All practice questions

Question 7781Question →

Which of the following observations demonstrates that cathode rays possess kinetic energy and momentum?

The rotation of a light paddle wheel placed directly in their path

The casting of a sharp shadow when an opaque Maltese cross is placed in their path

The production of fluorescence when the rays strike the glass walls of the discharge tube

The deflection of the rays toward a positively charged plate in an electric field

Show answer & explanation

Answer: The rotation of a light paddle wheel placed directly in their path

Answer

The rotation of a light paddle wheel placed directly in their path

Cathode rays consist of high-speed electrons possessing mass and kinetic energy. When these particles strike the vanes of a light paddle wheel inside a discharge tube, they transfer mechanical momentum to the wheel, causing it to roll along the glass rails. This directly proves that the rays carry kinetic energy and momentum.

Step-by-Step Solution

1

Recall the physical properties of cathode rays evidenced by classical discharge tube experiments.

Cathode rays travel in straight lines, carry negative charge, produce fluorescence, and possess mechanical momentum.

Identifying the specific experimental evidence associated with kinetic energy and momentum.

2

Analyze the mechanical effect of particle bombardment on a physical object in a vacuum tube.

When stream particles strike the blades of a lightweight paddle wheel, they exert a mechanical force that causes the wheel to turn.

Transfer of momentum (

\Delta p

) upon collision converts particle kinetic energy (

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

) into rotational kinetic energy of the paddle wheel.

Key Concept

Particle Nature and Mechanical Energy of Cathode Rays

Estimated Time:45s

Question 7782Question →

Arrange the following values related to arithmetic and geometric progressions in ascending order (from smallest to largest):

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct order from smallest to largest value is: the common ratio of the GP (4), the common difference of the AP (8), the 4th term of the AP (11), and the sum of the first 3 terms of the GP (13).

Evaluating each progression property gives numerical values of 4, 8, 11, and 13 respectively. Ordering these from least to greatest results in the order: common ratio of the GP (4), common difference of the AP (8), 4th term of the AP (11), and sum of the first 3 terms of the GP (13).

Step-by-Step Solution

1

Calculate the value for the first item (common ratio

r

)

Using

T_4 = T_2 \cdot r^2

, we have

80 = 5 r^2 \implies r^2 = 16 \implies r = 4

.

The terms of a GP follow

T_n = a r^{n-1}

.

2

Calculate the value for the second item (common difference

d

)

Using

T_5 = a + 4d

, we get

35 = 3 + 4d \implies 4d = 32 \implies d = 8

.

The nth term of an AP is given by

T_n = a + (n-1)d

.

3

Calculate the value for the third item (

4^{\text{th}}

term of AP)

T_4 = 2 + (4 - 1)(3) = 2 + 9 = 11

.

Direct application of the AP nth term formula.

4

Calculate the value for the fourth item (Sum of first

3

terms of GP)

S_3 = 1 + 3 + 9 = 13

(or using

S_n = \frac{a(r^n - 1)}{r - 1} = \frac{1(3^3 - 1)}{3 - 1} = 13

).

Sum of a finite geometric sequence.

5

Compare and arrange the calculated numerical values in ascending order

4 < 8 < 11 < 13

.

Ordering the numbers establishes the correct item sequence.

Key Concept

Evaluation of terms, differences, ratios, and sums in Arithmetic and Geometric Progressions.

Question 7783Question →

In a double-slit experiment, monochromatic light of wavelength $500\text{ nm}$ is incident normally on two narrow slits. If the third-order ( $m = 3$ ) bright fringe is observed at an angle of $30^\circ$ from the central maximum, what is the slit separation, $d$ , in micrometers ( $\mu\text{m}$ )?

Show answer & explanation

Answer: 3

Answer

The slit separation is

3.0\text{ }\mu\text{m}

.

For bright fringes in a double-slit setup, constructive interference occurs when

d \sin\theta = m\lambda

. Rearranging for slit separation yields

d = \frac{m\lambda}{\sin\theta}

. Substituting

m = 3

,

\lambda = 0.50\text{ }\mu\text{m}

, and

\theta = 30^\circ

gives

d = \frac{3 \times 0.50\text{ }\mu\text{m}}{\sin 30^\circ} = \frac{1.50\text{ }\mu\text{m}}{0.50} = 3.0\text{ }\mu\text{m}

.

Step-by-Step Solution

1

Identify the constructive interference equation for Young's double-slit experiment.

d \sin\theta = m\lambda

Bright fringes occur where waves from the two slits interfere constructively, which corresponds to path differences equal to integer multiples of the wavelength.

2

Convert the given wavelength to micrometers and substitute known values.

