Physics CSS Solved Papers

CSS Physics Paper-I 2026

Engr. Muhammad Yar Saqib

CSS Physics Paper-I 2026 Solved is a complete solved guide for CSS aspirants who need real answers, not only short topic hints. This post solves the subjective section question by question with definitions, formulas, derivations, numerical substitutions, final answers and exam-ready explanations. It covers divergence, gradient, curl, accelerated motion in the xy-plane, conservation of angular momentum, kinetic friction, work-energy theorem, normal force work, Kepler’s third law, satellite orbital speed, Bernoulli theorem, hydraulic lift, Young’s double-slit experiment, constructive and destructive interference, spring-block SHM period, diffraction of light, X-ray diffraction by crystals, diffraction grating maximum, first law of thermodynamics and work in isothermal expansion.

Central Argument: A CSS Physics solved paper should not only identify the topic asked in the paper. It should define the concept, derive the formula, substitute values carefully and explain the physical meaning. Therefore, this CSS Physics Paper-I 2026 Solved post gives the complete route to each answer so students can reproduce the solution in the examination hall.

Scan Note: The uploaded text is mostly readable. In Q8(b), the temperature written as 0micro C is interpreted as 0°C, because the standard thermodynamics problem requires temperature in Celsius and conversion to Kelvin.

CSS Physics Paper-I 2026 Solved Study Scope

This post covers CSS Physics Paper-I 2026 Solved in a complete, structured and WordPress-ready format. Each question includes the printed question, part-wise solution, formulas, mathematical working, final answer and examiner-friendly explanation.

Use this solved paper as a study document. First revise the formula sheet, then read Q2 to Q8 one by one. After reading, close the post and rewrite each solution from memory. CSS Physics rewards clean definitions, correct assumptions, dimensional accuracy, correct units and disciplined presentation.

Show Table of Contents
  1. Overview
  2. Study Scope
  3. Important Formula Sheet
  4. Question Map
  5. Question 2: Divergence, Gradient, Curl and Accelerated Motion
  6. Question 3: Angular Momentum, Kinetic Friction and Normal Force Work
  7. Question 4: Kepler’s Third Law and Satellite Speed
  8. Question 5: Bernoulli Theorem and Hydraulic Lift
  9. Question 6: Young Double Slit and Spring-Block Period
  10. Question 7: X-Ray Diffraction and Grating Angle
  11. Question 8: First Law and Isothermal Expansion Work
  12. Revision Plan
  13. Internal and External Resources
  14. FAQs

Important Formula Sheet for CSS Physics Paper-I 2026 Solved

Vector Calculus and Motion

grad φ = ∇φ

div A = ∇·A

curl A = ∇×A

v = u + at

Mechanics and Orbits

L = r × p

τ = dL/dt

Wnet = ΔK

T² = 4π²a³/(GM)

v = √[GM/(R+h)]

Fluids and Waves

P + 1/2ρv² + ρgh = constant

F1/A1 = F2/A2

d sinθ = mλ

T = 2π√(m/k)

Thermodynamics and Diffraction

ΔU = Q - Wby

Wby = nRT ln(Vf/Vi)

Won = -nRT ln(Vf/Vi)

2d sinθ = nλ

Question Map of CSS Physics Paper-I 2026 Solved

Question Main Area What Is Fully Solved
Q2 Vector calculus and kinematics Difference between divergence, gradient and curl with examples; velocity vector for accelerated xy-plane motion.
Q3 Rotational dynamics and work-energy Law of conservation of angular momentum, block speed with kinetic friction, work done by normal force.
Q4 Orbital mechanics Kepler’s third law proof and satellite speed at altitude h.
Q5 Fluid mechanics Bernoulli theorem derivation and hydraulic lift numerical.
Q6 Wave optics and SHM Young’s double-slit formulas for bright/dark fringes and spring-block oscillation period.
Q7 Diffraction and crystals Diffraction of light, X-ray diffraction by crystals and diffraction grating angle.
Q8 Thermodynamics First law of thermodynamics, isothermal expansion work derivation and numerical calculation.

