Tüm alıştırma soruları

Soru 941Soru →

A radioactive mixture initially contains two radioisotopes, $P$ and $Q$ , such that the initial number of undecayed nuclei of $P$ is $8$ times that of $Q$ . If the half-life of isotope $P$ is $2\text{ hours}$ and the half-life of isotope $Q$ is $6\text{ hours}$ , calculate the time, in hours, after which the number of undecayed nuclei of both isotopes will be equal.

Cevabı ve açıklamayı göster

Cevap: 9

Cevap

The time after which the number of undecayed nuclei of both isotopes will be equal is 9 hours.

By applying the radioactive decay law

N(t) = N_0(1/2)^{t/T_{1/2}}

to both isotopes with initial ratio

N_{P0} = 8N_{Q0}

and equating

N_P(t) = N_Q(t)

, we obtain

8 = 2^{t/3}

, which gives

t = 9\text{ hours}

.

Adım Adım Çözüm

1

Write the decay equations for isotopes P and Q based on their half-lives

N_P(t) = N_{P0}\left(\frac{1}{2}\right)^{t/2}

and

N_Q(t) = N_{Q0}\left(\frac{1}{2}\right)^{t/6}

Radioactive decay follows the exponential relationship

N(t) = N_0 \left(\frac{1}{2}\right)^{t/T_{1/2}}

.

2

Apply the initial condition

N_{P0} = 8 N_{Q0}

and set the two expressions equal

8 N_{Q0} \left(\frac{1}{2}\right)^{t/2} = N_{Q0} \left(\frac{1}{2}\right)^{t/6}

The problem asks for the time

t

when both isotopes have equal remaining undecayed nuclei.

3

Divide both sides by

N_{Q0} \left(\frac{1}{2}\right)^{t/2}

and simplify the exponents

8 = \frac{(1/2)^{t/6}}{(1/2)^{t/2}} = \left(\frac{1}{2}\right)^{-t/3} = 2^{t/3}

Applying exponent laws simplifies the ratio of powers of one-half into a single base-two exponent.

4

Solve for time

t

using powers of 2

2^3 = 2^{t/3} \implies \frac{t}{3} = 3 \implies t = 9\text{ hours}

Equating the exponents of identical base 2 gives the exact time.

Anahtar Kavram

Radioactive Decay Law and Half-life for Isotope Mixtures

Tahmini Süre:2m 0s

Soru 942Soru →

A $250\text{ dm}^3$ sample of hard water contains $0.012\text{ mol dm}^{-3}$ of dissolved magnesium tetraoxosulfate(VI), $\text{MgSO}_4$ . What mass, in grams, of anhydrous sodium trioxocarbonate(IV), $\text{Na}_2\text{CO}_3$ , is required to completely precipitate all the magnesium ions as magnesium trioxocarbonate(IV) and soften the water? $[\text{Molar mass of Na}_2\text{CO}_3 = 106\text{ g mol}^{-1}]$

Cevabı ve açıklamayı göster

Cevap: 318

Cevap

318 g of anhydrous sodium trioxocarbonate(IV) is required.

Permanent water hardness caused by soluble magnesium salts like magnesium tetraoxosulfate(VI) (

\text{MgSO}_4

) is removed by reaction with sodium trioxocarbonate(IV) (

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

). The balanced reaction

\text{MgSO}_4\text{(aq)} + \text{Na}_2\text{CO}_3\text{(aq)} \rightarrow \text{MgCO}_3\text{(s)} + \text{Na}_2\text{SO}_4\text{(aq)}

shows a 1:1 molar ratio. A

250\text{ dm}^3

volume at

0.012\text{ mol dm}^{-3}

contains

3.0\text{ moles}

of

\text{MgSO}_4

, which requires

3.0\text{ moles}

of

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

. Multiplying by its molar mass (

106\text{ g mol}^{-1}

) gives

318\text{ g}

.

Adım Adım Çözüm

1

Calculate the total number of moles of magnesium tetraoxosulfate(VI) in the water sample.

n(\text{MgSO}_4) = 250\text{ dm}^3 \times 0.012\text{ mol dm}^{-3} = 3.0\text{ mol}

Molar amount is calculated by multiplying the solution volume by its molar concentration.

2

Write the balanced chemical equation for softening permanent hardness with sodium trioxocarbonate(IV).

\text{MgSO}_4(\text{aq}) + \text{Na}_2\text{CO}_3(\text{aq}) \rightarrow \text{MgCO}_3(\text{s}) + \text{Na}_2\text{SO}_4(\text{aq})

Soluble magnesium ions causing permanent hardness are removed by precipitation as insoluble magnesium trioxocarbonate(IV).

3

Calculate the mass of anhydrous sodium trioxocarbonate(IV) needed.

\text{Mass} = 3.0\text{ mol} \times 106\text{ g mol}^{-1} = 318\text{ g}

From the 1:1 stoichiometric ratio, 3.0 moles of sodium trioxocarbonate(IV) is required.