\lambda = 0.50\text{ }\mu\text{m}

,

m = 3

,

\sin(30^\circ) = 0.50

Converting units to micrometers early simplifies direct calculation of

d

in

\mu\text{m}

.

3

Rearrange the equation and evaluate for

d

.

d = \frac{3 \times 0.50\text{ }\mu\text{m}}{0.50} = 3.0\text{ }\mu\text{m}

Dividing the numerator by

\sin(30^\circ) = 0.50

doubles the value of

m\lambda

.

Key Concept

Angular position condition for constructive interference in double-slit diffraction

Question 7784Question →

If $y = \frac{e^{3x}}{\cos x + \sin x}$ , what is the value of $\frac{dy}{dx}$ at $x = 0$ ?

Show answer & explanation

Answer: 2

Answer

The value of

\frac{dy}{dx}

at

x = 0

is

2

.

Applying the quotient rule

\frac{dy}{dx} = \frac{v u' - u v'}{v^2}

to

y = \frac{e^{3x}}{\cos x + \sin x}

yields

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

. Evaluating this expression at

x = 0

gives

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

.

Step-by-Step Solution

1

Identify numerator

u(x)

and denominator

v(x)

for the quotient rule.

u(x) = e^{3x}

and

v(x) = \cos x + \sin x

.

The function

y

is expressed as a quotient of exponential and trigonometric functions.

2

Find the first derivatives of

u(x)

and

v(x)

.

u'(x) = 3e^{3x}

and

v'(x) = -\sin x + \cos x

.

Derivative of

e^{kx}

is

k e^{kx}

, derivative of

\cos x

is

-\sin x

, and derivative of

\sin x

is

\cos x

.

3

Substitute

u(x)

,

v(x)

,

u'(x)

, and

v'(x)

into the quotient rule formula

\frac{dy}{dx} = \frac{u'v - uv'}{v^2}

.

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

.

The quotient rule is required to differentiate

\frac{u(x)}{v(x)}

.

4

Evaluate the derivative expression at

x = 0

.

\frac{dy}{dx}\Big|_{x=0} = \frac{3(1)(1 + 0) - 1(1 - 0)}{(1 + 0)^2} = \frac{3 - 1}{1} = 2

.

At

x = 0

,

e^0 = 1

,

\cos 0 = 1

, and

\sin 0 = 0

.

Key Concept

Differentiation of exponential and trigonometric functions using the quotient rule

Question 7785Question →

Given the matrices $A = \begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & -2 \\ 0 & 3 \end{pmatrix}$ , if $C = AB - 2I$ , where $I$ is the $2 \times 2$ identity matrix, what is the determinant of matrix $C$ ?

3

-3

0

15

Show answer & explanation

Answer: $3$

Answer

The determinant of matrix C is 3.

Multiplying matrices A and B using standard row-by-column multiplication yields \begin{pmatrix} 2 & -1 \\ 3 & 6 \end{pmatrix}. Subtracting the scaled identity matrix \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} produces matrix C = \begin{pmatrix} 0 & -1 \\ 3 & 4 \end{pmatrix}. Taking the determinant ad - bc gives (0)(4) - (-1)(3) = 3.

Step-by-Step Solution

1

Compute the matrix product AB.

AB = \begin{pmatrix} (2)(1) + (1)(0) & (2)(-2) + (1)(3) \\ (3)(1) + (4)(0) & (3)(-2) + (4)(3) \end{pmatrix} = \begin{pmatrix} 2 & -1 \\ 3 & 6 \end{pmatrix}

Matrix multiplication uses dot products of rows of the first matrix and columns of the second matrix.

2

Subtract 2I from AB to find matrix C.

C = \begin{pmatrix} 2 & -1 \\ 3 & 6 \end{pmatrix} - \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ 3 & 4 \end{pmatrix}

The identity matrix I multiplied by scalar 2 has 2 on its main diagonal and 0 elsewhere.

3

Calculate the determinant of matrix C.

\det(C) = (0)(4) - (-1)(3) = 0 + 3 = 3

For a 2x2 matrix \begin{pmatrix} a & b \\ c & d \end{pmatrix}, the determinant is ad - bc.

Key Concept

Matrix multiplication, matrix arithmetic operations, and evaluation of 2x2 determinants.

Estimated Time:1m 30s

Question 7786Question →

A progressive wave traveling through a uniform medium is described by the displacement equation $y = 0.04 \sin\left(150\pi t - 6\pi x\right)$ , where $x$ and $y$ are measured in meters and $t$ is in seconds. What is the speed of propagation of the wave?

Show answer & explanation

Answer: 25

Answer

The speed of propagation of the wave is 25 m/s.

By matching

y = 0.04 \sin(150\pi t - 6\pi x)

to the standard wave equation

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

, we obtain

\omega = 150\pi\text{ rad/s}

and

k = 6\pi\text{ rad/m}

. The speed of the wave

v

is calculated as

v = \frac{\omega}{k} = \frac{150\pi}{6\pi} = 25\text{ m/s}

.

Step-by-Step Solution

1

Compare the given wave equation with the general progressive wave equation

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

.

Angular frequency

\omega = 150\pi\text{ rad/s}

and wave number

k = 6\pi\text{ rad/m}

.

Matching coefficients allows direct extraction of angular frequency and spatial wave number.

2

Calculate the wave speed using the relationship

v = \frac{\omega}{k}

.

v = \frac{150\pi}{6\pi} = 25\text{ m/s}

.

The velocity of a progressive wave is equal to the ratio of its angular frequency to its wave number.

Key Concept

Determining wave velocity from progressive wave equation parameters

Question 7787Question →

What is the result of the subtraction $423_6 - 145_6$ expressed in base 6?

234_6

232_6

344_6

322_6

Show answer & explanation

Answer: $234_6$

Answer

234_6

Subtracting

145_6

from

423_6

column by column requires borrowing from higher place values. In the units column, borrowing 1 (worth 6) gives

3 + 6 - 5 = 4

. In the sixes column, borrowing 1 leaves 1 six, and

1 + 6 - 4 = 3

. In the thirty-sixes column,

3 - 1 = 2

. Thus, the correct result is

234_6

.

Step-by-Step Solution

1

Convert both base 6 numbers into base 10 (decimal).

423_6 = 4 \times 6^2 + 2 \times 6^1 + 3 \times 6^0 = 144 + 12 + 3 = 159_{10}

, and

145_6 = 1 \times 6^2 + 4 \times 6^1 + 5 \times 6^0 = 36 + 24 + 5 = 65_{10}

.

Converting both numbers to decimal provides a familiar base for subtraction.

2

Subtract the decimal numbers.

159_{10} - 65_{10} = 94_{10}

.

Perform standard decimal subtraction.

3

Convert the decimal result

94_{10}

back to base 6.

94 \div 6 = 15

R

4

;

15 \div 6 = 2

R

3

;

2 \div 6 = 0

R

2

. Arranging the remainders from bottom to top gives

234_6

.

Successive division by 6 extracts the base 6 digits.

Key Concept

Non-decimal base subtraction and conversions

Question 7788Question →

What is the value of $(-43 \times 15) \pmod{8}$ expressed in standard non-negative remainder form?

3

-5

5

-3

Show answer & explanation

Answer: $3$

Answer

3

Reducing

-43

modulo

8

gives

5

, and

15

modulo

8

gives

7

. Multiplying these results yields

35 \equiv 3 \pmod{8}

, which is the canonical non-negative remainder.

Step-by-Step Solution

1

Reduce individual factors modulo 8

-43 \equiv 5 \pmod{8}

(since

-43 = -6 \times 8 + 5

) and

15 \equiv 7 \pmod{8}

(since

15 = 1 \times 8 + 7

)

Simplifying terms before multiplication makes arithmetic easier.

2

Multiply the reduced remainders

5 \times 7 = 35

Modular arithmetic preserves multiplication.

3

Reduce the product modulo 8 to obtain the canonical non-negative remainder

35 = 4 \times 8 + 3 \implies 35 \equiv 3 \pmod{8}

The final answer in modular arithmetic must fall within the range

[0, n-1]

.

Key Concept

Modular Arithmetic and Canonical Non-Negative Remainders

Question 7789Question →

According to Bohr's model of the hydrogen atom, the total energy of an electron in the first excited state ( $n = 2$ ) is $-3.40\text{ eV}$ . Calculate the electric potential energy of the electron in this state, in electron-volts ( $\text{eV}$ ).

Show answer & explanation

Answer: -6.8

Answer

The electric potential energy of the electron in the first excited state is

-6.80\text{ eV}

.

In Bohr's model of the hydrogen atom, an electron held in a circular orbit by electrostatic attraction has a kinetic energy

K = -E

and an electric potential energy

U = 2E

. Given a total energy

E = -3.40\text{ eV}

, multiplying by 2 yields

U = -6.80\text{ eV}

.

Step-by-Step Solution

1

Identify the relationship between total energy and electric potential energy in the Bohr atomic model.

U = 2E

, where

U

is potential energy and

E

is total energy.

For an inverse-square electrostatic force binding an electron in a circular orbit, the virial theorem dictates that kinetic energy

K = -E

and potential energy

U = 2E

.

2

Substitute the given value of total energy

E = -3.40\text{ eV}

to solve for

U

.

U = 2 \times (-3.40\text{ eV}) = -6.80\text{ eV}

.

Direct multiplication yields the exact electric potential energy of the bound electron.

Key Concept

Energy components (kinetic, potential, and total energy) of an electron in Bohr's atomic model

Question 7790Question →

A $5.0\text{ }\mu\text{F}$ parallel-plate capacitor is connected across a potential difference of $20.0\text{ V}$ . What is the magnitude of the electric charge stored on either plate of the capacitor, in microcoulombs ( $\mu\text{C}$ )?

Show answer & explanation

Answer: 100

Answer

The magnitude of the electric charge stored on either plate of the capacitor is

100.0\text{ }\mu\text{C}

.

Using the capacitor charge equation

Q = C \times V

, substituting

C = 5.0\text{ }\mu\text{F}

and

V = 20.0\text{ V}

yields

Q = 5.0 \times 20.0 = 100.0\text{ }\mu\text{C}

.

Step-by-Step Solution

1

Identify the given physical quantities and the formula for electric charge on a capacitor.

Capacitance

C = 5.0\text{ }\mu\text{F}

, voltage

V = 20.0\text{ V}

. Formula:

Q = C V

.

The charge stored by a capacitor is directly proportional to the potential difference across its terminals.

2

Perform the multiplication to determine the charge magnitude in microcoulombs.

Q = 5.0 \times 20.0 = 100.0\text{ }\mu\text{C}

.

Multiplying capacitance in microfarads by potential difference in volts yields charge directly in microcoulombs.

Key Concept

Fundamental relationship between capacitance, charge, and potential difference (

Q = C V

)

Estimated Time:45s

Question 7791Question →

Three coplanar forces act simultaneously at a point $O$ . The first force of magnitude $10\text{ N}$ acts due East, and the second force of magnitude $10\text{ N}$ acts at an angle of $60^\circ$ North of East. If a third force keeps the system in static equilibrium, what is the magnitude of this third force?

10\sqrt{3}\text{ N}

20\text{ N}

10\sqrt{2}\text{ N}

10\text{ N}

Show answer & explanation

Answer: $10\sqrt{3}\text{ N}$

Answer

The magnitude of the third force required for equilibrium is

10\sqrt{3}\text{ N}

.

Resolving the forces along the horizontal (East) and vertical (North) axes gives components of

15\text{ N}

and

5\sqrt{3}\text{ N}

respectively. The magnitude of the resultant force is

\sqrt{15^2 + (5\sqrt{3})^2} = \sqrt{300} = 10\sqrt{3}\text{ N}

. Since the third force balances the system, its magnitude must equal that of the resultant, which is

10\sqrt{3}\text{ N}

.

Step-by-Step Solution

1

Resolve the two given forces into horizontal (

x

) and vertical (

y

) Cartesian components.

F_{1x} = 10\text{ N}

,

F_{1y} = 0\text{ N}

;

F_{2x} = 10 \cos(60^\circ) = 5\text{ N}

,

F_{2y} = 10 \sin(60^\circ) = 5\sqrt{3}\text{ N}

.

Vector addition requires breaking non-orthogonal vectors into perpendicular component directions.

2

Calculate the total horizontal (

R_x

) and vertical (

R_y

) components of the resultant of the first two forces.

R_x = 10 + 5 = 15\text{ N}

,

R_y = 0 + 5\sqrt{3} = 5\sqrt{3}\text{ N}

.

Adding aligned components yields the net component along each axis.

3

Compute the magnitude of the resultant force

R

.

R = \sqrt{R_x^2 + R_y^2} = \sqrt{15^2 + (5\sqrt{3})^2} = \sqrt{225 + 75} = \sqrt{300} = 10\sqrt{3}\text{ N}

.

Applying Pythagoras' theorem to orthogonal components determines the net magnitude.

4

Determine the magnitude of the equilibrant (third force).

Equibrant magnitude

= R = 10\sqrt{3}\text{ N}

.

For static equilibrium, the third force must be equal in magnitude and opposite in direction to the resultant of the first two forces.

Key Concept

Vector Addition and Equilibrium of Forces

Question 7792Question →

A uniform cylindrical metallic conductor has an initial resistance of $12.0\,\Omega$ . The conductor is stretched uniformly until its length increases by $50\%$ , while maintaining constant mass and density. If a constant potential difference of $27.0\,\text{V}$ is subsequently applied across the ends of the stretched conductor, what is the electric current passing through it?

1.0\,\text{A}

1.5\,\text{A}

2.25\,\text{A}

0.67\,\text{A}

Show answer & explanation

Answer: $1.0\,\text{A}$

Answer

The electric current passing through the stretched conductor is

1.0\,\text{A}

.

