All practice questions

Question 7961Question →

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 7962Question →

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 7963Question →

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 7964Question →

A cart of mass $40\text{ kg}$ carrying a package of mass $10\text{ kg}$ is coasting along a straight horizontal track at a constant velocity of $6.0\text{ m s}^{-1}$ . The package is suddenly ejected horizontally backward (opposite to the direction of motion of the cart) at a speed of $15\text{ m s}^{-1}$ relative to the ground. What is the new velocity of the cart after the package is ejected?

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

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

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

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

Show answer & explanation

Answer: $11.25\text{ m s}^{-1}$

Answer

The new velocity of the cart is

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

in the forward direction.

By the law of conservation of linear momentum, the total initial momentum of the system (cart plus package,

50\text{ kg}

moving at

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

) equals

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

. When the

10\text{ kg}

package is ejected backward at

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

, its momentum is

-150\text{ kg m s}^{-1}

. Setting

300 = -150 + 40 v_{\text{cart}}

yields

v_{\text{cart}} = 11.25\text{ m s}^{-1}

.

Step-by-Step Solution

1

Calculate the total initial momentum of the system before ejection.

Total mass

m_{\text{total}} = 40\text{ kg} + 10\text{ kg} = 50\text{ kg}

. Initial momentum

P_i = 50\text{ kg} \times 6.0\text{ m s}^{-1} = 300\text{ kg m s}^{-1}

.

The initial system consists of both the cart and the package moving together at

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

.

2

Set up the expression for final momentum considering direction.

Taking the forward direction as positive, the package's velocity is

v_{\text{pkg}} = -15\text{ m s}^{-1}

. Final momentum

P_f = (10 \times -15) + (40 \times v_{\text{cart}}) = -150 + 40 v_{\text{cart}}

.

Linear momentum is a vector quantity, so opposite motion must be assigned a negative sign.

3

Apply the law of conservation of linear momentum (

P_i = P_f

) to solve for the cart's final velocity.

300 = -150 + 40 v_{\text{cart}} \implies 450 = 40 v_{\text{cart}} \implies v_{\text{cart}} = 11.25\text{ m s}^{-1}

.

In the absence of external forces on the system, total linear momentum is conserved.

Key Concept

Conservation of Linear Momentum

Estimated Time:2m 0s

Question 7965Question →

A stone is projected horizontally with a speed of $15\text{ m/s}$ from the top of a vertical cliff of height $20\text{ m}$ . Taking acceleration due to gravity $g = 10\text{ m/s}^2$ , what is the magnitude of the velocity of the stone just before it strikes the ground?

15\text{ m/s}

20\text{ m/s}

25\text{ m/s}

35\text{ m/s}

Show answer & explanation

Answer: $25\text{ m/s}$

Answer

The magnitude of the velocity of the stone just before hitting the ground is

25\text{ m/s}

.

For a horizontally launched projectile, the horizontal velocity component remains constant at

v_x = 15\text{ m/s}

. The vertical component just before impact is found using

v_y^2 = u_y^2 + 2gh = 0 + 2(10)(20) = 400

, giving

v_y = 20\text{ m/s}

. Combining these perpendicular components yields a total speed of

v = \sqrt{15^2 + 20^2} = 25\text{ m/s}

.

Step-by-Step Solution

1

Identify horizontal velocity component

v_x = 15\text{ m/s}

Air resistance is neglected, so horizontal velocity remains constant throughout flight.

2

Calculate vertical velocity component just before impact using third equation of motion

v_y^2 = u_y^2 + 2gh = 0^2 + 2(10)(20) = 400 \implies v_y = 20\text{ m/s}

Initial vertical velocity

u_y = 0\text{ m/s}

for horizontal projection; stone falls through a vertical displacement of

20\text{ m}

under gravity.

3

Compute total resultant velocity magnitude using Pythagorean theorem

v = \sqrt{v_x^2 + v_y^2} = \sqrt{15^2 + 20^2} = \sqrt{225 + 400} = \sqrt{625} = 25\text{ m/s}

Horizontal and vertical velocity components are mutually perpendicular.

Key Concept

Horizontal Projection and Resultant Velocity Vector Synthesis

Estimated Time:2m 0s

Question 7966Question →

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 7967Question →

A sector of a circle of radius $21\text{ cm}$ subtends an angle of $60^\circ$ at the centre of the circle. What is the total perimeter of the sector? (Take $\pi = \frac{22}{7}$ )

22\text{ cm}

43\text{ cm}

64\text{ cm}

231\text{ cm}

Show answer & explanation

Answer: $64\text{ cm}$

Answer

The total perimeter of the sector is

64\text{ cm}

.