Question 2: Divergence, Gradient, Curl and Accelerated Motion

Full Question

Q.2. (a) What is the difference between divergence, gradient and curl? Give one example of each.

Q.2. (b) A particle moves in the xy-plane, starting from the origin at t=0 with an initial velocity having an x-component of 20 m/s and a y-component of -15 m/s. The particle experiences an acceleration in the x-direction given by ax=4 m/s². Determine the total velocity vector at any time.

Q2(a): Difference Between Gradient, Divergence and Curl

Gradient, divergence and curl are three important vector operators used in physics. They are built from the del operator , but each one has a different meaning and acts on a different kind of field.

Operator Acts On Result Meaning Formula
Gradient Scalar field Vector field Direction and rate of maximum increase of a scalar quantity. grad φ = ∇φ
Divergence Vector field Scalar field Measures source or sink strength of a vector field at a point. div A = ∇·A
Curl Vector field Vector field Measures local rotation or circulation of a vector field. curl A = ∇×A

Gradient

The gradient of a scalar function φ(x,y,z) is:

∇φ = (∂φ/∂x)i + (∂φ/∂y)j + (∂φ/∂z)k

It points in the direction in which φ increases most rapidly. Its magnitude gives the maximum rate of change of φ.

Example of Gradient: If T(x,y,z) is temperature, then ∇T points toward the direction where temperature increases most rapidly.

Divergence

The divergence of a vector field A = Axi + Ayj + Azk is:

∇·A = ∂Ax/∂x + ∂Ay/∂y + ∂Az/∂z

Divergence tells whether the field behaves like a source or sink at a point. Positive divergence means field lines are spreading outward. Negative divergence means field lines are converging inward.

Example of Divergence: The electric field of a positive point charge has positive divergence at the charge because field lines originate from it.

Curl

The curl of a vector field A is:

∇×A =
| i j k |
| ∂/∂x ∂/∂y ∂/∂z |
| Ax Ay Az |

Curl measures the local tendency of a vector field to rotate around a point.

Example of Curl: A rotating fluid has non-zero curl. In electromagnetism, a changing electric field can produce circulation of magnetic field.
Final Answer for Q2(a): Gradient changes a scalar field into a vector field, divergence changes a vector field into a scalar field, and curl changes a vector field into another vector field representing local rotation.

Q2(b): Total Velocity Vector at Any Time

The particle moves in the xy-plane. Initial velocity components are:

v0x = 20 m/s
v0y = -15 m/s

The acceleration is only in the x-direction:

ax = 4 m/s²
ay = 0

Velocity in the x-direction after time t is:

vx = v0x + axt
vx = 20 + 4t

Velocity in the y-direction is constant because there is no y-acceleration:

vy = v0y + ayt
vy = -15 + 0t
vy = -15

Therefore, the total velocity vector is:

v(t) = vx i + vy j
v(t) = (20 + 4t)i – 15j m/s

The speed at any time is the magnitude of velocity:

|v(t)| = √[(20+4t)² + (-15)²]
|v(t)| = √[(20+4t)² + 225]
Final Answer for Q2(b): The total velocity vector at any time is v(t)=(20+4t)i - 15j m/s. The speed is √[(20+4t)²+225].

Question 3: Angular Momentum, Kinetic Friction and Normal Force Work

Full Question

Q.3. (a) What is the law of conservation of angular momentum? Give at least one example.

Q.3. (b) A 6 kg block initially at rest is pulled to the right along a horizontal surface by a constant horizontal force of 12 N. Find the speed of the block after it has moved 3 m if the surfaces in contact have a coefficient of kinetic friction 0.15.

Q.3. (c) Can a normal force do work? If not, why not? If so, give an example.

