myclass24
myclass24your class. your pace.
Q&A BANK

every question.
answered.

32 answers across CBSE, ICSE, and State boards — Class 1 to 12, every subject.

CBSEICSEState
SUBJECT
CLASS

Filtered results

32 TOTAL
PhysicsClass 12CBSE

Bullet fires with velocity u; after 24 cm velocity = u/3. Then comes to rest at end of block. Find total length.

Step-1: Find retardation (v² = u² − 2as):(u/3)² = u² − 2a(0.24)u²/9 = u² − 0.48a0.48a = u²(1 − 1/9) = 8u²/9 ⇒ a = 50u²/27Step-2: Additional distance to stop from v = u/3:0 = (u/3)² − 2a·s₂s₂ = (u²/9) / (2 × 50u²/27) = (u²/9) × (27/100u²) = 0.03 m = 3 cmStep-3: Total length:L = 24 cm + 3 cm = 27 cmAnswer: Total length of the block = 27 cm

PhysicsClass 11CBSE

λ₁ = 350 nm, λ₂ = 450 nm; maximum velocities differ by factor 2 (v₁ = 2v₂). Find work function φ.

Einstein’s equations:hc/λ₁ = φ + ½mᵉv₁² ...(1)hc/λ₂ = φ + ½mᵉv₂² ...(2)Substitute v₁ = 2v₂ ⇒ v₁² = 4v₂² and subtract (2) from (1):hc(1/λ₁ − 1/λ₂) = (3/2)mᵉv₂²Calculate left side:1/350nm − 1/450nm = 100/(350×450) nm⁻¹ = 6.349×10⁵ m⁻¹hc × 6.349×10⁵ = 1.262×10⁻¹⁹ JSolve for v₂²:v₂² = (2/3) × 1.262×10⁻¹⁹ / 9.1×10⁻³¹ = 9.245×10¹⁰ m²/s²Substitute into equation (2):hc/λ₂ = (6.626×10⁻³⁴×3×10⁸) / (450×10⁻⁹) = 4.417×10⁻¹⁹ J½mᵉv₂² = 0.5 × 9.1×10⁻³¹ × 9.245×10¹⁰ = 4.206×10⁻²⁰ Jφ = 4.417×10⁻¹⁹ − 0.4206×10⁻¹⁹ ≈ 3.996×10⁻¹⁹ J ≈ 2.5 eVAnswer: Work function φ ≈ 2.5 eV

PhysicsClass 11CBSE

Charges −q, +q, −q at corners of equilateral triangle (side a). Find net force on +q at centroid O.

Distance from centroid to vertex: r = a/√3Force magnitude from each corner on +q at O:F = kq²/r² = kq²/(a²/3) = 3kq²/a²Directions (place +q corner at top, −q corners at bottom-left and bottom-right):From +q at top: repels downward (0, −F)From −q at bottom-L: attracts toward B-L (−F√3/2, −F/2)From −q at bottom-R: attracts toward B-R (+F√3/2, −F/2)Sum of components:Ex = 0 − F√3/2 + F√3/2 = 0Ey = −F − F/2 − F/2 = −2FF_net = 2F = 6kq²/a²Answer: F_net = 6kq²/a², directed away from the +q corner (toward midpoint of opposite side)Symmetry only guarantees x-cancellation here. The y-components all add never assume the force is zero without resolving components.

PhysicsClass 11CBSE

Two identical charges q₁ = q₂ = 5 μC. Find largest charge q to transfer so force decreases to 1/2.5 of original.

Step-1: After transferring charge q: new charges = (5−q) and (5+q) μCStep-2: New force = F₀ / 2.5:k(5−q)(5+q)/r² = k×25/(2.5×r²) = 10k/r²(5−q)(5+q) = 1025 − q² = 10 ⇒ q² = 15 ⇒ q = √15 μCAnswer: q = √15 μC ≈ 3.87 μCNote: (5−q)(5+q) = 25−q² is the difference-of-squares identity. Transferring charge always decreases the product q₁q₂ for charges of the same sign.

