All practice questions

Question 901Question →

A cylindrical metal rivet has a diameter of $2.50\text{ cm}$ at a room temperature of $25^\circ\text{C}$ . It needs to be inserted into a hole of diameter $2.49\text{ cm}$ in a structural frame. By how many kelvins must the rivet be cooled so that its diameter shrinks to just match the diameter of the hole? (Linear expansivity of the metal is $2.0 \times 10^{-5}\text{ K}^{-1}$ ).

Show answer & explanation

Answer: 200

Answer

The rivet must be cooled by 200 K.

Thermal expansion or contraction of a linear dimension (such as diameter) is governed by

\Delta d = d_0 \alpha \Delta T

. Substituting

\Delta d = -0.01\text{ cm}

,

d_0 = 2.50\text{ cm}

, and

\alpha = 2.0 \times 10^{-5}\text{ K}^{-1}

gives

-0.01 = 2.50 \times (2.0 \times 10^{-5}) \times \Delta T

, leading to

\Delta T = -200\text{ K}

. Hence, cooling by 200 K is required.

Step-by-Step Solution

1

Calculate the required change in diameter

\Delta d = 2.49\text{ cm} - 2.50\text{ cm} = -0.01\text{ cm}

The diameter of the rivet must decrease from 2.50 cm to 2.49 cm to fit into the hole.

2

Set up the linear expansion equation

\Delta d = d_0 \alpha \Delta T

Linear contraction/expansion applies directly to any linear dimension of a solid, including diameter.

3

Substitute given values into the equation

-0.01\text{ cm} = (2.50\text{ cm}) \times (2.0 \times 10^{-5}\text{ K}^{-1}) \times \Delta T

Substitute initial diameter, linear expansivity, and change in diameter.

4

Solve for the temperature change

\Delta T = \frac{-0.01}{5.0 \times 10^{-5}} = -200\text{ K}

Dividing the change in length by the product of initial length and linear expansivity yields the temperature change.

Key Concept

Thermal Contraction and Linear Expansivity of Solids

Question 902Question →

In an electrical power station, two independent backup transformers, $T_1$ and $T_2$ , operate simultaneously during power surges. The probability that $T_1$ fails during a surge is $0.15$ , and the probability that $T_2$ fails during the same surge is $0.20$ . What is the probability that at least one transformer functions correctly during a power surge?

Show answer & explanation

Answer: 0.97

Answer

0.97

The failure probabilities are

P(F_1) = 0.15

and

P(F_2) = 0.20

. By the multiplication law for independent events, the probability of both failing is

P(F_1 \cap F_2) = 0.15 \times 0.20 = 0.03

. Using the complement rule, the probability of at least one functioning correctly is

1 - P(F_1 \cap F_2) = 1 - 0.03 = 0.97

.

Step-by-Step Solution

1

Determine the joint probability of both transformers failing.

P(F_1 \cap F_2) = 0.15 \times 0.20 = 0.03

Since the operational failures of the two transformers are independent events, the probability of both failing together is the product of their individual failure probabilities.

2

Calculate the probability that at least one transformer functions correctly.

P(\text{at least one functions}) = 1 - 0.03 = 0.97

The event that at least one transformer functions is the complement of the event that both transformers fail simultaneously.

Key Concept

Independent Compound Events and the Complement Rule

Question 903Question →

A projectile is launched from level ground with an initial speed of $25\text{ m/s}$ at an angle $\theta$ to the horizontal such that $\sin\theta = 0.80$ . Calculate the maximum height reached by the projectile in meters. [Take $g = 10\text{ m/s}^2$ ]

Show answer & explanation

Answer: 20

Answer

The maximum height reached by the projectile is

20\text{ m}

.

The vertical component of the initial launch velocity is

u_y = u \sin\theta = 25 \times 0.80 = 20\text{ m/s}

. Using the equation for maximum height

H = \frac{u_y^2}{2g}

, we substitute

u_y = 20\text{ m/s}

and

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

to obtain

H = \frac{400}{20} = 20\text{ m}

.

Step-by-Step Solution

1

Calculate the initial vertical velocity component (

u_y

)

u_y = 25\text{ m/s} \times 0.80 = 20\text{ m/s}

Only the vertical component of velocity determines the maximum height reached.

2

Calculate the maximum height (

H

) using kinematic equations

H = \frac{u_y^2}{2g} = \frac{20^2}{2 \times 10} = 20\text{ m}

At maximum height, the vertical component of velocity becomes zero.

Key Concept

Maximum height of a projectile depends entirely on its initial vertical component of velocity and acceleration due to gravity.

Question 904Question →

A cylindrical brass sleeve has an internal diameter of $5.000\text{ cm}$ at a room temperature of $20^\circ\text{C}$ . It is to be shrink-fitted onto a solid shaft of diameter $5.012\text{ cm}$ (also at $20^\circ\text{C}$ ). Assuming the linear expansivity of brass is $2.0 \times 10^{-5}\text{ K}^{-1}$ , what is the minimum temperature, in $^\circ\text{C}$ , to which the brass sleeve must be heated so that it just slips over the shaft?

Show answer & explanation

Answer: 140

Answer

140 °C

The required expansion in internal diameter is

\Delta d = 5.012\text{ cm} - 5.000\text{ cm} = 0.012\text{ cm}

. Using the linear expansion relation

\Delta d = d_0 \alpha \Delta T

, the required temperature change is

\Delta T = \frac{0.012}{5.000 \times 2.0 \times 10^{-5}} = 120^\circ\text{C}

. Adding this to the initial temperature of

20^\circ\text{C}

gives a final minimum temperature of

140^\circ\text{C}

.

Step-by-Step Solution

1

Determine the required increase in internal diameter (

\Delta d

) of the brass sleeve

\Delta d = 5.012\text{ cm} - 5.000\text{ cm} = 0.012\text{ cm}

The sleeve's internal diameter must expand until it equals the shaft diameter.

2

Apply the linear expansion formula

\Delta d = d_0 \alpha \Delta T

to find the temperature rise

\Delta T

\Delta T = \frac{0.012\text{ cm}}{5.000\text{ cm} \times 2.0 \times 10^{-5}\text{ K}^{-1}} = \frac{0.012}{1.0 \times 10^{-4}} = 120\text{ K}

Linear dimensions such as diameter expand in direct proportion to the linear expansivity coefficient

\alpha

.

3

Calculate the final temperature

T_2

T_2 = T_1 + \Delta T = 20^\circ\text{C} + 120^\circ\text{C} = 140^\circ\text{C}

The final temperature is found by adding the temperature increase to the initial temperature.

Key Concept

Linear Expansivity and One-Dimensional Expansion of Curved Boundaries

Question 905Question →

Given the function $f(x) = 5x^2 - 4x + 3$ , what is the numerical value of its derivative at $x = 2$ when evaluated using the first principles limit definition $\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$ ?

Show answer & explanation

Answer: 16

Answer

The numerical value of the derivative of

f(x) = 5x^2 - 4x + 3

at

x = 2

is 16.

Evaluating the first principles definition

\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

for

f(x) = 5x^2 - 4x + 3

yields the derivative

f'(x) = 10x - 4

. Substituting

x = 2

yields

10(2) - 4 = 16

.

Step-by-Step Solution

1

Determine the expanded form of

f(x+h)

f(x+h) = 5(x+h)^2 - 4(x+h) + 3 = 5x^2 + 10xh + 5h^2 - 4x - 4h + 3

Evaluating the function at

x+h

requires expanding the square and distributing the constant factors.

2

Compute the difference

f(x+h) - f(x)

f(x+h) - f(x) = (5x^2 + 10xh + 5h^2 - 4x - 4h + 3) - (5x^2 - 4x + 3) = 10xh + 5h^2 - 4h

Subtracting

f(x)

cancels terms independent of

h

.

3

Form and simplify the difference quotient

\frac{f(x+h) - f(x)}{h}

\frac{10xh + 5h^2 - 4h}{h} = 10x + 5h - 4

Factoring

h

out of the numerator allows division by

h

for non-zero

h

.

4

Evaluate the limit as

h \to 0

f'(x) = \lim_{h \to 0} (10x + 5h - 4) = 10x - 4

Taking the limit produces the general derivative function

\frac{\mathrm{d}y}{\mathrm{d}x}

.

5

Substitute

x = 2

into the derivative

f'(2) = 10(2) - 4 = 16

Evaluating at

x = 2

gives the instantaneous rate of change at that point.

Key Concept

Differentiation from First Principles

Estimated Time:1m 30s

Question 906Question →

Given that $\int \left( k x^3 + 12\cos(3x) \right) dx = 4x^4 + 4\sin(3x) + C$ , where $C$ is the arbitrary constant of integration, what is the numerical value of the constant $k$ ?

Show answer & explanation

Answer: 16

Answer

The numerical value of the constant

k

is 16.

Integrating

kx^3 + 12\cos(3x)

with respect to

x

yields

\frac{k}{4}x^4 + 4\sin(3x) + C

. Comparing the coefficient of

x^4

with the given result

4x^4 + 4\sin(3x) + C

gives

\frac{k}{4} = 4

, which leads to

k = 16

.

Step-by-Step Solution

1

Integrate the polynomial and trigonometric terms separately using standard integration rules.

\int (kx^3 + 12\cos(3x)) dx = \frac{k}{4}x^4 + 4\sin(3x) + C

Applying the power rule

\int x^n dx = \frac{x^{n+1}}{n+1}

gives

\int kx^3 dx = \frac{k}{4}x^4

, and applying

\int \cos(ax) dx = \frac{\sin(ax)}{a}

gives

\int 12\cos(3x) dx = \frac{12}{3}\sin(3x) = 4\sin(3x)

.

2

Equate the integrated expression to the right-hand side of the given equation.

\frac{k}{4}x^4 + 4\sin(3x) + C = 4x^4 + 4\sin(3x) + C

Both sides represent the same antiderivative of the function.

3

Equate corresponding coefficients of

x^4

to solve for

k

.

k = 16

\frac{k}{4} = 4 \implies k = 4 \times 4 = 16.

Key Concept

Indefinite Integration of Polynomial and Trigonometric Functions

Question 907Question →

The $5^{\text{th}}$ term of an arithmetic progression (AP) is $18$ and its $11^{\text{th}}$ term is $42$ . Calculate the value of the $20^{\text{th}}$ term of this progression.

Show answer & explanation

Answer: 78

Answer

78

Using the AP general term formula

T_n = a + (n-1)d

, the equations

a + 4d = 18

and

a + 10d = 42

yield

d = 4

and

a = 2

. Evaluating

T_{20} = 2 + 19(4)

yields

78

.

Step-by-Step Solution

1

Set up simultaneous equations using the general term formula

T_n = a + (n-1)d

a + 4d = 18

and

a + 10d = 42

Relate given terms to the first term

a

and common difference

d

.

2

Solve for the common difference

d

6d = 24 \implies d = 4

Subtracting the 5th term equation from the 11th term equation eliminates

a

.

3

Solve for the first term

a

a + 16 = 18 \implies a = 2

Substitute the value of

d

back into the first equation.

4

Calculate the 20th term

T_{20} = 2 + 19(4) = 78

Apply the nth term formula for

n = 20

.

Key Concept

Determining terms of an Arithmetic Progression using simultaneous equations

Estimated Time:1m 30s

Question 908Question →

A potentiometer wire of length $100\text{ cm}$ has a resistance of $10\text{ }\Omega$ . It is connected in series with a driver cell of electromotive force $3.0\text{ V}$ and internal resistance $2.0\text{ }\Omega$ , alongside an external series resistor of $8.0\text{ }\Omega$ . A test cell of unknown electromotive force $E$ gives a balance point at a length of $60\text{ cm}$ from the zero end of the wire. What is the value of $E$ in volts?

Show answer & explanation

Answer: 0.9

Answer

The electromotive force of the test cell is

0.9\text{ V}

.

The e.m.f. of the test cell is balanced by the potential difference across a length of

60\text{ cm}

of the potentiometer wire. Accounting for the driver cell's internal resistance (

2.0\text{ }\Omega

), external series resistor (

8.0\text{ }\Omega

), and wire resistance (

10\text{ }\Omega

), the total resistance of the primary circuit is

20.0\text{ }\Omega

. This yields a primary current of

0.15\text{ A}

and a potential drop across the wire of

1.5\text{ V}

. The resulting potential gradient is

0.015\text{ V/cm}

, which when multiplied by the balance length of

60\text{ cm}

gives an e.m.f. of

0.9\text{ V}

.

Step-by-Step Solution

1

Calculate total resistance in the primary driver circuit

R_{total} = 10\text{ }\Omega + 2.0\text{ }\Omega + 8.0\text{ }\Omega = 20.0\text{ }\Omega

The driver cell's internal resistance, potentiometer wire, and external series resistor are connected in series.

2

Calculate the current flowing through the potentiometer wire

I = \frac{3.0\text{ V}}{20.0\text{ }\Omega} = 0.15\text{ A}

Apply Ohm's law to the complete primary circuit.

3

Find the voltage drop across the potentiometer wire

V_{wire} = 0.15\text{ A} \times 10\text{ }\Omega = 1.5\text{ V}

The potential difference across the wire depends on its resistance and the primary current.

4

Determine the potential gradient along the wire

k = \frac{1.5\text{ V}}{100\text{ cm}} = 0.015\text{ V/cm}

Potential gradient is the potential drop per unit length of the wire.

5

Calculate the e.m.f. of the unknown test cell

E = 0.015\text{ V/cm} \times 60\text{ cm} = 0.9\text{ V}

At the balance point, no current flows from the test cell, so its e.m.f. equals the potential drop across the balance length.

Key Concept

Potentiometer principle and potential gradient

Estimated Time:2m 0s

Question 909Question →

A right-angled triangle has a perimeter of $60\text{ cm}$ and a hypotenuse of length $25\text{ cm}$ . What is the area of the triangle in $\text{cm}^2$ ?

Show answer & explanation

Answer: 150

Answer

150

Let the perpendicular sides of the right-angled triangle be

a

and

b

, and the hypotenuse be

c = 25\text{ cm}

. From the perimeter,

a + b + 25 = 60

, so

a + b = 35\text{ cm}

. By the Pythagorean theorem,

a^2 + b^2 = 25^2 = 625

. Squaring both sides of

a + b = 35

yields

(a + b)^2 = a^2 + b^2 + 2ab = 35^2 = 1225

. Substituting

a^2 + b^2 = 625

gives

625 + 2ab = 1225 \implies 2ab = 600 \implies ab = 300

. The area of the right-angled triangle is

\frac{1}{2}ab = \frac{1}{2} \times 300 = 150\text{ cm}^2

.

Step-by-Step Solution

1

Determine the sum of the two legs from the perimeter

a + b = 35\text{ cm}

The perimeter of the triangle is

a + b + c = 60\text{ cm}

, where the hypotenuse

c = 25\text{ cm}

.

2

Use the Pythagorean theorem for the sum of squares of the legs

a^2 + b^2 = 625

In any right-angled triangle with hypotenuse

25\text{ cm}

,

a^2 + b^2 = 25^2 = 625

.

3

Expand

(a + b)^2

to find the product of the legs

ab

1225 = 625 + 2ab \implies 2ab = 600 \implies ab = 300

Using the algebraic identity

(a + b)^2 = a^2 + b^2 + 2ab

allows determining

ab

directly without solving for individual side lengths.

4

Calculate the area of the right-angled triangle

\text{Area} = 150\text{ cm}^2

The area of a right-angled triangle with perpendicular sides

a

and

b

is given by

\frac{1}{2}ab

.

Key Concept

Perimeter and Area of Right-Angled Triangles using Algebraic Identities

Question 910Question →

A uniform metallic conductor of length $200\,\text{m}$ and cross-sectional area $2.0 \times 10^{-6}\,\text{m}^2$ has a resistivity of $1.6 \times 10^{-8}\,\Omega\cdot\text{m}$ at an initial temperature of $20\,^\circ\text{C}$ . The temperature coefficient of resistivity for the material is $5.0 \times 10^{-3}\,^\circ\text{C}^{-1}$ . If the operating temperature of the conductor increases to $120\,^\circ\text{C}$ while it is connected across a constant potential difference of $12\,\text{V}$ , what is the magnitude of the electric current flowing through the conductor in amperes?

Show answer & explanation

Answer: 5

Answer

The electric current flowing through the conductor is

5.0\,\text{A}

.

The temperature change of

100\,^\circ\text{C}

increases the resistivity of the material from

1.6 \times 10^{-8}\,\Omega\cdot\text{m}

to

2.4 \times 10^{-8}\,\Omega\cdot\text{m}

via

\rho = \rho_0(1 + \alpha \Delta T)

. Substituting this updated resistivity into

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

gives a resistance of

2.4\,\Omega

. Applying Ohm's Law

I = \frac{V}{R}

with a potential difference of

12\,\text{V}

yields

5.0\,\text{A}

.

Step-by-Step Solution

1

Calculate the temperature difference

\Delta T = 100\,^\circ\text{C}

The temperature change relative to the reference temperature dictates the change in resistivity.

2

Calculate the resistivity at the final temperature

\rho = 2.4 \times 10^{-8}\,\Omega\cdot\text{m}

Resistivity depends on temperature according to

\rho = \rho_0(1 + \alpha \Delta T)

.

3

Calculate the total electrical resistance of the conductor

R = 2.4\,\Omega

Resistance is related to physical geometry and resistivity by

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

.

4

Apply Ohm's law to solve for the current

I = 5.0\,\text{A}

Electric current is determined by potential difference divided by resistance (

I = V / R

).

Key Concept

Temperature dependence of resistivity and Ohm's Law

Estimated Time:2m 0s

Question 911Question →

A sample of ideal gas in a syringe occupies a volume of $300\text{ cm}^3$ at a temperature of $27^\circ\text{C}$ . If the pressure of the gas remains constant, what is its volume in $\text{cm}^3$ when the temperature is raised to $127^\circ\text{C}$ ?

Show answer & explanation

Answer: 400

Answer

400 cm³

According to Charles's Law, at constant pressure, the volume of a given mass of gas is directly proportional to its absolute temperature (

V \propto T

). Converting the given temperatures to kelvins yields

T_1 = 300\text{ K}

and

T_2 = 400\text{ K}

. Applying

\frac{V_1}{T_1} = \frac{V_2}{T_2}

gives

V_2 = \frac{300 \times 400}{300} = 400\text{ cm}^3

.

Step-by-Step Solution

1

Convert the initial and final temperatures from degrees Celsius to kelvins.

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

and

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

Gas law calculations require absolute temperature measured on the kelvin scale.

2

Apply Charles's Law, which relates volume and absolute temperature at constant pressure.

\frac{V_1}{T_1} = \frac{V_2}{T_2}

The pressure of the gas is maintained constant.

3

Substitute the values into the equation and solve for the final volume

V_2

.

\frac{300\text{ cm}^3}{300\text{ K}} = \frac{V_2}{400\text{ K}} \implies V_2 = 400\text{ cm}^3

Cross-multiplying gives

V_2 = \frac{300 \times 400}{300} = 400\text{ cm}^3

.

Key Concept

Charles's Law

Question 912Question →

A uniform conductor of length $4.0\,\text{m}$ and cross-sectional area $2.0 \times 10^{-7}\,\text{m}^2$ has an electrical resistance of $0.50\,\Omega$ . What is the electrical resistivity of the material of the conductor in units of $10^{-8}\,\Omega\cdot\text{m}$ ?

Show answer & explanation

Answer: 2.5

Answer

The electrical resistivity of the material is

2.5 \times 10^{-8}\,\Omega\cdot\text{m}

, which gives a numerical value of

2.5

in units of

10^{-8}\,\Omega\cdot\text{m}

.

The electrical resistance

R

of a conductor is given by

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

, where

\rho

is the resistivity,

L

is the length, and

A

is the cross-sectional area. Rearranging to solve for resistivity yields

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

. Substituting

R = 0.50\,\Omega

,

A = 2.0 \times 10^{-7}\,\text{m}^2

, and

L = 4.0\,\text{m}

gives

\rho = \frac{0.50 \times 2.0 \times 10^{-7}}{4.0} = 2.5 \times 10^{-8}\,\Omega\cdot\text{m}

. Expressed in units of

10^{-8}\,\Omega\cdot\text{m}

, the answer is

2.5

.

Step-by-Step Solution

1

Identify the relevant formula linking resistance, resistivity, length, and area.

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

The resistance of a uniform conductor is directly proportional to its length and inversely proportional to its cross-sectional area.

2

Rearrange the equation to solve for resistivity

\rho

.

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

Isolating

\rho

allows direct evaluation using the given quantitative values.

3

Substitute the known numerical values into the equation.

\rho = \frac{0.50 \times (2.0 \times 10^{-7})}{4.0} = 2.5 \times 10^{-8}\,\Omega\cdot\text{m}

Performing the algebraic calculation gives the resistivity in SI units.

Key Concept

Electrical Resistivity and Conductor Dimensions

Question 913Question →

A point $P(x, y)$ moves in the Cartesian plane such that it is always equidistant from the two fixed points $A(2, 1)$ and $B(6, 5)$ . If the locus of $P$ intersects the line $2x + y = 14$ at the point $(x_0, y_0)$ , what is the value of $x_0$ ?

Show answer & explanation

Answer: 7

Answer

The value of

x_0

is

7

.

The locus of points equidistant from

A(2, 1)

and

B(6, 5)

is the perpendicular bisector of line segment

AB

. The midpoint of

AB

is

(4, 3)

and its slope is

1

, giving the perpendicular bisector a slope of

-1

. The equation of this locus is

y - 3 = -1(x - 4)

, or

x + y = 7

. Subtracting

x + y = 7

from

2x + y = 14

directly gives

x_0 = 7

.

Step-by-Step Solution

1

Determine the equation of the locus of point P

The locus of P is the perpendicular bisector of segment AB, represented by

x + y = 7

.

The set of all points equidistant from two fixed points forms the perpendicular bisector of the line segment connecting those points.

2

Find the point of intersection with the line

2x + y = 14

Solving

x + y = 7

and

2x + y = 14

simultaneously gives

x_0 = 7

.

The intersection point of two geometric lines must satisfy both equations simultaneously.

Key Concept

Perpendicular Bisector Locus and Line Intersections

Question 914Question →

A $6.0\text{ }\mu\text{F}$ capacitor is charged to a potential difference of $100\text{ V}$ using a direct-current source and then disconnected. It is subsequently connected in parallel across an uncharged $4.0\text{ }\mu\text{F}$ capacitor. What is the total electrostatic potential energy lost in the system during the redistribution of charge, in millijoules (mJ)?

