All practice questions

Question 221Question →

A coil consisting of $50$ turns is placed in a region of changing magnetic field. If the magnetic flux passing through the coil increases uniformly from $0.2\text{ Wb}$ to $0.6\text{ Wb}$ in $2.0\text{ s}$ , what is the magnitude of the induced electromotive force in the coil in volts?

Show answer & explanation

Answer: 10

Answer

The magnitude of the induced electromotive force in the coil is

10\text{ V}

.

According to Faraday's law of electromagnetic induction, the magnitude of the induced electromotive force

E

is proportional to the number of turns

N

and the rate of change of magnetic flux

\frac{\Delta \Phi}{\Delta t}

. Given

N = 50

,

\Delta \Phi = 0.6\text{ Wb} - 0.2\text{ Wb} = 0.4\text{ Wb}

, and

\Delta t = 2.0\text{ s}

, substituting these into

E = N \frac{\Delta \Phi}{\Delta t}

gives

E = 50 \times \frac{0.4}{2.0} = 10\text{ V}

.

Step-by-Step Solution

1

Determine the change in magnetic flux through the coil

\Delta \Phi = 0.6\text{ Wb} - 0.2\text{ Wb} = 0.4\text{ Wb}

Induction depends on the change in magnetic flux over time.

2

Apply Faraday's law of electromagnetic induction to solve for the induced e.m.f.

E = N \frac{\Delta \Phi}{\Delta t} = 50 \times \frac{0.4\text{ Wb}}{2.0\text{ s}} = 10\text{ V}

Faraday's law states that the induced e.m.f. is equal to the product of the number of turns and the rate of change of flux.

Key Concept

Faraday's Law of Electromagnetic Induction

Question 222Question →

A shell of total mass $5.0\text{ kg}$ is moving horizontally with a velocity of $20\text{ m s}^{-1}$ when an internal explosion splits it into two fragments. One fragment of mass $2.0\text{ kg}$ is propelled backward along the original path at a speed of $10\text{ m s}^{-1}$ . What is the magnitude of the velocity, in $\text{m s}^{-1}$ , of the second fragment immediately after the explosion?

Show answer & explanation

Answer: 40

Answer

The magnitude of the velocity of the second fragment immediately after the explosion is

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

.

According to the Law of Conservation of Linear Momentum, the total momentum of a system remains constant when no net external horizontal force acts on it. Taking the original direction of motion as positive, the initial momentum is

100\text{ kg m s}^{-1}

. Since the

2.0\text{ kg}

fragment moves backward at

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

, its momentum is

-20\text{ kg m s}^{-1}

. For total momentum to remain

+100\text{ kg m s}^{-1}

, the remaining

3.0\text{ kg}

fragment must carry a momentum of

+120\text{ kg m s}^{-1}

, which corresponds to a velocity of

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

.

Step-by-Step Solution

1

Calculate the initial momentum of the shell prior to the explosion.

p_i = 5.0\text{ kg} \times 20\text{ m s}^{-1} = 100\text{ kg m s}^{-1}

in the initial forward direction.

Before the internal explosion, the system consists of a single mass moving with a constant velocity.

2

Apply the Law of Conservation of Linear Momentum taking vector direction into account.

p_i = m_1 v_1 + m_2 v_2 \implies 100 = 2.0(-10) + 3.0 v_2

In the absence of external forces, total momentum is conserved. The fragment propelled backward takes a negative sign relative to the initial forward motion.

3

Solve the algebraic equation for the unknown velocity

v_2

.

100 + 20 = 3.0 v_2 \implies 120 = 3.0 v_2 \implies v_2 = 40\text{ m s}^{-1}

Isolating

v_2

gives the forward velocity magnitude of the remaining

3.0\text{ kg}

piece.

Key Concept

Conservation of Linear Momentum in Explosions (1D Vector Sign Convention)

Question 223Question →

A light ray strikes the first face of a glass prism of refracting angle $30^\circ$ at normal incidence ( $i = 0^\circ$ ). If the refractive index of the glass is $1.50$ , calculate the angle of emergence, in degrees, as the ray leaves the second face into air. (Take $\arcsin(0.75) = 48.6^\circ$ )

Show answer & explanation

Answer: 48.6

Answer

The angle of emergence of the light ray as it exits the prism is

48.6^\circ

.

Because the ray is incident normally at the first surface, it continues undeviated into the glass (

r_1 = 0^\circ

). By prism geometry, the angle of incidence at the second face is equal to the apex angle of the prism (

r_2 = A = 30^\circ

). Applying Snell's Law at the glass-air boundary gives

1.50 \sin(30^\circ) = 1.00 \sin(e)

, leading to

\sin(e) = 0.75

, which evaluates to an emergent angle of

48.6^\circ

.

Step-by-Step Solution

1

Determine the angle of refraction at the first surface

r_1 = 0^\circ

Light entering a surface normally (

i_1 = 0^\circ

) passes straight through without bending.

2

Find the angle of incidence at the second surface inside the prism using prism geometry

r_2 = 30^\circ

For any triangular prism, the refracting angle

A = r_1 + r_2

. Since

r_1 = 0^\circ

,

r_2 = A = 30^\circ

.

3

Apply Snell's Law at the second interface (glass to air)

\sin(e) = 0.75

n_{\text{glass}} \sin(r_2) = n_{\text{air}} \sin(e) \implies 1.50 \times \sin(30^\circ) = 1.00 \times \sin(e)

.

4

Calculate the emergent angle

e

e = 48.6^\circ

Taking the inverse sine of

0.75

yields

e = \arcsin(0.75) = 48.6^\circ

.

Key Concept

Prism Geometry and Snell's Law Refraction

Question 224Question →

A converging lens has a focal length of $25\text{ cm}$ . What is the power of the lens in dioptres?

Show answer & explanation

Answer: 4

Answer

The power of the lens is

4\text{ D}

.

The power

P

of a lens in dioptres is defined as the reciprocal of its focal length

f

expressed in meters (

P = \frac{1}{f}

). Expressing

25\text{ cm}

in meters gives

0.25\text{ m}

. Substituting this value yields

P = \frac{1}{0.25\text{ m}} = 4\text{ D}

.

Step-by-Step Solution

1

Convert the focal length from centimeters to meters

f = 25\text{ cm} = 0.25\text{ m}

The unit of dioptre (

D

) is defined as reciprocal meters (

m^{-1}

), so focal length must be in meters.

2

Apply the lens power formula

P = \frac{1}{f}

and compute the value

P = \frac{1}{0.25} = 4\text{ D}

The power of a converging lens is positive and equal to the reciprocal of its focal length in meters.

Key Concept

Power of a Lens

Question 225Question →

A metallic wire has a resistance of $10.0\,\Omega$ at $0^\circ\text{C}$ and a temperature coefficient of resistance $\alpha = 4.0 \times 10^{-3}\,\text{K}^{-1}$ . The wire is uniformly stretched at constant temperature until its length is doubled while maintaining constant mass and volume. It is subsequently heated to $50^\circ\text{C}$ . What is the electric current, in Amperes, that flows through the wire when a potential difference of $120.0\,\text{V}$ is applied across its ends?

Show answer & explanation

Answer: 2.5

Answer

The electric current passing through the heated, stretched wire is 2.5 A.

When a wire of initial resistance 10.0 ohms is stretched to twice its original length, conservation of volume requires its cross-sectional area to halve, which quadruples its resistance to 40.0 ohms at 0 °C. Heating the wire by 50 K increases its resistance by a factor of (1 + 0.004 * 50) = 1.2, producing a final resistance of 48.0 ohms. Applying a 120.0 V potential difference across 48.0 ohms results in a current of 2.5 A.

Step-by-Step Solution

1

Determine resistance change due to wire stretching

R_0' = 40.0 ohms at 0 °C

Uniform stretching conserves total volume (V = A * L). Doubling length halves area, making resistance increase by a factor of 2^2 = 4.

2

Calculate resistance at 50 °C using temperature coefficient

R(50 °C) = 48.0 ohms

Resistance increases linearly with temperature: R(T) = R_0'(1 + alpha * Delta T).

3

Apply Ohm's law to find current

I = 2.5 A

Current is given by potential difference divided by total resistance at the operational temperature.

Key Concept

Resistance variation with geometric stretching and temperature coefficient of resistance

Question 226Question →

A rectangular coil of $200$ turns has dimensions $0.15\text{ m}$ by $0.10\text{ m}$ . The coil is placed with its plane perpendicular to a uniform magnetic field of $0.40\text{ T}$ . If the direction of the magnetic field is completely reversed in a time interval of $0.06\text{ s}$ , calculate the magnitude of the average electromotive force (e.m.f.) induced in the coil in volts.

Show answer & explanation

Answer: 40

Answer

The magnitude of the average induced electromotive force is

40\text{ V}

.

According to Faraday's law of electromagnetic induction, the magnitude of induced electromotive force (e.m.f.) is given by

\mathcal{E} = N \left| \frac{\Delta \Phi}{\Delta t} \right|

. Since the coil is initially perpendicular to the magnetic field

B

, the initial flux per turn is

\Phi_i = B A

. When the field is completely reversed, the final flux becomes

\Phi_f = -B A

, giving a magnitude of flux change per turn of

|\Delta \Phi| = B A - (-B A) = 2 B A

. Substituting

N = 200

,

A = 0.015\text{ m}^2

,

B = 0.40\text{ T}

, and

\Delta t = 0.06\text{ s}

gives

\mathcal{E} = 200 \times \frac{2 \times 0.40 \times 0.015}{0.06} = 40\text{ V}

.

Step-by-Step Solution

1

Calculate the cross-sectional area of the rectangular coil

A = 0.15\text{ m} \times 0.10\text{ m} = 0.015\text{ m}^2

The area is required to determine the magnetic flux passing through the coil.

2

Compute the initial magnetic flux per turn

\Phi_i = B A = 0.40 \times 0.015 = 0.006\text{ Wb}

Magnetic flux is defined as the product of magnetic field strength and area when perpendicular.

3

Calculate the change in flux per turn when the magnetic field reverses direction

|\Delta \Phi| = \Phi_i - (-\Phi_i) = 2 \Phi_i = 0.012\text{ Wb}

Reversing the field flips the direction of the flux vectors, resulting in a net change equal to twice the magnitude of the initial flux.

4

Apply Faraday's Law of Electromagnetic Induction to find the induced e.m.f.

\mathcal{E} = N \frac{|\Delta \Phi|}{\Delta t} = 200 \times \frac{0.012}{0.06} = 40\text{ V}

The magnitude of induced e.m.f. equals the total rate of change of magnetic flux linkage across all turns.

Key Concept

Faraday's Law of Electromagnetic Induction (Magnetic Flux Reversal)

Question 227Question →

A rigid vessel of fixed volume contains an ideal gas at an initial pressure of $1.50 \times 10^5\text{ Pa}$ and a temperature of $27^\circ\text{C}$ . Additional gas is pumped into the vessel until the total number of moles of gas is doubled. If the temperature of the gas increases to $87^\circ\text{C}$ , what is the final pressure of the gas in units of $10^5\text{ Pa}$ ?

Show answer & explanation

Answer: 3.6

Answer

The final pressure of the gas is

3.6 \times 10^5\text{ Pa}

(which is

3.6

in units of

10^5\text{ Pa}

).

Using the ideal gas equation

PV = nRT

at fixed volume, the ratio of final to initial pressure is given by

P_2/P_1 = (n_2/n_1) \times (T_2/T_1)

. Converting temperatures to Kelvin yields

T_1 = 300\text{ K}

and

T_2 = 360\text{ K}

. Given that the number of moles doubles (

n_2/n_1 = 2

), substituting the values gives

P_2 = 1.50 \times 10^5 \times 2 \times (360/300) = 3.60 \times 10^5\text{ Pa}

.

Step-by-Step Solution

1

Convert initial and final temperatures from Celsius to absolute temperature in Kelvin.

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

and

T_2 = 87 + 273 = 360\text{ K}

.

Gas laws require absolute temperatures in Kelvin for thermodynamic calculations.

2

Formulate the pressure relation from the ideal gas equation

PV = nRT

.

Since volume

V

and the universal gas constant

R

are constant,

P_2 = P_1 \times \frac{n_2}{n_1} \times \frac{T_2}{T_1}

.

Pressure is directly proportional to both the number of moles and the absolute temperature when volume is fixed.

3

Substitute the mole ratio

n_2/n_1 = 2

, initial pressure, and Kelvin temperatures to compute

P_2

.

P_2 = 1.50 \times 10^5 \times 2 \times \frac{360}{300} = 3.60 \times 10^5\text{ Pa}

.

Evaluates the final gas pressure in the requested numerical units.

Key Concept

Ideal Gas Equation and Variable Moles under Constant Volume

Question 228Question →

Two point charges of $+2.0 \times 10^{-6}\text{ C}$ and $+4.0 \times 10^{-6}\text{ C}$ are placed in a vacuum at a distance of $0.3\text{ m}$ apart. Taking Coulomb's constant $k = 9.0 \times 10^9\text{ N}\cdot\text{m}^2\text{C}^{-2}$ , what is the magnitude of the electrostatic force exerted between them in Newtons?

Show answer & explanation

Answer: 0.8

Answer

The magnitude of the electrostatic force between the charges is 0.8 N.

According to Coulomb's Law, the force between two point charges is directly proportional to the product of the magnitude of the charges and inversely proportional to the square of the distance between them:

F = \frac{k q_1 q_2}{r^2}

. Substituting

q_1 = 2.0 \times 10^{-6}\text{ C}

,

q_2 = 4.0 \times 10^{-6}\text{ C}

,

r = 0.3\text{ m}

, and

k = 9.0 \times 10^9\text{ N}\cdot\text{m}^2\text{C}^{-2}

into the formula gives

F = 0.8\text{ N}

.

Step-by-Step Solution

1

Identify known quantities from the problem statement

q_1 = 2.0 \times 10^{-6}\text{ C}

,

q_2 = 4.0 \times 10^{-6}\text{ C}

,

r = 0.3\text{ m}

,

k = 9.0 \times 10^9\text{ N}\cdot\text{m}^2\text{C}^{-2}

Clear identification of parameters is required for substitution into Coulomb's Law.

2

Apply Coulomb's Law formula

F = \frac{k q_1 q_2}{r^2}

Coulomb's Law quantifies the electrostatic force between two stationary point charges.

3

Substitute values and perform arithmetic calculation

F = \frac{9.0 \times 10^9 \times (2.0 \times 10^{-6}) \times (4.0 \times 10^{-6})}{0.09} = \frac{0.072}{0.09} = 0.8\text{ N}

Squaring the separation distance

0.3\text{ m}

gives

0.09\text{ m}^2

, and evaluating the numerator gives

0.072\text{ N}\cdot\text{m}^2

.

Key Concept

Coulomb's Law

Question 229Question →

A hypermetropic eye has its unassisted near point situated at $75\text{ cm}$ from the eye. Calculate the focal length, in $\text{cm}$ , of the converging spectacle lens needed to enable the person to read a book comfortably at a distance of $25\text{ cm}$ from the eye.

Show answer & explanation

Answer: 37.5

Answer

The focal length of the required converging lens is

37.5\text{ cm}

.

To correct hypermetropia, the spectacle lens must form a virtual image of an object located at the desired near point (

u = +25\text{ cm}

) at the eye's actual, unassisted near point (

v = -75\text{ cm}

). Substituting

u = +25\text{ cm}

and

v = -75\text{ cm}

into the thin lens formula

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

gives

\frac{1}{f} = \frac{1}{25} - \frac{1}{75} = \frac{2}{75}\text{ cm}^{-1}

. Taking the reciprocal yields

f = 37.5\text{ cm}

.

Step-by-Step Solution

1

Determine the object distance and image distance with sign conventions

u = +25\text{ cm}

and

v = -75\text{ cm}

.

The object is placed at the normal reading distance (

25\text{ cm}

), and the spectacle lens creates a virtual image on the same side of the lens at the person's near point (

75\text{ cm}

).

2

Set up the thin lens formula

\frac{1}{f} = \frac{1}{25} - \frac{1}{75}.

The thin lens formula relates the focal length to object and image distances.

3

Calculate the focal length

f

f = 37.5\text{ cm}.

Simplifying

\frac{3 - 1}{75} = \frac{2}{75}

and taking the reciprocal yields

f = \frac{75}{2} = 37.5\text{ cm}

.

Key Concept

Correction of hypermetropia (farsightedness) using thin lens formula with virtual image sign convention

Question 230Question →

A variable force $F$ acts on a body moving along a straight horizontal path. The force increases linearly from $0\text{ N}$ at position $x = 0\text{ m}$ to $20\text{ N}$ at $x = 4\text{ m}$ , remains constant at $20\text{ N}$ from $x = 4\text{ m}$ to $x = 7\text{ m}$ , and then decreases linearly back to $0\text{ N}$ at $x = 10\text{ m}$ . What is the total work done by the force in Joules over the $10\text{ m}$ displacement?

Show answer & explanation

Answer: 130

Answer

130 J

The work done by a variable force is equal to the total area under the force-displacement (

F

-

x

) graph. The region under the graph forms a trapezoid bounded by parallel sides of length

10\text{ m}

(total displacement) and

3\text{ m}

(constant force interval), with a height of

20\text{ N}

. Calculating the area yields

W = \frac{1}{2}(10 + 3) \times 20 = 130\text{ J}

.

Step-by-Step Solution

1

Relate work done to the force-displacement graph

Work done

W

equals the total area under the

F

-

x

graph between

x = 0\text{ m}

and

x = 10\text{ m}

.

By definition,

W = \int F \, dx

, which corresponds to the geometric area under the force-displacement curve.

2

Calculate the geometric area under each section of the graph

First section (

0

to

4\text{ m}

): Triangle area =

\frac{1}{2} \times 4 \times 20 = 40\text{ J}

. Second section (

4

to

7\text{ m}

): Rectangle area =

(7 - 4) \times 20 = 60\text{ J}

. Third section (

7

to

10\text{ m}

): Triangle area =

\frac{1}{2} \times (10 - 7) \times 20 = 30\text{ J}

.

Breaking down a complex piecewise curve into simple geometric shapes allows straightforward area evaluation without calculus.

3

Sum the areas of all sections

Total Work

W = 40\text{ J} + 60\text{ J} + 30\text{ J} = 130\text{ J}

.

The total work done is the scalar sum of the work done across each contiguous segment of displacement.

Key Concept

Work Done by a Variable Force (Area under Force-Displacement Graph)

Question 231Question →

Two marksmen, Kemi and Chidi, independently shoot at a target once. The probability that Kemi hits the target is $0.7$ , and the probability that Chidi hits the target is $0.6$ . What is the probability that at least one of them hits the target?

Show answer & explanation

Answer: 0.88

Answer

The probability that at least one marksman hits the target is 0.88.

Since the two shooting attempts are independent, the probability that both marksmen miss is

(1 - 0.7) \times (1 - 0.6) = 0.3 \times 0.4 = 0.12

. Subtracting this probability from 1 yields

1 - 0.12 = 0.88

, which represents the probability that at least one marksman hits the target.

Step-by-Step Solution

1

Find the probability of each event not occurring (missing the target).

P(\text{Kemi misses}) = 0.3

and

P(\text{Chidi misses}) = 0.4

The sum of an event's probability and its complement is always 1.

2

Determine the probability of both events failing simultaneously.

P(\text{both miss}) = 0.3 \times 0.4 = 0.12

For independent events

A

and

B

,

P(A' \cap B') = P(A') \times P(B')

.

3

Use the complement rule to find the probability of at least one success.

P(\text{at least one hits}) = 1 - 0.12 = 0.88

The event 'at least one hits' is the exact complement of 'neither hits'.

Key Concept

Compound probability laws for independent events and complement of combined events

Question 232Question →

A crate on a smooth surface is simultaneously pulled by two horizontal forces. One force of magnitude $9.0\text{ N}$ acts towards the East, and another force of magnitude $12.0\text{ N}$ acts towards the North. What is the magnitude of the resultant force acting on the crate in newtons?

Show answer & explanation

Answer: 15

Answer

The magnitude of the resultant force acting on the crate is

15.0\text{ N}

.

Because the two pulling forces are perpendicular, their resultant magnitude is calculated using vector addition via the Pythagorean theorem:

R = \sqrt{9.0^2 + 12.0^2} = 15.0\text{ N}

.

Step-by-Step Solution

1

Determine the angle between the two given vectors

The forces are perpendicular (

90^\circ

).

East and North cardinal directions are orthogonal to each other.

2

Compute the resultant magnitude using vector synthesis

R = \sqrt{9.0^2 + 12.0^2} = \sqrt{81 + 144} = \sqrt{225} = 15.0\text{ N}

For perpendicular force vectors, the resultant magnitude is the hypotenuse of the right triangle formed by the vector components.

Key Concept

Vector addition of perpendicular forces using the Pythagorean theorem

Estimated Time:1m 15s

Question 233Question →

A stretched string of length $0.50\text{ m}$ fixed at both ends vibrates in its third harmonic mode at a frequency of $450\text{ Hz}$ . What is the speed of the transverse wave along the string in $\text{m/s}$ ?

Show answer & explanation

Answer: 150

Answer

The speed of the transverse wave along the string is

150\text{ m/s}

.

For a string fixed at both ends, standing wave modes produce harmonics given by

f_n = \frac{n v}{2L}

. Given

n = 3

,

L = 0.50\text{ m}

, and

f_3 = 450\text{ Hz}

, substituting these into the equation yields

450 = \frac{3v}{2(0.50)} = 3v

, leading to

v = 150\text{ m/s}

.

Step-by-Step Solution

1

Identify the standing wave frequency equation for a string fixed at both ends.

The frequency of the

n

-th harmonic is

f_n = \frac{n v}{2L}

, where

n

is the harmonic number,

v

is the wave speed, and

L

is the string length.

Fixed ends require nodes at both boundaries, producing standing wave modes with wavelengths

\lambda_n = \frac{2L}{n}

.

2

Substitute the given physical quantities into the harmonic equation.

450 = \frac{3 \times v}{2 \times 0.50}

.

The question specifies

L = 0.50\text{ m}

, third harmonic mode (

n = 3

), and frequency

f_3 = 450\text{ Hz}

.

3

Solve for the wave speed

v

.

v = 150\text{ m/s}

.

Simplifying

2 \times 0.50 = 1.0

gives

3v = 450

, so

v = \frac{450}{3} = 150\text{ m/s}

.

Key Concept

Standing Waves and Harmonics in Vibrating Strings

Question 234Question →

A thermometer calibrated on an arbitrary scale $X$ registers a lower fixed point of $-10^\circ\text{X}$ and an upper fixed point of $110^\circ\text{X}$ . What is the true temperature in degrees Celsius ( $^\circ\text{C}$ ) when this thermometer reads $20^\circ\text{X}$ ?

Show answer & explanation

Answer: 25

Answer

The true temperature on the Celsius scale is

25^\circ\text{C}

.

Using the relation

\frac{X - \text{LFP}_X}{\text{UFP}_X - \text{LFP}_X} = \frac{\theta}{100}

, substituting

X = 20

,

\text{LFP}_X = -10

, and

\text{UFP}_X = 110

gives

\frac{20 - (-10)}{110 - (-10)} = \frac{30}{120} = 0.25

. Multiplying

0.25

by

100

gives

25^\circ\text{C}

.

Step-by-Step Solution

1

Set up the linear relationship between the arbitrary temperature scale

X

and the Celsius scale

\frac{X - \text{LFP}_X}{\text{UFP}_X - \text{LFP}_X} = \frac{\theta - \text{LFP}_C}{\text{UFP}_C - \text{LFP}_C}

Thermometric properties vary linearly with temperature between fixed points.

2

Substitute the given numerical values into the formula

\frac{20 - (-10)}{110 - (-10)} = \frac{\theta - 0}{100 - 0} \implies \frac{30}{120} = \frac{\theta}{100}

The lower fixed point on scale

X

is

-10^\circ\text{X}

and the upper fixed point is

110^\circ\text{X}

.

3

Solve for the unknown temperature

\theta

in degrees Celsius

\theta = \frac{30}{120} \times 100 = 25^\circ\text{C}

Simplifying the fraction

\frac{30}{120}

yields

\frac{1}{4}

, and

\frac{1}{4} \times 100 = 25

.

Key Concept

Linear interpolation and conversion between thermometric temperature scales

Question 235Question →

The sum of the first three terms of an increasing geometric progression of positive real numbers is $21$ , and the sum of their squares is $189$ . What is the common ratio of this progression?

Show answer & explanation

Answer: 2

Answer

The common ratio of the geometric progression is 2.

Let the terms be

a

,

ar

, and

ar^2

. The given conditions yield

a(1+r+r^2) = 21

and

a^2(1+r^2+r^4) = 189

. Squaring the first equation gives

a^2(1+r+r^2)^2 = 441

. Dividing the sum of squares equation by this squared equation gives

\frac{1-r+r^2}{1+r+r^2} = \frac{189}{441} = \frac{3}{7}

. Simplifying

7(1-r+r^2) = 3(1+r+r^2)

results in

2r^2 - 5r + 2 = 0

, which factors into

(2r-1)(r-2) = 0

. Because the progression is increasing,

r > 1

, making

r = 2

the correct common ratio.

Step-by-Step Solution

1

Formulate algebraic expressions for the sum of terms and sum of squares.

a(1 + r + r^2) = 21

and

a^2(1 + r^2 + r^4) = 189

The first three terms of any geometric progression can be expressed as

a

,

ar

, and

ar^2

.

2

Eliminate the first term

a

by squaring the first equation and dividing.

\frac{a^2(1 + r^2 + r^4)}{a^2(1 + r + r^2)^2} = \frac{189}{441} \implies \frac{(1 + r + r^2)(1 - r + r^2)}{(1 + r + r^2)^2} = \frac{3}{7}

Using the algebraic factorization

1 + r^2 + r^4 = (1 + r + r^2)(1 - r + r^2)

allows cancellation of

a^2

and

(1 + r + r^2)

.

3

Solve the resulting equation for the common ratio

r

.

7(1 - r + r^2) = 3(1 + r + r^2) \implies 4r^2 - 10r + 4 = 0 \implies 2r^2 - 5r + 2 = 0

Cross-multiplying reduces the ratio to a standard quadratic equation.

4

Factor the quadratic equation and select the correct root.

(2r - 1)(r - 2) = 0 \implies r = 2 \text{ or } r = 0.5

Since the geometric progression is specified as increasing, the common ratio must be greater than 1 (

r = 2

).

Key Concept

Geometric progression term representations, sum formulas, and algebraic identity factorization.

Alternative Method

Find

a

and

r

by testing factors of

21

:

21 = 3 \times 7

, so the terms could be

3, 6, 12

(

a=3, r=2

). Check squares:

3^2 + 6^2 + 12^2 = 9 + 36 + 144 = 189

, which confirms

r = 2

.

Estimated Time:1m 30s

Question 236Question →

A transverse wave propagating through a stretched string has a frequency of $250\text{ Hz}$ and a wavelength of $1.4\text{ m}$ . What is the speed of propagation of the wave in $\text{m/s}$ ?

Show answer & explanation

Answer: 350

Answer

The speed of propagation of the wave is

350\text{ m/s}

.

The speed of propagation of a progressive wave is determined using the wave equation

v = f \lambda

. Substituting the given values

f = 250\text{ Hz}

and

\lambda = 1.4\text{ m}

gives

v = 250 \times 1.4 = 350\text{ m/s}

.

Step-by-Step Solution

1

Identify the given physical quantities

Frequency

f = 250\text{ Hz}

, Wavelength

\lambda = 1.4\text{ m}

.

These parameters are required to calculate the speed of propagation.

2

Apply the fundamental wave equation

v = f \lambda

The speed of a progressive wave equals the product of its frequency and its wavelength.

3

Perform the numerical calculation

v = 250 \times 1.4 = 350\text{ m/s}

Multiplying

250\text{ Hz}

by

1.4\text{ m}

yields the wave velocity in meters per second.

Key Concept

Wave Propagation Speed Equation (

v = f \lambda

)

Question 237Question →

A uniform metre rule of mass $120\text{ g}$ is balanced horizontally on a pivot placed at the $40\text{ cm}$ mark when an unknown mass $m$ is suspended at the $10\text{ cm}$ mark. What is the value of the mass $m$ in grams?

Show answer & explanation

Answer: 40

Answer

The mass

m

required to balance the metre rule horizontally is

40\text{ g}

.

According to the Principle of Moments, a system is in rotational equilibrium when the sum of anticlockwise moments equals the sum of clockwise moments about the pivot. The weight of the

120\text{ g}

uniform metre rule acts at its center of gravity (

50\text{ cm}

mark), which is

10\text{ cm}

to the right of the pivot at

40\text{ cm}

. This creates a clockwise moment of

120\text{ g} \times 10\text{ cm} = 1200\text{ g}\cdot\text{cm}

. The mass

m

is suspended at the

10\text{ cm}

mark, which is

30\text{ cm}

to the left of the pivot, creating an anticlockwise moment of

m \times 30\text{ cm}

. Setting

30m = 1200

yields

m = 40\text{ g}

.

Step-by-Step Solution

1

Determine the position of the center of gravity of the metre rule.

Center of gravity is at the

50\text{ cm}

mark.

A uniform metre rule has its weight concentrated at its geometric midpoint.

2

Calculate perpendicular distances from the pivot at

40\text{ cm}

to all acting forces.

Distance to rule weight =

50\text{ cm} - 40\text{ cm} = 10\text{ cm}

; Distance to mass

m

=

40\text{ cm} - 10\text{ cm} = 30\text{ cm}

.

Moments are calculated by multiplying force (or mass) by perpendicular distance from the turning point.

3

Apply the Principle of Moments for equilibrium.

Sum of anticlockwise moments = Sum of clockwise moments

\implies m \times 30 = 120 \times 10

.

For rotational equilibrium, the total clockwise moment must equal the total anticlockwise moment about the pivot.

4

Solve for the unknown mass

m

.

m = 40\text{ g}

.

Dividing

1200\text{ g}\cdot\text{cm}

by

30\text{ cm}

gives

40\text{ g}

.

Key Concept

Principle of Moments and Center of Gravity of a Uniform Body

Estimated Time:50s

Question 238Question →

A coil consisting of $150$ turns is placed in a magnetic field. The magnetic flux passing through the coil decreases uniformly from $4.0 \times 10^{-3}\text{ Wb}$ to $1.0 \times 10^{-3}\text{ Wb}$ over a time interval of $0.015\text{ s}$ . What is the magnitude of the average induced electromotive force in the coil in volts?

Show answer & explanation

Answer: 30

Answer

The magnitude of the average induced electromotive force in the coil is

30\text{ V}

.

According to Faraday's law of electromagnetic induction, the magnitude of the average induced electromotive force in a coil with

N

turns is given by

E = N \frac{\Delta \Phi}{\Delta t}

. Given

N = 150

,

\Delta \Phi = 3.0 \times 10^{-3}\text{ Wb}

, and

\Delta t = 0.015\text{ s}

, the magnitude of the induced e.m.f. is

E = 150 \times \frac{3.0 \times 10^{-3}}{0.015} = 30\text{ V}

.

Step-by-Step Solution

1

Calculate the magnitude of the change in magnetic flux through the coil

\Delta \Phi = 4.0 \times 10^{-3}\text{ Wb} - 1.0 \times 10^{-3}\text{ Wb} = 3.0 \times 10^{-3}\text{ Wb}

Faraday's law relates induced e.m.f. directly to the rate of change of magnetic flux.

2

Apply Faraday's Law of Electromagnetic Induction equation for an N-turn coil

E = N \frac{\Delta \Phi}{\Delta t}

The total induced electromotive force in a coil is proportional to the number of turns and the rate of change of flux.

3

Substitute the given numerical values to compute the magnitude of the induced e.m.f.

E = 150 \times \frac{3.0 \times 10^{-3}\text{ Wb}}{0.015\text{ s}} = 150 \times 0.2\text{ V} = 30\text{ V}

Performing clean calculation without needing a calculator.

Key Concept

Faraday's Law of Electromagnetic Induction

Question 239Question →

An atom in a gas discharge tube has a ground state energy level of $-10.4\text{ eV}$ and an excited energy level of $-3.8\text{ eV}$ . An electron absorbs a single photon to undergo a transition directly from the ground state to this excited level. If the frequency of the absorbed photon is expressed as $x \times 10^{15}\text{ Hz}$ , what is the numerical value of $x$ ? (Take Planck's constant $h = 6.6 \times 10^{-34}\text{ J}\cdot\text{s}$ and $1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$ )

Show answer & explanation

Answer: 1.6

Answer

The numerical value of

x

is 1.6.

The energy of the absorbed photon is equal to the difference between the excited state and ground state energies:

\Delta E = -3.8\text{ eV} - (-10.4\text{ eV}) = 6.6\text{ eV}

. Converting this to Joules yields

6.6 \times 1.6 \times 10^{-19}\text{ J} = 1.056 \times 10^{-18}\text{ J}

. Using Einstein's photon relation

E = hf

, the frequency is

f = \frac{1.056 \times 10^{-18}}{6.6 \times 10^{-34}} = 1.6 \times 10^{15}\text{ Hz}

, which gives

x = 1.6

.

Step-by-Step Solution

1

Determine the energy absorbed during transition

\Delta E = 6.6\text{ eV}

The energy of the absorbed photon equals the energy difference between the initial and final states:

\Delta E = -3.8\text{ eV} - (-10.4\text{ eV}) = 6.6\text{ eV}

.

2

Convert energy from eV to Joules

\Delta E = 1.056 \times 10^{-18}\text{ J}

Multiply by

1.6 \times 10^{-19}\text{ J/eV}

to convert energy to standard SI units.

3

Calculate photon frequency

f = 1.6 \times 10^{15}\text{ Hz}

Use Planck's energy equation

f = \frac{\Delta E}{h} = \frac{1.056 \times 10^{-18}\text{ J}}{6.6 \times 10^{-34}\text{ J}\cdot\text{s}} = 1.6 \times 10^{15}\text{ Hz}

.

Key Concept

Photon energy and atomic transition frequency

Estimated Time:2m 0s

Question 240Question →

An electric immersion heater rated at $200\text{ W}$ is used to heat $0.8\text{ kg}$ of a liquid contained in a vessel of heat capacity $120\text{ J K}^{-1}$ . The initial temperature of the liquid and vessel is $20^\circ\text{C}$ . If the heater is operated for $3.5\text{ minutes}$ and the final temperature reaches $50^\circ\text{C}$ , calculate the specific heat capacity of the liquid in $\text{J kg}^{-1}\text{ K}^{-1}$ , assuming no thermal energy is lost to the surroundings.

Show answer & explanation

Answer: 1600

Answer

The specific heat capacity of the liquid is

1600\text{ J kg}^{-1}\text{ K}^{-1}

.

Total energy supplied by the heater is

Q = 200\text{ W} \times 210\text{ s} = 42,000\text{ J}

. The temperature increase is

\Delta T = 30\text{ K}

. The energy absorbed by the container is

Q_{\text{vessel}} = C \Delta T = 120 \times 30 = 3,600\text{ J}

. The remaining energy

38,400\text{ J}

is absorbed by the liquid. Dividing this value by the product of the mass of liquid (

0.8\text{ kg}

) and temperature rise (

30\text{ K}

) gives the specific heat capacity

1600\text{ J kg}^{-1}\text{ K}^{-1}

.

Step-by-Step Solution

1

Convert the heating time to seconds and compute total heat energy supplied by the heater.

Q = P \times t = 200\text{ W} \times (3.5 \times 60\text{ s}) = 42,000\text{ J}

Heat energy supplied by an electric source equals electrical power multiplied by duration in seconds.

2

Determine the change in temperature of the system.

\Delta T = 50^\circ\text{C} - 20^\circ\text{C} = 30\text{ K}

Both the vessel and liquid experience the same initial and final temperatures.

3

Calculate the heat energy absorbed by the vessel.

Q_{\text{vessel}} = C \times \Delta T = 120\text{ J K}^{-1} \times 30\text{ K} = 3,600\text{ J}

Heat capacity

C

represents heat required per unit temperature rise for the container as a whole.

4

Subtract vessel absorption from total heat supplied to find heat absorbed by the liquid.

Q_{\text{liquid}} = 42,000\text{ J} - 3,600\text{ J} = 38,400\text{ J}

Conservation of energy dictates

Q_{\text{total}} = Q_{\text{vessel}} + Q_{\text{liquid}}

.

5

Calculate the specific heat capacity

c

of the liquid.

c = \frac{Q_{\text{liquid}}}{m \Delta T} = \frac{38,400}{0.8 \times 30} = 1600\text{ J kg}^{-1}\text{ K}^{-1}

Specific heat capacity isolates heat absorbed per unit mass per Kelvin.

Key Concept

Principle of conservation of thermal energy combining heat capacity of a vessel (

C

) and specific heat capacity of a liquid (

c

).

Estimated Time:1m 30s

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