All practice questions

Question 7921Question →

Points $P(k, 2)$ and $Q(3, 8)$ lie on a straight line $L_1$ . If $L_1$ is perpendicular to the line $L_2$ given by $4x + 3y - 12 = 0$ , what is the value of $k$ ?

Show answer & explanation

Answer: -5

Answer

The value of

k

is

-5

.

To determine

k

, the gradient of

L_2

(

4x + 3y - 12 = 0

) is found to be

-\frac{4}{3}

. Using the perpendicularity rule

m_1 \cdot m_2 = -1

, the gradient of

L_1

is

\frac{3}{4}

. Equating this to the slope formula

\frac{8 - 2}{3 - k}

gives

\frac{6}{3 - k} = \frac{3}{4}

, which simplifies to

k = -5

.

Step-by-Step Solution

1

Find the gradient

m_2

of line

L_2

m_2 = -\frac{4}{3}

Converting

4x + 3y - 12 = 0

to

y = mx + c

form gives

y = -\frac{4}{3}x + 4

.

2

Apply the perpendicular line condition to find

m_1

m_1 = \frac{3}{4}

Perpendicular lines have negative reciprocal gradients (

m_1 \cdot m_2 = -1

).

3

Express the gradient

m_1

using the coordinates of

P

and

Q

m_1 = \frac{6}{3 - k}

Using the slope formula

m = \frac{y_2 - y_1}{x_2 - x_1}

for points

P(k, 2)

and

Q(3, 8)

.

4

Solve for

k

k = -5

Equating

\frac{6}{3 - k} = \frac{3}{4}

yields

3(3 - k) = 24 \implies 9 - 3k = 24 \implies k = -5

.

Key Concept

Perpendicular Lines and Gradient Formula

Question 7922Question →

A ray of light travels from air into a liquid with a refractive index of $1.33$ . If the sine of the angle of incidence in air is $0.80$ , what is the sine of the angle of refraction in the liquid?

Show answer & explanation

Answer: 0.6

Answer

The sine of the angle of refraction in the liquid is 0.60.

According to Snell's law for light passing from air into a medium, the refractive index

n

is given by

n = \frac{\sin i}{\sin r}

. Rearranging this equation to solve for the sine of the angle of refraction yields

\sin r = \frac{\sin i}{n}

. Substituting

\sin i = 0.80

and

n = 1.33

(or

\frac{4}{3}

) gives

\sin r = \frac{0.80}{4/3} = 0.60

.

Step-by-Step Solution

1

Identify the given physical parameters and state Snell's law

Refractive index

n = 1.33

(or

\frac{4}{3}

),

\sin i = 0.80

. Snell's law:

n = \frac{\sin i}{\sin r}

Snell's law relates the ratio of the sines of the angles of incidence and refraction to the refractive index of the medium.

2

Rearrange the equation to express the sine of the angle of refraction

\sin r = \frac{\sin i}{n}

Algebraically isolating

\sin r

allows direct substitution of the known quantities.

3

Substitute the values and compute the result

\sin r = \frac{0.80}{4/3} = 0.60

Dividing

0.80

by

\frac{4}{3}

gives

0.60

.

Key Concept

Snell's Law of Refraction

Alternative Method

Convert decimal numbers into simple fractions:

n = \frac{4}{3}

and

\sin i = \frac{4}{5}

. Evaluating

\sin r = \frac{4/5}{4/3}

simplifies directly to

\frac{3}{5} = 0.60

.

Estimated Time:45s

Question 7923Question →

A team of $8$ identical excavators working together at a constant rate can clear a parcel of land in $15\text{ days}$ . All $8$ excavators work together for the first $3\text{ days}$ , after which $2$ excavators break down and are removed from the site. Assuming the remaining excavators continue working at the same constant rate, how many additional days will it take to complete the clearing of the land?

16\text{ days}

20\text{ days}

12\text{ days}

13\text{ days}

Show answer & explanation

Answer: $16\text{ days}$

Answer

The correct answer is

16\text{ days}

.

The total work required for the project is

120

excavator-days. During the first

3

days, the

8

excavators complete

24

excavator-days of work, leaving

96

excavator-days of work remaining. Since

2

excavators break down,

6

excavators remain. The additional time needed for these

6

excavators to clear the rest of the land is

96 \div 6 = 16

days.