PhysicsClass 11CBSE

Charges +Q, −Q, +Q, −Q at corners of a square (side 5 cm). Find field at centre. (Q = 1 μC)

With alternating charges, the two +Q charges are at diagonally opposite corners, and the two −Q charges are at diagonally opposite corners.Field at centre from each pair:Both +Q fields: exactly opposite directions ⇒ cancelBoth −Q fields: exactly opposite directions ⇒ cancelEvery field vector has an equal and opposite partner by symmetry.Answer: Electric field at centre = 0 (zero)

PhysicsClass 12CBSE

Magnetic Moment of a Current-Carrying Wire Bent into a CircleWire length = 2 m, current = 3.14 A (≈ π A). Find magnetic moment M.

Step-1: Radius of circle:2πr = 2 ⇒ r = 1/π mStep-2: Area:A = πr² = π × (1/π)² = 1/π m²Step 3 — Magnetic moment:M = I × A = π × (1/π) = 1 A·m²Answer: M = 1 A·m²

PhysicsClass 10CBSE

Causes of friction

Interlocking of microscopic surface irregularities (asperities) — primary causeMolecular adhesive forces (van der Waals forces between contact surfaces)Normal force — greater pressing force = greater interlocking = more friction (f = μN)Nature of materials — rubber/concrete = high friction; ice/ice = very lowSurface deformation — soft materials deform to increase contact areaLubricants — reduce friction by creating a separating film

PhysicsClass 10CBSE

What is are force?

Force is any push or pull that changes or tends to change the state of rest or uniform motion of a body.F = ma (Newton's Second Law) | SI unit: Newton (N) = 1 kg·m/s²Four fundamental forces of nature:GravitationalElectromagneticStrong NuclearWeak NuclearSir Isaac Newton formally defined force in Principia Mathematica (1687).Force in one word: Push or Pull (technically: "Interaction").

PhysicsClass 9CBSE

A ball is dropped from a height. Describe its motion in terms of velocity and acceleration, neglecting air resistance.

When a ball is dropped from a height, neglecting air resistance:Acceleration: The acceleration of the ball is constant and equal to the acceleration due to gravity (g), which is approximately 9.8 m/s². Its direction is always vertically downwards.Velocity: The ball starts with an initial velocity of zero (dropped from rest). As it falls, its velocity continuously increases in the downward direction due to the constant downward acceleration. This means its speed increases steadily.

PhysicsClass 9CBSE

Distinguish between instantaneous speed and average speed. When are they equal?

Instantaneous speed is the speed of an object at a particular instant in time. It is what a speedometer in a car measures. It can vary from moment to moment.Average speed is the total distance covered by an object divided by the total time taken for the entire journey.Instantaneous speed and average speed are equal when an object is moving with uniform speed (constant speed) throughout its motion. In this case, the speed at any instant is the same as the overall average speed.

PhysicsClass 9CBSE

A car travels from city A to city B at a speed of 40 km/h and returns from city B to city A at a speed of 60 km/h. What is the average velocity of the car for the entire journey?

For the entire journey from city A to city B and back to city A, the initial position is city A and the final position is also city A. Therefore, the total displacement of the car is zero.Average Velocity = Total Displacement / Total TimeSince Total Displacement = 0,Average Velocity = 0 / Total Time = 0 km/h.

PhysicsClass 9CBSE

Look at the given velocity-time graph. Describe the motion of the object in each segment (OA, AB, BC).

Segment OA: The velocity-time graph is a straight line passing through the origin with a positive slope. This indicates that the object is moving with uniform positive acceleration, starting from rest.Segment AB: The velocity-time graph is a straight line parallel to the time axis. This indicates that the object is moving with a constant velocity (uniform velocity), meaning its acceleration is zero.Segment BC: The velocity-time graph is a straight line with a negative slope, ending at zero velocity. This indicates that the object is moving with uniform negative acceleration (deceleration or retardation) and eventually comes to rest.

PhysicsClass 9CBSE

A car is traveling at 25 m/s. The driver applies the brakes, and the car decelerates uniformly at 5 m/s². How long does it take for the car to stop, and what distance does it cover during this time?