Q3(a): Law of Conservation of Angular Momentum

Angular momentum is the rotational analogue of linear momentum. For a particle, angular momentum about a point is:

L = r × p

where r is position vector and p is linear momentum.

Torque is the rate of change of angular momentum:

τ = dL/dt

If net external torque is zero:

τext = 0

then:

dL/dt = 0

Therefore:

L = constant

This is the law of conservation of angular momentum.

Law: If the net external torque acting on a system is zero, the total angular momentum of the system remains constant.

Examples of Conservation of Angular Momentum

  1. Spinning skater: A skater spins faster when pulling arms inward because moment of inertia decreases and angular velocity increases to keep constant.
  2. Planetary motion: A planet speeds up near perihelion and slows down near aphelion because angular momentum is conserved under the Sun’s central gravitational force.
  3. Diver in air: A diver curls the body to rotate faster and stretches out to rotate slower.
Final Answer for Q3(a): Angular momentum remains conserved when net external torque is zero. A spinning skater pulling arms inward is a common example.

Q3(b): Speed of Block With Kinetic Friction

Given:

Mass, m = 6 kg
Applied force, F = 12 N
Distance, s = 3 m
Coefficient of kinetic friction, μk = 0.15
Initial speed, u = 0
g = 9.8 m/s²

Step 1: Normal Force

Since the block is on a horizontal surface and there is no vertical acceleration:

N = mg
N = 6 × 9.8
N = 58.8 N

Step 2: Kinetic Friction

fk = μkN
fk = 0.15 × 58.8
fk = 8.82 N

Step 3: Net Force

Fnet = F – fk
Fnet = 12 – 8.82
Fnet = 3.18 N

Step 4: Net Work

Wnet = Fnet s
Wnet = 3.18 × 3
Wnet = 9.54 J

Step 5: Work-Energy Theorem

According to work-energy theorem:

Wnet = ΔK

The block starts from rest, so:

Wnet = 1/2 mv²

Therefore:

v = √(2Wnet/m)
v = √[(2)(9.54)/6]
v = √3.18
v ≈ 1.78 m/s
Final Answer for Q3(b): The speed of the block after moving 3 m is approximately 1.78 m/s.

Q3(c): Can a Normal Force Do Work?

Work is defined as:

W = F · s = Fs cosθ

A normal force does no work when it is perpendicular to displacement. For example, when a block slides along a fixed horizontal table, the normal force is vertical while displacement is horizontal. The angle is 90°, so:

W = Ns cos90° = 0

When Normal Force Can Do Work

A normal force can do work if the surface itself moves and the displacement has a component along the normal force.

Example: A person standing in an accelerating elevator is pushed upward by the floor. If the person moves upward, the normal force and displacement are in the same direction, so the normal force does positive work.
Final Answer for Q3(c): A normal force usually does no work when it is perpendicular to displacement, but it can do work if displacement occurs along the direction of the normal force, such as in an elevator.

Question 4: Kepler’s Third Law and Satellite Speed

Full Question

Q.4. (a) What is Kepler’s third law? Prove that the square of the period is proportional to the cube of the semimajor axis.

Q.4. (b) Consider a satellite of mass m moving in a circular orbit around Earth at a constant speed v and at an altitude h above Earth’s surface. Determine the speed of the satellite in terms of G, h, R and M.

Q4(a): Kepler’s Third Law

Kepler’s third law states that the square of the orbital period of a planet is directly proportional to the cube of the semimajor axis of its orbit.

T² ∝ a³

For a circular orbit, the semimajor axis a becomes the orbital radius r.

Proof of Kepler’s Third Law

For a planet or satellite moving in a circular orbit, gravitational force provides the centripetal force.