The perimeter of a sector consists of its curved arc length plus its two bounding radii (

2r

). Using

\theta = 60^\circ

,

r = 21\text{ cm}

, and

\pi = \frac{22}{7}

, the arc length is

\frac{60}{360} \times 2 \times \frac{22}{7} \times 21 = 22\text{ cm}

. Adding

2 \times 21\text{ cm} = 42\text{ cm}

yields

64\text{ cm}

.

Step-by-Step Solution

1

Calculate the arc length (

s

) of the sector using the formula

s = \frac{\theta}{360^\circ} \times 2\pi r

.

s = \frac{60^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 21 = \frac{1}{6} \times 132 = 22\text{ cm}

.

The arc length represents the curved boundary of the sector.

2

Calculate the total perimeter of the sector by adding the arc length and the two bounding radii:

P = s + 2r

.

P = 22\text{ cm} + 2(21\text{ cm}) = 22 + 42 = 64\text{ cm}

.

A sector boundary consists of the curved arc plus the two straight radial edges.

Key Concept

Perimeter of a circular sector

Estimated Time:1m 30s

Question 7968Question →

A clean metal surface with threshold frequency $f_0$ is illuminated by monochromatic radiation of frequency $2f_0$ and light intensity $I$ , causing emission of photoelectrons with maximum kinetic energy $K_1$ and stopping potential $V_1$ . If the source is changed to radiate light of frequency $3f_0$ while its intensity is simultaneously doubled to $2I$ , what are the new maximum kinetic energy $K_2$ and stopping potential $V_2$ in terms of $K_1$ and $V_1$ ?

K_2 = 2K_1

and

V_2 = 2V_1

K_2 = 4K_1

and

V_2 = 4V_1

K_2 = 4K_1

and

V_2 = 2V_1

K_2 = 3K_1

and

V_2 = 3V_1

Show answer & explanation

Answer: $K_2 = 2K_1$ and $V_2 = 2V_1$

Answer

K_2 = 2K_1

and

V_2 = 2V_1

By Einstein's photoelectric law,

K_{\max} = hf - W_0

. For incident frequency

2f_0

and work function

W_0 = hf_0

, the initial kinetic energy is

K_1 = 2hf_0 - hf_0 = hf_0

. For incident frequency

3f_0

, the new kinetic energy is

K_2 = 3hf_0 - hf_0 = 2hf_0 = 2K_1

. Since stopping potential is related by

eV_s = K_{\max}

,

V_2 = 2V_1

. Doubling the light intensity from

I

to

2I

doubles the rate of photoelectron emission but does not alter maximum kinetic energy or stopping potential.

Step-by-Step Solution

1

Apply Einstein's photoelectric equation to the initial condition.

K_1 = h(2f_0) - hf_0 = hf_0

, and

e V_1 = K_1 = hf_0

.

The maximum kinetic energy is the energy of the incident photon minus the work function of the metal plate.

2

Apply Einstein's photoelectric equation to the new frequency condition.

K_2 = h(3f_0) - hf_0 = 2hf_0 = 2K_1

, and

e V_2 = K_2 = 2hf_0 = 2e V_1 \implies V_2 = 2V_1

.

The new photon energy is

3hf_0

, leaving

2hf_0

of excess energy as photoelectron maximum kinetic energy.

3

Evaluate the effect of doubling light intensity to

2I

.

Light intensity has zero impact on individual photoelectron kinetic energy or stopping potential; it only increases the emission rate (saturation photocurrent).

Photoelectric emission is a one-to-one photon-electron interaction process where photon frequency dictates individual energy.

Key Concept

Independence of photoelectron kinetic energy and stopping potential from light intensity

Question 7969Question →

A spring balance calibrated in newtons has a zero error of $+2.0\text{ N}$ (it displays $+2.0\text{ N}$ before any load is attached). When an object is suspended from the balance in a location where the acceleration due to gravity is $10\text{ m s}^{-2}$ , the scale displays a reading of $42.0\text{ N}$ . What is the true mass of the object?

4.2\text{ kg}

4.0\text{ kg}

4.4\text{ kg}

40.0\text{ kg}

Show answer & explanation

Answer: $4.0\text{ kg}$

Answer

4.0\text{ kg}

To obtain the true weight from a spring balance with a positive zero error, the zero offset must be subtracted from the indicated reading:

\text{True Weight} = 42.0\text{ N} - 2.0\text{ N} = 40.0\text{ N}

. Dividing this true weight by the gravitational acceleration (

10\text{ m s}^{-2}

) gives the true mass of

4.0\text{ kg}

.

