & explanation

Answer: 9

Answer

The constant of integration

C

is equal to 9.

Integrating

f'(x) = 3x^2 - 4x + 6\sin(3x)

gives

f(x) = x^3 - 2x^2 - 2\cos(3x) + C

. Substituting

x = 0

and

f(0) = 7

leads to

7 = 0 - 0 - 2(1) + C

, which simplifies to

C = 9

.

Step-by-Step Solution

1

Integrate the rate of change function

f'(x) = 3x^2 - 4x + 6\sin(3x)

with respect to

x

.

f(x) = x^3 - 2x^2 - 2\cos(3x) + C

Using the power rule

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

and trigonometric integration rule

\int \sin(kx) dx = -\frac{1}{k}\cos(kx)

.

2

Substitute the initial condition

x = 0

and

f(0) = 7

into the expression for

f(x)

.

7 = (0)^3 - 2(0)^2 - 2\cos(3 \cdot 0) + C

The curve passes through

x = 0

with value

y = 7

.

3

Evaluate the trigonometric term at zero and solve for

C

.

7 = -2(1) + C \implies C = 9

Since

\cos(0) = 1

, the expression simplifies to

7 = -2 + C

, yielding

C = 9

.

Key Concept

Indefinite Integration of Polynomial and Trigonometric Functions with Boundary Conditions

Question 874Question →

A sound transmitter and a projectile launcher are co-located at a distance of $210\text{ m}$ directly in front of a tall, flat vertical cliff. At time $t = 0\text{ s}$ , a projectile is launched directly away from the cliff at a constant speed of $60\text{ m s}^{-1}$ , while a sound pulse is emitted simultaneously towards the cliff. The sound wave reflects off the cliff face and travels back to overtake the moving projectile. Assuming the speed of sound in air is $340\text{ m s}^{-1}$ , calculate the distance from the cliff face, in meters, to the position where the reflected sound wave intercepts the projectile.

Show answer & explanation

Answer: 300

Answer

The distance of the projectile from the cliff face at the instant of interception is

300\text{ m}

.

The correct calculation accounts for both the two-part path of the sound wave (forward to cliff + back to projectile) and the displacement of the projectile moving away from the cliff over the same time interval, yielding an interception distance of

300\text{ m}

from the cliff.

Step-by-Step Solution

1

Set up the total distance expression for the sound wave from the cliff face

Total sound path =

210\text{ m} + x

The sound pulse must travel

210\text{ m}

forward to hit the cliff face, plus an additional distance

x

away from the cliff face after reflection to reach the projectile.

2

Set up the distance expression for the projectile from its starting point

Projectile path =

x - 210\text{ m}

The projectile starts

210\text{ m}

away from the cliff and moves farther away to position

x

.

3

Equate the time elapsed for both sound propagation and projectile movement

\frac{210 + x}{340} = \frac{x - 210}{60}

Both events happen simultaneously over the exact same time interval

t

.

4

Solve the linear equation for

x

x = 300\text{ m}

Cross-multiplying yields

60(210 + x) = 340(x - 210)

, which simplifies to

28x = 8400

, giving

x = 300\text{ m}

.

Key Concept

Echo reflection path combined with relative linear kinematics

Question 875Question →

A progressive wave traveling along a stretched string is represented by the equation $y = 0.02 \sin(120\pi t - 3\pi x)$ , where $x$ and $y$ are in meters and $t$ is in seconds. What is the velocity of the wave in $\text{m s}^{-1}$ ?

Show answer & explanation

Answer: 40

Answer

The velocity of the wave is

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

.

The standard progressive wave equation is

y = A \sin(\omega t - kx)

, where

\omega

is the angular frequency and

k

is the wave number (wave vector). Comparing the given equation

y = 0.02 \sin(120\pi t - 3\pi x)

with the standard form yields

\omega = 120\pi \text{ rad s}^{-1}

and

k = 3\pi \text{ m}^{-1}

. The velocity of propagation of the wave is given by

v = \frac{\omega}{k} = \frac{120\pi}{3\pi} = 40 \text{ m s}^{-1}

.

Step-by-Step Solution

1

Compare given equation with the standard progressive wave equation

Matching

y = 0.02 \sin(120\pi t - 3\pi x)

to

y = A \sin(\omega t - kx)

gives

\omega = 120\pi \text{ rad s}^{-1}

and

k = 3\pi \text{ rad m}^{-1}

.