Anahtar Kavram

Quantitative removal of permanent water hardness using sodium trioxocarbonate(IV) (washing soda)

Soru 943Soru →

A non-uniform wooden pole of length $6.0\text{ m}$ and weight $150\text{ N}$ is balanced horizontally on a pivot placed $2.4\text{ m}$ from its heavy end $P$ . The system achieves rotational equilibrium when a load of $50\text{ N}$ is hung directly from end $P$ . What is the distance of the center of gravity of the pole from end $P$ ?

Cevabı ve açıklamayı göster

Cevap: 3.2

Cevap

The distance of the center of gravity of the pole from end

P

is

3.2\text{ m}

.

Taking moments about the pivot at

2.4\text{ m}

from end

P

, the counter-clockwise moment created by the

50\text{ N}

load (

50\text{ N} \times 2.4\text{ m} = 120\text{ N}\cdot\text{m}

) must balance the clockwise moment created by the

150\text{ N}

weight of the pole acting at its center of gravity (

150\text{ N} \times (d - 2.4\text{ m})

). Equating these gives

120 = 150(d - 2.4)

, leading to

d - 2.4 = 0.8\text{ m}

, so

d = 3.2\text{ m}

.

Adım Adım Çözüm

1

Identify force positions relative to the pivot

The

50\text{ N}

load is

2.4\text{ m}

to the left of the pivot. The

150\text{ N}

weight acts at the center of gravity, which is

(d - 2.4\text{ m})

to the right of the pivot.

Moments are evaluated relative to the fulcrum to eliminate the unknown normal reaction force at the pivot.

2

Apply the Principle of Moments

Anti-clockwise moment =

50 \times 2.4 = 120\text{ N}\cdot\text{m}

. Clockwise moment =

150 \times (d - 2.4)

. Setting them equal:

120 = 150(d - 2.4)

.

For a body in rotational equilibrium, the total clockwise moment about any pivot equals the total anti-clockwise moment.

3

Solve the equation for distance

d

d - 2.4 = 0.8 \implies d = 3.2\text{ m}

.

Adding the displacement from the pivot (

0.8\text{ m}

) to the pivot position from end

P

(

2.4\text{ m}

) yields the position of the center of gravity from end

P

.

Anahtar Kavram

Rotational equilibrium and Principle of Moments for non-uniform rigid bodies

Tahmini Süre:1m 30s

Soru 944Soru →

A $4.00\text{ g}$ sample of a copper oxide is completely reduced by dry hydrogen gas to yield $3.20\text{ g}$ of metallic copper. According to the Law of Definite Proportions, what mass of this same copper oxide, in grams, will be produced when $5.00\text{ g}$ of pure copper is completely oxidized?

Cevabı ve açıklamayı göster

Cevap: 6.25

Cevap

6.25 g

According to the Law of Definite Proportions (or Constant Composition), a chemical compound always contains its component elements in fixed mass ratios. In the first sample, copper makes up

3.20\text{ g} / 4.00\text{ g} = 0.80

or

80\%

of the total mass. Therefore, in any sample of this oxide,

5.00\text{ g}

of copper represents

80\%

of the total mass. Dividing

5.00\text{ g}

by

0.80

gives

6.25\text{ g}

of copper oxide.

Adım Adım Çözüm

1

Determine the mass percentage (or mass fraction) of copper in the compound from the first experiment

Mass fraction of Cu = 3.20 / 4.00 = 0.80 (80%)

The first experiment provides quantitative data regarding the mass of copper contained in a known mass of oxide.

2

Apply the Law of Definite Proportions to calculate the required mass of copper oxide for 5.00 g of copper

Mass of Copper Oxide = 5.00 / 0.80 = 6.25 g

The Law of Definite Proportions dictates that the mass composition ratio remains constant regardless of the sample source or size.

Anahtar Kavram

Law of Definite Proportions

Soru 945Soru →

A particle of mass $0.50\text{ kg}$ executes simple harmonic motion along a straight line. When its displacement from the equilibrium position is $0.06\text{ m}$ , its speed is $0.80\text{ m/s}$ and its potential energy is $0.09\text{ J}$ . What is the magnitude of the maximum acceleration of the particle in $\text{m/s}^2$ ?

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Cevap: 10

Cevap

The magnitude of the maximum acceleration of the particle is

10\text{ m/s}^2

.

Using potential energy

E_p = \frac{1}{2}m\omega^2 x^2

, the angular frequency

\omega

is

10\text{ rad/s}

. Using

v = \omega\sqrt{A^2 - x^2}

, the amplitude

A

is

0.10\text{ m}

. Substituting these values into

a_{\max} = \omega^2 A

yields

10\text{ m/s}^2

.

Adım Adım Çözüm

1

Find angular frequency

\omega

from potential energy.

\omega = 10\text{ rad/s}

Using potential energy

E_p = \frac{1}{2}m\omega^2 x^2

:

0.09 = \frac{1}{2}(0.50)\omega^2 (0.06)^2 \implies 0.09 = 0.0009 \omega^2 \implies \omega^2 = 100 \implies \omega = 10\text{ rad/s}

.