Step-by-Step Solution

1

Calculate the total work required in excavator-days.

\text{Total Work} = 8 \text{ excavators} \times 15 \text{ days} = 120 \text{ excavator-days}

.

Work done is directly proportional to the product of rate (number of machines) and time.

2

Calculate the work completed in the first 3 days and the remaining work.

\text{Work Done} = 8 \times 3 = 24 \text{ excavator-days}

;

\text{Remaining Work} = 120 - 24 = 96 \text{ excavator-days}

.

Subtracting completed work from total work yields the portion left to be done.

3

Determine the number of remaining excavators and compute the additional days required.

\text{Remaining Excavators} = 8 - 2 = 6

;

\text{Additional Days} = \frac{96 \text{ excavator-days}}{6 \text{ excavators}} = 16 \text{ days}

.

Dividing the remaining work by the active rate gives the time needed to finish the project.

Key Concept

Inverse proportion and work-rate problem involving partial completion and workforce changes.

Question 7924Question →

How many integer values of $x$ satisfy the quadratic inequality $2x^2 - 7x - 4 \le 0$ ?

Show answer & explanation

Answer: 5

Answer

There are 5 integer values of x that satisfy the inequality.

Solving the quadratic inequality yields

-\frac{1}{2} \le x \le 4

. The integer solutions within this interval are

0, 1, 2, 3,

and

4

. Counting them gives a total of

5

valid integer values.

Step-by-Step Solution

1

Factor the quadratic expression.

(2x + 1)(x - 4) \le 0

Factoring allows determination of the critical boundary points.

2

Find the critical values (roots of the equation).

x = -\frac{1}{2}

and

x = 4

The roots divide the number line into test intervals.

3

Determine the solution set of the inequality.

-\frac{1}{2} \le x \le 4

Since the quadratic coefficient is positive, the quadratic expression is non-positive between its roots.

4

List and count the integers within the range.

The integers are

0, 1, 2, 3, 4

, making a total of

5

integers.

Counting only whole numbers in the closed interval

[ -0.5, 4 ]

.

Key Concept

Quadratic Inequalities and Integer Solution Counting

Question 7925Question →

A metallic wire has a resistance of $12.0\,\Omega$ at $0\,^\circ\text{C}$ and a temperature coefficient of resistance of $4.0 \times 10^{-3}\,^\circ\text{C}^{-1}$ . The wire is uniformly stretched until its length increases by $25\%$ . Assuming the density and total volume of the wire remain constant during stretching, what is the resistance of the stretched wire at $50\,^\circ\text{C}$ ?

Show answer & explanation

Answer: 22.5

Answer

The resistance of the stretched wire at

50\,^\circ\text{C}

is

22.5\,\Omega

.

Stretching a wire by

25\%

increases its length by a factor of

1.25

and reduces its cross-sectional area by a factor of

1.25

(since volume is conserved). The resistance at

0\,^\circ\text{C}

scales as

(1.25)^2 = 1.5625

, giving

18.75\,\Omega

. Accounting for the temperature increase to

50\,^\circ\text{C}

via

R(T) = R'_0(1 + \alpha T)

yields

18.75 \times (1 + 4.0 \times 10^{-3} \times 50) = 18.75 \times 1.20 = 22.5\,\Omega

.

Step-by-Step Solution

1

Calculate the resistance of the wire at

0\,^\circ\text{C}

after uniform stretching.

R'_0 = 18.75\,\Omega

Uniform stretching by

25\%

increases length to

L' = 1.25 L_0

. Volume conservation (

V = A L

) requires area to decrease to

A' = A_0 / 1.25

. Since

R = \rho L / A

,

R'_0 = R_0 (L'/L_0)^2 = 12.0 \times (1.25)^2 = 18.75\,\Omega

.

2

Apply the temperature coefficient formula to calculate resistance at

50\,^\circ\text{C}

.

