All practice questions

Question 4461Question →

A uniform conductor of length $50\,\text{m}$ and cross-sectional area $2.0 \times 10^{-6}\,\text{m}^2$ is made of a material with a resistivity of $4.0 \times 10^{-7}\,\Omega\cdot\text{m}$ . What is the electrical resistance of the conductor in ohms?

Show answer & explanation

Answer: 10

Answer

The resistance of the conductor is

10\,\Omega

.

The resistance of a uniform conductor is given by

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

. Substituting the values

L = 50\,\text{m}

,

A = 2.0 \times 10^{-6}\,\text{m}^2

, and

\rho = 4.0 \times 10^{-7}\,\Omega\cdot\text{m}

yields

R = \frac{(4.0 \times 10^{-7})(50)}{2.0 \times 10^{-6}} = 10\,\Omega

.

Step-by-Step Solution

1

Identify given physical quantities

L = 50\,\text{m}

,

A = 2.0 \times 10^{-6}\,\text{m}^2

,

\rho = 4.0 \times 10^{-7}\,\Omega\cdot\text{m}

Extracting given parameter values clearly sets up the mathematical relationship.

2

Apply the resistivity formula for resistance

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

Resistance varies directly with length and resistivity, and inversely with cross-sectional area.

3

Perform the calculation

R = \frac{(4.0 \times 10^{-7}\,\Omega\cdot\text{m})(50\,\text{m})}{2.0 \times 10^{-6}\,\text{m}^2} = 10\,\Omega

Multiplying the numerator gives

2.0 \times 10^{-5}\,\Omega\cdot\text{m}^2

; dividing by

2.0 \times 10^{-6}\,\text{m}^2

yields

10\,\Omega

.

Key Concept

Direct computation of electrical resistance using resistivity, length, and cross-sectional area

Estimated Time:45s

Question 4462Question →

Two point charges, $Q_1 = +3.0 \times 10^{-6}\text{ C}$ and $Q_2 = +4.0 \times 10^{-6}\text{ C}$ , are positioned in a vacuum at coordinates $(0\text{ m}, 3.0\text{ m})$ and $(3.0\text{ m}, 0\text{ m})$ respectively on a Cartesian plane. Taking the electrostatic constant $k = 9.0 \times 10^9\text{ N m}^2\text{ C}^{-2}$ , what is the magnitude of the net electric field intensity at the origin $(0,0)$ ?

1.0 \times 10^3\text{ N C}^{-1}

5.0 \times 10^3\text{ N C}^{-1}

7.0 \times 10^3\text{ N C}^{-1}

1.5 \times 10^4\text{ N C}^{-1}

Show answer & explanation

Answer: $5.0 \times 10^3\text{ N C}^{-1}$

Answer

The magnitude of the net electric field intensity at the origin is

5.0 \times 10^3\text{ N C}^{-1}

.

The electric field intensity at the origin is a vector sum of the individual electric fields created by each point charge. The field due to the charge on the y-axis points downward along the y-axis with a magnitude of

3.0 \times 10^3\text{ N C}^{-1}

, while the field due to the charge on the x-axis points leftward along the x-axis with a magnitude of

4.0 \times 10^3\text{ N C}^{-1}

. Because these two fields act at right angles to each other, their vector sum magnitude is given by

\sqrt{(3.0 \times 10^3)^2 + (4.0 \times 10^3)^2} = 5.0 \times 10^3\text{ N C}^{-1}

.

Step-by-Step Solution

1

Calculate the electric field intensity

E_1

at the origin due to charge

Q_1

E_1 = \frac{k |Q_1|}{r_1^2} = \frac{(9.0 \times 10^9)(3.0 \times 10^{-6})}{3.0^2} = 3.0 \times 10^3\text{ N C}^{-1}

, directed along the negative y-axis.

Electric field magnitude follows Coulomb's law for field strength, and positive charges create fields directed away from themselves.

2

Calculate the electric field intensity

E_2

at the origin due to charge

Q_2

E_2 = \frac{k |Q_2|}{r_2^2} = \frac{(9.0 \times 10^9)(4.0 \times 10^{-6})}{3.0^2} = 4.0 \times 10^3\text{ N C}^{-1}

, directed along the negative x-axis.

The charge is located at

(3.0, 0)

on the x-axis, producing a field pointing toward the origin.

3

Calculate the net electric field vector magnitude at the origin

E_{\text{net}} = \sqrt{E_1^2 + E_2^2} = \sqrt{(3.0 \times 10^3)^2 + (4.0 \times 10^3)^2} = 5.0 \times 10^3\text{ N C}^{-1}

.

Since the two component fields are perpendicular along orthogonal axes (x and y), their resultant is found using the Pythagorean theorem.

Key Concept

Vector addition of electric field intensities from multiple point charges

Question 4463Question →

An alternating current (AC) circuit consists of a $40\ \Omega$ resistor, an inductor with a reactance of $70\ \Omega$ , and a capacitor with a reactance of $40\ \Omega$ connected in series across an AC supply. What is the power factor of the circuit?

0.80

0.60

0.27

1.00

Show answer & explanation

Answer: 0.80

Answer

0.80

The total impedance of the series RLC circuit is found using phasor addition:

Z = \sqrt{40^2 + (70 - 40)^2} = 50\ \Omega

. The power factor is the ratio of resistance to impedance:

\cos \theta = \frac{40}{50} = 0.80

.

Step-by-Step Solution

1

Calculate the net reactance (

X_{\text{net}}

) of the circuit

X_{\text{net}} = X_L - X_C = 70\ \Omega - 40\ \Omega = 30\ \Omega

In a series RLC circuit, inductive and capacitive reactances oppose each other in phase.

2

Calculate total impedance (

Z

) using phasor addition

Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{40^2 + 30^2} = \sqrt{1600 + 900} = 50\ \Omega

Resistance and net reactance act at right angles in the impedance phasor diagram.

3

Calculate the power factor (

\cos \theta

)

\cos \theta = \frac{R}{Z} = \frac{40}{50} = 0.80

The power factor is defined as the cosine of the phase angle, which equals the ratio of resistance to total impedance.

Key Concept

Power factor of a series RLC AC circuit

Estimated Time:1m 30s

Question 4464Question →

A progressive wave traveling along a stretched string is represented by the equation $y = 0.02 \sin(50\pi t - 4\pi x)$ , where $x$ and $y$ are in meters and $t$ is in seconds. What is the speed of the wave in meters per second ( $\text{m/s}$ )?

Show answer & explanation

Answer: 12.5

Answer

The speed of the wave is

12.5\text{ m/s}

.

Comparing the wave equation

y = 0.02 \sin(50\pi t - 4\pi x)

with the standard form

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

shows that the angular frequency is

\omega = 50\pi\text{ rad/s}

and the wave number is

k = 4\pi\text{ rad/m}

. Using the relation

v = \frac{\omega}{k}

, the speed of the wave is calculated as

v = \frac{50\pi}{4\pi} = 12.5\text{ m/s}

.

Step-by-Step Solution

1

Extract angular frequency and wave number from the equation

Comparing

y = 0.02 \sin(50\pi t - 4\pi x)

with

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

gives

\omega = 50\pi\text{ rad/s}

and

k = 4\pi\text{ rad/m}

.

Matching corresponding terms yields the wave parameters.

2

Calculate the wave propagation speed

v = \frac{\omega}{k} = \frac{50\pi}{4\pi} = 12.5\text{ m/s}

.

Wave speed equals the ratio of angular frequency to wave number.

Key Concept

Calculating wave speed from a mathematical wave equation

Question 4465Question →

The interior angles of a convex pentagon are given as $(2x + 10)^\circ$ , $(x + 25)^\circ$ , $(3x - 15)^\circ$ , $(2x + 30)^\circ$ , and $(2x + 10)^\circ$ . What is the value of $x$ ?

30

48

60

66

Show answer & explanation

Answer: 48

Answer

The value of

x

is 48.

For a pentagon (

n = 5

), the sum of interior angles is

(5 - 2) \times 180^\circ = 540^\circ

. Summing the five given interior angle expressions yields

10x + 60

. Equating

10x + 60 = 540

gives

10x = 480

, which solves to

x = 48

.

Step-by-Step Solution

1

Calculate the sum of interior angles of a 5-sided polygon (pentagon)

Sum

= (5 - 2) \times 180^\circ = 3 \times 180^\circ = 540^\circ

The formula for the sum of interior angles of a polygon with

n

sides is

(n - 2) \times 180^\circ

.

2

Sum the given algebraic expressions for the interior angles

(2x + 10) + (x + 25) + (3x - 15) + (2x + 30) + (2x + 10) = 10x + 60

Combine like terms for the

x

terms and constant terms.

3

Equate the sum of expressions to

540^\circ

and solve for

x

10x + 60 = 540 \implies 10x = 480 \implies x = 48

Subtract 60 from both sides and divide by 10 to isolate

x

.

Key Concept

Interior Angle Sum of Polygons

Estimated Time:1m 15s

Question 4466Question →

A car initially traveling at a constant speed of $15\text{ m/s}$ accelerates uniformly at $2\text{ m/s}^2$ for $5\text{ s}$ . It then maintains the acquired maximum speed for $10\text{ s}$ before coming to rest under uniform retardation in $4\text{ s}$ . What is the total distance covered by the car during the entire motion?

Show answer & explanation

Answer: 400

Answer

The total distance covered by the car during the entire motion is

400\text{ m}

.

The total distance is calculated by summing the distances covered in the three distinct phases of motion: acceleration (

100\text{ m}

), uniform velocity (

250\text{ m}

), and uniform retardation (

50\text{ m}

), yielding a total distance of

400\text{ m}

.

Step-by-Step Solution

1

Calculate final speed and distance for the acceleration phase.

Final speed

v = 25\text{ m/s}

and distance

s_1 = 100\text{ m}

.

Using kinematic equations

v = u + a t_1 = 15 + (2)(5) = 25\text{ m/s}

and

s_1 = u t_1 + \frac{1}{2}a t_1^2 = 15(5) + \frac{1}{2}(2)(5^2) = 75 + 25 = 100\text{ m}

.

2

Calculate the distance covered during the constant speed phase.

Distance

s_2 = 250\text{ m}

.

The car maintains the acquired speed of

25\text{ m/s}

for

10\text{ s}

, giving

s_2 = v \cdot t_2 = 25 \times 10 = 250\text{ m}

.

3

Calculate the distance covered during the retardation phase.

Distance

s_3 = 50\text{ m}

.

Using average velocity for uniform retardation to rest:

s_3 = \frac{v + 0}{2} t_3 = \frac{25}{2} \times 4 = 50\text{ m}

.

4

Sum the distances from all three stages to determine total distance.

Total distance

S = 400\text{ m}

.

S = s_1 + s_2 + s_3 = 100 + 250 + 50 = 400\text{ m}

.

Key Concept

Multi-stage linear motion and equations of uniform acceleration

Question 4467Question →

Under the same conditions of temperature and pressure, Gas X has a molar mass of $4\text{ g/mol}$ and Gas Y has a molar mass of $64\text{ g/mol}$ . How many times faster will Gas X diffuse compared to Gas Y?

4 times as fast

16 times as fast

2 times as fast

0.25 times as fast

Show answer & explanation

Answer: 4 times as fast

Answer

Gas X diffuses 4 times as fast as Gas Y.

According to Graham's Law of Diffusion, the rate of diffusion

r

of a gas is inversely proportional to the square root of its molar mass

M

. Therefore,

\frac{r_X}{r_Y} = \sqrt{\frac{M_Y}{M_X}} = \sqrt{\frac{64}{4}} = \sqrt{16} = 4

. This means Gas X diffuses 4 times faster than Gas Y.

Step-by-Step Solution

1

State Graham's Law formula for relative rates of diffusion

\frac{r_X}{r_Y} = \sqrt{\frac{M_Y}{M_X}}

The rate of diffusion is inversely proportional to the square root of molar mass.

2

Substitute the given molar masses into the formula

\frac{r_X}{r_Y} = \sqrt{\frac{64}{4}} = \sqrt{16}

Molar mass of Gas Y (

M_Y

) is

64\text{ g/mol}

and Gas X (

M_X

) is

4\text{ g/mol}

.

3

Simplify the square root expression

\frac{r_X}{r_Y} = 4

The square root of 16 is 4, showing Gas X diffuses 4 times as fast as Gas Y.

Key Concept

Graham's Law of Diffusion

Estimated Time:45s

Question 4468Question →

A radioactive isotope has a half-life of $4\text{ hours}$ . If a sample initially contains $80\text{ g}$ of the isotope, what mass of the isotope, in grams, will remain undecayed after $12\text{ hours}$ ?

Show answer & explanation

Answer: 10

Answer

The mass of the radioactive isotope remaining undecayed after

12\text{ hours}

is

10\text{ g}

.

After

3

half-lives (

12\text{ hours}

total elapsed time with a half-life of

4\text{ hours}

), the fraction of the initial sample remaining is

\left(\frac{1}{2}\right)^3 = \frac{1}{8}

. Multiplying this fraction by the initial mass of

80\text{ g}

gives

10\text{ g}

.

Step-by-Step Solution

1

Determine the number of elapsed half-lives (

n

)

n = \frac{12\text{ hours}}{4\text{ hours}} = 3

half-lives

Dividing the total time elapsed by the half-life period gives the number of decay cycles.

2

Calculate the mass remaining after

3

half-lives

N = 80 \times \left(\frac{1}{2}\right)^3 = 80 \times \frac{1}{8} = 10\text{ g}

The remaining mass halves during each half-life interval according to the exponential decay rule

N = N_0 (1/2)^n

.

Key Concept

Radioactive Decay Law and Half-life

Question 4469Question →

Based on the postulates of the Kinetic Molecular Theory, a gas is considered ideal when intermolecular forces are negligible and the actual volume of the gas molecules is insignificant compared to the container volume. Under high pressure and low temperature conditions, real gases deviate markedly from this ideal behavior and undergo liquefaction. Which of the following best explains the microscopic behavior of gas particles under these extreme conditions?