When a metallic conductor of fixed mass and volume is stretched, increasing its length by a factor of

n = 1.5

causes its cross-sectional area to decrease by a factor of

1.5

. Because resistance is directly proportional to length and inversely proportional to cross-sectional area (

R = \rho L / A

), the new resistance becomes

n^2

times the initial resistance (

1.5^2 \times 12.0\,\Omega = 27.0\,\Omega

). By Ohm's law,

I = V / R = 27.0\,\text{V} / 27.0\,\Omega = 1.0\,\text{A}

.

Step-by-Step Solution

1

Determine the new length and cross-sectional area of the stretched conductor

L_2 = 1.5 L_1

and

A_2 = \frac{A_1}{1.5}

Increasing the length by

50\%

means

L_2 = L_1 + 0.5 L_1 = 1.5 L_1

. Since the volume

V = A \cdot L

remains constant during stretching,

A_1 L_1 = A_2 L_2

, which gives

A_2 = A_1 / 1.5

.

2

Calculate the new resistance of the conductor

R_2 = 27.0\,\Omega

Resistance is given by

R = \rho \frac{L}{A}

. Substituting the new length and area gives

R_2 = \rho \frac{1.5 L_1}{A_1 / 1.5} = (1.5)^2 \rho \frac{L_1}{A_1} = 2.25 R_1 = 2.25 \times 12.0\,\Omega = 27.0\,\Omega

.

3

Apply Ohm's law to find the current

I = 1.0\,\text{A}

Using

I = \frac{V}{R_2}

, substitute

V = 27.0\,\text{V}

and

R_2 = 27.0\,\Omega

to get

I = \frac{27.0\,\text{V}}{27.0\,\Omega} = 1.0\,\text{A}

.

Key Concept

Resistance variation with length and cross-sectional area under constant volume constraint (

R \propto L^2

when stretched)

Estimated Time:2m 0s

Question 7793Question →

If $y = e^{-2x} \sin(3x)$ , find $\frac{dy}{dx}$ .

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

e^{-2x} (\cos(3x) - 2\sin(3x))

e^{-2x} (3\cos(3x) - 2\sin(3x))

-6e^{-2x} \cos(3x)

Show answer & explanation

Answer: $e^{-2x} (3\cos(3x) - 2\sin(3x))$

Answer

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

Applying the product rule

\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}

to

u = e^{-2x}

and

v = \sin(3x)

yields

\frac{du}{dx} = -2e^{-2x}

and

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

. Substituting these terms gives

e^{-2x}(3\cos(3x)) + \sin(3x)(-2e^{-2x}) = e^{-2x}(3\cos(3x) - 2\sin(3x))

.

Step-by-Step Solution

1

Identify the component functions for the product rule

Let

u = e^{-2x}

and

v = \sin(3x)

.

The function

y

is a product of an exponential function and a trigonometric function.

2

Differentiate each component using the chain rule

\frac{du}{dx} = -2e^{-2x}

and

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

.

\frac{d}{dx}(e^{kx}) = k e^{kx}

and

\frac{d}{dx}(\sin(kx)) = k \cos(kx)

where

k

is a constant.

3

Apply the product rule formula

\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}

and factor out

e^{-2x}

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

.

Combining terms correctly gives the exact derivative.

Key Concept

Differentiation of Product of Transcendental Functions (Exponential and Trigonometric)

Question 7794Question →

An electric iron draws a steady current of $2.5\,\text{A}$ when connected to a mains supply. What total electric charge passes through the heating element of the appliance in $2.0\,\text{minutes}$ ?

5.0\,\text{C}

48\,\text{C}

300\,\text{C}

600\,\text{C}

Show answer & explanation

Answer: $300\,\text{C}$

Answer

The total electric charge passing through the heating element is

300\,\text{C}

.

Electric charge

Q

is calculated using the formula

Q = I \times t

. Converting

2.0\,\text{minutes}

into seconds gives

120\,\text{s}

. Multiplying the current of

2.5\,\text{A}

by

120\,\text{s}

gives

300\,\text{C}

.

Step-by-Step Solution

1

Convert the given time duration from minutes into seconds.

t = 2.0\,\text{minutes} \times 60\,\text{s/min} = 120\,\text{s}

.

Standard units require time to be measured in seconds when evaluating charge in coulombs.

2

Apply the electric charge formula

Q = I \times t

.

Q = 2.5\,\text{A} \times 120\,\text{s} = 300\,\text{C}

.

Electric current is defined as the total charge passing a given cross-section per unit time.

Key Concept

Electric Current and Charge Relationship

Question 7795Question →

A motorcycle traveling at a constant speed of $18\text{ m/s}$ passes a landmark. Exactly $4.0\text{ s}$ after passing the landmark, the rider accelerates uniformly at a rate of $2.5\text{ m/s}^2$ until reaching a speed of $28\text{ m/s}$ . What is the total distance, in meters, traveled by the motorcycle from the instant it passed the landmark to the moment it reaches $28\text{ m/s}$ ?

Show answer & explanation

Answer: 164

Answer

The total distance traveled by the motorcycle is

164\text{ m}

.

The total distance is obtained by finding the displacement during the

4.0\text{ s}

of constant speed at

18\text{ m/s}

(

72\text{ m}

) and adding the displacement during uniform acceleration from

18\text{ m/s}

to

28\text{ m/s}

at

2.5\text{ m/s}^2

(

92\text{ m}

), giving a sum of

164\text{ m}

.

Step-by-Step Solution

1

Calculate the distance covered during the initial constant speed stage.

The distance covered in the first

4.0\text{ s}

is

72\text{ m}

.

At a constant velocity

v = 18\text{ m/s}

, the distance

s_1 = v \times t = 18 \times 4.0 = 72\text{ m}

.

2

Calculate the distance covered during the accelerated motion stage.

The distance covered while accelerating from

18\text{ m/s}

to

28\text{ m/s}

is

92\text{ m}

.

Using the kinematic equation

v^2 = u^2 + 2as_2

, substitute

u = 18\text{ m/s}

,

v = 28\text{ m/s}

, and

a = 2.5\text{ m/s}^2

to get

28^2 = 18^2 + 2(2.5)s_2 \implies 784 = 324 + 5s_2 \implies s_2 = 92\text{ m}

.

3

Find the total distance traveled.

Total distance

s_{\text{total}} = 164\text{ m}

.

The total distance is the sum of the distance covered during the constant speed period (

72\text{ m}

) and during uniform acceleration (

92\text{ m}

).

Key Concept

Kinematics equations for multi-stage motion combining uniform speed and uniform acceleration.

Question 7796Question →

The line $(k+1)x + 3y - 5 = 0$ is perpendicular to the line passing through the points $(2, 1)$ and $(4, 5)$ . What is the value of $k$ ?

-7

\frac{1}{2}

-\frac{5}{2}

\frac{5}{2}

Show answer & explanation

Answer: $\frac{1}{2}$

Answer

The value of

k

is

\frac{1}{2}

.

The line passing through

(2, 1)

and

(4, 5)

has a gradient

m_1 = \frac{5-1}{4-2} = 2

. Rearranging

(k+1)x + 3y - 5 = 0

into

y = -\frac{k+1}{3}x + \frac{5}{3}

gives its gradient

m_2 = -\frac{k+1}{3}

. Using the perpendicular condition

m_1 \cdot m_2 = -1

, we get

2 \cdot \left(-\frac{k+1}{3}\right) = -1

, which simplifies to

2k + 2 = 3

and yields

k = \frac{1}{2}

.

Step-by-Step Solution

1

Calculate the gradient of the line passing through the points

(2, 1)

and

(4, 5)

.

m_1 = \frac{5 - 1}{4 - 2} = \frac{4}{2} = 2

The gradient between two points

(x_1, y_1)

and

(x_2, y_2)

is given by

m = \frac{y_2 - y_1}{x_2 - x_1}

.

2

Express the line

(k+1)x + 3y - 5 = 0

in slope-intercept form

y = mx + c

to find its gradient

m_2

.

3y = -(k+1)x + 5 \implies y = -\left(\frac{k+1}{3}\right)x + \frac{5}{3}

, so

m_2 = -\frac{k+1}{3}

The coefficient of

x

when solved for

y

represents the gradient of the line.

3

Apply the perpendicularity condition

m_1 \cdot m_2 = -1

and solve for

k

.

2 \cdot \left(-\frac{k+1}{3}\right) = -1 \implies -\frac{2(k+1)}{3} = -1 \implies 2(k+1) = 3 \implies 2k + 2 = 3 \implies k = \frac{1}{2}

Two non-vertical lines are perpendicular if and only if the product of their gradients is

-1

.

Key Concept

Perpendicular Lines and Gradients

Estimated Time:1m 30s

Question 7797Question →

A solid metal block of mass $4.0\text{ kg}$ has a density of $8000\text{ kg m}^{-3}$ at $20^\circ\text{C}$ . If the linear expansivity of the metal is $2.0 \times 10^{-5}\text{ K}^{-1}$ , calculate the increase in volume of the block, in $\text{cm}^3$ , when its temperature is raised to $120^\circ\text{C}$ .

Show answer & explanation

Answer: 3

Answer

The increase in volume of the block is

3.0\text{ cm}^3

.