2

Find amplitude

A

from speed.

A = 0.10\text{ m}

Using speed

v = \omega\sqrt{A^2 - x^2}

:

0.80 = 10\sqrt{A^2 - 0.06^2} \implies 0.08 = \sqrt{A^2 - 0.0036} \implies 0.0064 = A^2 - 0.0036 \implies A^2 = 0.0100 \implies A = 0.10\text{ m}

.

3

Calculate maximum acceleration.

a_{\max} = 10\text{ m/s}^2

Using maximum acceleration formula

a_{\max} = \omega^2 A

:

a_{\max} = 100 \times 0.10 = 10\text{ m/s}^2

.

Anahtar Kavram

Simple Harmonic Motion Energy and Kinematic Relations

Soru 946Soru →

When a $2.0\text{ g}$ sample of a solid solute is completely dissolved in $100.0\text{ g}$ of distilled water initially at $25.0^\circ\text{C}$ , an exothermic process occurs and the temperature of the water rises to $30.0^\circ\text{C}$ . Assuming the specific heat capacity of water is $4.2\text{ J g}^{-1}\text{ K}^{-1}$ and neglecting the heat capacity of the calorimeter container, what is the amount of heat energy, in Joules, absorbed by the water?

Cevabı ve açıklamayı göster

Cevap: 2100

Cevap

The amount of heat energy absorbed by the water is

2100\text{ J}

.

In an exothermic process, heat is transferred to the surroundings (water). Using the calorimetric relation

Q = m c \Delta T

with

m = 100.0\text{ g}

,

c = 4.2\text{ J g}^{-1}\text{ K}^{-1}

, and

\Delta T = (30.0 - 25.0)\text{ K} = 5.0\text{ K}

, the heat gained by the water is

Q = 100.0 \times 4.2 \times 5.0 = 2100\text{ J}

.

Adım Adım Çözüm

1

Determine the change in temperature (

\Delta T

) of the water.

\Delta T = T_{\text{final}} - T_{\text{initial}} = 30.0^\circ\text{C} - 25.0^\circ\text{C} = 5.0^\circ\text{C} = 5.0\text{ K}

The heat calculation requires the temperature difference resulting from the exothermic dissolution.

2

Apply the heat formula

Q = m c \Delta T

.

Q = 100.0\text{ g} \times 4.2\text{ J g}^{-1}\text{ K}^{-1} \times 5.0\text{ K}

Heat absorbed depends directly on the mass of water, its specific heat capacity, and the temperature rise.

3

Calculate the total heat energy absorbed.

Q = 2100\text{ J}

Multiplying the mass, heat capacity, and temperature change yields

2100\text{ J}

.

Anahtar Kavram

Calorimetric Heat Calculation (

Q = m c \Delta T

)

Soru 947Soru →

Heavy water is water in which the hydrogen atoms are replaced by the hydrogen isotope deuterium ( $^{2}_{1}\text{H}$ or $\text{D}$ ). Given that the atomic mass of deuterium is $2\text{ g mol}^{-1}$ and oxygen ( $^{16}_{8}\text{O}$ ) is $16\text{ g mol}^{-1}$ , what is the molar mass of heavy water ( $\text{D}_2\text{O}$ ) in $\text{g mol}^{-1}$ ?

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Cevap: 20

Cevap

The molar mass of heavy water (

\text{D}_2\text{O}

) is

20\text{ g mol}^{-1}

.

Heavy water (

\text{D}_2\text{O}

) contains two atoms of deuterium (

^2\text{H}

, atomic mass =

2

) and one atom of oxygen (

^{16}\text{O}

, atomic mass =

16

). The molar mass is calculated as

(2 \times 2) + 16 = 20\text{ g mol}^{-1}

.

Adım Adım Çözüm

1

Determine the molecular composition of heavy water

Heavy water (

\text{D}_2\text{O}

) contains 2 deuterium atoms and 1 oxygen atom.

Deuterium is an isotope of hydrogen containing one proton and one neutron, giving it a mass number of 2.

2

Sum the relative atomic masses of all constituent atoms

(2 \times 2) + 16 = 20\text{ g mol}^{-1}

Multiplying the mass of deuterium by two and adding the mass of one oxygen atom yields the molar mass of the compound.

Anahtar Kavram

Heavy Water and Deuterium Molar Mass

Soru 948Soru →

A particle of mass $0.20\text{ kg}$ executes simple harmonic motion along a straight line. When its displacement from the equilibrium position is $0.03\text{ m}$ , its speed is $0.16\text{ m/s}$ . When its displacement is $0.04\text{ m}$ , its speed is $0.12\text{ m/s}$ . What is the total mechanical energy of the particle in millijoules ( $\text{mJ}$ )?

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

The total mechanical energy of the particle is

4\text{ mJ}

.