The average kinetic energy of the molecules decreases sufficiently for intermolecular attractive forces to become dominant, while the actual volume of the gas particles becomes a significant fraction of the total container volume.

The average kinetic energy of the molecules increases dramatically, causing collisions against container walls to become completely inelastic and resulting in rapid loss of momentum.

The intermolecular attractive forces completely vanish as particles draw closer together, forcing the gas volume to shrink linearly to zero at absolute zero temperature.

High pressure increases the speed of the gas molecules, causing repulsive forces to dominate over attractive forces and preventing condensation.

Show answer & explanation

Answer: The average kinetic energy of the molecules decreases sufficiently for intermolecular attractive forces to become dominant, while the actual volume of the gas particles becomes a significant fraction of the total container volume.

Answer

The average kinetic energy of the molecules decreases sufficiently for intermolecular attractive forces to become dominant, while the actual volume of the gas particles becomes a significant fraction of the total container volume.

According to the kinetic theory of matter, ideal gases assume no intermolecular attraction and zero molecular volume. At low temperatures, the decreased kinetic energy allows attractive forces between gas particles to overcome thermal motion. Simultaneously, at high pressures, gas particles are packed closely, making their individual volume significant relative to the total volume. These combined effects violate the ideal gas postulates and cause the gas to condense into a liquid.

Step-by-Step Solution

1

Analyze the effect of low temperature on particle kinetic energy

Decreasing temperature lowers the average kinetic energy of gas molecules (

E_k \propto T

), slowing them down.

Slower-moving molecules spend more time in proximity, allowing weak intermolecular attractive forces (Van der Waals forces) to overcome kinetic energy and pull particles together.

2

Analyze the effect of high pressure on gas particle volume

High pressure compresses the total volume of the container, forcing gas particles close together.

As the space between particles diminishes, the finite volume of the gas molecules themselves is no longer negligible relative to the reduced total container volume.

3

Synthesize the breakdown of Kinetic Molecular Theory postulates leading to liquefaction

Both key assumptions of ideal behavior (zero intermolecular forces and negligible particle volume) fail simultaneously.

The dominant attractive forces and significant molecular volume cause real gases to deviate from ideal gas laws and condense into a liquid state.

Key Concept

Deviations of Real Gases from Kinetic Molecular Theory Postulates

Estimated Time:1m 30s

Question 4470Question →

A dentist uses a small concave mirror with a focal length of $20\text{ mm}$ to inspect a patient's tooth. If the mirror produces an upright image that is magnified $4$ times, how far from the tooth is the mirror placed?

15\text{ mm}

25\text{ mm}

80\text{ mm}

5\text{ mm}

Show answer & explanation

Answer: $15\text{ mm}$

Answer

The mirror must be placed

15\text{ mm}

from the tooth.

For a concave mirror, an upright image is virtual. Linear magnification

m = 4

implies

v = -4u

. Substituting

f = 20\text{ mm}

into the mirror formula

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

gives

\frac{1}{20} = \frac{1}{u} - \frac{1}{4u} = \frac{3}{4u}

, solving to

u = 15\text{ mm}

.

Step-by-Step Solution

1

Identify given values and apply sign conventions.

Focal length of concave mirror

f = +20\text{ mm}

. Magnification

m = +4

because the image is upright (virtual).

Upright images formed by single optical mirrors are always virtual, requiring a negative image distance.

2

Express image distance

v

in terms of object distance

u

.

Linear magnification

m = -\frac{v}{u} \implies +4 = -\frac{v}{u} \implies v = -4u

.

The linear magnification formula relates orientation, object distance, and image distance.

3

Substitute

f

and

v

into the mirror formula.

\frac{1}{f} = \frac{1}{u} + \frac{1}{v} \implies \frac{1}{20} = \frac{1}{u} - \frac{1}{4u} = \frac{3}{4u}.

The mirror equation determines object and image locations relative to the focal length.

4

Solve for the object distance

u

.

4u = 60 \implies u = 15\text{ mm}.

Cross-multiplying gives the required distance between the mirror and the tooth.

Key Concept

Reflection and image formation by concave spherical mirrors (virtual magnified image)

Question 4471Question →

An arithmetic progression (AP) has a first term of $2$ and a common difference of $3$ . A geometric progression (GP) has a first term of $1$ and a common ratio of $2$ . Arrange the following quantities in ascending order of their numerical values:

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct ascending order is the 4th term of the GP (8), followed by the 4th term of the AP (11), the sum of the first 3 terms of the AP (15), and the 5th term of the GP (16).

Evaluating each term individually gives: the 4th term of the GP equals 8, the 4th term of the AP equals 11, the sum of the first 3 terms of the AP equals 15, and the 5th term of the GP equals 16. Arranging these calculated values from smallest to largest gives the order 8, 11, 15, 16.

Step-by-Step Solution

1

Calculate the 4th term of the AP

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

Using the AP

n^{\text{th}}

term formula

T_n = a + (n-1)d

.

2

Calculate the 5th term of the GP

T_5 = 1 \times 2^{5-1} = 2^4 = 16

Using the GP

n^{\text{th}}

term formula

T_n = a r^{n-1}

.

3

Calculate the sum of the first 3 terms of the AP

S_3 = \frac{3}{2}[2(2) + (3-1)3] = \frac{3}{2}[4 + 6] = 15

Using the AP sum formula

S_n = \frac{n}{2}[2a + (n-1)d]

.

4

Calculate the 4th term of the GP

T_4 = 1 \times 2^{4-1} = 2^3 = 8

Using the GP

n^{\text{th}}

term formula

T_n = a r^{n-1}

.

5

Compare and arrange the values in ascending order

8 < 11 < 15 < 16

Arranging from smallest to largest numerical value.

Key Concept

Nth term and sum formulas of Arithmetic and Geometric Progressions

Question 4472Question →

A $5.00\text{ g}$ sample of impure limestone ( $\text{CaCO}_3$ ) is strongly heated until decomposition is complete. If the loss in mass due to the escape of carbon dioxide ( $\text{CO}_2$ ) gas is $1.76\text{ g}$ , what is the percentage purity of the limestone sample? $[\text{Relative atomic masses: } \text{Ca} = 40, \text{C} = 12, \text{O} = 16]$

Show answer & explanation

Answer: 80

Answer

The percentage purity of the limestone sample is 80%.

Thermal decomposition of calcium carbonate yields calcium oxide and carbon dioxide. The mass loss of 1.76 g corresponds to the evolved CO2. From the molar masses (CaCO3 = 100 g/mol, CO2 = 44 g/mol), 44 g of CO2 is released by 100 g of pure CaCO3. Thus, 1.76 g of CO2 is released by 4.00 g of pure CaCO3. Dividing the pure mass (4.00 g) by the original sample mass (5.00 g) and multiplying by 100 yields a percentage purity of 80%.

Step-by-Step Solution

1

Write the balanced equation for the decomposition reaction.

\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)

The decrease in mass is entirely due to the evolved carbon dioxide gas.

2

Calculate the relative formula mass of calcium carbonate and carbon dioxide.

\text{Molar mass of } \text{CaCO}_3 = 100\text{ g/mol}, \quad \text{Molar mass of } \text{CO}_2 = 44\text{ g/mol}

Required to relate the mass of evolved gas to the mass of reacting calcium carbonate.

3

Calculate the mass of pure calcium carbonate in the sample.

\text{Mass of pure } \text{CaCO}_3 = \left(\frac{100}{44}\right) \times 1.76\text{ g} = 4.00\text{ g}

Direct stoichiometric ratio derived from 1 mol CaCO3 producing 1 mol CO2.

4

Calculate percentage purity.

\text{Percentage purity} = \left(\frac{4.00\text{ g}}{5.00\text{ g}}\right) \times 100 = 80\%

Ratio of pure reactant mass to total sample mass expressed as a percentage.

Key Concept

Calculating percentage purity using stoichiometry and gravimetric decomposition data.

Question 4473Question →

What is the correct sequential order of laboratory steps required to obtain pure, dry copper(II) sulfate crystals from a mixture of solid copper(II) sulfate and insoluble sand?

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence begins with dissolving the soluble component in water, filtering to remove insoluble sand, heating the filtrate to obtain a saturated solution, cooling gradually to precipitate pure crystals, and finally filtering and drying the isolated crystals.

The purification of a mixture containing a soluble salt and an insoluble impurity requires dissolving the salt in water first, followed by filtration to remove the insoluble residue. To obtain pure crystals, the filtrate must be heated only until saturated and then allowed to cool slowly, as rapid or total evaporation to dryness destroys hydrated crystals. Finally, filtering and blotting the crystals isolates pure, dry hydrated salt.

Step-by-Step Solution

1

Dissolve the mixture in water.

Copper(II) sulfate dissolves into solution while sand remains solid.

Exploits the differential solubility of the two components.

2

Filter the suspension.

Sand is retained on the filter paper as residue; clear copper(II) sulfate solution passes through as filtrate.

Filtration physically separates an insoluble solid from a liquid.

3

Evaporate partially to crystallizing point.

A hot saturated solution is formed.

Evaporating to dryness would decompose the hydrated salt into an anhydrous powder.

4

Cool the saturated solution.

Pure copper(II) sulfate crystals precipitate from the solution.

Solubility of most solid solutes decreases as temperature falls.

5

Isolate and dry the crystals.

Pure, dry crystals of hydrated copper(II) sulfate are obtained.

Filtration collects the solid crystals, and drying with filter paper removes residual moisture without removing water of crystallization.

Key Concept

Multi-step purification of soluble and insoluble solid mixtures via selective dissolution, filtration, evaporation to saturation, and crystallization.

Question 4474Question →

A light wave of frequency $5.00 \times 10^{14}\text{ Hz}$ travels from air into a glass medium with a refractive index of $1.50$ . What is the frequency of the light wave inside the glass medium?

3.33 \times 10^{14}\text{ Hz}

5.00 \times 10^{14}\text{ Hz}

7.50 \times 10^{14}\text{ Hz}

2.25 \times 10^{14}\text{ Hz}

Show answer & explanation

Answer: $5.00 \times 10^{14}\text{ Hz}$

Answer

The frequency of the light wave inside the glass medium remains

5.00 \times 10^{14}\text{ Hz}

.

When light passes from one medium into another, its speed and wavelength change, but its frequency remains constant because it depends solely on the source producing the wave.

Step-by-Step Solution

1

Identify the given physical parameters.

Initial frequency

f = 5.00 \times 10^{14}\text{ Hz}

and refractive index

n = 1.50

.

Understanding provided quantities is the starting point for wave refraction analysis.

2

Apply the principle of frequency conservation across optical boundaries.

Frequency in glass medium

f_{\text{glass}} = f_{\text{air}} = 5.00 \times 10^{14}\text{ Hz}

.

When a wave passes into a different medium, its speed and wavelength alter in proportion to the refractive index, but its frequency is determined solely by the periodic source and remains constant.

Key Concept

Invariance of Wave Frequency across Refracting Media

Question 4475Question →

In a regular polygon, the measure of each interior angle is $132^\circ$ greater than the measure of each exterior angle. How many sides does this polygon have?

Show answer & explanation

Answer: 15

Answer

The polygon has 15 sides.

Since the interior angle

I

and exterior angle

E

of a regular polygon sum to

180^\circ

(

I + E = 180^\circ

) and their given difference is

I - E = 132^\circ

, subtracting the difference equation from the sum equation gives

2E = 48^\circ

, which simplifies to

E = 24^\circ

. The number of sides is

n = \frac{360^\circ}{E} = \frac{360^\circ}{24^\circ} = 15

.

Step-by-Step Solution

1

Set up the linear pair equation for interior and exterior angles

I + E = 180^\circ

An interior angle and its adjacent exterior angle at any vertex of a polygon lie on a straight line and sum to

180^\circ

.

2

Set up the given condition equation

I - E = 132^\circ

The question states that each interior angle is

132^\circ

greater than each exterior angle.

3

Solve for the exterior angle

E

E = 24^\circ

Subtracting

I - E = 132^\circ

from

I + E = 180^\circ

yields

2E = 48^\circ

, giving

E = 24^\circ

.

4

Calculate the number of sides

n

n = 15

The sum of all exterior angles of any convex polygon is

360^\circ

, so

n = \frac{360^\circ}{E} = \frac{360^\circ}{24^\circ} = 15

.

Key Concept

Interior and Exterior Angle Properties of Regular Polygons

Question 4476Question →

A heavy radioactive parent nucleus undergoes a decay series emitting five $\alpha$ -particles and four $\beta^{-}$ -particles to form a stable daughter nucleus. What is the net change in the mass number ( $A$ ) and atomic number ( $Z$ ) of the nucleus?

Mass number decreases by

20

, and atomic number decreases by

6

Mass number decreases by

20

, and atomic number decreases by

14

Mass number decreases by

9

, and atomic number decreases by

6

Mass number decreases by

20

, and atomic number decreases by

10

Show answer & explanation

Answer: Mass number decreases by $20$ , and atomic number decreases by $6$

Answer

The mass number decreases by

20

, and the atomic number decreases by

6

.

An alpha particle emission removes 2 protons and 2 neutrons (

^{4}_{2}\text{He}

), reducing the atomic mass number by 4 and the atomic number by 2. Five alpha emissions reduce

A

by

20

and

Z

by

10

. A beta-minus emission (

^{0}_{-1}e

) converts a neutron into a proton, leaving

A

unchanged while increasing

Z

by 1. Four beta-minus emissions increase

Z

by 4. Combining these gives a net decrease in mass number of 20 and a net decrease in atomic number of 6.

Step-by-Step Solution

1

Determine the change in mass number (

A

) and atomic number (

Z

) due to

\alpha

-emissions.

Each

\alpha

-particle (

^{4}_{2}\text{He}

) decreases

A

by

4

and

Z

by

2

. For

5

\alpha

-particles:

\Delta A_{\alpha} = 5 \times (-4) = -20

,

\Delta Z_{\alpha} = 5 \times (-2) = -10

.