R(50) = 22.5\,\Omega

Using

R(T) = R'_0 (1 + \alpha T)

, substitute

R'_0 = 18.75\,\Omega

,

\alpha = 4.0 \times 10^{-3}\,^\circ\text{C}^{-1}

, and

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

to find

R(50) = 18.75 \times (1 + 0.20) = 22.5\,\Omega

.

Key Concept

Combined effects of dimensional deformation and temperature on electrical resistance

Question 7926Question →

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 7927Question →

What is the equation of the locus of a point $P(x, y)$ that is always equidistant from the point $(0, 4)$ and the line $y = -4$ ?

Show answer & explanation

Answer: x^2 = 16y; x^2 - 16y = 0; x^2-16y=0; y = x^2/16; y = \frac{x^2}{16}; x^{2}=16y; x^{2}-16y=0

Answer

The equation of the locus is

x^2 = 16y

(or

x^2 - 16y = 0

).

Equating the distance from

P(x, y)

to

(0, 4)

, which is

\sqrt{x^2 + (y-4)^2}

, to the perpendicular distance to the line

y = -4

, which is

|y + 4|

, and squaring both sides gives

x^2 + (y-4)^2 = (y+4)^2

. Expanding yields

x^2 + y^2 - 8y + 16 = y^2 + 8y + 16

, which simplifies directly to

x^2 = 16y

or

x^2 - 16y = 0

.

Step-by-Step Solution

1

Formulate the distance expressions from point

P(x, y)

to the given point

(0, 4)

and line

y = -4

.

Distance to

(0, 4) = \sqrt{(x - 0)^2 + (y - 4)^2}

. Perpendicular distance to

y = -4

is

|y - (-4)| = |y + 4|

.

By definition of geometric locus, the distance from

P(x, y)

to the fixed point must equal its distance to the fixed line.

2

Equate the two distance expressions.

\sqrt{x^2 + (y - 4)^2} = |y + 4|

The point

P(x, y)

is equidistant from both geometric entities.

3

Square both sides of the equation to clear the square root and absolute value.

x^2 + (y - 4)^2 = (y + 4)^2

Squaring eliminates radical and absolute value signs while preserving algebraic equality.

4

Expand both squared binomial expressions and simplify.

x^2 + y^2 - 8y + 16 = y^2 + 8y + 16 \implies x^2 = 16y

Subtracting

y^2 + 16

from both sides leaves

x^2 - 8y = 8y

, which simplifies to

x^2 = 16y

.

Key Concept

The locus of points equidistant from a fixed point (focus) and a fixed straight line (directrix) forms a parabola.

Estimated Time:1m 30s

Question 7928Question →

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 7929Question →

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 7930Question →

Given that $(x + 2)$ is a factor of the polynomial $P(x) = 2x^3 - x^2 + ax + b$ , and that dividing $P(x)$ by $(2x - 1)$ leaves a remainder of $15$ , what is the value of $a + b$ ?

6

11

14

-1

Show answer & explanation

Answer: $14$

Answer

The value of

a + b

is

14

.

Applying the Factor Theorem with

x = -2

gives the equation

-2a + b = 20

. Applying the Remainder Theorem with

x = \frac{1}{2}

gives the equation

a + 2b = 30

. Solving these simultaneous equations yields

a = -2

and

b = 16

. Adding

a

and

b

gives

14

.

Step-by-Step Solution

1

Apply the Factor Theorem for

(x + 2)

-2a + b = 20

Since

(x + 2)

is a factor of

P(x)

,

P(-2) = 0

. Substituting

x = -2

yields

2(-2)^3 - (-2)^2 + a(-2) + b = 0

, which simplifies to

-16 - 4 - 2a + b = 0

or

-2a + b = 20

.

2

Apply the Remainder Theorem for

(2x - 1)

a + 2b = 30

Dividing

P(x)

by

(2x - 1)

leaves a remainder of

15

, so

P\left(\frac{1}{2}\right) = 15

. Substituting

x = \frac{1}{2}

yields

2\left(\frac{1}{8}\right) - \frac{1}{4} + \frac{a}{2} + b = 15

, which simplifies to

\frac{a}{2} + b = 15

or

a + 2b = 30

.

3

Solve the system of linear equations for

a

and

b

a = -2

and

b = 16

From step 1,

b = 2a + 20

. Substituting this into step 2 gives

a + 2(2a + 20) = 30 \implies 5a + 40 = 30 \implies 5a = -10 \implies a = -2

. Substituting

a = -2

into

b = 2a + 20

gives

b = 16

.

4

Calculate

a + b

14

Summing the calculated values gives

a + b = -2 + 16 = 14

.

Key Concept

Factor Theorem and Remainder Theorem for Polynomials

Question 7931Question →

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 7932Question →

A micrometer screw gauge has a pitch of $0.5\text{ mm}$ and $100$ equal divisions on its circular thimble scale. When the anvil and spindle are fully closed without any object, the zero mark on the circular scale lies $4$ divisions below the datum line. The instrument is then used to measure the total thickness of a tightly bound stack of $10$ identical metal sheets. The main scale reading is $2.5\text{ mm}$ and the $38\text{th}$ division on the circular scale aligns with the datum line. What is the actual mean thickness of a single metal sheet?

0.269\text{ mm}

0.271\text{ mm}

0.267\text{ mm}

0.284\text{ mm}

Show answer & explanation

Answer: $0.267\text{ mm}$

Answer

The actual mean thickness of a single metal sheet is

0.267\text{ mm}

.

The least count of the micrometer is

\frac{0.5\text{ mm}}{100} = 0.005\text{ mm}

. With the zero mark 4 divisions below the datum line, there is a positive zero error of

+0.020\text{ mm}

. The observed total reading for 10 sheets is

2.5\text{ mm} + (38 \times 0.005\text{ mm}) = 2.690\text{ mm}

. Subtracting the positive zero error yields an actual thickness of

2.690\text{ mm} - 0.020\text{ mm} = 2.670\text{ mm}

for 10 sheets, which gives

0.267\text{ mm}

per sheet.

Step-by-Step Solution

1

Calculate the least count of the micrometer screw gauge

\text{Least count} = \frac{\text{Pitch}}{\text{Number of circular scale divisions}} = \frac{0.5\text{ mm}}{100} = 0.005\text{ mm}

The least count determines the value of each division on the circular scale.

2

Determine the zero error

\text{Zero error} = +4 \times 0.005\text{ mm} = +0.020\text{ mm}

Since the zero mark on the thimble is below the main scale datum line when closed, the error is positive.

3

Calculate the observed reading for 10 sheets

\text{Observed reading} = \text{Main scale reading} + (\text{Circular scale division} \times \text{Least count}) = 2.5\text{ mm} + (38 \times 0.005\text{ mm}) = 2.5\text{ mm} + 0.190\text{ mm} = 2.690\text{ mm}

Combines the main scale and circular scale readings to find the raw measured value.

4

Calculate the corrected reading for 10 sheets

\text{Corrected reading} = \text{Observed reading} - \text{Zero error} = 2.690\text{ mm} - 0.020\text{ mm} = 2.670\text{ mm}

Subtracting a positive zero error gives the true thickness of the 10 sheets.

5

Find the thickness of a single sheet

\text{Thickness per sheet} = \frac{2.670\text{ mm}}{10} = 0.267\text{ mm}

Dividing the total corrected thickness by the total number of sheets gives the average thickness per sheet.

Key Concept

Micrometer Screw Gauge Least Count and Zero Error Correction

Question 7933Question →

A sealed rigid canister contains a fixed mass of nitrogen gas at an initial pressure of $1.50 \times 10^5\text{ Pa}$ and a temperature of $27^\circ\text{C}$ . If the gas is heated at constant volume until its temperature reaches $127^\circ\text{C}$ , what is the new pressure of the gas?

2.00 \times 10^5\text{ Pa}

7.06 \times 10^5\text{ Pa}

1.13 \times 10^5\text{ Pa}

3.19 \times 10^4\text{ Pa}

Show answer & explanation

Answer: $2.00 \times 10^5\text{ Pa}$

Answer

The new pressure of the gas is

2.00 \times 10^5\text{ Pa}

.

At constant volume, the pressure of a gas is directly proportional to its absolute temperature (

P_1/T_1 = P_2/T_2

). Converting temperatures to Kelvin gives

300\text{ K}

and

400\text{ K}

. Substituting these values yields

P_2 = 1.50 \times 10^5 \times (400/300) = 2.00 \times 10^5\text{ Pa}

.

Step-by-Step Solution

1

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

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

and

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

.

Gas laws require absolute temperature in Kelvin for proportional reasoning.

2

Apply Pressure Law (Gay-Lussac's Law) for constant volume.

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

.

At constant volume, the pressure of a fixed mass of gas is directly proportional to its absolute temperature.

3

Substitute the known values into the equation and solve for

P_2

.

P_2 = 1.50 \times 10^5\text{ Pa} \times \frac{400\text{ K}}{300\text{ K}} = 2.00 \times 10^5\text{ Pa}

.

Multiplying initial pressure by the temperature ratio gives the final pressure.

Key Concept

Pressure Law (Gay-Lussac's Law) states that at constant volume,

P \propto T

where

T

must be in Kelvin.

Question 7934Question →

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 7935Question →

A beam of cathode rays is accelerated from rest through an electric potential difference $V$ before entering a uniform magnetic field $B$ applied perpendicular to the direction of motion, causing the rays to bend into a circular arc of radius $r$ . If the accelerating potential difference is increased to $2V$ and the magnetic field intensity is increased to $2B$ , what is the new radius of curvature of the cathode ray path?

\frac{r}{\sqrt{2}}

r

\sqrt{2}r

2r

Show answer & explanation

Answer: $\frac{r}{\sqrt{2}}$

Answer

The new radius of curvature is

\frac{r}{\sqrt{2}}

.

The kinetic energy gained by an electron of mass

m

and charge

e

accelerated through potential difference

V

is

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

, giving

v = \sqrt{\frac{2eV}{m}}

. When entering a perpendicular magnetic field

B

, centripetal force gives

e v B = \frac{m v^2}{r}

, leading to

r = \frac{m v}{e B} = \frac{1}{B}\sqrt{\frac{2m V}{e}}

. Replacing

V

with

2V

and

B

with

2B

gives

r' = \frac{\sqrt{2}}{2}r = \frac{r}{\sqrt{2}}

.

Step-by-Step Solution

1

Relate electron velocity to accelerating potential difference

V

v = \sqrt{\frac{2eV}{m}}

The electrical potential energy lost equals the kinetic energy gained by the cathode ray electrons:

eV = \frac{1}{2}mv^2

.

2

Express the radius of curvature

r

in terms of

V

and

B

r = \frac{mv}{eB} = \frac{1}{B}\sqrt{\frac{2mV}{e}}

The magnetic force

evB

provides the necessary centripetal force

\frac{mv^2}{r}

.

3

Substitute the scaled values

V' = 2V

and

B' = 2B

into the radius expression

r' = \frac{1}{2B}\sqrt{\frac{2m(2V)}{e}} = \frac{\sqrt{2}}{2}\left(\frac{1}{B}\sqrt{\frac{2mV}{e}}\right) = \frac{r}{\sqrt{2}}

Increasing

V

by a factor of 2 increases

v

by

\sqrt{2}

, while doubling

B

increases the denominator by 2.

Key Concept

Deflection of cathode rays in magnetic fields and energy conversion of accelerated charges

Question 7936Question →

A Vernier caliper with a least count of $0.01\text{ cm}$ is used to measure the thickness of a uniform metal plate. When the jaws are closed together without the plate inserted, the zero line of the Vernier scale lies to the right of the main scale zero, with the $3\text{rd}$ Vernier division coinciding with a main scale mark. When clamped around the metal plate, the main scale reading is $1.8\text{ cm}$ and the $4\text{th}$ Vernier division aligns perfectly with a main scale mark. What is the correct thickness of the metal plate?

1.81\text{ cm}

1.84\text{ cm}

1.87\text{ cm}

1.77\text{ cm}

Show answer & explanation

Answer: $1.81\text{ cm}$

Answer

The correct thickness of the metal plate is

1.81\text{ cm}

.

When the Vernier zero is positioned to the right of the main scale zero when jaws are closed, it indicates a positive zero error equal to

+ (3 \times 0.01\text{ cm}) = +0.03\text{ cm}

. The observed measurement with the metal plate is

1.8\text{ cm} + (4 \times 0.01\text{ cm}) = 1.84\text{ cm}

. Applying the standard correction formula

\text{Correct Reading} = \text{Observed Reading} - \text{Zero Error}

gives

1.84\text{ cm} - 0.03\text{ cm} = 1.81\text{ cm}

.

Step-by-Step Solution

1

Determine the zero error of the Vernier caliper

Zero error

= +3 \times 0.01\text{ cm} = +0.03\text{ cm}

Because the Vernier zero lies to the right of the main scale zero, the instrument has a positive zero error.

2

Calculate the observed reading

Observed reading

= 1.8\text{ cm} + (4 \times 0.01\text{ cm}) = 1.84\text{ cm}

The total observed value is the sum of the main scale reading and the Vernier scale reading.

3

Apply the zero error correction to obtain the actual reading

Correct reading

= 1.84\text{ cm} - (+0.03\text{ cm}) = 1.81\text{ cm}

True Reading = Observed Reading - Zero Error (with proper sign).

Key Concept

Vernier Caliper Zero Error Correction

Estimated Time:1m 15s

Question 7937Question →

A $2.0\text{ }\mu\text{F}$ capacitor and a $3.0\text{ }\mu\text{F}$ capacitor are connected in series across a $100\text{ V}$ d.c. power supply. What is the magnitude of the electric charge stored on the $2.0\text{ }\mu\text{F}$ capacitor?

500\text{ }\mu\text{C}

120\text{ }\mu\text{C}

200\text{ }\mu\text{C}

80\text{ }\mu\text{C}

Show answer & explanation

Answer: $120\text{ }\mu\text{C}$

Answer

The electric charge stored on the

2.0\text{ }\mu\text{F}

capacitor is

120\text{ }\mu\text{C}

.

The correct answer is

120\text{ }\mu\text{C}

. For capacitors connected in series, the equivalent capacitance

C_{eq}

is determined using

\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}

, giving

C_{eq} = \frac{2.0 \times 3.0}{2.0 + 3.0} = 1.2\text{ }\mu\text{F}

. Multiplying by the source voltage of

100\text{ V}

yields a total charge

Q = C_{eq} V = 120\text{ }\mu\text{C}

. Because capacitors in series store equal amounts of charge, the charge on the

2.0\text{ }\mu\text{F}

capacitor is

120\text{ }\mu\text{C}

.

Step-by-Step Solution

1

Calculate the equivalent capacitance

C_{eq}

of the series combination.

C_{eq} = \frac{C_1 C_2}{C_1 + C_2} = \frac{2.0 \times 3.0}{2.0 + 3.0} = 1.2\text{ }\mu\text{F}

Capacitors in series combine according to the reciprocal formula

\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}

.

2

Determine the total charge

Q

supplied by the

100\text{ V}

source.

Q = C_{eq} V = 1.2\text{ }\mu\text{F} \times 100\text{ V} = 120\text{ }\mu\text{C}

The total charge is the product of the equivalent capacitance and the total voltage.

3

Identify the charge on the individual

2.0\text{ }\mu\text{F}

capacitor.

Q_1 = Q = 120\text{ }\mu\text{C}

Components connected in series carry the exact same electric charge.

Key Concept

Equivalent Capacitance and Charge Distribution in Series Circuits

Estimated Time:1m 30s

Question 7938Question →

In an X-ray tube, non-relativistic electrons accelerated from rest hit a target anode, producing continuous X-radiation with a minimum cut-off wavelength of $\lambda_0$ . If the operating potential difference across the tube is adjusted such that the maximum momentum of the colliding electrons increases by $50\%$ , what is the new cut-off wavelength of the emitted X-rays in terms of $\lambda_0$ ?

\frac{2}{3}\lambda_0

\frac{4}{9}\lambda_0

\frac{9}{4}\lambda_0

\frac{3}{2}\lambda_0

Show answer & explanation

Answer: $\frac{4}{9}\lambda_0$

Answer

The new cut-off wavelength of the emitted X-rays is

\frac{4}{9}\lambda_0

.

According to the Duane-Hunt law, the maximum photon energy produced in continuous X-radiation equals the maximum kinetic energy of the striking electrons:

E_{\text{max}} = \frac{hc}{\lambda_{\min}} = E_k

. Expressing kinetic energy in terms of momentum yields

E_k = \frac{p^2}{2m}

, which gives

\lambda_{\min} = \frac{2mhc}{p^2}

. Therefore,

\lambda_{\min}

is inversely proportional to

p^2

. When momentum increases by

50\%

(

p_2 = 1.5 p_1 = \frac{3}{2}p_1

),

p^2

increases by a factor of

\frac{9}{4}

. Consequently, the new cut-off wavelength becomes

\frac{4}{9}\lambda_0

.

Step-by-Step Solution

1

Relate electron momentum to electron kinetic energy

The kinetic energy

E_k

of non-relativistic electrons of mass

m

with momentum

p

is

E_k = \frac{p^2}{2m}

.

Electrons are accelerated through potential difference

V

, gaining kinetic energy

E_k = e V = \frac{p^2}{2m}

.

2

Apply the Duane-Hunt law for minimum X-ray wavelength

\lambda_0 = \frac{hc}{E_k} = \frac{2mhc}{p_1^2}

.

The maximum energy of an X-ray photon corresponds to the shortest (cut-off) wavelength

\lambda_{\min} = \frac{hc}{E_k}

.

3

Calculate the updated momentum and new kinetic energy ratio

New momentum

p_2 = 1.5 p_1 = \frac{3}{2} p_1

, so

p_2^2 = \frac{9}{4} p_1^2

.

An increase of

50\%

means multiplying the initial momentum by

1 + 0.5 = 1.5 = \frac{3}{2}

.

4

Determine the new cut-off wavelength

\lambda_{\text{new}}

\lambda_{\text{new}} = \frac{2mhc}{p_2^2} = \frac{2mhc}{\frac{9}{4}p_1^2} = \frac{4}{9} \left(\frac{2mhc}{p_1^2}\right) = \frac{4}{9}\lambda_0

.

Since

\lambda_{\min}

is inversely proportional to

p^2

, scaling momentum by

\frac{3}{2}

reduces the cut-off wavelength by a factor of

\left(\frac{2}{3}\right)^2 = \frac{4}{9}

.

Key Concept

Duane-Hunt Law and Electron Kinetics

Question 7939Question →

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 7940Question →

Ultraviolet radiation of fixed frequency $f$ , which exceeds the threshold frequency $f_0$ of a zinc emitter plate, causes photoelectron emission. If the intensity of the incident radiation is multiplied by two without altering its frequency, what is the effect on the maximum kinetic energy of the photoelectrons and the emission current?

The maximum kinetic energy is unchanged, but the emission current is doubled.

The maximum kinetic energy is doubled, and the emission current is also doubled.

The maximum kinetic energy is doubled, while the emission current remains unchanged.

The maximum kinetic energy and emission current both remain unchanged.

Show answer & explanation

Answer: The maximum kinetic energy is unchanged, but the emission current is doubled.

Answer

The maximum kinetic energy is unchanged, but the emission current is doubled.

In the photoelectric effect, the maximum kinetic energy of emitted electrons (

K_{\text{max}} = hf - W_0

) depends only on the frequency of the incident radiation and the work function of the metal. Changing light intensity changes only the number of photons arriving per second, which doubles the rate of photoelectron emission and hence doubles the photoelectric current while leaving maximum kinetic energy unchanged.

Step-by-Step Solution

1

Analyze the dependence of maximum kinetic energy on radiation parameters using Einstein's photoelectric equation.

K_{\text{max}} = hf - W_0

. Since frequency

f

and work function

W_0

remain constant,

K_{\text{max}}

remains unchanged.

Individual photon energy is determined strictly by frequency (

E = hf

). Intensity does not alter individual photon energy.

2

Analyze the relationship between light intensity and photoelectric current.

Intensity

I

is directly proportional to the number of incident photons per second (

N

). Since one photon liberates one electron, doubling intensity doubles the photoelectron emission rate, thereby doubling the photoelectric current.

Photoelectric current measures the rate of charge emission, which depends directly on photon flux.

Key Concept

Independence of photoelectron kinetic energy from light intensity

Answer

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