Step-by-Step Solution

1

Correct the scale reading for instrument zero error to obtain true weight

True Weight

W = 42.0\text{ N} - 2.0\text{ N} = 40.0\text{ N}

A positive zero error means the balance overcounts force, so the initial reading must be subtracted from the scale display.

2

Calculate the true mass using the formula

W = mg

Mass

m = \frac{W}{g} = \frac{40.0\text{ N}}{10\text{ m s}^{-2}} = 4.0\text{ kg}

Mass is found by dividing the true force of gravity (weight) by the acceleration due to gravity.

Key Concept

Instrument zero error correction and mass-weight relationship

Estimated Time:1m 0s

Question 7970Question →

Two fixed points in a Cartesian plane are given as $A(-1, 2)$ and $B(3, 4)$ . A point $P(x, y)$ moves in the plane such that $PA^2 + PB^2 = 26$ . Which of the following equations represents the locus of $P$ ?

x^2 + y^2 - 2x - 6y + 2 = 0

x^2 + y^2 + 2x + 6y + 2 = 0

x^2 + y^2 - 4x - 12y + 4 = 0

x^2 + y^2 - 2x - 6y + 30 = 0

Show answer & explanation

Answer: $x^2 + y^2 - 2x - 6y + 2 = 0$

Answer

The equation representing the locus of

P

is

x^2 + y^2 - 2x - 6y + 2 = 0

.

Using the Cartesian coordinate distance formula,

PA^2 = (x+1)^2 + (y-2)^2

and

PB^2 = (x-3)^2 + (y-4)^2

. Adding these together yields

2x^2 + 2y^2 - 4x - 12y + 30

. Setting this equal to 26 gives

2x^2 + 2y^2 - 4x - 12y + 4 = 0

. Dividing the entire equation by 2 yields

x^2 + y^2 - 2x - 6y + 2 = 0

, which is the correct locus equation.

Step-by-Step Solution

1

Express

PA^2

using the distance formula between

P(x, y)

and

A(-1, 2)

.

PA^2 = (x - (-1))^2 + (y - 2)^2 = (x + 1)^2 + (y - 2)^2 = x^2 + 2x + 1 + y^2 - 4y + 4 = x^2 + y^2 + 2x - 4y + 5

The square of the distance between two points

(x_1, y_1)

and

(x_2, y_2)

is

(x_2 - x_1)^2 + (y_2 - y_1)^2

.

2

Express

PB^2

using the distance formula between

P(x, y)

and

B(3, 4)

.

PB^2 = (x - 3)^2 + (y - 4)^2 = x^2 - 6x + 9 + y^2 - 8y + 16 = x^2 + y^2 - 6x - 8y + 25

Expanding the distance squared formula for point

B

.

3

Sum

PA^2

and

PB^2

and equate to the given constant 26.

(x^2 + y^2 + 2x - 4y + 5) + (x^2 + y^2 - 6x - 8y + 25) = 26 \implies 2x^2 + 2y^2 - 4x - 12y + 30 = 26

Substitute the algebraic expressions into the locus condition

PA^2 + PB^2 = 26

.

4

Rearrange into general circle form and divide through by 2.

2x^2 + 2y^2 - 4x - 12y + 4 = 0 \implies x^2 + y^2 - 2x - 6y + 2 = 0

Subtract 26 from both sides and divide all terms by 2 to obtain the standard Cartesian equation.

Key Concept

Locus of a point with sum of squared distances to two fixed points equal to a constant

Question 7971Question →

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 7972Question →

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 7973Question →

A rigid solid object with negligible thermal expansivity is completely submerged in a container of water initially at $10^\circ\text{C}$ . The water is then uniformly cooled down to $0^\circ\text{C}$ . How does the upthrust (buoyant force) exerted by the water on the submerged object vary during this cooling process?

It increases to a maximum value at

4^\circ\text{C}

and then decreases as the temperature falls to

0^\circ\text{C}

.

It decreases continuously from

10^\circ\text{C}

down to

0^\circ\text{C}

as the liquid cools.

It remains constant throughout the cooling process because the volume of the submerged object does not change.

It decreases to a minimum value at

4^\circ\text{C}

and then increases as the temperature reaches

0^\circ\text{C}

.