Alpha particles consist of two protons and two neutrons.

2

Determine the change in mass number (

A

) and atomic number (

Z

) due to

\beta^{-}

-emissions.

Each

\beta^{-}

-particle (

^{0}_{-1}e

) leaves

A

unchanged and increases

Z

by

1

. For

4

\beta^{-}

-particles:

\Delta A_{\beta} = 0

,

\Delta Z_{\beta} = 4 \times (+1) = +4

.

Beta-minus decay involves a neutron converting into a proton, emitting an electron.

3

Sum the total changes for

A

and

Z

.

Net

\Delta A = -20 + 0 = -20

(decrease by

20

). Net

\Delta Z = -10 + 4 = -6

(decrease by

6

).

Combining independent particle emission effects yields the total net change.

Key Concept

Natural radioactive radiation emissions change nuclear composition: alpha emission (

^{4}_{2}\text{He}

) reduces

A

by

4

and

Z

by

2

, whereas beta-minus emission (

^{0}_{-1}e

) keeps

A

constant and increases

Z

by

1

.

Question 4477Question →

A ray of light passes symmetrically through an equilateral glass prism of refractive index $\sqrt{2}$ . What is the angle of minimum deviation of the ray, in degrees?

Show answer & explanation

Answer: 30

Answer

The angle of minimum deviation of the ray is

30^\circ

.

For an equilateral triangular prism, the apex angle

A

is

60^\circ

. At minimum deviation, light travels symmetrically through the prism and obeys the exact relation

n = \frac{\sin\left(\frac{A + D_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}

. Substituting

n = \sqrt{2}

and

A = 60^\circ

gives

\sin\left(\frac{60^\circ + D_m}{2}\right) = \frac{\sqrt{2}}{2} = \sin(45^\circ)

. Equating arguments yields

\frac{60^\circ + D_m}{2} = 45^\circ

, leading directly to

D_m = 30^\circ

.

Step-by-Step Solution

1

Determine the refracting angle of the prism

A = 60^\circ

An equilateral prism has interior angles of

60^\circ

each, so the apex angle

A = 60^\circ

.

2

Set up the prism formula for minimum deviation

n = \frac{\sin\left(\frac{A + D_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}

When light passes symmetrically through a prism, the deviation is at its minimum value

D_m

.

3

Substitute known values into the equation

\sqrt{2} = \frac{\sin\left(\frac{60^\circ + D_m}{2}\right)}{\sin(30^\circ)}

Given

n = \sqrt{2}

and

A = 60^\circ

, with

\sin(30^\circ) = 0.5

.

4

Calculate the sine of the half-angle

\sin\left(\frac{60^\circ + D_m}{2}\right) = \sqrt{2} \times 0.5 = \frac{\sqrt{2}}{2}

Multiplying both sides by

\sin(30^\circ) = 0.5

.

5

Solve for the minimum deviation angle

D_m

D_m = 30^\circ

Since

\arcsin\left(\frac{\sqrt{2}}{2}\right) = 45^\circ

, we set

\frac{60^\circ + D_m}{2} = 45^\circ \implies 60^\circ + D_m = 90^\circ \implies D_m = 30^\circ

.

Key Concept

Refraction of light through a prism at the angle of minimum deviation

Question 4478Question →

A wheel and axle machine having a wheel radius of $25\text{ cm}$ and an axle radius of $5\text{ cm}$ is used to raise a load of mass $80\text{ kg}$ vertically through a height of $10\text{ m}$ . If the efficiency of the machine is $80\%$ , what is the work done against friction during this operation? (Take $g = 10\text{ m s}^{-2}$ )

1,600\text{ J}

2,000\text{ J}

8,000\text{ J}

10,000\text{ J}

Show answer & explanation

Answer: $2,000\text{ J}$

Answer

2,000\text{ J}

The useful work done on the load is

W_{\text{out}} = mgh = 80 \times 10 \times 10 = 8000\text{ J}

. Since efficiency is

80\%

, the total work input required is

W_{\text{in}} = \frac{8000}{0.80} = 10,000\text{ J}

. Therefore, the work lost to friction is

W_{\text{in}} - W_{\text{out}} = 10,000\text{ J} - 8,000\text{ J} = 2,000\text{ J}

.

Step-by-Step Solution

1

Calculate the useful work output of the machine

W_{\text{out}} = mgh = 80\text{ kg} \times 10\text{ m s}^{-2} \times 10\text{ m} = 8,000\text{ J}

Useful work output is the gravitational potential energy gained by raising the load.

2

Calculate the total work input using the efficiency equation

W_{\text{in}} = \frac{W_{\text{out}}}{\text{Efficiency}} = \frac{8,000\text{ J}}{0.80} = 10,000\text{ J}

Efficiency is defined as

\frac{\text{Work Output}}{\text{Work Input}}

, so Work Input =

\frac{\text{Work Output}}{\text{Efficiency}}

.

