All practice questions

Question 4361Question →

A satellite orbits the Earth at an altitude equal to three times the radius of the Earth, $R$ . If the acceleration due to gravity at the Earth's surface is $g$ , what is the gravitational field strength experienced by the satellite at its orbit?

\frac{g}{16}

\frac{g}{9}

\frac{g}{4}

\frac{g}{3}

Show answer & explanation

Answer: $\frac{g}{16}$

Answer

The gravitational field strength at the orbital position is

\frac{g}{16}

.

The total distance from the center of the Earth to the satellite is

r = R + 3R = 4R

. Since gravitational field strength is inversely proportional to the square of the distance from the center of mass, increasing the distance by a factor of

4

reduces the field strength by a factor of

4^2 = 16

, giving

\frac{g}{16}

.

Step-by-Step Solution

1

Determine the total distance from the center of the Earth to the satellite.

The orbital radius is

r = R + h = R + 3R = 4R

.

Gravitational calculations must be measured from the center of mass of the attracting body, not its surface.

2

Apply Newton's law of universal gravitation for gravitational field strength.

At the surface (

r = R

),

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

. At orbit (

r = 4R

),

g' = \frac{GM}{(4R)^2} = \frac{GM}{16R^2}

.

Gravitational field strength follows an inverse-square law with respect to distance from the center of mass.

3

Express the orbital gravitational field strength in terms of the surface gravity

g

.

g' = \frac{1}{16} \left(\frac{GM}{R^2}\right) = \frac{g}{16}

.

Substituting the surface value

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

yields the simplified ratio.

Key Concept

Gravitational Field Strength and Inverse-Square Law

Question 4362Question →

A rectangular metal sheet has an initial area of $2.0\text{ m}^2$ at $20^\circ\text{C}$ . If the linear expansivity of the metal is $2.5 \times 10^{-5}\text{ K}^{-1}$ , what is the final temperature required for its area to increase by $0.005\text{ m}^2$ ?

70^\circ\text{C}

120^\circ\text{C}

50^\circ\text{C}

53.3^\circ\text{C}

Show answer & explanation

Answer: $70^\circ\text{C}$

Answer

The final temperature required is

70^\circ\text{C}

.

Area expansion is governed by

\Delta A = A_0 \beta \Delta T

, where the area expansivity

\beta = 2\alpha

. Substituting

A_0 = 2.0\text{ m}^2

,

\Delta A = 0.005\text{ m}^2

, and

\beta = 5.0 \times 10^{-5}\text{ K}^{-1}

gives a temperature rise

\Delta T = 50^\circ\text{C}

. Adding the initial temperature of

20^\circ\text{C}

yields a final temperature of

70^\circ\text{C}

.

Step-by-Step Solution

1

Determine the area expansivity (superficial expansivity)

\beta

from the linear expansivity

\alpha

.

\beta = 2\alpha = 2 \times 2.5 \times 10^{-5}\text{ K}^{-1} = 5.0 \times 10^{-5}\text{ K}^{-1}

Area expansion depends on area expansivity, which is twice the linear expansivity.

2

Calculate the temperature change

\Delta T

using the formula

\Delta A = A_0 \beta \Delta T

.

$\Delta T = \frac{\Delta A}{A_0 \beta} = \frac{0.005\text{ m}^2}{2.0\text{ m}^2 \times 5.0 \times 10^{-5}\text{ K}^{-1}} = \frac{5 \times 10^{-3}}{1.0 \times 10^{-4}} = 50\text{ K}$

Rearranging the expansion formula isolates the temperature change variable.

3

Calculate the final temperature

T_2

by adding

\Delta T

to the initial temperature

T_1

.

T_2 = T_1 + \Delta T = 20^\circ\text{C} + 50^\circ\text{C} = 70^\circ\text{C}

The final temperature is the sum of the initial temperature and the rise in temperature.

Key Concept

Thermal Expansion of Solids (Area Expansivity)

Estimated Time:1m 30s

Question 4363Question →

A motorist travels from Town A to Town B at a constant speed of $60\text{ km/h}$ for $2\text{ hours}$ , and then continues from Town B to Town C at a constant speed of $90\text{ km/h}$ for $3\text{ hours}$ . What is the average speed of the motorist for the entire journey in $\text{km/h}$ ?

Show answer & explanation

Answer: 78

Answer

The average speed for the entire journey is

78\text{ km/h}

.

The correct average speed is

78\text{ km/h}

, obtained by dividing the total distance of

390\text{ km}

(

120\text{ km} + 270\text{ km}

) by the total duration of

5\text{ hours}

(

2\text{ h} + 3\text{ h}

).

Step-by-Step Solution

1

Find the distance traveled during each segment of the journey

First segment distance =

120\text{ km}

, second segment distance =

270\text{ km}

Distance equals speed multiplied by time (

D = v \times t

).