Using the relation

v^2 = \omega^2(A^2 - x^2)

for the two given state points

(0.03\text{ m}, 0.16\text{ m/s})

and

(0.04\text{ m}, 0.12\text{ m/s})

forms a set of simultaneous equations. Subtracting them yields

\omega^2 = 16\text{ rad}^2/\text{s}^2

, leading to

A^2 = 0.0025\text{ m}^2

. Substituting these values into

E = \frac{1}{2}m\omega^2 A^2

gives

E = 0.004\text{ J}

, which converts to

4\text{ mJ}

.

Adım Adım Çözüm

1

Set up kinematic equations for both displacement points using

v^2 = \omega^2(A^2 - x^2)

.

0.0256 = \omega^2(A^2 - 0.0009)

and

0.0144 = \omega^2(A^2 - 0.0016)

.

The equation relates linear speed, angular frequency, amplitude, and instantaneous displacement in SHM.

2

Subtract the two simultaneous equations to eliminate

A^2

and find

\omega^2

.

0.0112 = 0.0007\omega^2 \implies \omega^2 = 16\text{ rad}^2/\text{s}^2

.

Eliminating amplitude isolates the angular frequency squared.

3

Determine

A^2

by substituting

\omega^2 = 16

back into one of the state equations.

A^2 = 0.0025\text{ m}^2 \implies A = 0.05\text{ m}

.

Amplitude is required to calculate the maximum potential or total mechanical energy.

4

Calculate total mechanical energy

E = \frac{1}{2}m\omega^2 A^2

and convert to millijoules.

E = \frac{1}{2} \times 0.20 \times 16 \times 0.0025 = 0.004\text{ J} = 4\text{ mJ}

.

Total energy in SHM is constant and proportional to mass, square of angular frequency, and square of amplitude.

Anahtar Kavram

Conservation of energy and phase-space relationship between velocity and displacement in simple harmonic motion.

Soru 949Soru →

A small spherical lead shot of radius $3.0 \times 10^{-3}\text{ m}$ and density $8000\text{ kg/m}^3$ falls vertically through a viscous oil of density $800\text{ kg/m}^3$ and dynamic viscosity $0.18\text{ Pa}\cdot\text{s}$ . What is the magnitude of the terminal velocity of the sphere in $\text{m/s}$ ? (Take $g = 10\text{ m/s}^2$ )

Cevabı ve açıklamayı göster

Cevap: 0.8

Cevap

The terminal velocity of the lead shot is

0.8\text{ m/s}

.

At terminal velocity, acceleration is zero because the upward forces (viscous drag

6\pi \eta r v_t

plus buoyant upthrust

\frac{4}{3}\pi r^3 \rho_f g

) completely balance the downward weight of the sphere (

\frac{4}{3}\pi r^3 \rho_s g

). Solving for velocity gives

v_t = \frac{2 r^2 (\rho_s - \rho_f) g}{9 \eta} = 0.8\text{ m/s}

.

Adım Adım Çözüm

1

Formulate the force balance equation at terminal velocity.

Weight (

W

) = Upthrust (

U

) + Viscous drag (

F_v

), which simplifies to

v_t = \frac{2 r^2 (\rho_s - \rho_f) g}{9 \eta}

.

When terminal velocity is attained, the net acceleration of the sphere is zero, so upward forces balance downward force.

2

Substitute the physical parameters into the terminal velocity formula.

v_t = \frac{2 \times (3.0 \times 10^{-3})^2 \times (8000 - 800) \times 10}{9 \times 0.18} = 0.8\text{ m/s}

.

Direct calculation using Stokes' law and Archimedes' principle.

Anahtar Kavram

Terminal Velocity and Stokes' Law in a Viscous Medium

Soru 950Soru →

A uniform horizontal rod $AB$ of length $2.0\text{ m}$ and mass $6.0\text{ kg}$ is suspended horizontally by two light vertical strings attached at end $A$ and at a point $C$ located $0.5\text{ m}$ from end $B$ . Taking $g = 10\text{ m/s}^2$ , what is the tension in the string at point $C$ in Newtons?

Cevabı ve açıklamayı göster

Cevap: 40

Cevap

The tension in the string at point C is 40 N.

The weight of the rod (

60\text{ N}

) acts at its midpoint (

1.0\text{ m}

from end

A

). Point

C

is located

1.5\text{ m}

from end

A

. Taking moments about end

A

gives

60\text{ N} \times 1.0\text{ m} = T_C \times 1.5\text{ m}

, which evaluates to

T_C = 40\text{ N}

.

Adım Adım Çözüm

1

Calculate total weight and identify the center of gravity position

Weight

W = 60\text{ N}

acting at

1.0\text{ m}

from end

A

For a uniform rod, the weight acts vertically downwards at its geometric center.

2

Set up the moment equilibrium equation taking end A as pivot

60\text{ N} \times 1.0\text{ m} = T_C \times 1.5\text{ m}

Taking moments about point

A

eliminates the force at

A

and equates clockwise moment from weight to counter-clockwise moment from tension at

C

.

3

Solve for the tension force at point C

T_C = 40\text{ N}

Dividing the total moment of

60\text{ N}\cdot\text{m}

by the moment arm of

1.5\text{ m}

yields

40\text{ N}

.