3

Determine the work done against friction

W_{\text{friction}} = W_{\text{in}} - W_{\text{out}} = 10,000\text{ J} - 8,000\text{ J} = 2,000\text{ J}

The energy wasted as heat and friction is the difference between total work input and useful work output.

Key Concept

Efficiency of Simple Machines and Work Done Against Friction

Question 4479Question →

A traffic officer on a stationary motorcycle spots a car passing at a constant speed of $15\text{ m/s}$ . The officer immediately pursues the car, accelerating uniformly from rest at $4\text{ m/s}^2$ for $5\text{ s}$ , after which the motorcycle continues at the constant speed attained. How long after setting off does the motorcycle overtake the car?

10\text{ s}

7.5\text{ s}

5\text{ s}

15\text{ s}

Show answer & explanation

Answer: $10\text{ s}$

Answer

The total time required for the motorcycle to overtake the car is

10\text{ s}

.

During the first

5\text{ s}

, the motorcycle accelerates from rest at

4\text{ m/s}^2

to

20\text{ m/s}

, covering

50\text{ m}

. In that same interval, the car moving at

15\text{ m/s}

covers

75\text{ m}

, creating a

25\text{ m}

separation gap. Beyond

5\text{ s}

, the motorcycle maintains

20\text{ m/s}

and closes in on the car (

15\text{ m/s}

) at a relative speed of

5\text{ m/s}

. It takes

5\text{ s}

to close the

25\text{ m}

gap, making the total pursuit time

10\text{ s}

.

Step-by-Step Solution

1

Calculate the state of both vehicles at the end of the motorcycle's acceleration phase (

t_1 = 5\text{ s}

).

Motorcycle speed

v = u + at = 0 + 4(5) = 20\text{ m/s}

. Motorcycle distance

s_1 = \frac{1}{2} a t^2 = \frac{1}{2}(4)(5^2) = 50\text{ m}

. Car distance

s_{\text{car}} = v_{\text{car}} \times t = 15 \times 5 = 75\text{ m}

.

Determine the position gap and relative speed after the acceleration phase ends.

2

Determine the remaining distance gap and relative speed between the vehicles.

Separation gap

d = 75\text{ m} - 50\text{ m} = 25\text{ m}

. Relative speed

v_{\text{rel}} = 20\text{ m/s} - 15\text{ m/s} = 5\text{ m/s}

.

Both vehicles now move at constant velocities, so relative speed determines how quickly the gap closes.

3

Calculate the time taken in the second phase and sum for total time.

Phase 2 time

t_2 = \frac{d}{v_{\text{rel}}} = \frac{25}{5} = 5\text{ s}

. Total time

t = t_1 + t_2 = 5\text{ s} + 5\text{ s} = 10\text{ s}

.

The total duration of pursuit is the acceleration duration plus the constant velocity catch-up duration.

Key Concept

Multi-stage linear motion involving uniform acceleration followed by constant velocity.

Estimated Time:1m 30s

Question 4480Question →

The $2^{\text{nd}}$ term of a geometric progression (GP) is $6$ and the $5^{\text{th}}$ term is $48$ . What is the sum of the first $6$ terms of the progression?

189

93

162

192

Show answer & explanation

Answer: $189$

Answer

The sum of the first 6 terms of the geometric progression is 189.

By using the GP term formula

T_n = a r^{n-1}

, we establish

a r = 6

and

a r^4 = 48

. Dividing the fifth term by the second term yields

r^3 = 8

, giving

r = 2

. Substituting

r = 2

into

a r = 6

gives

a = 3

. Finally, using the sum formula

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

, we evaluate

S_6 = \frac{3(2^6 - 1)}{2 - 1} = 3(63) = 189

.

Step-by-Step Solution

1

Set up equations for the given terms using the nth term formula

T_n = a r^{n-1}

.

T_2 = a r = 6

and

T_5 = a r^4 = 48

.

The nth term of a GP is defined by

T_n = a r^{n-1}

.

2

Solve for the common ratio

r

by dividing

T_5

by

T_2

.

\frac{a r^4}{a r} = \frac{48}{6} \implies r^3 = 8 \implies r = 2

.

Dividing the terms eliminates the first term

a

.

3

Find the first term

a

.

a(2) = 6 \implies a = 3

.

Substitute

r = 2

back into the equation for

T_2

.

4

Calculate the sum of the first 6 terms using

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

.

S_6 = \frac{3(2^6 - 1)}{2 - 1} = 3(64 - 1) = 3 \times 63 = 189

.

Apply the sum formula for a GP with

r > 1

.

Key Concept

Geometric Progression nth term and sum formulas

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