Direct parameter identification from the wave function gives the angular frequency and wave number.

2

Calculate wave velocity

Wave velocity

v = \frac{\omega}{k} = \frac{120\pi}{3\pi} = 40 \text{ m s}^{-1}

.

The ratio of angular frequency to wave number equals the phase velocity of the wave.

Key Concept

Extracting wave parameters (angular frequency and wave number) from the mathematical wave equation to determine wave velocity.

Question 876Question →

The sum of the first $4$ terms of a geometric progression (GP) with a common ratio of $2$ is $45$ . What is the $6^{\text{th}}$ term of the progression?

Show answer & explanation

Answer: 96

Answer

The

6^{\text{th}}

term of the geometric progression is

96

.

Using the sum formula

S_4 = \frac{a(2^4 - 1)}{2 - 1} = 45

gives

15a = 45

, so the first term

a

is

3

. Substituting

a = 3

and

r = 2

into the term formula

T_6 = a r^5

gives

3 \times 32 = 96

.

Step-by-Step Solution

1

Express the sum of the first 4 terms using the GP sum formula to find the first term

a

.

Setting up

45 = \frac{a(2^4 - 1)}{2 - 1}

yields

15a = 45

, so

a = 3

.

The sum formula

S_n = \frac{a(r^n - 1)}{r - 1}

allows us to isolate the unknown initial term

a

when

S_4

and

r

are given.

2

Calculate the

6^{\text{th}}

term

T_6

using the

n^{\text{th}}

term formula

T_n = a r^{n-1}

.

T_6 = 3 \times 2^{6-1} = 3 \times 32 = 96

.

The exponent for the common ratio in the

n^{\text{th}}

term formula is

n - 1

, giving

5

as the exponent.

Key Concept

Sum and

n^{\text{th}}

term of a Geometric Progression

Question 877Question →

The energy of an electron in the first excited state of a hydrogen atom is $-3.4\text{ eV}$ . What is the minimum energy, in joules ( $\text{J}$ ), required to completely remove the electron from this state to ionize the atom? ( $1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$ )

Show answer & explanation

Answer: 5.44e-19

Answer

The minimum energy required to ionize the atom from its first excited state is

5.44 \times 10^{-19}\text{ J}

.

Ionization energy is defined as the minimum energy necessary to completely remove an electron from its bound atomic energy level to infinity (

E_{\infty} = 0\text{ eV}

). For an electron at

E = -3.4\text{ eV}

, the required energy change is

\Delta E = 0 - (-3.4\text{ eV}) = 3.4\text{ eV}

. Converting this to joules using

1\text{ eV} = 1.6 \times 10^{-19}\text{ J}

gives

3.4 \times 1.6 \times 10^{-19} = 5.44 \times 10^{-19}\text{ J}

.

Step-by-Step Solution

1

Determine the energy required for ionization in electron-volts (eV)

\Delta E = E_{\infty} - E_2 = 0\text{ eV} - (-3.4\text{ eV}) = 3.4\text{ eV}

Ionization requires supplying sufficient energy to raise the electron from its bound energy state (

E_2 = -3.4\text{ eV}

) to the ionization limit where it is free (

E_{\infty} = 0\text{ eV}

).

2

Convert the ionization energy from electron-volts to joules

E = 3.4\text{ eV} \times 1.6 \times 10^{-19}\text{ J/eV} = 5.44 \times 10^{-19}\text{ J}

To convert energy from electron-volts (eV) to joules (J), multiply the value in eV by the elementary charge conversion factor

1.6 \times 10^{-19}\text{ J/eV}

.

Key Concept

Ionization Energy and Energy Level Transitions

Estimated Time:1m 30s

Question 878Question →

A bag contains red, blue, and yellow marbles. The theoretical probability of selecting a red marble at random from the bag is $\frac{2}{5}$ . In a probability experiment, a marble is drawn from the bag, its color recorded, and then replaced. This procedure is repeated $250$ times, resulting in a red marble being drawn $115$ times. What is the absolute difference between the observed experimental frequency of red marbles and the theoretical expected frequency?

Show answer & explanation

Answer: 15

Answer

15

The theoretical expected frequency of selecting a red marble over

250

trials is found by multiplying the total trials by the theoretical probability:

250 \times \frac{2}{5} = 100

. The experimental frequency observed was

115

. Taking the absolute difference gives

|115 - 100| = 15

.