First, the initial volume of the metal block is found using

V_0 = \frac{m}{\rho} = \frac{4.0\text{ kg}}{8000\text{ kg m}^{-3}} = 5.0 \times 10^{-4}\text{ m}^3 = 500\text{ cm}^3

. Next, because a solid expands in three dimensions, the cubic expansivity is

\gamma = 3\alpha = 3 \times (2.0 \times 10^{-5}\text{ K}^{-1}) = 6.0 \times 10^{-5}\text{ K}^{-1}

. The temperature change is

\Delta T = 120^\circ\text{C} - 20^\circ\text{C} = 100\text{ K}

. Finally, the increase in volume is computed as

\Delta V = V_0 \gamma \Delta T = 500\text{ cm}^3 \times (6.0 \times 10^{-5}\text{ K}^{-1}) \times 100\text{ K} = 3.0\text{ cm}^3

.

Step-by-Step Solution

1

Calculate the initial volume of the metal block from its mass and density.

V_0 = \frac{m}{\rho_0} = \frac{4.0\text{ kg}}{8000\text{ kg m}^{-3}} = 5.0 \times 10^{-4}\text{ m}^3 = 500\text{ cm}^3

Volume is equal to mass divided by density.

2

Convert linear expansivity (

\alpha

) to volume expansivity (

\gamma

).

\gamma = 3\alpha = 3 \times (2.0 \times 10^{-5}\text{ K}^{-1}) = 6.0 \times 10^{-5}\text{ K}^{-1}

For an isotropic solid, cubic expansivity is three times its linear expansivity.

3

Determine the change in temperature.

\Delta T = 120^\circ\text{C} - 20^\circ\text{C} = 100\text{ K}

Thermal expansion is driven by the change in temperature.

4

Calculate the total volume expansion.

\Delta V = V_0 \gamma \Delta T = 500\text{ cm}^3 \times (6.0 \times 10^{-5}\text{ K}^{-1}) \times 100\text{ K} = 3.0\text{ cm}^3

Applying the thermal volume expansion formula

\Delta V = V_0 \gamma \Delta T

.

Key Concept

Thermal Volume Expansion of Solids

Estimated Time:2m 0s

Question 7798Question →

A traffic officer on a stationary motorcycle spots a car moving past at a constant speed of $30\text{ m/s}$ . The officer takes $2\text{ s}$ of reaction time before accelerating uniformly at $4\text{ m/s}^2$ to a maximum cruise speed of $40\text{ m/s}$ , after which the motorcycle continues at this constant speed. What is the total distance traveled by the motorcycle from its initial position to the point where it catches up with the car?

360\text{ m}

510\text{ m}

600\text{ m}

840\text{ m}

Show answer & explanation

Answer: $840\text{ m}$

Answer

840\text{ m}

The correct answer is

840\text{ m}

. Over the first

2\text{ s}

, the motorcycle is stationary while the car covers

60\text{ m}

. Over the next

10\text{ s}

, the motorcycle accelerates to

40\text{ m/s}

, covering

200\text{ m}

, while the car travels another

300\text{ m}

(totaling

360\text{ m}

). To close the remaining

160\text{ m}

gap at a relative speed of

10\text{ m/s}

requires an additional

16\text{ s}

. The total elapsed time of

28\text{ s}

yields a total catch-up distance of

30\text{ m/s} \times 28\text{ s} = 840\text{ m}

.

Step-by-Step Solution

1

Calculate car position and motorcycle position at the end of the reaction time period (

t = 2\text{ s}

).

During the reaction time

t_r = 2\text{ s}

, the motorcycle remains stationary (

s_m = 0\text{ m}

). The car travels a distance

s_c = 30\text{ m/s} \times 2\text{ s} = 60\text{ m}

.

The motorcycle does not begin accelerating until after the officer's reaction time elapses.

2

Determine the time and distance required for the motorcycle to reach its maximum speed of

40\text{ m/s}

.

Time to reach maximum speed:

t_{acc} = \frac{v_{max} - u}{a} = \frac{40 - 0}{4} = 10\text{ s}

. Distance during acceleration:

s_{acc} = \frac{v_{max}^2 - u^2}{2a} = \frac{40^2 - 0}{2(4)} = 200\text{ m}

.

The motorcycle accelerates uniformly from rest until reaching its capped maximum velocity.

3

Calculate total positions at

t_{total1} = 2\text{ s} + 10\text{ s} = 12\text{ s}

from the instant the car passed.

Motorcycle position:

x_m(12) = 200\text{ m}

. Car position:

x_c(12) = 30\text{ m/s} \times 12\text{ s} = 360\text{ m}

. Distance gap remaining:

\Delta x = 360\text{ m} - 200\text{ m} = 160\text{ m}

.

Comparing positions at

12\text{ s}

establishes the remaining distance gap to be closed during the constant speed phase.