Show answer & explanation

Answer: 12

Answer

The total electrostatic potential energy lost in the system during the redistribution of charge is 12 mJ.

The initial energy stored in the charged capacitor is

U_i = \frac{1}{2} C_1 V_1^2 = \frac{1}{2}(6.0 \times 10^{-6}\text{ F})(100\text{ V})^2 = 30\text{ mJ}

. When connected in parallel to the uncharged capacitor, the total charge

Q = 600\text{ }\mu\text{C}

is conserved across an equivalent capacitance of

C_{eq} = 6.0\text{ }\mu\text{F} + 4.0\text{ }\mu\text{F} = 10.0\text{ }\mu\text{F}

. The common potential becomes

V_f = \frac{Q}{C_{eq}} = 60\text{ V}

, leading to a final stored energy

U_f = \frac{1}{2} C_{eq} V_f^2 = 18\text{ mJ}

. The energy lost is

\Delta U = U_i - U_f = 30\text{ mJ} - 18\text{ mJ} = 12\text{ mJ}

.

Step-by-Step Solution

1

Calculate the initial stored charge

Q

and initial energy

U_i

in the charged

6.0\text{ }\mu\text{F}

capacitor.

Q = 6.0 \times 10^{-4}\text{ C} = 600\text{ }\mu\text{C}

and

U_i = 3.0 \times 10^{-2}\text{ J} = 30\text{ mJ}

.

Before connection, all charge and energy reside solely on the first capacitor.

2

Find the equivalent capacitance

C_{eq}

when the two capacitors are connected in parallel.

C_{eq} = 6.0\text{ }\mu\text{F} + 4.0\text{ }\mu\text{F} = 10.0\text{ }\mu\text{F}

.

Capacitances add directly when connected in parallel.

3

Determine the common final potential difference

V_f

across the combination.

V_f = \frac{Q}{C_{eq}} = \frac{600\text{ }\mu\text{C}}{10.0\text{ }\mu\text{F}} = 60\text{ V}

.

Total electric charge is conserved during redistribution between connected capacitors.

4

Calculate the final total energy

U_f

stored in the combined system.

U_f = \frac{1}{2} C_{eq} V_f^2 = \frac{1}{2} (10.0 \times 10^{-6}\text{ F})(60\text{ V})^2 = 18\text{ mJ}

.

Both capacitors now store energy under the new common potential difference.

5

Compute the total energy lost

\Delta U = U_i - U_f

.

\Delta U = 30\text{ mJ} - 18\text{ mJ} = 12\text{ mJ}

.

The difference in energy is dissipated as heat in connecting wires and spark/radiation.

Key Concept

Charge Conservation and Energy Dissipation during Charge Sharing in Capacitors

Estimated Time:2m 0s

Question 915Question →

An electron inside an excited gas atom undergoes a transition from an upper energy level of $-2.40\text{ eV}$ to a lower energy level of $-5.15\text{ eV}$ . What is the wavelength, in nanometers ( $\text{nm}$ ), of the emitted photon? (Take Planck's constant $h = 6.6 \times 10^{-34}\text{ J s}$ , speed of light $c = 3.0 \times 10^8\text{ m s}^{-1}$ , and $1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$ )

Show answer & explanation

Answer: 450

Answer

The wavelength of the emitted photon is 450 nm.

When an electron transitions from a higher energy level to a lower energy level, a photon is emitted with energy equal to the difference between the two energy states. Converting 2.75 eV to 4.40 x 10^-19 J and applying lambda = hc / E yields a wavelength of 4.50 x 10^-7 m, which equals 450 nm.

Step-by-Step Solution

1

Calculate the energy change of the transition in eV

\Delta E = -2.40\text{ eV} - (-5.15\text{ eV}) = 2.75\text{ eV}

The energy of the emitted photon equals the difference between the higher and lower electronic energy states.

2

Convert the transition energy into SI units (Joules)

\Delta E = 2.75 \times 1.6 \times 10^{-19}\text{ J} = 4.40 \times 10^{-19}\text{ J}

Standard physics constants h and c require energy to be expressed in Joules.

3

Calculate photon wavelength and convert to nanometers

\lambda = \frac{hc}{\Delta E} = \frac{1.98 \times 10^{-25}\text{ J m}}{4.40 \times 10^{-19}\text{ J}} = 4.50 \times 10^{-7}\text{ m} = 450\text{ nm}

Applying the de Broglie/Einstein relation connecting photon energy and wavelength.