Gravitational force is:

Fg = GMm/r²

Centripetal force is:

Fc = mv²/r

Equate both:

GMm/r² = mv²/r

Cancel m:

GM/r² = v²/r

Therefore:

v² = GM/r

For circular motion:

v = 2πr/T

Substitute into v²=GM/r:

(2πr/T)² = GM/r
4π²r²/T² = GM/r

Rearrange:

T² = 4π²r³/(GM)

Since 4π²/(GM) is constant for a given central mass:

T² ∝ r³

For an elliptical orbit, r is replaced by semimajor axis a:

T² = 4π²a³/(GM)
Final Answer for Q4(a): Kepler’s third law is T²∝a³. It follows from gravitational force providing centripetal force.

Q4(b): Speed of Satellite at Altitude h

The satellite is at altitude h above Earth’s surface. If Earth’s radius is R, orbital radius is:

r = R + h

For circular orbit:

GMm/r² = mv²/r

Cancel m:

GM/r² = v²/r

Therefore:

v² = GM/r

Substitute r=R+h:

v² = GM/(R+h)

So:

v = √[GM/(R+h)]
Final Answer for Q4(b): The orbital speed of the satellite is v = √[GM/(R+h)].

Question 5: Bernoulli Theorem and Hydraulic Lift

Full Question

Q.5. (a) What is Bernoulli’s theorem? Show that P + 1/2ρv² + ρgh = constant, where P is pressure, ρ is density, v is velocity, g is acceleration due to gravity and h is height above a reference level.

Q.5. (b) In a car lift used in a service station, compressed air exerts a force on a small piston of circular cross-section radius 5 cm. Pressure is transmitted by liquid to a piston of radius 15 cm. What force must compressed air exert to lift a car weighing 1000 N?

Q5(a): Bernoulli’s Theorem

Bernoulli’s theorem states that for steady, incompressible, non-viscous flow along a streamline, the sum of pressure energy, kinetic energy and potential energy per unit volume remains constant.

P + 1/2ρv² + ρgh = constant

Assumptions of Bernoulli’s Theorem

  1. The fluid is incompressible.
  2. The fluid is non-viscous.
  3. The flow is steady.
  4. The equation applies along a streamline.
  5. No energy is added or removed by pumps or turbines between the points considered.

Derivation of Bernoulli’s Equation

Consider a fluid element moving along a streamline from point 1 to point 2. According to work-energy principle, work done by pressure forces equals change in kinetic energy plus change in gravitational potential energy.

Pressure work done on volume ΔV at point 1 is:

W1 = P1ΔV

Pressure work done by the fluid at point 2 is:

W2 = P2ΔV

Net pressure work is:

W = P1ΔV – P2ΔV
W = (P1 – P2)ΔV

Mass of the fluid element is:

m = ρΔV

Change in kinetic energy is:

ΔK = 1/2 m v2² – 1/2 m v1²
ΔK = 1/2 ρΔV(v2² – v1²)

Change in potential energy is:

ΔU = mg(h2 – h1)
ΔU = ρΔV g(h2 – h1)

Using work-energy principle:

(P1 – P2)ΔV = 1/2ρΔV(v2²-v1²) + ρΔVg(h2-h1)

Divide by ΔV:

P1 – P2 = 1/2ρ(v2²-v1²) + ρg(h2-h1)

Rearrange:

P1 + 1/2ρv1² + ρgh1 = P2 + 1/2ρv2² + ρgh2

Therefore:

P + 1/2ρv² + ρgh = constant
Final Answer for Q5(a): Bernoulli’s theorem is an energy conservation equation for ideal fluid flow: P + 1/2ρv² + ρgh = constant.

Q5(b): Hydraulic Lift Numerical

A hydraulic lift works on Pascal’s law. Pressure applied to an enclosed liquid is transmitted equally in all directions.