4

Calculate the time required during the constant-speed phase to close the remaining gap and find the total meeting distance.

Relative speed:

v_{rel} = 40\text{ m/s} - 30\text{ m/s} = 10\text{ m/s}

. Additional time needed:

\Delta t = \frac{160\text{ m}}{10\text{ m/s}} = 16\text{ s}

. Total elapsed time:

t = 12\text{ s} + 16\text{ s} = 28\text{ s}

. Catch-up distance:

s_{total} = 30\text{ m/s} \times 28\text{ s} = 840\text{ m}

.

With both vehicles moving at constant speeds, the relative velocity determines how quickly the remaining separation is eliminated.

Key Concept

Multi-stage linear motion involving delayed reaction time, uniform acceleration, and constant speed pursuit.

Estimated Time:3m 0s

Question 7799Question →

A parallel-plate capacitor with air between its plates has a capacitance of $15\text{ }\mu\text{F}$ . If the plate separation is reduced to one-third of its initial value and a dielectric material of relative permittivity $\varepsilon_r = 4.0$ is completely inserted between the plates, what is the new capacitance of the capacitor?

20\text{ }\mu\text{F}

45\text{ }\mu\text{F}

60\text{ }\mu\text{F}

180\text{ }\mu\text{F}

Show answer & explanation

Answer: $180\text{ }\mu\text{F}$

Answer

The new capacitance of the capacitor is

180\text{ }\mu\text{F}

.

The capacitance of a parallel-plate capacitor is given by

C = \frac{\varepsilon_r \varepsilon_0 A}{d}

. Reducing plate separation to one-third increases capacitance by a factor of 3. Adding a dielectric with relative permittivity

\varepsilon_r = 4.0

increases capacitance by a factor of 4. Combining both effects increases capacitance by a total factor of

3 \times 4 = 12

, yielding

12 \times 15\text{ }\mu\text{F} = 180\text{ }\mu\text{F}

.

Step-by-Step Solution

1

Express the initial capacitance

C_1

using the formula for a parallel-plate air capacitor.

C_1 = \frac{\varepsilon_0 A}{d} = 15\text{ }\mu\text{F}

Air has a relative permittivity of 1.

2

Write the formula for the modified capacitance

C_2

with plate separation

d' = \frac{d}{3}

and dielectric constant

\varepsilon_r = 4.0

.

C_2 = \frac{\varepsilon_r \varepsilon_0 A}{d'} = \frac{4.0 \varepsilon_0 A}{\frac{d}{3}} = 4.0 \times 3 \times \left(\frac{\varepsilon_0 A}{d}\right) = 12 C_1

Reducing distance to

d/3

increases capacitance by a factor of 3, and inserting the dielectric increases capacitance by a factor of 4.

3

Calculate the value of the new capacitance.

C_2 = 12 \times 15\text{ }\mu\text{F} = 180\text{ }\mu\text{F}

Multiplying the combined scaling factor by the initial capacitance gives the final answer.

Key Concept

Parallel Plate Capacitance and Dielectrics

Estimated Time:1m 30s

Question 7800Question →

The cumulative frequency distribution below shows the monthly electricity consumption (in kWh) recorded for $50$ households in a residential estate:

Electricity Usage (kWh)	Cumulative Frequency
$\leq 50$	5
$\leq 100$	15
$\leq 150$	35
$\leq 200$	45
$\leq 250$	50

Using linear interpolation, calculate the 60th percentile ( $P_{60}$ ) of electricity consumption in kWh.

Show answer & explanation

Answer: 137.5

Answer

137.5 kWh

The 60th percentile corresponds to the 30th observation out of 50 (

0.60 \times 50 = 30

). From the cumulative frequency table, the 30th value falls in the class interval between

100

and

150

. Using the lower boundary of

100

, previous cumulative frequency of

15

, interval frequency of

20

, and interval width of

50

, linear interpolation gives

100 + \frac{30 - 15}{20} \times 50 = 137.5

kWh.

Step-by-Step Solution

1

Determine the rank for the 60th percentile

Rank position = 30th household

The 60th percentile rank corresponds to

60\%

of the total sample size (

N = 50

), which is

0.60 \times 50 = 30

.

2

Identify the parameters of the percentile class interval

Lower boundary

L = 100

,

CF_{\text{prev}} = 15

, class frequency

f = 20

, width

w = 50

The cumulative frequency rises from 15 to 35 in the class interval

(100, 150]

, so the 30th value lies within this interval.

3

Calculate the value using linear interpolation

137.5 kWh

Substitute values into

P_{60} = 100 + \left(\frac{30 - 15}{20}\right) \times 50 = 100 + 37.5 = 137.5

kWh.

Key Concept

Linear interpolation for percentiles from a cumulative frequency distribution

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