2

Determine the total distance traveled and total time taken

Total distance =

390\text{ km}

, Total time =

5\text{ hours}

Average rate requires the sum of all distances and the sum of all durations.

3

Calculate the average speed

Average speed =

78\text{ km/h}

Average speed is defined as total distance divided by total time.

Key Concept

Average Rate and Speed

Question 4364Question →

A compound microscope in normal adjustment consists of an objective lens with a focal length of $1.5\text{ cm}$ and an eyepiece with a focal length of $5.0\text{ cm}$ . An object is placed at a distance of $1.6\text{ cm}$ in front of the objective lens. Calculate the distance, in centimeters, between the objective lens and the eyepiece.

Show answer & explanation

Answer: 29

Answer

The distance between the objective lens and the eyepiece is 29.0 cm.

Applying the thin lens formula to the objective lens yields

\frac{1}{1.5} = \frac{1}{1.6} + \frac{1}{v_o}

, giving an image distance

v_o = 24.0\text{ cm}

. Under normal adjustment, the intermediate image falls on the focal point of the eyepiece, making

u_e = f_e = 5.0\text{ cm}

. The total separation between the two lenses is

L = v_o + u_e = 24.0 + 5.0 = 29.0\text{ cm}

.

Step-by-Step Solution

1

Apply the thin lens formula to the objective lens to find the intermediate image position

v_o

.

\frac{1}{1.5} = \frac{1}{1.6} + \frac{1}{v_o} \Rightarrow \frac{1}{v_o} = \frac{2}{3} - \frac{5}{8} = \frac{1}{24}\text{ cm}^{-1} \Rightarrow v_o = 24.0\text{ cm}

The objective lens forms a real, inverted, magnified image at distance

v_o

from the objective.

2

Identify the object distance for the eyepiece

u_e

under normal adjustment.

u_e = f_e = 5.0\text{ cm}

For normal adjustment of a optical instrument, the final image is formed at infinity, requiring the intermediate image to sit exactly at the principal focus of the eyepiece.

3

Sum the intermediate image distance and eyepiece object distance to obtain total lens separation

L

.

L = v_o + u_e = 24.0\text{ cm} + 5.0\text{ cm} = 29.0\text{ cm}

The separation of lenses in a compound microscope is the distance from the objective to the intermediate image plus the distance from the intermediate image to the eyepiece.

Key Concept

Compound microscope optics and lens separation in normal adjustment

Question 4365Question →

A convex security mirror installed in a store has a focal length of magnitude $20\text{ cm}$ . If an upright image of a customer is formed with a linear magnification of $0.25$ , at what distance from the mirror is the customer standing?

60\text{ cm}

100\text{ cm}

15\text{ cm}

25\text{ cm}

Show answer & explanation

Answer: $60\text{ cm}$

Answer

The customer is standing at a distance of

60\text{ cm}

from the convex mirror.

By sign convention, a convex mirror has a negative focal length (

f = -20\text{ cm}

). An upright image formed by a mirror has a positive magnification

m = +0.25

. Using

m = -v/u

, we get

v = -u/4

. Substituting these into the mirror equation

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

gives

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

, which yields

u = 60\text{ cm}

.

Step-by-Step Solution

1

Identify the optical properties and apply sign conventions for a convex mirror

Focal length

f = -20\text{ cm}

(convex mirror focal length is virtual/negative). Magnification

m = +0.25

(upright image).

Convex mirrors always form virtual, upright, diminished images behind the mirror.

2

Relate image distance

v

to object distance

u

using the linear magnification formula

Since

m = -v/u = 1/4

, we obtain

v = -u/4

.

The negative sign in the magnification definition accounts for virtual image distance.

3

Substitute

f = -20\text{ cm}

and

v = -u/4

into the mirror formula

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

\frac{1}{-20} = \frac{1}{u} + \frac{1}{-u/4} \implies -\frac{1}{20} = \frac{1}{u} - \frac{4}{u} = -\frac{3}{u}

.

Combines fractions with a common denominator

u

to solve for the unknown object distance.

4

Solve for the object distance

u

\frac{1}{20} = \frac{3}{u} \implies u = 60\text{ cm}

.

Cross-multiplying yields the real object distance in front of the mirror.

Key Concept

Reflection at Convex Mirrors and Optical Sign Conventions

Question 4366Question →

A flat circular coil consisting of $50$ turns and enclosing an area of $0.02\text{ m}^2$ is placed in a uniform magnetic field directed vertically upwards. The magnitude of the magnetic field decreases steadily from $0.5\text{ T}$ to $0.1\text{ T}$ in a time interval of $0.2\text{ s}$ . If the total electrical resistance of the coil is $5.0\text{ }\Omega$ , what is the magnitude and direction of the induced current in the coil when viewed from above?

0.4\text{ A}

in an anticlockwise direction

0.4\text{ A}

in a clockwise direction

1.0\text{ A}

in an anticlockwise direction

1.0\text{ A}

in a clockwise direction

Show answer & explanation

Answer: $0.4\text{ A}$ in an anticlockwise direction

Answer

0.4\text{ A}

in an anticlockwise direction

According to Faraday's law, the induced e.m.f. is given by

E = N A \frac{\Delta B}{\Delta t} = 50 \times 0.02 \times \frac{0.4}{0.2} = 2.0\text{ V}

. By Ohm's law, the current magnitude is

I = \frac{2.0\text{ V}}{5.0\text{ }\Omega} = 0.4\text{ A}

. By Lenz's law, because the upward magnetic field is decreasing, the coil opposes this decrease by generating an upward magnetic field. By the right-hand grip rule, an upward magnetic field corresponds to an anticlockwise current flow when viewed from above.

Step-by-Step Solution

1

Calculate the magnitude of the rate of change of magnetic field strength

\frac{\Delta B}{\Delta t} = \frac{0.5\text{ T} - 0.1\text{ T}}{0.2\text{ s}} = \frac{0.4\text{ T}}{0.2\text{ s}} = 2.0\text{ T/s}

Faraday's law depends on the rate at which the magnetic flux changes over time.

2

Calculate the magnitude of the induced electromotive force (e.m.f.)

E = N A \left(\frac{\Delta B}{\Delta t}\right) = 50 \times 0.02\text{ m}^2 \times 2.0\text{ T/s} = 2.0\text{ V}

The total induced e.m.f. is proportional to the number of turns and the enclosed area.

3

Determine the magnitude of the induced current using Ohm's law

I = \frac{E}{R} = \frac{2.0\text{ V}}{5.0\text{ }\Omega} = 0.4\text{ A}

Current equals induced voltage divided by total coil resistance.

4

Determine the direction of the induced current using Lenz's law and the right-hand rule

Anticlockwise direction when viewed from above

The upward magnetic field is decreasing, so the induced current must produce its own upward magnetic field to oppose the reduction in magnetic flux.

Key Concept

Faraday's Law and Lenz's Law of Electromagnetic Induction

Estimated Time:2m 0s

Question 4367Question →

A box contains $30$ tickets numbered $1$ to $30$ . In a probability experiment, a ticket is drawn at random from the box and its number recorded before being replaced. This trial is performed $150$ times, and a ticket with a number that is a multiple of $4$ is recorded $45$ times. What is the absolute difference between the experimental probability and the theoretical probability of selecting a ticket bearing a multiple of $4$ ?

\frac{1}{15}

\frac{3}{10}

\frac{7}{30}

\frac{8}{15}

Show answer & explanation

Answer: $\frac{1}{15}$

Answer

The absolute difference between the experimental probability and the theoretical probability is

\frac{1}{15}

.

The theoretical probability of drawing a multiple of 4 is

\frac{7}{30}

since there are 7 favorable tickets (

4, 8, 12, 16, 20, 24, 28

) out of 30 total tickets. The experimental probability from 150 draws is

\frac{45}{150} = \frac{9}{30}

. Taking the absolute difference gives

\frac{9}{30} - \frac{7}{30} = \frac{2}{30} = \frac{1}{15}

.

Step-by-Step Solution

1

Determine the theoretical probability

Theoretical probability

P(T) = \frac{7}{30}

The multiples of

4

between

1

and

30

are

4, 8, 12, 16, 20, 24, 28

, giving

7

favorable outcomes out of

30

possible outcomes.

2

Determine the experimental probability

Experimental probability

P(E) = \frac{45}{150} = \frac{3}{10}

The event occurred

45

times out of

150

experimental trials.

3

Calculate the absolute difference between

P(E)

and

P(T)

|P(E) - P(T)| = |\frac{9}{30} - \frac{7}{30}| = \frac{2}{30} = \frac{1}{15}

Express both fractions with a common denominator of

30

and subtract the smaller probability from the larger.

Key Concept

Experimental probability is calculated from trial data (favorable trials divided by total trials), whereas theoretical probability is calculated from expected sample space outcomes (favorable outcomes divided by total sample space size).

Estimated Time:1m 30s

Question 4368Question →

A function

f(x)

is defined by

f(x) = \begin{cases} \frac{x^2 + x - 6}{x - 2}, & x \neq 2 \\ 2k - 1, & x = 2 \end{cases}

If

f(x)

is continuous at

x = 2

, what is the value of the constant

k

?

3

2

5

-1

Show answer & explanation

Answer: 3

Answer

The value of the constant

k

is

3

.

By definition, a function

f(x)

is continuous at

x = c

if

\lim_{x \to c} f(x) = f(c)

. Factoring the numerator gives

x^2 + x - 6 = (x - 2)(x + 3)

. Canceling the common factor

(x - 2)

for

x \neq 2

, the limit as

x \to 2

is

2 + 3 = 5

. Equating

f(2) = 2k - 1

to 5 yields

2k - 1 = 5

, which solves to

k = 3

.

Step-by-Step Solution

1

Evaluate the limit of

f(x)

as

x

approaches

2

.

\lim_{x \to 2} \frac{x^2 + x - 6}{x - 2} = \lim_{x \to 2} \frac{(x - 2)(x + 3)}{x - 2} = \lim_{x \to 2} (x + 3) = 5

Direct substitution gives the indeterminate form

\frac{0}{0}

, so factor the numerator to simplify.

2

Apply the definition of continuity at a point.

f(2) = \lim_{x \to 2} f(x) \implies 2k - 1 = 5

For

f(x)

to be continuous at

x = 2

, the value of the function at

x = 2

must equal its limit as

x \to 2

.

3

Solve the linear equation for

k

.

2k = 6 \implies k = 3

Add 1 to both sides and divide by 2.

Key Concept

Continuity of a Piecewise Function at a Point

Alternative Method

Alternatively, use L'Hôpital's rule to evaluate the limit:

\lim_{x \to 2} \frac{\frac{d}{dx}(x^2+x-6)}{\frac{d}{dx}(x-2)} = \lim_{x \to 2} \frac{2x+1}{1} = 5

. Then set

2k - 1 = 5

to find

k = 3

.

Estimated Time:1m 30s

Question 4369Question →

A bar magnet is moved with its north pole approaching one end of a closed circular wire coil. According to Lenz's law, which of the following correctly describes the induced magnetic polarity at that face of the coil and the direction of the induced current as viewed from the side of the magnet?

Anticlockwise current, developing a North pole

Clockwise current, developing a South pole

Clockwise current, developing a North pole

Anticlockwise current, developing a South pole

Show answer & explanation

Answer: Anticlockwise current, developing a North pole

Answer

Anticlockwise current, developing a North pole

According to Lenz's law, an induced current flows in a direction such that its magnetic field opposes the change in magnetic flux that produced it. As the North pole of a magnet approaches the coil face, the coil must oppose this motion by establishing a North pole on that face. By the right-hand grip rule, looking at a North pole corresponds to an anticlockwise flow of current.

Step-by-Step Solution

1

Apply Lenz's law to determine the induced magnetic polarity

The near face of the coil must develop a North pole to repel the approaching North pole of the bar magnet

Lenz's law states that the direction of an induced current always opposes the magnetic flux change causing it.

2

Determine the direction of induced current corresponding to a North magnetic pole

Looking directly at a North magnetic pole, the induced current flows in an anticlockwise direction

By the right-hand rule (or N-S rule for coils), an anticlockwise current produces a magnetic field directed out of the face (North polarity).

Key Concept

Lenz's Law

Estimated Time:45s

Question 4370Question →

In a mathematics examination, a student is required to answer $5$ questions out of $8$ available questions. If the first $2$ questions are compulsory, in how many different ways can the student select the questions to answer?

Show answer & explanation

Answer: 20

Answer

The student can select the questions in 20 different ways.

With 2 compulsory questions, the student only has to select 3 more questions from the remaining 6 questions. The number of ways to select 3 items from 6 without regard to order is given by ^6C_3 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20.

Step-by-Step Solution

1

Deduct compulsory questions from both the required total and available total

The student must choose 3 additional questions from the remaining 6 questions.

Compulsory questions are fixed and provide only 1 selection choice.

2

Apply the combination formula ^6C_3 to calculate selection ways

^6C_3 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20

The order in which the student chooses the examination questions does not alter the group of questions selected.

Key Concept

Combinations with restricted/compulsory choices

Question 4371Question →

Evaluate the limit $\lim_{x \to 0} \frac{\sqrt{x + 4} - 2}{x}$ . What is the numerical value of this limit?

Show answer & explanation

Answer: 0.25

Answer

The numerical value of the limit is 0.25 (or

\frac{1}{4}

).

When direct substitution into

\frac{\sqrt{x + 4} - 2}{x}

yields the indeterminate form

\frac{0}{0}

, rationalizing the numerator by multiplying by its conjugate

\sqrt{x + 4} + 2

simplifies the expression to

\frac{1}{\sqrt{x + 4} + 2}

. Taking the limit as

x \to 0

gives

\frac{1}{4} = 0.25

.

Step-by-Step Solution

1

Check direct substitution

Indeterminate form

\frac{0}{0}

Directly evaluating at

x = 0

yields zero in both numerator and denominator.

2

Multiply by the conjugate of the numerator

\frac{(\sqrt{x + 4} - 2)(\sqrt{x + 4} + 2)}{x(\sqrt{x + 4} + 2)} = \frac{x}{x(\sqrt{x + 4} + 2)}

Rationalizing the radical in the numerator allows cancellation of the term causing the zero denominator.

3

Cancel common factors and evaluate limit

\frac{1}{\sqrt{0 + 4} + 2} = 0.25

Canceling

x

removes the zero factor, permitting direct evaluation.

Key Concept

Limits of indeterminate algebraic expressions using surd rationalization

Question 4372Question →

A body of mass $4\text{ kg}$ moves with a constant velocity of $5\text{ m s}^{-1}$ . Calculate the kinetic energy of the body in Joules.

Show answer & explanation

Answer: 50

Answer

The kinetic energy of the body is

50\text{ J}

.

The kinetic energy of an object in linear motion is calculated using

E_k = \frac{1}{2} m v^2

. Substituting

m = 4\text{ kg}

and

v = 5\text{ m s}^{-1}

gives

E_k = \frac{1}{2} \times 4 \times 5^2 = 50\text{ J}

.

Step-by-Step Solution

1

Identify the given physical quantities

Mass

m = 4\text{ kg}

and velocity

v = 5\text{ m s}^{-1}

Extract values provided in the problem statement.

2

Apply the formula for kinetic energy

E_k = \frac{1}{2} m v^2

Kinetic energy is defined as half the product of mass and the square of velocity.

3

Substitute values and solve

E_k = \frac{1}{2} \times 4 \times (5)^2 = 50\text{ J}

Perform arithmetic evaluation to get the final energy in Joules.

Key Concept

Kinetic Energy

Question 4373Question →

Match each physical phenomenon or calculation involving magnetic forces on the left with its corresponding rule, equation, or physical principle on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

1 matches with Fleming's Left-Hand Rule; 2 matches with the Right-Hand Grip Rule; 3 matches with the ratio

r = \frac{mv}{qB}

; 4 matches with the parallel current interaction law

\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}

.

Each electromagnetic phenomenon correctly aligns with its governing physical rule or formula: force direction on a moving charge is determined by Fleming's Left-Hand Rule, magnetic field orientation around a wire by the Right-Hand Grip Rule, circular orbital radius by balancing magnetic force with centripetal force (

r = \frac{mv}{qB}

), and force between parallel conductors by Ampere's force law.

Step-by-Step Solution

1

Identify the directional rule for magnetic force on a moving charge.

Magnetic force direction is perpendicular to both particle velocity and magnetic field, given by Fleming's Left-Hand Rule.

Fleming's Left-Hand Rule relates thrust/force (thumb), magnetic field (forefinger), and current/positive charge motion (middle finger).

2

Identify the field mapping rule for a current-carrying wire.

Concentric magnetic field lines around a straight wire are mapped using the Right-Hand Grip Rule.

Pointing the right thumb along conventional current causes fingers to curl in the direction of the magnetic field vector.

3

Derive the trajectory equation for a charge in a magnetic field.

Equating magnetic force

qvB

to centripetal force

\frac{mv^2}{r}

yields

r = \frac{mv}{qB}

.

Because the magnetic force acts as a pure centripetal force, the charge follows a circular trajectory of fixed radius

r

.

4

Identify the force law between parallel currents.

The attractive force per length is given by

\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}

.

Current

I_1

sets up a magnetic field

B_1 = \frac{\mu_0 I_1}{2\pi d}

at wire 2, producing force per length

B_1 I_2

.

Key Concept

Magnetic Force and Electromagnetism Rules and Equations

Question 4374Question →

Radioactive nuclei emit different types of emissions ( $\alpha$ , $\beta^{-}$ , $\beta^{+}$ , and $\gamma$ ) characterized by distinct deflection behaviors, energy spectra, and physical properties when passing through electric fields. Match each type of radiation emission listed on the left with its corresponding physical properties and field response on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Alpha particle emission matches weak deflection towards the negative electrode with discrete energy levels; Beta-minus particle emission matches strong deflection towards the positive electrode with a continuous spectrum; Beta-plus particle emission matches strong deflection towards the negative electrode accompanied by a neutrino; Gamma ray emission matches zero deflection in electric fields and high speed.

Matching each emission type correctly relies on evaluating electric field deflection (determined by charge sign and

q/m

ratio) and spectral characteristics. Alpha particles are heavy positive ions showing slight deflection toward the negative electrode with discrete energy states. Beta-minus emissions are light negative particles deflecting strongly toward the positive electrode in a continuous spectrum. Beta-plus emissions are light positive particles deflecting strongly toward the negative electrode alongside a neutrino. Gamma radiation is uncharged photon energy showing zero deflection.

Step-by-Step Solution

1

Analyze charge and mass ratios of radioactive emissions to determine magnetic and electric field deflection direction and magnitude.

Alpha particles (

+2e

, mass

4\text{ u}

) deflect slightly toward negative plate; Beta-minus (

-e

, negligible mass) deflects strongly toward positive plate; Beta-plus (

+e

, negligible mass) deflects strongly toward negative plate; Gamma rays (

0

charge,

0

mass) do not deflect.

Deflection angle in an electric field depends directly on the charge-to-mass ratio (

q/m

) and the sign of the charge.

2

Evaluate energy spectra characteristics and secondary particle emissions for decay modes.

Alpha decay produces discrete kinetic energy peaks. Beta decay produces a continuous spectrum due to three-body decay sharing kinetic energy with a neutrino or antineutrino. Gamma photons carry discrete transition energy.

Conservation of momentum and energy in three-body beta decay requires kinetic energy sharing with the (anti)neutrino.

3

Pair each radiation type with its full physical description.

Alpha matches weak deflection to negative plate with discrete energy; Beta-minus matches strong deflection to positive plate with continuous spectrum; Beta-plus matches strong deflection to negative plate with neutrino co-emission; Gamma matches no deflection and lowest specific ionization.

Combines field deflection, charge-to-mass ratio, and spectral traits into unique matching pairings.

Key Concept

Deflection characteristics, charge-to-mass ratios, and energy spectrum nature of alpha, beta, and gamma radiation emissions

Question 4375Question →

A high-pressure storage vessel contains a sample of gas at an initial pressure of $2.00 \times 10^5\text{ Pa}$ and a temperature of $27^\circ\text{C}$ . A relief valve releases one-third of the total mass of the gas while maintaining a constant internal volume. After the valve closes, the vessel and remaining gas are heated to $127^\circ\text{C}$ . What is the final pressure of the gas inside the vessel?

1.78 \times 10^5\text{ Pa}

6.27 \times 10^5\text{ Pa}

8.89 \times 10^4\text{ Pa}

2.67 \times 10^5\text{ Pa}

Show answer & explanation

Answer: $1.78 \times 10^5\text{ Pa}$

Answer

The final pressure of the gas inside the vessel is

1.78 \times 10^5\text{ Pa}

.

According to the ideal gas equation

PV = \frac{m}{M}RT

, pressure is directly proportional to mass and absolute temperature at fixed volume (

P \propto m T

). Converting temperatures to Kelvin gives

T_1 = 300\text{ K}

and

T_2 = 400\text{ K}

. Since one-third of the gas escaped, two-thirds remains (

m_2 = \frac{2}{3}m_1

). The final pressure is therefore

P_2 = P_1 \times \frac{2}{3} \times \frac{400}{300} = 2.00 \times 10^5\text{ Pa} \times \frac{8}{9} \approx 1.78 \times 10^5\text{ Pa}

.

Step-by-Step Solution

1

Convert given initial and final temperatures from Celsius to Kelvin.

T_1 = 27^\circ\text{C} + 273 = 300\text{ K}

and

T_2 = 127^\circ\text{C} + 273 = 400\text{ K}

.

Absolute temperature in Kelvin is required for all gas law calculations.

2

Determine the remaining mass fraction of the gas.

m_2 = m_1 - \frac{1}{3}m_1 = \frac{2}{3}m_1

.

The pressure depends on the quantity of gas remaining in the rigid container after venting.

3

Apply the ideal gas equation

PV = nRT = \frac{m}{M}RT

for constant volume

V

and molar mass

M

.

\frac{P_2}{P_1} = \left(\frac{m_2}{m_1}\right) \left(\frac{T_2}{T_1}\right)

.

Pressure is directly proportional to both mass and absolute temperature when volume is constant.

4

Substitute the known values to compute

P_2

.

P_2 = (2.00 \times 10^5\text{ Pa}) \times \left(\frac{2}{3}\right) \times \left(\frac{400\text{ K}}{300\text{ K}}\right) = 2.00 \times 10^5 \times \frac{8}{9} \approx 1.78 \times 10^5\text{ Pa}

.

Evaluates the combined effects of mass reduction and temperature elevation.

Key Concept

Ideal Gas Law variations involving changing gas mass and absolute temperature at constant volume.

Question 4376Question →

A particle is projected from horizontal ground at an initial angle $\theta$ to the horizontal. At its maximum height $H = 20\text{ m}$ , its kinetic energy is exactly half of its initial kinetic energy at launch. Taking acceleration due to gravity $g = 10\text{ m/s}^2$ , what is the horizontal range of the projectile?

40\text{ m}

80\text{ m}

20\text{ m}

0\text{ m}

Show answer & explanation

Answer: $80\text{ m}$

Answer

The horizontal range of the projectile is

80\text{ m}

.

At the trajectory's highest point, the vertical component of velocity becomes zero while the horizontal component

u\cos\theta

remains unchanged. The kinetic energy at the apex is therefore

\frac{1}{2}m(u\cos\theta)^2 = E_{k,i}\cos^2\theta

. Setting

\cos^2\theta = \frac{1}{2}

gives

\theta = 45^\circ

. For a

45^\circ

launch angle, the horizontal range

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

is related to the maximum height

H = \frac{u^2}{4g}

by

R = 4H

. Substituting

H = 20\text{ m}

yields

R = 80\text{ m}

.

Step-by-Step Solution

1

Relate kinetic energy at maximum height to initial kinetic energy.

Initial kinetic energy

E_k(0) = \frac{1}{2}m u^2

. At peak height, vertical velocity component

v_y = 0

, so velocity is

v_x = u\cos\theta

. Thus,

E_k(\text{peak}) = \frac{1}{2}m (u\cos\theta)^2 = E_k(0)\cos^2\theta

.

At maximum height, only the horizontal component of velocity remains.

2

Calculate the launch angle

\theta

.

Given

E_k(\text{peak}) = \frac{1}{2} E_k(0)

, we have

\cos^2\theta = \frac{1}{2}

, giving

\theta = 45^\circ

.

Setting the energy expression equal to the given condition enables solving for the angle.

3

Relate maximum height

H

to horizontal range

R

for

\theta = 45^\circ

.

For

\theta = 45^\circ

,

H = \frac{u^2 \sin^2(45^\circ)}{2g} = \frac{u^2}{4g}

and

R = \frac{u^2 \sin(90^\circ)}{g} = \frac{u^2}{g}

. Therefore,

R = 4H

.

Standard formulas for maximum height and range express both quantities in terms of launch speed

u

and acceleration due to gravity

g

.

4

Substitute the maximum height value into the range formula.

R = 4 \times 20\text{ m} = 80\text{ m}

.

Multiplying the given peak height of

20\text{ m}

by

4

yields the exact horizontal range.

Key Concept

Kinetic Energy Conservation and Range-Height Relationship in Projectile Motion

Estimated Time:2m 0s

Question 4377Question →

A metal block of mass $2.5\text{ kg}$ absorbs $1200\text{ J}$ of thermal energy, causing its temperature to rise by $15\text{ K}$ . What is the heat capacity of the block in $\text{J K}^{-1}$ ?

Show answer & explanation

Answer: 80

Answer

The heat capacity of the block is

80\text{ J K}^{-1}

.

Heat capacity

C

represents the quantity of heat energy required to raise the temperature of the entire object by

1\text{ K}

. Using the relation

C = \frac{Q}{\Delta T}

, substituting

Q = 1200\text{ J}

and

\Delta T = 15\text{ K}

gives

C = \frac{1200}{15} = 80\text{ J K}^{-1}

.

Step-by-Step Solution

1

Identify the given quantities from the problem.

Heat energy

Q = 1200\text{ J}

, temperature change

\Delta T = 15\text{ K}

, and mass

m = 2.5\text{ kg}

.

Clear identification of parameters is necessary to choose the correct formula.

2

Apply the formula for heat capacity.

C = \frac{Q}{\Delta T}

Heat capacity

C

measures the heat required to change the temperature of the entire body by

1\text{ K}

, regardless of mass per unit quantity.

3

Substitute the values to calculate the heat capacity.

C = \frac{1200}{15} = 80\text{ J K}^{-1}

Dividing total thermal energy absorbed by the resulting temperature rise yields the heat capacity.

Key Concept

Heat Capacity (

C = \frac{Q}{\Delta T}

)

Estimated Time:45s

Question 4378Question →

Complete the sentence below by filling in the blank with the appropriate word to complete the idiomatic expression.

Fill in the blanks below

After months of failing to resolve their internal conflict, the board members decided to call in an independent auditor to help them clear the .

Show answer & explanation

Answer

The word 'air' completes the idiomatic expression 'clear the air', which means to remove tension, misunderstandings, or doubts between people.

The idiom 'clear the air' means to discuss feelings or resolve misunderstandings openly so that a tense situation improves. In this sentence context, bringing in an external mediator aims to resolve the underlying friction.

Step-by-Step Solution

1

Analyze the sentence context

The context describes board members seeking to resolve lingering internal conflict and misunderstandings.

An independent auditor is brought in specifically to facilitate open discussion and eliminate tension.

2

Identify the correct idiomatic term

The standard English idiom for eliminating tense feelings or resolving unspoken grievances is to 'clear the air'.

'Air' is the fixed word in this established idiomatic structure.

Key Concept

Contextual Completion of Fixed Idiomatic Expressions

Question 4379Question →

A metallic container with an initial volume of $400 \text{ cm}^3$ at $15^\circ\text{C}$ is filled completely with paraffin. Upon heating the container and its contents to $65^\circ\text{C}$ , a volume of $18 \text{ cm}^3$ of paraffin spills over. Given that the linear expansivity of the metal container is $2.0 \times 10^{-5} \text{ K}^{-1}$ , what is the real cubic expansivity of the paraffin, expressed in units of $10^{-4} \text{ K}^{-1}$ ?