Therefore:

F1/A1 = F2/A2

Given:

Small piston radius, r1 = 5 cm
Large piston radius, r2 = 15 cm
Weight of car, F2 = 1000 N

Since area is proportional to radius squared:

A1/A2 = r1²/r2²

So:

F1 = F2(r1²/r2²)

Substitute values:

F1 = 1000 × (5²/15²)
F1 = 1000 × (25/225)
F1 = 1000/9
F1 = 111.1 N
Final Answer for Q5(b): The compressed air must exert a force of approximately 111 N on the small piston.

Question 6: Young Double Slit and Spring-Block Period

Full Question

Q.6. (a) What is Young’s double-slit experiment? Derive the mathematical formula for constructive and destructive interference of light.

Q.6. (b) A 200 g block connected to a light spring with force constant 5 N/m is free to oscillate on a frictionless horizontal surface. The block is displaced 5 cm from equilibrium and released from rest. Find the period of its motion.

Q6(a): Young’s Double-Slit Experiment

Young’s double-slit experiment demonstrates the wave nature of light through interference. A monochromatic light source illuminates two narrow slits separated by a small distance d. The two slits behave as coherent sources. Their waves overlap on a distant screen and produce alternating bright and dark fringes.

Path Difference

At a point on the screen making angle θ with the central axis, the path difference between waves from the two slits is:

Δ = d sinθ

For small angles:

sinθ ≈ tanθ = y/D

where D is the distance between slits and screen, and y is distance of the fringe from the central maximum.

So:

Δ = dy/D

Constructive Interference

Constructive interference occurs when path difference is an integral multiple of wavelength:

Δ = mλ

Therefore:

d sinθ = mλ

For small angles:

dy/D = mλ

So position of bright fringe is:

y_m = mλD/d

Destructive Interference

Destructive interference occurs when path difference is an odd half-integral multiple of wavelength:

Δ = (m + 1/2)λ

Therefore:

d sinθ = (m + 1/2)λ

For small angles:

y_m = (m + 1/2)λD/d

Fringe Width

The distance between two consecutive bright or dark fringes is:

β = λD/d

Fringe Pattern

Screen intensity
^
| bright bright bright
| /\ /\ /\
| / \ / \ / \
|______/____\______/____\______/____\____> y
dark bright dark bright dark

Bright: d sinθ = mλ
Dark: d sinθ = (m+1/2)λ

Final Answer for Q6(a): Young’s experiment proves interference of light. Bright fringes occur at d sinθ=mλ, dark fringes occur at d sinθ=(m+1/2)λ, and fringe width is β=λD/d.

Q6(b): Period of Spring-Block Oscillation

Given:

Mass, m = 200 g = 0.2 kg
Spring constant, k = 5 N/m
Amplitude = 5 cm

For a mass-spring oscillator, the period is:

T = 2π√(m/k)

Substitute values:

T = 2π√(0.2/5)
T = 2π√0.04
T = 2π(0.2)
T = 1.257 s
Final Answer for Q6(b): The period of oscillation is approximately 1.26 s. The amplitude does not affect the period for ideal simple harmonic motion.

Question 7: X-Ray Diffraction and Grating Angle

Full Question

Q.7. (a) What is diffraction of light? Explain diffraction of X-rays by crystals.

Q.7. (b) Monochromatic light from a helium-neon laser, λ = 632 nm, is incident normally on a diffraction grating containing 4000 grooves per centimeter. Find the angles at which the second-order maximum is observed.

Q7(a): Diffraction of Light

Diffraction is the bending or spreading of waves when they pass through an aperture or around an obstacle. Diffraction becomes prominent when the size of the aperture or obstacle is comparable to the wavelength of the wave.

Light shows diffraction because it has wave nature. Diffraction patterns consist of bright and dark regions due to constructive and destructive interference of secondary wavelets.

Types of Diffraction

Type Explanation
Fresnel diffraction Source or screen is at finite distance from the aperture or obstacle.
Fraunhofer diffraction Source and screen are effectively at infinite distance, often achieved using lenses.

Diffraction of X-Rays by Crystals

X-rays have wavelengths comparable to the spacing between atoms in crystals. Therefore, a crystal can act as a three-dimensional diffraction grating for X-rays.

Crystal planes reflect X-rays. Constructive interference occurs when the path difference between rays reflected from adjacent crystal planes equals an integral multiple of wavelength.

Bragg’s Law

The condition for constructive interference is:

2d sinθ = nλ

Here:

  • d is spacing between crystal planes.
  • θ is glancing angle.
  • n is order of reflection.
  • λ is wavelength of X-rays.

Derivation of Bragg’s Law

Consider X-rays reflected from two adjacent parallel crystal planes separated by distance d. The ray reflected from the lower plane travels an extra path:

extra path = d sinθ + d sinθ
extra path = 2d sinθ

For constructive interference:

2d sinθ = nλ

Uses of X-Ray Diffraction

  1. Determination of crystal structure.
  2. Measurement of interplanar spacing.
  3. Identification of unknown crystalline substances.
  4. Study of metals, minerals and semiconductors.
  5. Protein crystallography and molecular structure analysis.
Final Answer for Q7(a): Diffraction is spreading of light around apertures or obstacles. X-rays are diffracted by crystal planes according to Bragg’s law 2d sinθ=nλ.

Q7(b): Diffraction Grating Angle for Second-Order Maximum

Given:

Wavelength, λ = 632 nm = 632×10⁻⁹ m
Grooves = 4000 per cm
Order, m = 2

Step 1: Convert Grooves per Centimeter to Grooves per Meter

4000 grooves/cm = 4000 × 100 grooves/m
N = 4.0×10⁵ grooves/m

Step 2: Grating Spacing

Grating spacing is reciprocal of number of grooves per meter:

d = 1/N
d = 1/(4.0×10⁵)
d = 2.5×10⁻⁶ m

Step 3: Grating Equation

For normal incidence:

d sinθ = mλ

Therefore:

sinθ = mλ/d

Substitute values:

sinθ = [2(632×10⁻⁹)]/(2.5×10⁻⁶)
sinθ = 1264×10⁻⁹ / 2.5×10⁻⁶
sinθ = 0.5056

Therefore:

θ = sin⁻¹(0.5056)
θ ≈ 30.36°

Since maxima appear symmetrically on both sides of the central maximum:

θ = ±30.36°
Final Answer for Q7(b): The second-order maximum is observed at approximately ±30.4°.

Question 8: First Law and Isothermal Expansion Work

Full Question

Q.8. (a) What is the first law of thermodynamics? Show that work done on the gas during isothermal expansion of an ideal gas is W=nRT ln(Vf/Vi), where R is universal gas constant, T is temperature, n is number of moles, and Vi and Vf are initial and final volumes.

Q.8. (b) A 1 mole sample of an ideal gas is kept at 0°C during expansion from 3 L to 10 L. How much work is done on the gas during the expansion?

Q8(a): First Law of Thermodynamics

The first law of thermodynamics is the law of conservation of energy applied to heat and work. It states that heat supplied to a system is used to increase its internal energy and to do work.

Using the sign convention where Wby is work done by the gas:

ΔU = Q – Wby

or:

Q = ΔU + Wby

Isothermal Expansion of an Ideal Gas

In an isothermal process, temperature remains constant:

T = constant

For an ideal gas, internal energy depends only on temperature. Therefore, in an isothermal process:

ΔU = 0

From the first law:

Q = Wby

Derivation of Work Done by Gas

Small work done by gas is:

dWby = P dV

For an ideal gas:

PV = nRT

So:

P = nRT/V

Substitute into work expression:

dWby = (nRT/V)dV

Integrate from initial volume Vi to final volume Vf:

Wby = ∫Vi^Vf (nRT/V)dV

Since nRT is constant in isothermal process:

Wby = nRT ∫Vi^Vf dV/V
Wby = nRT [lnV]Vi^Vf
Wby = nRT ln(Vf/Vi)

Work Done on Gas

Work done on the gas is opposite in sign to work done by the gas:

Won = -Wby

Therefore:

Won = -nRT ln(Vf/Vi)

Sign Note: Some textbooks write W=nRT ln(Vf/Vi) for work done by the gas. If the question specifically says “work done on the gas,” the correct sign for expansion is negative because the gas does work on the surroundings.