Anahtar Kavram

Principle of Moments and Rotational Equilibrium

Tahmini Süre:1m 30s

Soru 951Soru →

A sample of a radioactive isotope has an initial activity of $8000\text{ Bq}$ . After an elapsed time of $18\text{ minutes}$ , its activity reduces to $1000\text{ Bq}$ . Calculate the decay constant $\lambda$ of the isotope in $\text{min}^{-1}$ .

Cevabı ve açıklamayı göster

Cevap: 0.1155

Cevap

The decay constant of the radioactive isotope is

0.1155\text{ min}^{-1}

.

The activity decreases from

8000\text{ Bq}

to

1000\text{ Bq}

, which is a reduction to

\frac{1}{8}

of its initial value. Since

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

, exactly

3

half-lives have elapsed in

18\text{ minutes}

, meaning

T_{1/2} = 6\text{ minutes}

. Using the relationship

\lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.69315}{6\text{ min}}

, we obtain

\lambda \approx 0.1155\text{ min}^{-1}

.

Adım Adım Çözüm

1

Determine the number of elapsed half-lives from the activity reduction ratio.

The fraction of remaining activity is

\frac{A}{A_0} = \frac{1000}{8000} = \frac{1}{8} = \left(\frac{1}{2}\right)^3

, giving

n = 3

half-lives.

Radioactive decay follows the relation

A = A_0 (1/2)^n

.

2

Determine the half-life

T_{1/2}

of the isotope.

T_{1/2} = \frac{t}{n} = \frac{18\text{ min}}{3} = 6\text{ minutes}

.

Total elapsed time is equal to the number of half-lives multiplied by the duration of one half-life.

3

Compute the decay constant

\lambda

in

\text{min}^{-1}

.

\lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.69315}{6\text{ min}} = 0.1155\text{ min}^{-1}

.

The decay constant is fundamental to decay rate and related to half-life via

\lambda = \frac{\ln 2}{T_{1/2}}

.

Anahtar Kavram

Radioactive Decay Law and Decay Constant

Tahmini Süre:2m 0s

Soru 952Soru →

A composite cylindrical bar consists of two uniform sections of equal length joined end-to-end. Section A has a radius of $2.0\text{ cm}$ and a thermal conductivity of $300\text{ W m}^{-1}\text{K}^{-1}$ . Section B has a radius of $4.0\text{ cm}$ and a thermal conductivity of $150\text{ W m}^{-1}\text{K}^{-1}$ . The outer end of Section A is maintained at a constant temperature of $120^\circ\text{C}$ , while the outer end of Section B is held at $0^\circ\text{C}$ . Assuming the curved surfaces of both sections are perfectly insulated and heat flow is steady, what is the temperature at the junction between the two sections in $^\circ\text{C}$ ?

Cevabı ve açıklamayı göster

Cevap: 40

Cevap

The steady-state temperature at the junction between Section A and Section B is 40.0°C.

At steady state, the rate of heat conduction through Section A equals that through Section B. Because Section B has double the radius of Section A, its cross-sectional area is four times as large. Equating the heat flow rates gives 300 * A_A * (120 - T_J) = 150 * (4 A_A) * T_J, which simplifies directly to 120 - T_J = 2 T_J, yielding a junction temperature of 40.0°C.

Adım Adım Çözüm

1

Determine the relationship between the cross-sectional areas of Section A and Section B.

The area ratio A_B / A_A = (r_B / r_A)² = (4.0 cm / 2.0 cm)² = 4.

The cross-sectional area of a cylinder is proportional to the square of its radius.

2

Write the steady-state heat flow equation for each section.

H_A = (k_A * A_A * (120 - T_J)) / L and H_B = (k_B * A_B * (T_J - 0)) / L.

According to Fourier's law of thermal conduction, the heat transfer rate through a uniform layer is proportional to thermal conductivity, cross-sectional area, and temperature difference, and inversely proportional to length.

3

Equate the heat transfer rates H_A and H_B and solve for the junction temperature T_J.

300 * A_A * (120 - T_J) = 150 * (4 A_A) * T_J => 300 * (120 - T_J) = 600 * T_J => 120 - T_J = 2 T_J => 3 T_J = 120 => T_J = 40.0°C.

At steady state with insulated sides, heat does not accumulate or escape, so the rate of heat conduction through both sections must be identical.

Anahtar Kavram

Steady-state thermal conduction through composite conductors with differing cross-sectional areas and thermal conductivities

Soru 953Soru →

An alternating current (AC) circuit contains an inductor of inductance $L = \frac{0.2}{\pi^2}\ \text{H}$ , a resistor of resistance $R = 40\ \Omega$ , and a variable capacitor $C$ connected in series across a $50\ \text{Hz}$ voltage supply. What capacitance $C$ , in microfarads ( $\mu\text{F}$ ), is required for the circuit to operate at electrical resonance?

Cevabı ve açıklamayı göster

Cevap: 500

Cevap

The capacitance required to achieve electrical resonance is 500 μF.

At electrical resonance in a series RLC circuit, the inductive reactance (

X_L

) equals the capacitive reactance (

X_C

). Setting

2\pi f L = \frac{1}{2\pi f C}

yields

f_0 = \frac{1}{2\pi\sqrt{LC}}

. Substituting

f_0 = 50\ \text{Hz}

and

L = \frac{0.2}{\pi^2}\ \text{H}

into this equation gives

C = 5 \times 10^{-4}\ \text{F}

, which equals

500\ \mu\text{F}

.

Adım Adım Çözüm

1

Recall the resonant frequency formula for a series RLC circuit.

f_0 = \frac{1}{2\pi\sqrt{LC}}

At resonance, inductive reactance equals capacitive reactance (

X_L = X_C

).

2

Substitute the given numerical parameters into the equation.

50 = \frac{1}{2\pi \sqrt{\frac{0.2}{\pi^2} \cdot C}}

Given frequency

f_0 = 50\ \text{Hz}

and inductance

L = \frac{0.2}{\pi^2}\ \text{H}

.

3

Isolate the square root term and simplify.

\sqrt{0.2 C} = 0.01

Simplifying

2\pi \cdot \frac{1}{\pi} = 2

and rearranging

2 \sqrt{0.2 C} = \frac{1}{50} = 0.02

.

4

Square both sides and solve for

C

in farads.

C = 5 \times 10^{-4}\ \text{F}

0.2 C = (0.01)^2 = 10^{-4}

, so

C = \frac{10^{-4}}{0.2} = 5 \times 10^{-4}\ \text{F}

.

5

Convert capacitance from farads to microfarads.

C = 500\ \mu\text{F}

Multiply farads by

10^6

to express the result in microfarads.

Anahtar Kavram

Resonant Frequency in Series AC Circuits

Soru 954Soru →

An RLC series circuit operating at resonance contains an inductor of inductance $0.1\text{ H}$ and a capacitor of capacitance $10\ \mu\text{F}$ . What is the resonant angular frequency of the circuit in radians per second ( $\text{rad/s}$ )?

Cevabı ve açıklamayı göster

Cevap: 1000

Cevap

The resonant angular frequency of the circuit is

1000\text{ rad/s}

.

The resonant angular frequency

\omega_0

of an AC circuit is determined by the formula

\omega_0 = \frac{1}{\sqrt{LC}}

. Substituting the given values

L = 0.1\text{ H}

and

C = 10^{-5}\text{ F}

gives

\omega_0 = \frac{1}{\sqrt{10^{-6}}} = 1000\text{ rad/s}

.

Adım Adım Çözüm

1

Convert given parameters to standard SI units

L = 0.1\text{ H}

and

C = 10 \times 10^{-6}\text{ F} = 10^{-5}\text{ F}

.

Calculations must be performed in base SI units for dimensional consistency.

2

Calculate the resonant angular frequency using

\omega_0 = \frac{1}{\sqrt{LC}}

\omega_0 = \frac{1}{\sqrt{0.1 \times 10^{-5}}} = \frac{1}{\sqrt{10^{-6}}} = 1000\text{ rad/s}

.

At resonance, inductive reactance equals capacitive reactance (

X_L = X_C

), which yields

\omega_0 = \frac{1}{\sqrt{LC}}

.

Anahtar Kavram

Resonant Angular Frequency

Soru 955Soru →

A U-tube open to the atmosphere at both ends contains water of density $1000\text{ kg m}^{-3}$ . An immiscible liquid is poured into one arm of the tube until it forms a column of height $15.0\text{ cm}$ . If the interface between the two liquids lies $12.0\text{ cm}$ below the surface of the water in the opposite arm, what is the density of the liquid in $\text{kg m}^{-3}$ ?

Cevabı ve açıklamayı göster

Cevap: 800

Cevap

The density of the immiscible liquid is

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

.

At the level of the liquid interface, the hydrostatic pressure exerted by the

15.0\text{ cm}

liquid column must equal the hydrostatic pressure exerted by the

12.0\text{ cm}

water column above it. Equating

\rho_1 h_1 = \rho_2 h_2

yields

\rho_{\text{liquid}} = 1000 \times (12.0 / 15.0) = 800\text{ kg m}^{-3}

.

Adım Adım Çözüm

1

Apply the principal of equal pressure at the same horizontal level in a continuous fluid at rest.

Pressure due to liquid column equals pressure due to water column above the interface line.

Hydrostatic pressure at depth

h

is given by

P = \rho g h

, and points at equal depth in a connected liquid body share identical pressure.

2

Set up the density-height ratio relationship:

\rho_{\text{liquid}} \cdot h_{\text{liquid}} = \rho_{\text{water}} \cdot h_{\text{water}}

.

\rho_{\text{liquid}} = \rho_{\text{water}} \times \frac{h_{\text{water}}}{h_{\text{liquid}}}

.

Acceleration due to gravity

g

cancels out from both sides of the pressure balance equation.

3

Substitute

h_{\text{liquid}} = 15.0\text{ cm}

,

h_{\text{water}} = 12.0\text{ cm}

, and

\rho_{\text{water}} = 1000\text{ kg m}^{-3}

.

\rho_{\text{liquid}} = 1000 \times \frac{12.0}{15.0} = 800\text{ kg m}^{-3}

.

Heights can remain in centimeters since their unit ratio is dimensionless.

Anahtar Kavram

Hydrostatic Pressure Balance in U-Tube Manometers

Tahmini Süre:1m 30s

Soru 956Soru →

Element $X$ exists naturally as three isotopes with mass numbers $24$ , $25$ , and $26$ . The relative atomic mass of element $X$ is $24.32$ . If the natural abundance of the isotope $^{25}X$ is $10\%$ , what is the percentage abundance of the isotope $^{26}X$ ?

Cevabı ve açıklamayı göster

Cevap: 11

Cevap

The percentage abundance of

^{26}X

is

11\%

.

By setting the percentage abundance of

^{24}X

as

x

and

^{26}X

as

z

, given

^{25}X = 10\%

, we have

x + z = 90\%

. Using the relative atomic mass formula

24x + 25(10) + 26z = 2432

and substituting

x = 90 - z

yields

2410 + 2z = 2432

, solving to

z = 11\%

.

Adım Adım Çözüm

1

Formulate the abundance relation for the three isotopes.

The sum of abundances is

x + 10 + z = 100\%

, which gives

x = 90 - z

.

The sum of percentage abundances of all naturally occurring isotopes of an element must equal 100%.

2

Set up the weighted relative atomic mass equation.

24x + 25(10) + 26z = 2432

.

Relative atomic mass is the weighted average of isotopic masses based on fractional abundance.

3

Substitute

x = 90 - z

and solve for

z

.

24(90 - z) + 250 + 26z = 2432 \implies 2410 + 2z = 2432 \implies z = 11\%

.

Substituting

x

reduces the equation to a single variable

z

, representing the percentage abundance of

^{26}X

.

Anahtar Kavram

Calculation of Isotopic Abundances from Relative Atomic Mass

Soru 957Soru →

Element $M$ has a relative atomic mass of $24.32$ . It exists naturally as three isotopes: $^{24}M$ with a relative abundance of $79\%$ , $^{25}M$ with a relative abundance of $10\%$ , and an isotope $^{x}M$ with a relative abundance of $11\%$ . What is the mass number ( $x$ ) of the third isotope?

Cevabı ve açıklamayı göster

Cevap: 26

Cevap

The mass number of the third isotope is 26.

The relative atomic mass is the weighted average of isotopic mass numbers:

24.32 = \frac{(24 \times 79) + (25 \times 10) + (x \times 11)}{100}

. Simplifying gives

2432 = 1896 + 250 + 11x

, which simplifies to

11x = 286

, yielding

x = 26

.

Adım Adım Çözüm

1

Set up the relative atomic mass equation based on percentage abundances.

\text{RAM} = (24 \times 0.79) + (25 \times 0.10) + (x \times 0.11)

The relative atomic mass of an element is the weighted average of the mass numbers of its naturally occurring isotopes.

2

Calculate the mass contributions of the first two isotopes.

24 \times 0.79 = 18.96

and

25 \times 0.10 = 2.50

, giving a combined sum of

21.46

Evaluating the contribution of known isotopes isolates the variable term.

3

Subtract the combined contribution from the given relative atomic mass to find the contribution of the third isotope.

0.11x = 24.32 - 21.46 = 2.86

The remaining mass contribution must come entirely from the third isotope.

4

Divide by the fractional abundance of the third isotope to solve for

x

.

x = \frac{2.86}{0.11} = 26

Dividing the mass contribution by fractional abundance yields the integer mass number.

Anahtar Kavram

Relative Atomic Mass Calculation from Isotopic Abundances

Tahmini Süre:1m 15s

Soru 958Soru →

A Geiger-Müller counter records a total count rate of $340\text{ counts per minute}$ near a radioactive source. The background radiation in the laboratory produces a steady count rate of $20\text{ counts per minute}$ . If the total count rate recorded by the counter drops to $60\text{ counts per minute}$ after an elapsed time of $15\text{ minutes}$ , what is the half-life of the radioactive source in minutes?

Cevabı ve açıklamayı göster

Cevap: 5

Cevap

The half-life of the radioactive source is

5\text{ minutes}

.

To find the true activity of the radioactive source, the constant background radiation of

20\text{ cpm}

must be subtracted from all detector readings. The initial source activity is

340 - 20 = 320\text{ cpm}

and the activity after

15\text{ minutes}

is

60 - 20 = 40\text{ cpm}

. The fraction of source activity remaining is

40 / 320 = 1/8 = (1/2)^3

, which means

3

half-lives have elapsed in

15\text{ minutes}

. Dividing the total time by the number of half-lives (

15 / 3

) gives a half-life of

5\text{ minutes}

.

Adım Adım Çözüm

1

Calculate the initial activity of the radioactive source by subtracting the background count rate.