Step-by-Step Solution

1

Calculate the theoretical expected frequency of red marble outcomes.

Theoretical expected frequency =

250 \times \frac{2}{5} = 100

.

Expected frequency is determined by multiplying the number of trials (

N = 250

) by the theoretical probability (

P = \frac{2}{5}

).

2

Find the absolute difference between the observed experimental frequency and the theoretical expected frequency.

|115 - 100| = 15

.

The observed experimental frequency is

115

and the theoretical expected frequency is

100

, so the positive difference is

15

.

Key Concept

Experimental Frequency vs. Theoretical Expected Frequency

Estimated Time:1m 30s

Question 879Question →

An unpolarized light beam with an initial intensity of $80\text{ W/m}^2$ passes through an ideal linear polarizer. What is the intensity of the transmitted light in $\text{W/m}^2$ ?

Show answer & explanation

Answer: 40

Answer

The intensity of the transmitted light is

40\text{ W/m}^2

.

When unpolarized light of initial intensity

I_0

encounters an ideal linear polarizing filter, the transmitted intensity

I

is always equal to half of the incident intensity (

I = \frac{1}{2}I_0

). Substituting

80\text{ W/m}^2

gives

I = 40\text{ W/m}^2

.

Step-by-Step Solution

1

Determine the fraction of unpolarized light intensity transmitted by a polarizer.

Transmitted intensity formula

I = \frac{1}{2} I_0

.

Unpolarized light consists of randomly oriented electric field vectors, resulting in an average transmission factor of one-half.

2

Substitute the incident intensity value into the formula.

I = \frac{80}{2} = 40\text{ W/m}^2

.

Direct mathematical calculation.

Key Concept

Polarization and intensity reduction of unpolarized light upon passing through a linear polarizer.

Question 880Question →

A liquid has a density of $840\text{ kg m}^{-3}$ at $10^\circ\text{C}$ . It is heated inside a container whose material has a linear expansivity of $2.0 \times 10^{-5}\text{ K}^{-1}$ . If the measured apparent cubic expansivity of the liquid in this container is $4.4 \times 10^{-4}\text{ K}^{-1}$ , what is the density of the liquid at $110^\circ\text{C}$ in $\text{kg m}^{-3}$ ?

Show answer & explanation

Answer: 800

Answer

800 kg m^-3

To find the density of the liquid at the elevated temperature, we must use its real cubic expansivity

\gamma_r

. The vessel expands with a cubic expansivity of

\gamma_v = 3\alpha = 6.0 \times 10^{-5}\text{ K}^{-1}

. The real cubic expansivity of the liquid is

\gamma_r = \gamma_a + \gamma_v = 4.4 \times 10^{-4} + 0.6 \times 10^{-4} = 5.0 \times 10^{-4}\text{ K}^{-1}

. Using the density variation formula

\rho_2 = \frac{\rho_1}{1 + \gamma_r \Delta T}

, we evaluate

\frac{840}{1 + (5.0 \times 10^{-4} \times 100)} = \frac{840}{1.05} = 800\text{ kg m}^{-3}

.

Step-by-Step Solution

1

Calculate the cubic expansivity of the container material

\gamma_v = 6.0 \times 10^{-5}\text{ K}^{-1}

The volume (cubic) expansivity of a solid container is three times its linear expansivity (

\gamma_v = 3\alpha

).

2

Determine the real cubic expansivity of the liquid

\gamma_r = 5.0 \times 10^{-4}\text{ K}^{-1}

The real cubic expansivity is the sum of the apparent cubic expansivity and the cubic expansivity of the container (

\gamma_r = \gamma_a + \gamma_v

).

3

Determine the change in temperature

\Delta T = 100\text{ K}

Subtract the initial temperature from the final temperature:

110^\circ\text{C} - 10^\circ\text{C} = 100\text{ K}

.

4

Calculate the final density of the liquid at

110^\circ\text{C}

\rho_2 = 800\text{ kg m}^{-3}

Density varies inversely with volumetric expansion according to

\rho_2 = \frac{\rho_1}{1 + \gamma_r \Delta T}

.

Key Concept

Relationship between real expansivity, apparent expansivity, container expansion, and density change in fluids

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept