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Question 1401Question →

A closed rigid container holds a mixture of dry air and water vapour at a temperature of $27^\circ\text{C}$ under a total pressure of $740\text{ mmHg}$ . The relative humidity of the air inside the container is $80\%$ , and the saturated vapour pressure of water at $27^\circ\text{C}$ is $25\text{ mmHg}$ . If the container is heated at constant volume to $127^\circ\text{C}$ , what is the partial pressure of the dry air in $\text{mmHg}$ at this higher temperature?

Show answer & explanation

Answer: 960

Answer

The partial pressure of dry air inside the container at

127^\circ\text{C}

is

960\text{ mmHg}

.

First, the partial pressure of water vapour at

27^\circ\text{C}

is determined by multiplying relative humidity (

80\%

) by the saturated vapour pressure (

25\text{ mmHg}

), yielding

20\text{ mmHg}

. Next, subtracting this vapour pressure from the total pressure of

740\text{ mmHg}

gives the partial pressure of dry air alone as

720\text{ mmHg}

at

27^\circ\text{C}

(

300\text{ K}

). Finally, applying the Pressure Law (

\frac{P_1}{T_1} = \frac{P_2}{T_2}

) for the dry air between

300\text{ K}

and

400\text{ K}

(

127^\circ\text{C}

) yields

P_2 = 720 \times \frac{400}{300} = 960\text{ mmHg}

.

Step-by-Step Solution

1

Calculate the partial pressure of water vapour at

27^\circ\text{C}

P_{\text{vapour}, 1} = 0.80 \times 25\text{ mmHg} = 20\text{ mmHg}

Relative humidity is the ratio of actual partial vapour pressure to the saturated vapour pressure at that temperature.

2

Determine the initial partial pressure of the dry air at

27^\circ\text{C}

using Dalton's Law

P_{\text{dry air}, 1} = 740\text{ mmHg} - 20\text{ mmHg} = 720\text{ mmHg}

Total pressure of a gas mixture is the sum of the partial pressures of its individual components.

3

Convert temperatures to Kelvin and apply Gay-Lussac's Pressure Law for dry air at constant volume

T_1 = 27 + 273 = 300\text{ K}

,

T_2 = 127 + 273 = 400\text{ K}

.

P_{\text{dry air}, 2} = 720\text{ mmHg} \times \left(\frac{400\text{ K}}{300\text{ K}}\right) = 960\text{ mmHg}

For a fixed mass of gas at constant volume, pressure is directly proportional to absolute temperature.

Key Concept

Dalton's Law of Partial Pressures and Gay-Lussac's Pressure Law applied to gas-vapour mixtures

Question 1402Question →

A sample of hydrated sodium trioxocarbonate(IV), $\text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O}$ , is heated strongly in a crucible until a constant mass is reached. If the salt loses $62.94\%$ of its initial mass as water vapor during heating, what is the integer value of $x$ ?

[Relative atomic masses: $\text{Na} = 23$ , $\text{C} = 12$ , $\text{O} = 16$ , $\text{H} = 1$ ]

Show answer & explanation

Answer: 10

Answer

The integer value of

x

is 10.

Heating hydrated sodium trioxocarbonate(IV) drives off all water of crystallization as steam. Since

62.94\%

of the total mass is lost, water accounts for

62.94\%

of the molar mass of

\text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O}

. Solving

\frac{18x}{106 + 18x} = 0.6294

yields

x = 10

, representing decahydrate crystals,

\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}

.

Step-by-Step Solution

1

Calculate the formula masses of the anhydrous salt

\text{Na}_2\text{CO}_3

and water

\text{H}_2\text{O}

.

Molar mass of

\text{Na}_2\text{CO}_3 = 106\text{ g/mol}

and molar mass of

\text{H}_2\text{O} = 18\text{ g/mol}

.

Establishing the molar component masses is required to determine the percentage ratio of water of crystallization to the total mass of the hydrated salt.

2

Set up an algebraic ratio relating the mass of water lost to the total hydrated mass using the given percentage.

\frac{18x}{106 + 18x} = 0.6294

.

Heating to constant mass removes all water of crystallization, meaning the mass lost corresponds to the

x\text{H}_2\text{O}

component of the hydrated crystal.

3

Solve the algebraic equation for

x

.

x = 10

.

Isolating

x

yields the exact stoichiometric coefficient of water molecules per mole of hydrated salt.

Key Concept

Quantitative determination of water of crystallization using percentage mass loss.

Question 1403Question →

A light, rigid horizontal bar $AB$ of length $1.5\text{ m}$ is smoothly pivoted at end $A$ . A vertical downward load of $40\text{ N}$ is hung from end $B$ . The bar is kept in horizontal equilibrium by a light string attached at point $C$ , located $1.0\text{ m}$ from $A$ . The string exerts a tension force $T$ pulling upwards at an angle of $30^\circ$ relative to the horizontal bar. What is the magnitude of the tension $T$ in newtons?

Show answer & explanation

Answer: 120

Answer

The magnitude of the tension

T

in the string is

120\text{ N}

.

For the bar to maintain rotational equilibrium, the clockwise moment about pivot

A

must equal the counterclockwise moment about

A

. The

40\text{ N}

load exerts a clockwise moment of

40\text{ N} \times 1.5\text{ m} = 60\text{ N}\cdot\text{m}

. The string tension

T

exerts a counterclockwise moment given by its vertical component multiplied by the distance from the pivot:

(T \sin 30^\circ) \times 1.0\text{ m} = 0.5T \text{ N}\cdot\text{m}

. Equating the two moments gives

0.5T = 60\text{ N}\cdot\text{m}

, yielding

T = 120\text{ N}

.

Step-by-Step Solution

1

Calculate the clockwise moment about the pivot at end

A

τ_{\text{clockwise}} = 40\text{ N} \times 1.5\text{ m} = 60\text{ N}\cdot\text{m}

The weight at

B

acts vertically downward at a perpendicular distance of

1.5\text{ m}

from pivot

A

.

2

Determine the perpendicular component of tension

T

relative to the bar

F_{\perp} = T \sin 30^\circ = 0.5T

Only the component of force perpendicular to the bar produces a moment about the pivot.

3

Set up the counterclockwise moment about pivot

A

τ_{\text{counterclockwise}} = (0.5T) \times 1.0\text{ m} = 0.5T \text{ N}\cdot\text{m}

The string is attached at point

C

, which is

1.0\text{ m}

away from pivot

A

.

4

Apply the principle of moments for rotational equilibrium and solve for

T

0.5T = 60 \implies T = 120\text{ N}

For rotational equilibrium, total clockwise moments must equal total counterclockwise moments about any pivot.

Key Concept

Principle of moments and rotational equilibrium for forces acting at non-perpendicular angles.

Estimated Time:1m 30s

Question 1404Question →

A Uranium-235 nucleus undergoes nuclear fission after absorbing a thermal neutron according to the reaction equation:

{^{235}_{92}\text{U}} + {^{1}_{0}\text{n}} \rightarrow {^{140}_{54}\text{Xe}} + {^{94}_{38}\text{Sr}} + x\ {^{1}_{0}\text{n}}

Given the rest masses:
- Mass of

{^{235}_{92}\text{U}} = 235.0439\text{ u}

- Mass of

{^{140}_{54}\text{Xe}} = 139.9216\text{ u}

- Mass of

{^{94}_{38}\text{Sr}} = 93.9154\text{ u}

- Mass of

{^{1}_{0}\text{n}} = 1.0087\text{ u}

-

1\text{ u} = 931.5\text{ MeV}

Calculate the total energy released during this fission process in MeV.

Show answer & explanation

Answer: 184.62

Answer

184.62 MeV

The total energy released is calculated by first balancing the nuclear equation to determine that 2 neutrons are emitted (

x = 2

). The total mass of reactants is

236.0526\text{ u}

and the products is

235.8544\text{ u}

. The mass defect of

0.1982\text{ u}

multiplied by

931.5\text{ MeV/u}

yields

184.62\text{ MeV}

.

Step-by-Step Solution

1

Balance the mass numbers in the nuclear equation to find the number of emitted neutrons (

x

)

x = 2

neutrons

Conservation of mass number requires

235 + 1 = 140 + 94 + x

, giving

236 = 234 + x

, so

x = 2

.

2

Calculate the total mass of the reactants before fission

236.0526\text{ u}

Summing the mass of U-235 (

235.0439\text{ u}

) and the incident neutron (

1.0087\text{ u}

).

3

Calculate the total mass of the products after fission

235.8544\text{ u}

Summing masses of Xe-140 (

139.9216\text{ u}

), Sr-94 (

93.9154\text{ u}

), and 2 neutrons (

2 \times 1.0087\text{ u} = 2.0174\text{ u}

).

4

Determine the mass defect (mass difference)

\Delta m = 0.1982\text{ u}

Mass defect

\Delta m = m_{\text{reactants}} - m_{\text{products}} = 236.0526 - 235.8544 = 0.1982\text{ u}

.

5

Convert mass defect to energy using

1\text{ u} = 931.5\text{ MeV}

184.62\text{ MeV}

Multiplying mass defect

0.1982\text{ u}

by

931.5\text{ MeV/u}

gives

184.6233\text{ MeV}

, which rounds to

184.62\text{ MeV}

.

Key Concept

Mass defect and mass-energy equivalence in nuclear fission reactions

Estimated Time:2m 30s

Question 1405Question →

A metal block of mass $4.0\text{ kg}$ absorbs $12\text{ kJ}$ of thermal energy, causing its temperature to rise by $15\text{ K}$ . What is the specific heat capacity of the metal block in $\text{J kg}^{-1}\text{K}^{-1}$ ?

Show answer & explanation

Answer: 200

Answer

The specific heat capacity of the metal block is

200\text{ J kg}^{-1}\text{K}^{-1}

.

Specific heat capacity

c

is determined using

c = \frac{Q}{m \Delta T}

. Substituting

Q = 12,000\text{ J}

,

m = 4.0\text{ kg}

, and

\Delta T = 15\text{ K}

yields

c = \frac{12000}{4.0 \times 15} = 200\text{ J kg}^{-1}\text{K}^{-1}

.

Step-by-Step Solution

1

Convert given energy to standard SI units

Q = 12\text{ kJ} = 12,000\text{ J}

Calculation of specific heat capacity requires heat energy in Joules.

2

Relate heat energy, mass, specific heat capacity, and temperature change

Q = m c \Delta T \implies c = \frac{Q}{m \Delta T}

Specific heat capacity

c

represents heat energy per unit mass per Kelvin temperature change.

3

Substitute the values and compute the result

c = \frac{12000}{4.0 \times 15} = 200\text{ J kg}^{-1}\text{K}^{-1}

Evaluating the expression gives

200\text{ J kg}^{-1}\text{K}^{-1}

.

Key Concept

Specific heat capacity is the quantity of heat required to raise the temperature of

1\text{ kg}

of a substance by

1\text{ K}

, expressed as

c = \frac{Q}{m \Delta T}

.

Question 1406Question →

A concave shaving mirror has a radius of curvature of $60\text{ cm}$ . A person places their face in front of the mirror such that an upright image magnified $3$ times is formed. What is the distance of the face from the mirror, in centimeters?

Show answer & explanation

Answer: 20

Answer

The distance of the person's face from the mirror is

20\text{ cm}

.

For a concave mirror with a radius of curvature of

60\text{ cm}

, the focal length is

f = +30\text{ cm}

. An upright image is virtual, corresponding to a positive magnification

m = +3

. Since

m = -\frac{v}{u}

, the image distance is

v = -3u

. Substituting these into the mirror formula

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

gives

\frac{1}{30} = \frac{1}{u} - \frac{1}{3u} = \frac{2}{3u}

, solving to

u = 20\text{ cm}

.

Step-by-Step Solution

1

Determine the focal length of the concave mirror.

f = 30\text{ cm}

The focal length is half the radius of curvature (

f = \frac{r}{2} = \frac{60\text{ cm}}{2} = 30\text{ cm}

).

2

Express the image distance

v

in terms of the object distance

u

using the magnification relationship.

v = -3u

An upright image produced by a spherical mirror is virtual, so linear magnification

m = +3

. Using

m = -\frac{v}{u} = +3

, we obtain

v = -3u

.

3

Substitute

f

and

v

into the mirror equation to solve for

u

.

u = 20\text{ cm}

Applying the mirror formula

\frac{1}{f} = \frac{1}{u} + \frac{1}{v} \implies \frac{1}{30} = \frac{1}{u} - \frac{1}{3u} = \frac{2}{3u} \implies 3u = 60 \implies u = 20\text{ cm}

.

Key Concept

Mirror equation and sign conventions for virtual images formed by concave mirrors

Question 1407Question →

Two long, straight parallel wires separated by a distance of $5.0\text{ cm}$ in air carry equal currents in opposite directions. If the repulsive force per unit length between the wires is $1.6 \times 10^{-3}\text{ N/m}$ , determine the magnitude of the current flowing through each wire in amperes. (Take $\mu_0 = 4\pi \times 10^{-7}\text{ T}\cdot\text{m/A}$ )

Show answer & explanation

Answer: 20

Answer

The magnitude of the current flowing through each wire is

20\text{ A}

.

Using the parallel conductor force formula

\frac{F}{L} = \frac{\mu_0 I^2}{2\pi d}

, we substitute

\frac{F}{L} = 1.6 \times 10^{-3}\text{ N/m}

,

d = 0.05\text{ m}

, and

\mu_0 = 4\pi \times 10^{-7}\text{ T}\cdot\text{m/A}

. Simplifying yields

1.6 \times 10^{-3} = 4 \times 10^{-6} I^2

, giving

I^2 = 400

and

I = 20\text{ A}

.

Step-by-Step Solution

1

Recall the expression for force per unit length between two current-carrying parallel wires

\frac{F}{L} = \frac{\mu_0 I^2}{2\pi d}

The magnetic field generated by one wire exerts a magnetic force on the current in the adjacent wire.

2

Convert distance to meters and substitute all given values into the formula

d = 0.05\text{ m}

, leading to

1.6 \times 10^{-3} = \frac{(4\pi \times 10^{-7}) I^2}{2\pi (0.05)}

Standard SI unit for distance is meters, necessary for dimensional consistency.

3

Simplify the equation and compute the current magnitude

1.6 \times 10^{-3} = 4 \times 10^{-6} I^2 \implies I^2 = 400 \implies I = 20\text{ A}

Solving the quadratic term gives the scalar current magnitude in amperes.

Key Concept

Force per unit length between parallel current-carrying conductors

Question 1408Question →

A car of mass $1200\text{ kg}$ ascends a straight road inclined at an angle $\theta$ to the horizontal, where $\sin\theta = 0.1$ , at a steady speed of $15\text{ m s}^{-1}$ . If the total frictional resistance to motion is $400\text{ N}$ , what is the useful mechanical power output of the engine in kilowatts? (Take $g = 10\text{ m s}^{-2}$ )

Show answer & explanation

Answer: 24

Answer

24 kW

To maintain a constant ascending speed, the engine must supply a force equal to the sum of the component of weight parallel to the incline (

mg\sin\theta = 1200\text{ N}

) and the frictional force (

400\text{ N}

), resulting in a total force of

1600\text{ N}

. Multiplying this total force by the constant speed of

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

yields a power output of

24000\text{ W}

, which corresponds to

24\text{ kW}

.

Step-by-Step Solution

1

Calculate the gravitational force component acting down the slope

1200 N

The component of the car's weight parallel to the incline opposes upward motion:

F_g = m g \sin\theta = 1200 \text{ kg} \times 10 \text{ m s}^{-2} \times 0.1 = 1200 \text{ N}

.

2

Determine the total tractive force required from the car engine

1600 N

Because the velocity is constant, the net force is zero; hence, the engine force must balance both the gravitational slope component and the frictional resistance:

F_{\text{engine}} = F_g + F_{\text{friction}} = 1200 \text{ N} + 400 \text{ N} = 1600 \text{ N}

.

3

Calculate the mechanical power delivered by the engine

24 kW

Power is the product of tractive force and constant speed:

P = F_{\text{engine}} \times v = 1600 \text{ N} \times 15 \text{ m s}^{-1} = 24000 \text{ W} = 24 \text{ kW}

.

Key Concept

Power required to maintain motion against opposing forces on an inclined plane

Question 1409Question →

A small object lies at the bottom of a transparent vessel containing two immiscible liquid layers, $A$ and $B$ . Layer $A$ (top) has a real thickness of $7.0\text{ cm}$ and a refractive index of $1.40$ . Layer $B$ (bottom) has a real thickness of $8.0\text{ cm}$ and a refractive index of $1.60$ . Calculate the apparent displacement of the object, in centimeters, when viewed vertically from directly above.

Show answer & explanation

Answer: 5

Answer

The apparent displacement of the object is

5.0\text{ cm}

.

For multiple refractive layers viewed normally, the total apparent depth is the sum of individual layer apparent depths:

\frac{7.0}{1.40} + \frac{8.0}{1.60} = 5.0 + 5.0 = 10.0\text{ cm}

. Subtracting this total apparent depth from the total real depth of

15.0\text{ cm}

gives an apparent vertical shift (displacement) of

5.0\text{ cm}

.

Step-by-Step Solution

1

Calculate the apparent depth of the top liquid layer (Layer A)

Apparent depth of Layer A =

5.0\text{ cm}

Apparent depth for a single medium is obtained by dividing real depth by its refractive index:

\frac{7.0}{1.40} = 5.0\text{ cm}

.

2

Calculate the apparent depth of the bottom liquid layer (Layer B)

Apparent depth of Layer B =

5.0\text{ cm}

Apparent depth for Layer B is

\frac{8.0}{1.60} = 5.0\text{ cm}

.

3

Calculate total real depth and total apparent depth

Total real depth =

15.0\text{ cm}

; Total apparent depth =

10.0\text{ cm}

Depths in composite media are additive.

4

Calculate vertical apparent displacement

Apparent displacement =

5.0\text{ cm}

Displacement is the difference between total real depth and total apparent depth:

15.0\text{ cm} - 10.0\text{ cm} = 5.0\text{ cm}

.

Key Concept

Refraction through composite media and vertical apparent displacement

Question 1410Question →

An electric water heater operating at a voltage of $240\,\text{V}$ has a heating element of resistance $48\,\Omega$ . It is used to heat $1.5\,\text{kg}$ of water from $20^\circ\text{C}$ to $100^\circ\text{C}$ . If the thermal efficiency of the heating process is $80\%$ , calculate the total electrical energy consumed by the heater in kilojoules ( $\text{kJ}$ ). [Take specific heat capacity of water = $4200\,\text{J}\cdot\text{kg}^{-1}\cdot\text{K}^{-1}$ ]

Show answer & explanation

Answer: 630

Answer

The total electrical energy consumed by the heater is

630\,\text{kJ}

.

The thermal energy required to raise the temperature of

1.5\,\text{kg}

of water by

80^\circ\text{C}

is

Q = 1.5 \times 4200 \times 80 = 504,000\,\text{J} = 504\,\text{kJ}

. Taking into account the

80\%

thermal efficiency, the total electrical energy consumed is

E_{\text{elec}} = \frac{504\,\text{kJ}}{0.80} = 630\,\text{kJ}

.

Step-by-Step Solution

1

Calculate the useful heat energy needed to heat the water.

Q = m c (T_2 - T_1) = 1.5 \times 4200 \times (100 - 20) = 504,000\,\text{J} = 504\,\text{kJ}

.

The thermal energy transferred to the water depends on its mass, specific heat capacity, and temperature increase.

2

Account for the efficiency of the heating element to find total electrical energy input.

E_{\text{elec}} = \frac{Q}{\text{Efficiency}} = \frac{504\,\text{kJ}}{0.80} = 630\,\text{kJ}

.

Since only

80\%

of the electrical energy is converted into useful heat energy for the water, the input electrical energy must be greater than the output heat energy.

Key Concept

Conversion of electrical energy to thermal energy and application of thermal efficiency.

Question 1411Question →

The mass defect of a nitrogen nucleus $^{14}_{7}\text{N}$ is calculated to be $0.112\text{ u}$ . Taking $1\text{ u} = 931.5\text{ MeV}$ , what is the total binding energy of the nucleus in $\text{MeV}$ ?

Show answer & explanation

Answer: 104.328

Answer

The total binding energy of the

^{14}_{7}\text{N}

nucleus is

104.328\text{ MeV}

.

The total binding energy of a nucleus is equal to the mass defect multiplied by the energy equivalent of one atomic mass unit (

931.5\text{ MeV/u}

). Calculating

0.112\text{ u} \times 931.5\text{ MeV/u}

gives

104.328\text{ MeV}

.

Step-by-Step Solution

1

Apply the binding energy formula

E_b = \Delta m \times 931.5\text{ MeV/u}

.

E_b = 0.112\text{ u} \times 931.5\text{ MeV/u} = 104.328\text{ MeV}

.

The binding energy is obtained by converting the mass defect directly into its energy equivalent.

Key Concept

Mass Defect and Binding Energy Conversion

Question 1412Question →

An object is placed at a distance $u$ in front of a concave mirror of focal length $12\text{ cm}$ . A plane mirror is placed perpendicular to the principal axis at a distance of $32\text{ cm}$ in front of the concave mirror, between the object and the concave mirror. If the real image formed by the concave mirror coincides in space with the virtual image formed by the plane mirror, what is the value of $u$ in centimeters?

Show answer & explanation

Answer: 48

Answer

The correct object distance

u

is

48\text{ cm}

.

The object is located at distance

u

from the concave mirror. With the plane mirror at

32\text{ cm}

from the concave mirror, the object distance from the plane mirror is

u - 32

. The plane mirror forms an image at distance

u - 32

behind itself, which corresponds to

32 - (u - 32) = 64 - u

from the concave mirror. Setting

v = 64 - u

in the mirror formula

\frac{1}{12} = \frac{1}{u} + \frac{1}{64 - u}

gives

u^2 - 64u + 768 = 0

. Factoring yields

u = 48\text{ cm}

or

u = 16\text{ cm}

. Because the plane mirror is between the object and the concave mirror,

u > 32\text{ cm}

, so

u = 48\text{ cm}

.

Step-by-Step Solution

1

Find the position of the image formed by the plane mirror in terms of

u

.

The object is at a distance

(u - 32)\text{ cm}

in front of the plane mirror. Its virtual image is formed

(u - 32)\text{ cm}

behind the plane mirror, which places it at

32 - (u - 32) = (64 - u)\text{ cm}

in front of the concave mirror.

A plane mirror forms an image behind it at a distance equal to the object distance in front of it.

2

Equate the image distance of the concave mirror

v

to the position of the plane mirror image.

v = 64 - u

The question states that the image formed by the concave mirror coincides in position with the image formed by the plane mirror.

3

Substitute

f = 12\text{ cm}

and

v = 64 - u

into the mirror equation.

\frac{1}{12} = \frac{1}{u} + \frac{1}{64 - u}

The standard mirror formula relates focal length, object distance, and image distance.

4

Solve the algebraic equation for

u

.

\frac{1}{12} = \frac{(64 - u) + u}{u(64 - u)} \implies 64u - u^2 = 768 \implies u^2 - 64u + 768 = 0

Combining fractions and multiplying across gives a quadratic equation in standard form.

5

Factor the quadratic equation and select the physical root.

(u - 48)(u - 16) = 0 \implies u = 48\text{ cm}

or

u = 16\text{ cm}

. Since

u > 32\text{ cm}

,

u = 48\text{ cm}

.

The plane mirror is situated between the object and the concave mirror at

32\text{ cm}

, so the object distance

u

must be greater than

32\text{ cm}

.

Key Concept

Image coincidence in combined plane and curved optical systems

Estimated Time:3m 0s

Question 1413Question →

A spring balance is attached to the ceiling of an elevator accelerating downwards at $2.0\text{ m s}^{-2}$ . Suspended from the hook of the spring balance is a light, frictionless pulley carrying two masses of $3.0\text{ kg}$ and $1.0\text{ kg}$ connected by a light inextensible string. Taking the acceleration due to gravity $g = 10.0\text{ m s}^{-2}$ , what is the reading registered by the spring balance in newtons?

Show answer & explanation

Answer: 24

Answer

The reading registered by the spring balance is 24 N.

In a frame accelerating downwards at

a = 2.0\text{ m s}^{-2}

, the effective acceleration due to gravity is reduced to

g' = g - a = 8.0\text{ m s}^{-2}

. Within this frame, the tension in the Atwood machine string is

T = \frac{2(3.0)(1.0)}{3.0 + 1.0} \times 8.0 = 12.0\text{ N}

. Since two string segments act downward on the light pulley suspended from the spring balance, the total tension force registered by the balance scale is

2T = 24.0\text{ N}

.

Step-by-Step Solution

1

Determine the effective local acceleration due to gravity inside the accelerating elevator

g' = 8.0\text{ m s}^{-2}

Because the elevator accelerates downward at

a = 2.0\text{ m s}^{-2}

, objects inside experience an apparent gravitational acceleration of

g' = g - a

.

2

Compute the tension in the string supporting the two masses in the modified gravitational field

T = 12.0\text{ N}

For an Atwood machine system in effective gravity

g'

, string tension is

T = \frac{2 m_1 m_2}{m_1 + m_2} g' = \frac{2(3.0)(1.0)}{4.0} \times 8.0 = 12.0\text{ N}

.

3

Calculate the downward pull on the spring balance

F = 24.0\text{ N}

The spring balance supports the frictionless pulley, which experiences a downward force from two upward string segments, making the total measured weight force equal to

2T = 24.0\text{ N}

.

Key Concept

Apparent weight measurement and tension forces in accelerating frames

Question 1414Question →

Two long, straight, parallel horizontal conductors are separated vertically by a distance of $2.0\text{ cm}$ . The upper conductor has a mass per unit length of $0.04\text{ kg/m}$ and carries a steady current of $50\text{ A}$ . Assuming the currents in the two conductors flow in opposite directions so that the resulting magnetic force is repulsive, what current (in amperes) must flow through the lower conductor to magnetically levitate and balance the weight of the upper conductor? (Take $g = 9.8\text{ m/s}^2$ and $\mu_0 = 4\pi \times 10^{-7}\text{ T}\cdot\text{m/A}$ )

Show answer & explanation

Answer: 784

Answer

The required current in the lower conductor is

784\text{ A}

.

Equating magnetic repulsion per unit length

\frac{\mu_0 I_1 I_2}{2\pi d}

to weight per unit length

\lambda g

gives

\frac{(2 \times 10^{-7}) \times 50 \times I_2}{0.02} = 0.04 \times 9.8

, which simplifies to

5 \times 10^{-4} I_2 = 0.392

, yielding

I_2 = 784\text{ A}

.

Step-by-Step Solution

1

Equate the upward repulsive magnetic force per unit length to the downward gravitational weight per unit length.

\frac{\mu_0 I_1 I_2}{2\pi d} = \lambda g

For the upper conductor to levitate in vertical static equilibrium, the upward magnetic force per meter must exactly balance its weight per meter.

2

Substitute all given physical values in SI units into the force balance equation.

\frac{(4\pi \times 10^{-7}) \times 50 \times I_2}{2\pi \times 0.02} = 0.04 \times 9.8

Converting distance

d = 2.0\text{ cm} = 0.02\text{ m}

and using mass density

\lambda = 0.04\text{ kg/m}

sets up a single equation with unknown

I_2

.

3

Simplify both sides of the equation.

5 \times 10^{-4} I_2 = 0.392

Calculating

\frac{2 \times 10^{-7} \times 50}{0.02} = 5 \times 10^{-4}\text{ N/(A}\cdot\text{m)}

and

0.04 \times 9.8 = 0.392\text{ N/m}

.

4

Solve for the unknown current

I_2

.

I_2 = \frac{0.392}{5 \times 10^{-4}} = 784\text{ A}

Dividing the weight per meter by the magnetic force coefficient yields the exact required current magnitude.

Key Concept

Interaction force between parallel current-carrying conductors and mechanical equilibrium

Question 1415Question →

A nitrogen nucleus $^{14}_{7}\text{N}$ has a nuclear mass of $13.9992\text{ u}$ . Given that the mass of a proton is $1.0073\text{ u}$ and the mass of a neutron is $1.0087\text{ u}$ , what is the total binding energy of the nucleus in $\text{MeV}$ ? (Take $1\text{ u} = 931\text{ MeV}$ )

Show answer & explanation

Answer: 105.02

Answer

The total binding energy of the

^{14}_{7}\text{N}

nucleus is

105.02\text{ MeV}

.

The total binding energy is computed by finding the total rest mass of 7 protons and 7 neutrons (14.1120 u), subtracting the actual nuclear mass of Nitrogen-14 (13.9992 u) to obtain a mass defect of 0.1128 u, and converting this mass defect to energy by multiplying by 931 MeV/u, yielding 105.02 MeV.

Step-by-Step Solution

1

Determine the number of constituent protons and neutrons

Z = 7

protons and

N = 14 - 7 = 7

neutrons

The atomic number is 7 and the mass number is 14.

2

Calculate the total mass of the constituent nucleons

M_{\text{nucleons}} = (7 \times 1.0073\text{ u}) + (7 \times 1.0087\text{ u}) = 7.0511\text{ u} + 7.0609\text{ u} = 14.1120\text{ u}

The sum of the individual rest masses of all isolated protons and neutrons.

3

Compute the mass defect

\Delta m = 14.1120\text{ u} - 13.9992\text{ u} = 0.1128\text{ u}

Mass defect is the difference between total mass of isolated nucleons and the bound nuclear mass.

4

Convert the mass defect into energy in MeV

E_b = 0.1128\text{ u} \times 931\text{ MeV/u} = 105.0168\text{ MeV} \approx 105.02\text{ MeV}

Applying the mass-energy equivalence factor

1\text{ u} = 931\text{ MeV}

.

Key Concept

Mass Defect and Binding Energy

Estimated Time:2m 0s

Question 1416Question →

A simple pendulum suspended in a physics laboratory has a length of $0.64\text{ m}$ . Given that acceleration due to gravity $g = 10\text{ m/s}^2$ and taking $\pi^2 = 10$ , what is the period of oscillation of the pendulum in seconds?

Show answer & explanation

Answer: 1.6

Answer

The period of oscillation of the simple pendulum is

1.6\text{ s}

.

Using the pendulum period relationship

T = 2\pi \sqrt{\frac{L}{g}}

, substituting

L = 0.64\text{ m}

,

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

, and

\pi = \sqrt{10}

yields

T = 2\sqrt{10}\sqrt{\frac{0.64}{10}} = 2\sqrt{0.64} = 1.6\text{ s}

.

Step-by-Step Solution

1

Identify the formula for the period of a simple pendulum.

The period formula is

T = 2\pi \sqrt{\frac{L}{g}}

.

The period depends on the length of the pendulum

L

and acceleration due to gravity

g

.

2

Substitute the given values into the formula.

T = 2\pi \sqrt{\frac{0.64\text{ m}}{10\text{ m/s}^2}}

.

Given parameters are

L = 0.64\text{ m}

and

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

.

3

Simplify the equation using

\pi = \sqrt{10}

.

T = 2\sqrt{10} \times \sqrt{0.064} = 2 \sqrt{10 \times 0.064} = 2 \sqrt{0.64} = 2 \times 0.8 = 1.6\text{ s}

.

Using

\pi^2 = 10

simplifies the calculation cleanly without requiring a calculator.

Key Concept

Simple Pendulum Period of Oscillation

Question 1417Question →

A particle moves along a circular track of radius $r$ with a constant speed $v$ . If the radius of the track is measured with a percentage error of $1.2\%$ , and the speed is measured as $(10.0 \pm 0.3)\text{ m s}^{-1}$ , what is the percentage error in the calculated centripetal acceleration of the particle?

Show answer & explanation

Answer: 7.2

Answer

The percentage error in the calculated centripetal acceleration is 7.2%.

The centripetal acceleration is

a_c = \frac{v^2}{r}

. The percentage error in

v

is

\frac{0.3}{10.0} \times 100\% = 3.0\%

. According to error propagation rules, the percentage error in

a_c

is

2 \times (\text{percentage error in } v) + (\text{percentage error in } r) = 2(3.0\%) + 1.2\% = 7.2\%

.

Step-by-Step Solution

1

Calculate the percentage error in the speed measurement

v

.

\text{Percentage error in } v = \frac{0.3}{10.0} \times 100\% = 3.0\%

Percentage error is given by dividing the absolute error by the measured value and multiplying by 100%.

2

Write the relationship for centripetal acceleration

a_c

in terms of

v

and

r

.

a_c = \frac{v^2}{r}

Centripetal acceleration is directly proportional to the square of speed and inversely proportional to radius.

3

Apply the fractional error propagation law for a quantity involving powers and quotients.

\frac{\Delta a_c}{a_c} = 2\left(\frac{\Delta v}{v}\right) + \frac{\Delta r}{r}

When a measured quantity is raised to a power

n

, its fractional error contribution is multiplied by

n

. Errors accumulate additively.

4

Substitute the individual percentage errors to find the total percentage error in

a_c

.

\text{Percentage error in } a_c = 2(3.0\%) + 1.2\% = 7.2\%

Multiplying the speed's percentage error by 2 and adding the radius percentage error yields the total relative error.

Key Concept

Propagation of percentage errors in physical formulas containing powers and ratios.

Question 1418Question →

A nation's economic records for a given fiscal year provide the following national income figures:

- Gross Domestic Product ( $\text{GDP}$ ): $\text{N}5,400\text{ million}$
- Income earned by domestic citizens working abroad: $\text{N}450\text{ million}$
- Income earned by foreign nationals operating domestically: $\text{N}600\text{ million}$
- Capital Consumption Allowance ( $\text{CCA}$ ): $\text{N}380\text{ million}$

Calculate the Net National Product ( $\text{NNP}$ ) of the country in millions of Naira.

Show answer & explanation

Answer: 4870

Answer

The Net National Product (NNP) of the country is N4,870 million.

Net National Product (NNP) is obtained by adding Net Factor Income from Abroad (NFIA) to Gross Domestic Product (GDP) to get Gross National Product (GNP), and then subtracting Capital Consumption Allowance (CCA). Here, NFIA = N450 million - N600 million = -N150 million. Thus, GNP = N5,400 million - N150 million = N5,250 million. Finally, NNP = N5,250 million - N380 million = N4,870 million.

Step-by-Step Solution

1

Calculate Net Factor Income from Abroad (NFIA)

NFIA = N450 million - N600 million = -N150 million

NFIA measures the net flow of factor payments between domestic citizens abroad and foreign residents domestically.

2

Calculate Gross National Product (GNP)

GNP = N5,400 million + (-N150 million) = N5,250 million

GNP adjusts GDP for net factor receipts from abroad.

3

Calculate Net National Product (NNP)

NNP = N5,250 million - N380 million = N4,870 million

NNP reflects the net output available to an economy after accounting for capital depreciation.

Key Concept

Basic National Income Aggregates (GDP, GNP, NNP, NDP)

Question 1419Question →

A sample of a radioactive isotope used in medical imaging has an initial activity of $320\text{ MBq}$ . If its activity decreases to $20\text{ MBq}$ after an elapsed time of $15\text{ hours}$ , what is the half-life of the isotope in hours?

Show answer & explanation

Answer: 3.75

Answer

The half-life of the radioactive isotope is

3.75\text{ hours}

.

The initial activity of

320\text{ MBq}

drops to

20\text{ MBq}

, which is a reduction to

\frac{20}{320} = \frac{1}{16}

of its original value. Since

\frac{1}{16} = \left(\frac{1}{2}\right)^4

, exactly 4 half-lives have elapsed over the period of

15\text{ hours}

. Therefore, one half-life is

\frac{15\text{ hours}}{4} = 3.75\text{ hours}

.

Step-by-Step Solution

1

Calculate the fraction of initial activity remaining

Fraction remaining =

\frac{20\text{ MBq}}{320\text{ MBq}} = \frac{1}{16}

The ratio of current activity to initial activity determines the fraction of un-decayed nuclei remaining.

2

Determine the number of half-lives (

n

) that have elapsed

n = 4

half-lives, since

\left(\frac{1}{2}\right)^4 = \frac{1}{16}

Each half-life reduces the remaining sample activity by half.

3

Calculate the half-life duration (

T_{1/2}

)

T_{1/2} = \frac{t}{n} = \frac{15\text{ hours}}{4} = 3.75\text{ hours}

Dividing total elapsed time by the number of half-lives gives the duration of one half-life.

Key Concept

Radioactive Decay Law and relationship between remaining activity fraction, number of half-lives, and total elapsed time.

Question 1420Question →

A mixture of $15\text{ cm}^3$ of ethene ( $C_2H_4$ ) and $50\text{ cm}^3$ of oxygen gas was exploded at constant temperature and pressure. If the resulting mixture was cooled to room temperature, what is the total volume of the residual gas in $\text{cm}^3$ ?

Show answer & explanation

Answer: 35

Answer

35 cm³

According to Gay-Lussac's Law,

15\text{ cm}^3

of ethene reacts with

45\text{ cm}^3

of oxygen to produce

30\text{ cm}^3

of carbon(IV) oxide gas. Since

50\text{ cm}^3

of oxygen was initially present,

5\text{ cm}^3

of unreacted oxygen gas remains. Upon cooling to room temperature, water condenses to liquid, leaving a total residual gaseous volume of

5\text{ cm}^3 + 30\text{ cm}^3 = 35\text{ cm}^3

.

Step-by-Step Solution

1

Write the balanced equation for the complete combustion of ethene gas.

C_2H_4(g) + 3O_2(g) \rightarrow 2CO_2(g) + 2H_2O(l)

By Gay-Lussac's Law of Combining Volumes, gases react in simple whole-number volume ratios equal to their stoichiometric coefficients at constant temperature and pressure.

2

Determine the volume of oxygen required and the unreacted excess volume.

Volume of

O_2

consumed =

3 \times 15\text{ cm}^3 = 45\text{ cm}^3

. Remaining unreacted

O_2 = 50\text{ cm}^3 - 45\text{ cm}^3 = 5\text{ cm}^3

.

Since

15\text{ cm}^3

of

C_2H_4

requires only

45\text{ cm}^3

of

O_2

, oxygen is in excess and ethene is the limiting reactant.

3

Calculate the volume of carbon(IV) oxide gas produced.

Volume of

CO_2

produced =

2 \times 15\text{ cm}^3 = 30\text{ cm}^3

.

1 volume of

C_2H_4

produces 2 volumes of

CO_2

gas.

4

Calculate the total volume of the residual gaseous mixture after cooling to room temperature.

Total residual gaseous volume =

5\text{ cm}^3 \text{ (excess } O_2\text{)} + 30\text{ cm}^3 \text{ (produced } CO_2\text{)} = 35\text{ cm}^3

.

Water formed is in the liquid state at room temperature and contributes negligible volume to the gaseous mixture.

Key Concept

Gay-Lussac's Law of Combining Volumes and Residual Gas Volume Calculations

Estimated Time:1m 30s

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