Show answer & explanation

Answer: It increases to a maximum value at $4^\circ\text{C}$ and then decreases as the temperature falls to $0^\circ\text{C}$ .

Answer

The upthrust increases to a maximum value at

4^\circ\text{C}

and then decreases as cooling continues to

0^\circ\text{C}

.

According to Archimedes' principle, upthrust is given by

U = \rho_{\text{fluid}} V g

. For a submerged object of constant volume, upthrust is directly proportional to the density of water. As water cools from

10^\circ\text{C}

to

4^\circ\text{C}

, it contracts normally, increasing its density to a maximum at

4^\circ\text{C}

(

1000\text{ kg m}^{-3}

). Upon further cooling from

4^\circ\text{C}

to

0^\circ\text{C}

, water undergoes anomalous expansion, increasing in volume and decreasing in density. Consequently, the upthrust reaches a maximum at

4^\circ\text{C}

and decreases down to

0^\circ\text{C}

.

Step-by-Step Solution

1

Identify the expression for upthrust (buoyant force)

U = \rho_{\text{water}} \cdot V_{\text{object}} \cdot g

By Archimedes' principle, upthrust depends directly on the density of the displaced liquid

\rho_{\text{water}}

when the volume

V_{\text{object}}

and gravitational acceleration

g

are constant.

2

Analyze the density behavior of water from

10^\circ\text{C}

down to

4^\circ\text{C}

Density increases, reaching its maximum value

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

at

4^\circ\text{C}

.

Between

10^\circ\text{C}

and

4^\circ\text{C}

, water contracts normally upon cooling.

3

Analyze the density behavior of water from

4^\circ\text{C}

down to

0^\circ\text{C}

Density decreases as temperature drops from

4^\circ\text{C}

to

0^\circ\text{C}

.

Water exhibits anomalous expansion between

4^\circ\text{C}

and

0^\circ\text{C}

, expanding in volume and consequently decreasing in density.

4

Relate density variation directly to upthrust variation

Upthrust increases from

10^\circ\text{C}

up to a peak at

4^\circ\text{C}

, then decreases from

4^\circ\text{C}

to

0^\circ\text{C}

.

Since upthrust is directly proportional to density, its value mirrors the density profile of water.

Key Concept

Anomalous expansion of water and its effect on density and upthrust

Question 7974Question →

A ray of light strikes a plane mirror at an angle of incidence of $35^\circ$ . What is the angle of deviation of the reflected ray?

110^\circ

70^\circ

55^\circ

145^\circ

Show answer & explanation

Answer: $110^\circ$

Answer

The angle of deviation of the reflected ray is

110^\circ

.

The angle of deviation

d

represents the angle through which a ray of light is turned from its original path. For a plane mirror,

d = 180^\circ - 2i

. Substituting

i = 35^\circ

yields

d = 180^\circ - 70^\circ = 110^\circ

.

Step-by-Step Solution

1

Identify the given angle of incidence

i = 35^\circ

The angle of incidence is measured relative to the normal line.

2

Apply the law of reflection

Angle of reflection

r = i = 35^\circ

The angle of reflection equals the angle of incidence.

3

Calculate the angle of deviation

d = 180^\circ - (i + r) = 180^\circ - 2(35^\circ) = 110^\circ

The angle of deviation measures how much the light ray is turned from its original initial straight path.

Key Concept

Angle of deviation for reflection at a plane surface

Estimated Time:45s

Question 7975Question →

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

Question 7976Question →

An X-ray tube operates at an accelerating potential difference of $20\text{ kV}$ . What is the maximum energy of the produced X-ray photons in Joules? (Take elementary charge $e = 1.6 \times 10^{-19}\text{ C}$ )

3.2 \times 10^{-15}\text{ J}

3.2 \times 10^{-18}\text{ J}

1.25 \times 10^{23}\text{ J}

2.0 \times 10^{4}\text{ J}

Show answer & explanation

Answer: $3.2 \times 10^{-15}\text{ J}$

Answer

3.2 \times 10^{-15}\text{ J}

According to the Duane-Hunt law, the maximum energy of an emitted X-ray photon equals the maximum kinetic energy gained by an electron accelerated through potential difference

V

, which is

E_{\text{max}} = e V

. Substituting

e = 1.6 \times 10^{-19}\text{ C}

and

V = 20,000\text{ V}

yields

3.2 \times 10^{-15}\text{ J}

.

Step-by-Step Solution

1

Convert the accelerating potential difference from kilovolts (kV) to volts (V).

V = 20\text{ kV} = 20,000\text{ V} = 2.0 \times 10^{4}\text{ V}

Standard SI units require potential difference in volts.

2

Apply the Duane-Hunt relationship for maximum photon energy

E_{\text{max}} = e V

.

E_{\text{max}} = 1.6 \times 10^{-19}\text{ C} \times 2.0 \times 10^{4}\text{ V} = 3.2 \times 10^{-15}\text{ J}

The kinetic energy acquired by accelerated electrons is completely converted into the maximum energy of an emitted X-ray photon.

Key Concept

Duane-Hunt Law and Maximum X-ray Photon Energy

Estimated Time:45s

Question 7977Question →

A international conference delegation of $6$ members is to be selected from $5$ diplomats and $4$ translators. In how many distinct ways can the delegation be formed if it must include at least $4$ diplomats?

Show answer & explanation

Answer: 34

Answer

The total number of distinct ways to form the delegation is 34.

To form a delegation of 6 with at least 4 diplomats out of 5 diplomats and 4 translators, we evaluate two mutually exclusive cases: selecting 4 diplomats and 2 translators (5C4 * 4C2 = 30 ways) and selecting 5 diplomats and 1 translator (5C5 * 4C1 = 4 ways). Summing these gives 34 distinct ways.

Step-by-Step Solution

1

Determine the valid combinations of diplomats and translators.

Two valid cases exist: (4 diplomats, 2 translators) and (5 diplomats, 1 translator).

The total delegation size is 6 and it must contain at least 4 diplomats out of the 5 available.

2

Compute the selection ways for Case 1 (4 diplomats and 2 translators).

5 * 6 = 30 ways

Selecting 4 diplomats out of 5 is 5C4 = 5 ways, and selecting 2 translators out of 4 is 4C2 = 6 ways.

3

Compute the selection ways for Case 2 (5 diplomats and 1 translator).

1 * 4 = 4 ways

Selecting 5 diplomats out of 5 is 5C5 = 1 way, and selecting 1 translator out of 4 is 4C1 = 4 ways.

4

Sum the combinations from both mutually exclusive cases.

30 + 4 = 34 ways

By the addition principle of counting, mutually exclusive scenarios are added together.

Key Concept

Combinations with constraints and mutually exclusive cases

Question 7978Question →

A wooden block of mass $4.0\text{ kg}$ is suspended vertically at rest. A bullet of mass $0.05\text{ kg}$ travelling horizontally at $400\text{ m s}^{-1}$ strikes the block, passes completely through it, and emerges on the opposite side with a reduced speed of $100\text{ m s}^{-1}$ . If a constant retarding force brings the moving block to rest in $0.25\text{ s}$ after the bullet emerges, calculate the magnitude of this retarding force in newtons.

Show answer & explanation

Answer: 60

Answer

The magnitude of the retarding force acting on the block is

60\text{ N}

.

During the impact, the bullet loses momentum equal to

\Delta p = 0.05\text{ kg} \times (400\text{ m s}^{-1} - 100\text{ m s}^{-1}) = 15\text{ N s}

. By the conservation of linear momentum, this exact amount of momentum is gained by the block. Applying Newton's second law (

F = \frac{\Delta p}{\Delta t}

), the magnitude of the constant retarding force needed to reduce the block's momentum to zero in

0.25\text{ s}

is

F = \frac{15\text{ N s}}{0.25\text{ s}} = 60\text{ N}

.

Step-by-Step Solution

1

Calculate the momentum lost by the bullet during penetration.

\Delta p_{\text{bullet}} = 0.05\text{ kg} \times (400\text{ m s}^{-1} - 100\text{ m s}^{-1}) = 15\text{ N s}

The momentum lost by the bullet equals its mass multiplied by the change in its horizontal velocity vector.

2

Determine the initial momentum imparted to the wooden block using the law of conservation of linear momentum.

p_{\text{block}} = \Delta p_{\text{bullet}} = 15\text{ N s}

Since no external horizontal force acts during the collision impact, the momentum lost by the bullet is fully transferred to the block.

3

Calculate the retarding force required to bring the block to rest using the impulse-momentum theorem.

F = \frac{\Delta p_{\text{block}}}{\Delta t} = \frac{15\text{ N s}}{0.25\text{ s}} = 60\text{ N}

According to Newton's second law of motion, the net force acting on a body equals the rate of change of momentum.

Key Concept

Conservation of Linear Momentum and Newton's Second Law

Question 7979Question →

Complete the statement below regarding the chemical reaction between ammonia and boron trifluoride.