A_0 = 340\text{ cpm} - 20\text{ cpm} = 320\text{ cpm}

Background radiation contributes to the detector reading and must be isolated from the source activity.

2

Calculate the activity of the source after 15 minutes by subtracting the background count rate.

A(t) = 60\text{ cpm} - 20\text{ cpm} = 40\text{ cpm}

The background count remains constant at

20\text{ cpm}

, so the actual count due to the source is

40\text{ cpm}

.

3

Determine the remaining fraction of the radioactive source.

\frac{A(t)}{A_0} = \frac{40}{320} = \frac{1}{8}

Radioactive decay follows an exponential decay law based on the fraction of initial undecayed nuclei.

4

Calculate the number of elapsed half-lives

n

.

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

The remaining fraction equals

(1/2)^n

where

n

is the number of half-lives.

5

Compute the half-life

T_{1/2}

.

T_{1/2} = \frac{t}{n} = \frac{15\text{ minutes}}{3} = 5\text{ minutes}

The total elapsed time is the product of the number of half-lives and the duration of one half-life.

Anahtar Kavram

Radioactive Decay Law and Half-life with Background Radiation Correction

Soru 959Soru →

A sample of a radioactive substance has a decay constant of $0.0154\text{ h}^{-1}$ . What is the elapsed time, in hours, required for $87.5\%$ of the original sample to decay? (Take $\ln 2 = 0.693$ )

Cevabı ve açıklamayı göster

Cevap: 135

Cevap

135 hours

First, calculate the half-life of the substance using the relation

T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{0.0154\text{ h}^{-1}} = 45\text{ hours}

. Since

87.5\%

of the sample has decayed, the remaining fraction of the sample is

100\% - 87.5\% = 12.5\% = \frac{1}{8}

. Expressing

\frac{1}{8}

as a power of

\frac{1}{2}

gives

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

, indicating that

3

half-lives have passed. The total elapsed time is therefore

3 \times 45\text{ hours} = 135\text{ hours}

.

Adım Adım Çözüm

1

Calculate the half-life from the given decay constant

Half-life

T_{1/2} = 45\text{ hours}

The decay constant

\lambda

and half-life

T_{1/2}

are related by the formula

T_{1/2} = \frac{\ln 2}{\lambda}

2

Find the remaining percentage and fraction of the sample

Remaining fraction is

12.5\%

or

\frac{1}{8}

Radioactive decay equations use the undecayed remaining amount, which is

100\% - 87.5\% = 12.5\%

3

Calculate the number of half-lives elapsed

Number of half-lives

n = 3

Since

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

, solving for

n

gives

n = 3

4

Compute the total elapsed time

Total elapsed time

t = 135\text{ hours}

Total elapsed time is the product of the number of half-lives and the half-life duration (

3 \times 45\text{ hours}

)

Anahtar Kavram

Radioactive Decay Law and Half-life Relationship

Soru 960Soru →

A uniform beam $PQ$ of length $4.0\text{ m}$ and mass $20\text{ kg}$ is supported horizontally on a pivot at end $P$ and by a vertical wire attached at end $Q$ . A load of mass $30\text{ kg}$ is placed on the beam at a distance of $1.0\text{ m}$ from $P$ . Taking the acceleration due to gravity $g = 10\text{ m/s}^2$ , what is the tension in the vertical wire attached at $Q$ in newtons?

Cevabı ve açıklamayı göster

Cevap: 175

Cevap

The tension in the vertical wire attached at end

Q

is

175\text{ N}

.

By applying the principle of moments about the pivot at

P

, the sum of downward clockwise moments produced by the

30\text{ kg}

load (

300\text{ N} \times 1.0\text{ m} = 300\text{ N}\cdot\text{m}

) and the beam's center of gravity (

200\text{ N} \times 2.0\text{ m} = 400\text{ N}\cdot\text{m}

) equals

700\text{ N}\cdot\text{m}

. Equating this to the counterclockwise moment of the tension force (

T \times 4.0\text{ m}

) gives

T = 175\text{ N}

.

Adım Adım Çözüm

1

Determine forces and their perpendicular distances from the pivot at

P

.

The load exerts a downward force of

300\text{ N}

at

1.0\text{ m}

from

P

. The uniform beam's weight of

200\text{ N}

acts at its midpoint (

2.0\text{ m}

from

P

). The vertical tension

T

acts upward at

Q

(

4.0\text{ m}

from

P

).

Before applying the principle of moments, all force magnitudes and their distance arms relative to the pivot point must be identified.

2

Equate total clockwise moments to total counterclockwise moments about

P

.

(300\text{ N} \times 1.0\text{ m}) + (200\text{ N} \times 2.0\text{ m}) = T \times 4.0\text{ m}

For rotational equilibrium, the sum of clockwise moments about any pivot must equal the sum of counterclockwise moments about that same pivot.

3

Calculate the value of the tension force

T

.

T = \frac{700}{4.0} = 175\text{ N}

Simplifying the moment equation gives the magnitude of the upward supporting force.

Anahtar Kavram

Principle of Moments and Rotational Equilibrium

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