Initial velocity (u) = 25 m/sFinal velocity (v) = 0 m/s (comes to stop)Acceleration (a) = -5 m/s² (deceleration)Part 1: Time to stop (t)Using v = u + at:0 = 25 + (-5) * t0 = 25 - 5t5t = 25t = 5 seconds.Part 2: Distance covered (s)Using v² = u² + 2as:0² = 25² + 2 * (-5) * s0 = 625 - 10s10s = 625s = 62.5 meters.Alternatively, using s = ut + (1/2)at²:s = (25 * 5) + (1/2) * (-5) * (5)²s = 125 + (1/2) * (-5) * 25s = 125 - 62.5s = 62.5 meters.

PhysicsClass 9CBSE

Does an object in uniform circular motion have acceleration? If yes, in which direction does it act?

Yes, an object in uniform circular motion does have acceleration. This is because, even though its speed is constant, its velocity is continuously changing due to the continuous change in direction. This acceleration is called centripetal acceleration, and it always acts towards the center of the circular path.

PhysicsClass 9CBSE

In uniform circular motion, is the speed constant? Is the velocity constant? Explain your answer.

In uniform circular motion, the speed is constant because the magnitude of the velocity remains the same. However, the velocity is not constant because its direction continuously changes as the object moves along the circular path. Since velocity is a vector quantity (magnitude and direction), a change in direction means a change in velocity.

PhysicsClass 9CBSE

A boy is sitting on a merry-go-round which is moving. With respect to whom is the boy at rest, and with respect to whom is he in motion?

The boy is at rest with respect to the merry-go-round itself (e.g., the seat he is sitting on). He is in motion with respect to an observer standing on the ground, or with respect to the trees and buildings around the merry-go-round.

PhysicsClass 9CBSE

A bus moving at 36 km/h is brought to rest by applying brakes. If the brakes produce a uniform retardation of 0.5 m/s², how far does the bus travel before coming to rest?

Initial velocity (u) = 36 km/hConvert u to m/s: 36 km/h * (1000 m / 1 km) * (1 h / 3600 s) = 10 m/sFinal velocity (v) = 0 m/s (comes to rest)Acceleration (a) = -0.5 m/s² (retardation is negative acceleration)Using the third equation of motion: v² = u² + 2as0² = 10² + 2 * (-0.5) * s0 = 100 - ss = 100 meters.

PhysicsClass 9CBSE

A car accelerates uniformly from 10 m/s to 20 m/s in 5 seconds. What is the distance covered by the car during this time?

Initial velocity (u) = 10 m/sFinal velocity (v) = 20 m/sTime (t) = 5 sFirst, calculate acceleration (a) using v = u + at:20 = 10 + a * 510 = 5aa = 2 m/s²Now, calculate displacement (s) using s = ut + (1/2)at²:s = (10 * 5) + (1/2) * 2 * (5)²s = 50 + (1/2) * 2 * 25s = 50 + 25s = 75 meters.Alternatively, using v² = u² + 2as:20² = 10² + 2 * 2 * s400 = 100 + 4s300 = 4ss = 300 / 4 = 75 meters.

PhysicsClass 9CBSE

A train starting from rest attains a velocity of 72 km/h in 5 minutes. Assuming the acceleration is uniform, calculate the acceleration in m/s².

Initial velocity (u) = 0 m/s (since it starts from rest)Final velocity (v) = 72 km/hConvert v to m/s: 72 km/h * (1000 m / 1 km) * (1 h / 3600 s) = 20 m/sTime (t) = 5 minutesConvert t to seconds: 5 minutes * 60 seconds/minute = 300 secondsUsing the first equation of motion: v = u + at20 = 0 + a * 30020 = 300aa = 20 / 300 = 1 / 15 m/s² ≈ 0.0667 m/s².

PhysicsClass 9CBSE

A car travels a distance of 100 km in 2 hours and then another 150 km in 3 hours. Calculate the average speed of the car for the entire journey.

Total Distance = 100 km + 150 km = 250 kmTotal Time = 2 hours + 3 hours = 5 hoursAverage Speed = Total Distance / Total Time = 250 km / 5 hours = 50 km/h.

STILL STUCK?

Get a tutor for just your question.

One-on-one help, verified tutors. Matched within 24 hours.

Get Started →