Final Answer for Q8(a): For isothermal expansion, work done by gas is Wby=nRT ln(Vf/Vi), while work done on gas is Won=-nRT ln(Vf/Vi).

Q8(b): Work Done During Expansion from 3 L to 10 L

Given:

n = 1 mol
T = 0°C = 273.15 K
Vi = 3 L
Vf = 10 L
R = 8.314 J/mol K

Work done by the gas is:

Wby = nRT ln(Vf/Vi)

Substitute values:

Wby = (1)(8.314)(273.15) ln(10/3)

Now:

ln(10/3) = ln(3.333) ≈ 1.204

Therefore:

Wby = 8.314 × 273.15 × 1.204
Wby ≈ 2733 J
Wby ≈ 2.73 kJ

Work done on the gas is:

Won = -Wby
Won ≈ -2.73 kJ
Final Answer for Q8(b): Work done by the gas is approximately +2.73 kJ. Therefore, work done on the gas during expansion is approximately -2.73 kJ.

Revision Plan for CSS Physics Paper-I 2026 Solved

After reading this complete CSS Physics Paper-I 2026 Solved guide, revise it in three rounds. In the first round, learn the definitions and formulas. In the second round, reproduce each derivation without looking. In the third round, solve the numerical questions again with units and compare your answer with the final result.

Question Revision Task
Q2 Compare gradient, divergence and curl; then derive v(t)=(20+4t)i-15j.
Q3 State angular momentum conservation, solve the friction block again and explain when normal force does work.
Q4 Prove T²∝a³ and derive v=√[GM/(R+h)].
Q5 Derive Bernoulli equation and solve the hydraulic lift force using area ratio.
Q6 Write bright/dark fringe conditions and solve spring-block period.
Q7 Explain X-ray diffraction by crystals and solve grating angle for second order.
Q8 Derive isothermal work and solve expansion from 3 L to 10 L.

Related Resources for CSS Physics Paper-I 2026 Solved

Exam Note: For CSS Physics, every numerical answer should show the given data, formula, substitution, calculation and final unit. Every derivation should start from a basic law and end with a boxed result. This improves both readability and marks.

FAQs About CSS Physics Paper-I 2026 Solved

What does CSS Physics Paper-I 2026 Solved include?

CSS Physics Paper-I 2026 Solved includes complete solved answers for Q2 to Q8 with definitions, derivations, numerical calculations, formulas, final answers and exam-ready notes.

What is the answer for the velocity vector in Q2?

The total velocity vector at any time is v(t)=(20+4t)i - 15j m/s.

What is the speed of the block in Q3?

The speed of the block after moving 3 m is approximately 1.78 m/s.

What is the satellite speed formula in Q4?

The satellite speed at altitude h above Earth is v=√[GM/(R+h)].

What is the hydraulic lift answer in Q5?

The compressed air must exert a force of approximately 111 N on the small piston to lift a car weighing 1000 N.

What is the period of the spring-block system in Q6?

For m=0.2 kg and k=5 N/m, the period is approximately 1.26 s.

What is the grating angle in Q7?

The second-order maximum for λ=632 nm and 4000 grooves/cm occurs at approximately ±30.4°.

What is the work done in Q8?

Work done by the gas is approximately +2.73 kJ. Work done on the gas during expansion is approximately -2.73 kJ.

Can I paste this HTML into WordPress?

Yes. The post avoids an H1 heading so your WordPress post title can remain the only H1. The internal structure begins with H2 and continues with H3/H4 headings.

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