Show answer & explanation

Answer: 9.6

Answer

The real cubic expansivity of the paraffin is

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

, which corresponds to a numerical value of

9.6

in units of

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

.

The real cubic expansivity of a liquid accounts for both the observed (apparent) expansion of the liquid and the expansion of the container holding it. By applying

\gamma_a = \frac{\Delta V_a}{V_0 \Delta T} = 9.0 \times 10^{-4} \text{ K}^{-1}

and

\gamma_v = 3\alpha = 0.6 \times 10^{-4} \text{ K}^{-1}

, we sum them to obtain the real cubic expansivity

\gamma_r = 9.6 \times 10^{-4} \text{ K}^{-1}

.

Step-by-Step Solution

1

Determine the temperature change (

\Delta T

) and the apparent change in volume (

\Delta V_a

).

\Delta T = 65^\circ\text{C} - 15^\circ\text{C} = 50 \text{ K}

and

\Delta V_a = 18 \text{ cm}^3

.

The overflow volume represents the apparent expansion of the liquid relative to the expanding container over the temperature rise.

2

Calculate the apparent cubic expansivity (

\gamma_a

) of the paraffin.

\gamma_a = \frac{\Delta V_a}{V_0 \Delta T} = \frac{18}{400 \times 50} = \frac{18}{20000} = 9.0 \times 10^{-4} \text{ K}^{-1}

.

Apparent expansivity relates the apparent volume expansion to the original volume and temperature increase.

3

Calculate the cubic expansivity of the metallic vessel (

\gamma_v

).

\gamma_v = 3 \alpha = 3 \times 2.0 \times 10^{-5} \text{ K}^{-1} = 6.0 \times 10^{-5} \text{ K}^{-1} = 0.6 \times 10^{-4} \text{ K}^{-1}

.

Cubic expansivity of an isotropic solid container is three times its linear expansivity.

4

Calculate the real cubic expansivity of the paraffin (

\gamma_r

).

\gamma_r = \gamma_a + \gamma_v = 9.0 \times 10^{-4} \text{ K}^{-1} + 0.6 \times 10^{-4} \text{ K}^{-1} = 9.6 \times 10^{-4} \text{ K}^{-1}

.

The real expansion of a liquid is the sum of its apparent expansion and the expansion of the containing vessel.

Key Concept

Real vs. Apparent Expansion of Liquids (

\gamma_r = \gamma_a + \gamma_v

where

\gamma_v = 3\alpha

)

Question 4380Question →

A uniform wooden cube of edge length $0.20\text{ m}$ floats in water of density $1000\text{ kg/m}^3$ with $0.15\text{ m}$ of its vertical height submerged. Taking acceleration due to gravity $g = 10\text{ m/s}^2$ , what minimum mass, in kilograms, must be placed on the top surface of the cube so that its upper face becomes just flush with the water surface?

Show answer & explanation

Answer: 2

Answer

The minimum mass required to submerge the cube completely flush with the water surface is

2.0\text{ kg}

.

By the Law of Flotation, a floating body displaces its own weight of fluid. Initially, the cube displaces a volume of

0.20\text{ m} \times 0.20\text{ m} \times 0.15\text{ m} = 0.006\text{ m}^3

of water, corresponding to an upthrust of

60\text{ N}

(or mass of

6.0\text{ kg}

). When completely submerged, the total volume displaced is

0.20^3 = 0.008\text{ m}^3

, providing a total upthrust of

80\text{ N}

(or mass equivalent of

8.0\text{ kg}

). The additional mass required on top is therefore the difference:

8.0\text{ kg} - 6.0\text{ kg} = 2.0\text{ kg}

.

Step-by-Step Solution

1

Calculate the cross-sectional area of the cube

A = (0.20\text{ m})^2 = 0.04\text{ m}^2

The base area is needed to find the volume of the block submerged and unsubmerged.

2

Calculate the volume of the cube above the water surface

V_{\text{above}} = 0.04\text{ m}^2 \times (0.20\text{ m} - 0.15\text{ m}) = 0.002\text{ m}^3

To push the cube level with the surface, the additional weight added on top must balance the extra upthrust created by submerging this remaining volume.

3

Calculate the additional mass required

m = \rho_{\text{water}} \times V_{\text{above}} = 1000\text{ kg/m}^3 \times 0.002\text{ m}^3 = 2.0\text{ kg}

By Archimedes' principle, the additional downward mass must equal the mass of the extra water displaced when fully submerged.

Key Concept

Archimedes' Principle and Law of Flotation

Answer

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