Key Concept

Energy Level Transitions and Atomic Emission Spectra

Question 916Question →

A galvanometer has an internal resistance of $19.0\text{ }\Omega$ and produces a full-scale deflection for a current of $50\text{ mA}$ . What value of shunt resistance, in ohms ( $\Omega$ ), must be connected in parallel with the galvanometer to convert it into an ammeter capable of measuring currents up to $1.0\text{ A}$ ?

Show answer & explanation

Answer: 1

Answer

The required shunt resistance is

1.0\text{ }\Omega

.

To convert a sensitive galvanometer into an ammeter, a low-resistance resistor called a shunt (

R_s

) is connected in parallel with the galvanometer. This provides an alternative path for the bulk of the total current. Since the potential difference across parallel branches is equal,

I_s R_s = I_g R_g

. Substituting

I_g = 0.05\text{ A}

,

R_g = 19.0\text{ }\Omega

, and

I_s = 1.0\text{ A} - 0.05\text{ A} = 0.95\text{ A}

yields

R_s = \frac{0.05 \times 19.0}{0.95} = 1.0\text{ }\Omega

.

Step-by-Step Solution

1

Convert the galvanometer full-scale deflection current

I_g

to amperes.

I_g = 50\text{ mA} = 0.05\text{ A}

Standard SI units must be used for electrical calculations.

2

Calculate the current

I_s

that must bypass the galvanometer through the shunt resistor.

I_s = I - I_g = 1.0\text{ A} - 0.05\text{ A} = 0.95\text{ A}

By Kirchhoff's current law, the total maximum current splits into galvanometer current and shunt current.

3

Calculate the required shunt resistance

R_s

using the parallel voltage relation.

R_s = \frac{I_g R_g}{I_s} = \frac{0.05\text{ A} \times 19.0\text{ }\Omega}{0.95\text{ A}} = 1.0\text{ }\Omega

Because the galvanometer and shunt resistor are connected in parallel, they share the exact same potential difference.

Key Concept

Galvanometer Conversion to Ammeter using a Shunt Resistor

Question 917Question →

A cargo ship departs from port $M$ and sails $24\text{ km}$ on a bearing of $050^\circ$ to reach point $N$ . From point $N$ , the ship changes course and sails $10\text{ km}$ on a bearing of $140^\circ$ to reach point $P$ . What is the direct distance, in kilometers, from port $M$ to point $P$ ?

Show answer & explanation

Answer: 26

Answer

The direct distance from port M to point P is 26 km.

The back bearing from N to M is 230°, and the bearing from N to P is 140°. The interior angle at N is 230° - 140° = 90°. Using the Pythagorean theorem for the right triangle formed by M, N, and P, the direct distance is √(24² + 10²) = √676 = 26 km.

Step-by-Step Solution

1

Calculate the interior angle

\angle MNP

at point

N

\angle MNP = (050^\circ + 180^\circ) - 140^\circ = 230^\circ - 140^\circ = 90^\circ

The back bearing from

N

to

M

is

230^\circ

. Subtracting the forward bearing to

P

(

140^\circ

) gives the enclosed interior angle.

2

Apply the Pythagorean theorem to right-angled triangle

MNP

MP = \sqrt{MN^2 + NP^2} = \sqrt{24^2 + 10^2} = \sqrt{576 + 100} = \sqrt{676} = 26\text{ km}

Because

\angle MNP = 90^\circ

, triangle

MNP

is a right-angled triangle with hypotenuse

MP

.

Key Concept

Bearings and Right-Angled Triangles

Question 918Question →

A Vernier caliper has $10$ divisions on its Vernier scale that coincide with $9$ main scale divisions of $1\text{ mm}$ each. When the measuring jaws are fully closed without any object between them, the zero mark of the Vernier scale lies to the left of the main scale zero mark, and the $7\text{th}$ Vernier division coincides precisely with a main scale mark. When used to measure the internal diameter of a hollow brass ring, the main scale reads $3.5\text{ cm}$ and the $4\text{th}$ Vernier division coincides with a main scale mark. What is the actual corrected internal diameter of the ring in cm?

Show answer & explanation

Answer: 3.57

Answer

The corrected internal diameter of the hollow brass ring is

3.57\text{ cm}

.

The least count of the Vernier caliper is

0.01\text{ cm}

. The negative zero error is

-(10 - 7) \times 0.01\text{ cm} = -0.03\text{ cm}

. The observed reading is

3.5\text{ cm} + (4 \times 0.01\text{ cm}) = 3.54\text{ cm}

. Correcting for zero error gives

3.54\text{ cm} - (-0.03\text{ cm}) = 3.57\text{ cm}

.

Step-by-Step Solution

1

Calculate the least count of the Vernier caliper.

Least count =

0.01\text{ cm}

(

0.1\text{ mm}

).