Fill in the blanks below

In the formation of the adduct between ammonia (

NH_3

) and boron trifluoride (

BF_3

), the nitrogen atom contributes both electrons to form the chemical bond with boron. This type of linkage is known as a bond, and nitrogen functions as the electron pair .

Show answer & explanation

Answer

Blank 1 should be filled with 'dative covalent' (or 'coordinate covalent'), and Blank 2 should be filled with 'donor' (or 'Lewis base').

When

NH_3

reacts with

BF_3

, nitrogen donates its unshared lone pair of electrons into the empty p-orbital of the electron-deficient boron atom. A covalent bond where one atom contributes both bonding electrons is a dative (coordinate covalent) bond, and the species donating the electron pair is the donor (or Lewis base).

Step-by-Step Solution

1

Examine the valence electron configurations of the reacting species.

Ammonia (

NH_3

) has a non-bonding lone pair on the nitrogen atom, whereas boron trifluoride (

BF_3

) has an incomplete octet with only six valence electrons around the central boron atom.

Determining the presence of unshared pairs and electron deficiency identifies the roles of each atom.

2

Identify the nature of the bond formed between nitrogen and boron.

Nitrogen shares its unshared lone pair to complete the octet of the boron atom, forming a bond where both shared electrons originate from nitrogen.

A covalent bond in which one atom provides both shared electrons is defined as a dative or coordinate covalent bond.

3

Assign the appropriate terminology to the electron-donating species.

The atom or molecule supplying the shared electron pair is termed the electron pair donor or Lewis base.

By definition in Lewis acid-base theory, the species donating the electron pair is the donor.

Key Concept

Dative (Coordinate) Covalent Bonding and Lewis Acid-Base Theory

Question 7980Question →

A micrometer screw gauge with a pitch of $0.5\text{ mm}$ and $50$ circular scale divisions is used to measure the diameter of a uniform brass sphere. When the anvil and spindle are brought into contact without the sphere, the $45\text{th}$ division on the thimble scale aligns with the main scale index line. When the sphere is clamped between the anvil and spindle, the main scale reads $3.5\text{ mm}$ and the $28\text{th}$ division on the thimble scale coincides with the index line. What is the actual diameter of the brass sphere?

3.83\text{ mm}

3.78\text{ mm}

3.73\text{ mm}

3.33\text{ mm}

Show answer & explanation

Answer: $3.83\text{ mm}$

Answer

The actual diameter of the brass sphere is

3.83\text{ mm}

.

The correct answer of

3.83\text{ mm}

is derived by first establishing the least count (

0.01\text{ mm}

). When the jaws are closed, the

45\text{th}

mark lies below the reference line, giving a negative zero error of

-0.05\text{ mm}

. Adding the main scale (

3.5\text{ mm}

) to the thimble reading (

0.28\text{ mm}

) gives an observed value of

3.78\text{ mm}

. Subtracting the negative zero error yields

3.78\text{ mm} - (-0.05\text{ mm}) = 3.83\text{ mm}

.

Step-by-Step Solution

1

Determine the least count (precision) of the micrometer screw gauge.

\text{Least Count (LC)} = \frac{\text{Pitch}}{\text{Number of circular scale divisions}} = \frac{0.5\text{ mm}}{50} = 0.01\text{ mm}

.

Least count defines the minimum measurement value represented by one thimble scale division.

2

Calculate the zero error of the instrument.

Since the

45\text{th}

division is aligned when closed, it is

5

divisions below the zero line (

45 - 50 = -5

). Thus,

\text{Zero Error (ZE)} = -5 \times 0.01\text{ mm} = -0.05\text{ mm}

.

When the zero mark on the thimble lies below the index line, the instrument has a negative zero error.

3

Calculate the observed reading of the sphere.

\text{Observed Reading (OR)} = 3.5\text{ mm} + (28 \times 0.01\text{ mm}) = 3.5\text{ mm} + 0.28\text{ mm} = 3.78\text{ mm}.

The total observed reading combines the main scale reading and the circular thimble reading.

4

Apply the zero error correction to find the true diameter.

\text{True Diameter} = \text{Observed Reading} - \text{Zero Error} = 3.78\text{ mm} - (-0.05\text{ mm}) = 3.78\text{ mm} + 0.05\text{ mm} = 3.83\text{ mm}.

True measurement is always obtained by subtracting the zero error (including its sign) from the observed reading.

Key Concept

Negative zero error correction in micrometer screw gauge measurements

Estimated Time:2m 0s

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