One main scale division is

1\text{ mm} = 0.1\text{ cm}

. Ten Vernier divisions equal nine main scale divisions (

0.9\text{ mm}

), so one Vernier division =

0.09\text{ cm}

. Least count =

0.1\text{ cm} - 0.09\text{ cm} = 0.01\text{ cm}

.

2

Determine the zero error of the instrument.

Zero Error =

-0.03\text{ cm}

.

For a negative zero error where the Vernier zero lies to the left of the main scale zero, Zero Error =

-(N - n) \times \text{least count}

, where

N = 10

and

n = 7

. Thus, Zero Error =

-(10 - 7) \times 0.01\text{ cm} = -0.03\text{ cm}

.

3

Calculate the uncorrected observed reading.

Observed Reading =

3.54\text{ cm}

.

Observed Reading = Main scale reading + (Vernier coincided division \times Least count) =

3.5\text{ cm} + (4 \times 0.01\text{ cm}) = 3.54\text{ cm}

.

4

Apply zero error correction to obtain the actual reading.

Corrected Reading =

3.57\text{ cm}

.

Corrected Reading = Observed Reading - Zero Error =

3.54\text{ cm} - (-0.03\text{ cm}) = 3.57\text{ cm}

.

Key Concept

Measurement of length using a Vernier caliper with negative zero error correction

Question 919Question →

A battery with an electromotive force (e.m.f.) of $12.0\text{ V}$ and an internal resistance of $1.0\text{ }\Omega$ is connected across a load resistor of $5.0\text{ }\Omega$ . What is the terminal potential difference across the battery in volts?

Show answer & explanation

Answer: 10

Answer

The terminal potential difference across the battery is

10.0\text{ V}

.

The total resistance in the circuit is the sum of the external load resistance and the battery's internal resistance (

R_{\text{total}} = 5.0\text{ }\Omega + 1.0\text{ }\Omega = 6.0\text{ }\Omega

). The current drawn from the battery is

I = \frac{12.0\text{ V}}{6.0\text{ }\Omega} = 2.0\text{ A}

. The terminal potential difference available to the external load is

V = I \cdot R = 2.0\text{ A} \times 5.0\text{ }\Omega = 10.0\text{ V}

.

Step-by-Step Solution

1

Find the total circuit resistance including internal resistance

R_{\text{total}} = 5.0\text{ }\Omega + 1.0\text{ }\Omega = 6.0\text{ }\Omega

The internal resistance of the cell acts in series with the external load resistor.

2

Determine total current in the circuit

I = \frac{E}{R_{\text{total}}} = \frac{12.0\text{ V}}{6.0\text{ }\Omega} = 2.0\text{ A}

By Ohm's law, current equals total electromotive force divided by total circuit resistance.

3

Compute the terminal potential difference

V = I \times R = 2.0\text{ A} \times 5.0\text{ }\Omega = 10.0\text{ V}

Terminal voltage is the voltage drop across the load resistor, equivalent to

E - I \cdot r

.

Key Concept

Terminal Potential Difference and Internal Resistance

Question 920Question →

In a cathode-ray tube, electrons of mass $9.10 \times 10^{-31}\text{ kg}$ and elementary charge $1.60 \times 10^{-19}\text{ C}$ are accelerated from rest by the electric field between the cathode and the anode. What potential difference, in volts, is required to accelerate these electrons to a speed of $8.00 \times 10^{6}\text{ m s}^{-1}$ ?

Show answer & explanation

Answer: 182

Answer

The potential difference required to accelerate the electrons to the specified speed is

182\text{ V}

.

The work done by an accelerating potential difference

V

on an electron of charge

e

is converted entirely into kinetic energy

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

. Solving

e V = \frac{1}{2} m v^2

for

V

yields

V = \frac{m v^2}{2 e} = 182\text{ V}

.

Step-by-Step Solution

1

Equate the work done by the electric field to the kinetic energy gained by an electron.

W = e V = \frac{1}{2} m v^2

Work done on a charged particle moving through an electric potential difference equals its gain in kinetic energy.

2

Isolate the potential difference

V

on one side of the equation.

V = \frac{m v^2}{2 e}

Algebraic rearrangement to solve for the target variable.

3

Substitute the given numerical parameters into the equation.

V = \frac{(9.10 \times 10^{-31}\text{ kg}) \times (8.00 \times 10^{6}\text{ m s}^{-1})^2}{2 \times (1.60 \times 10^{-19}\text{ C})}

Populating the formula with the specified values for electron mass, speed, and charge.

4

Perform the final calculation.

V = 182\text{ V}

Simplifying the numerical expression gives

182\text{ V}

.

Key Concept

Acceleration of charged particles in electric fields

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