Topic-wise Physics Formulas Class 11th - Download PDF

The list of topic-wise Physics Formulas Class 11th includes all important formulas from Physical World and Measurement, Kinematics, Laws of Motion, Work, Energy and Power, Motion of System of Particles and Rigid Body, Gravitation, and several other topics. You can easily download the PDF for topic-wise Physics Formulas Class 11th to revise all the topics in the CBSE Class 11 Physics syllabus 2025-26. 

There is a high weight of 20 to 40 marks given to the questions based on topic-wise Physics formulas. Read more to know about the most important formulas for Physics class 11 to score higher in the final examination.

Topic-wise Physics Formulas Class 11th: Download PDF

Downloading PDF for Class 11 Topic-wise Physics Formulas will help you keep a record of all topic-wise Physics Formulas altogether, making it easier to learn and revise for the exam day. You can download the Class 11 Physics Formulas PDF from the table given below. 

Also Read: Class 11th Physics Practical Experiments and Activities

Important Formulas for Physics Class 11

Here are some of the most important Physics formulas for class 11 that you must know:

Some Important Questions Based on Physics Formulas Class 11th 

Some of the important questions based on the Topic-wise Physics Formulas Class 11th are given below. You can practise Physics Formulas for Class 11 to get an idea of what questions are asked in the exam. 

Question 1: A car accelerates uniformly from rest to a speed of 25 m/s in 10 seconds. Calculate its acceleration.
Solution: Given:

• Initial velocity, u=0u = 0u=0 m/s
• Final velocity, v=25v = 25v=25 m/s
• Time, t=10t = 10t=10 s

Using the formula a=v−uta = frac{v - u}{t}a=tv−u​:
a=25−010=2.5 m/s2a = frac{25 - 0}{10} = 2.5 text{ m/s}^2a=1025−0​=2.5 m/s2

Question 2: A stone is thrown vertically upwards with a velocity of 20 m/s. Calculate the maximum height reached by the stone.
Solution: Given:

• Initial velocity, u=20u = 20u=20 m/s (upwards)
• Acceleration due to gravity, g=9.8g = 9.8g=9.8 m/s2^22

At maximum height, final velocity v=0v = 0v=0 m/s. Use the formula v2=u2+2asv^2 = u^2 + 2asv2=u2+2as, where a=−ga = -ga=−g:
0=(20)2−2⋅9.8⋅s0 = (20)^2 - 2 cdot 9.8 cdot s0=(20)2−2⋅9.8⋅s
Solving for sss:
s=(20)22⋅9.8=20.41 ms = frac{(20)^2}{2 cdot 9.8} = 20.41 text{ m}s=2⋅9.8(20)2​=20.41 m

Question 3: A ball is dropped from a height of 50 m. Calculate its velocity just before it hits the ground.
Solution: Given:

• Initial height, h=50h = 50h=50 m
• Acceleration due to gravity, g=9.8g = 9.8g=9.8 m/s2^22

Using the formula v2=u2+2asv^2 = u^2 + 2asv2=u2+2as, where u=0u = 0u=0:
v2=2⋅9.8⋅50v^2 = 2 cdot 9.8 cdot 50v2=2⋅9.8⋅50 v=2⋅9.8⋅50=980=31.3 m/sv = sqrt{2 cdot 9.8 cdot 50} = sqrt{980} = 31.3 text{ m/s}v=2⋅9.8⋅50​=980​=31.3 m/s

Question 4: A force of 10 N acts on a body of mass 5 kg for 2 seconds. Calculate the change in momentum of the body.
Solution: Given:

• Force, F=10F = 10F=10 N
• Mass, m=5m = 5m=5 kg
• Time, t=2t = 2t=2 s

Change in momentum Δp=F⋅tDelta p = F cdot tΔp=F⋅t:
Δp=10⋅2=20 kg m/sDelta p = 10 cdot 2 = 20 text{ kg m/s}Δp=10⋅2=20 kg m/s

Question 5: An object of mass 2 kg is moving with a velocity of 3 m/s. Calculate its kinetic energy.
Solution: Given:

• Mass, m=2m = 2m=2 kg
• Velocity, v=3v = 3v=3 m/s

Kinetic energy K.E.=12mv2K.E. = frac{1}{2}mv^2K.E.=21​mv2:
K.E.=12⋅2⋅(3)2=9 JK.E. = frac{1}{2} cdot 2 cdot (3)^2 = 9 text{ J}K.E.=21​⋅2⋅(3)2=9 J

Question 6: Calculate the gravitational force between two objects of masses 5 kg and 3 kg placed 10 m apart. Given G=6.67×10−11G = 6.67 times 10^{-11}G=6.67×10−11 N m2^22/kg2^22.
Solution: Given:

• Mass 1, m1=5m_1 = 5m1​=5 kg
• Mass 2, m2=3m_2 = 3m2​=3 kg
• Distance, r=10r = 10r=10 m
• Gravitational constant, G=6.67×10−11G = 6.67 times 10^{-11}G=6.67×10−11 N m2^22/kg2^22

Gravitational force F=G⋅m1⋅m2r2F = frac{G cdot m_1 cdot m_2}{r^2}F=r2G⋅m1​⋅m2​​:
F=6.67×10−11⋅5⋅3(10)2=1.001×10−10 NF = frac{6.67 times 10^{-11} cdot 5 cdot 3}{(10)^2} = 1.001 times 10^{-10} text{ N}F=(10)26.67×10−11⋅5⋅3​=1.001×10−10 N

Question 7: A spring with spring constant k=100k = 100k=100 N/m is compressed by 0.1 m. Calculate the potential energy stored in the spring.
Solution: Given:

• Spring constant, k=100k = 100k=100 N/m
• Compression, x=0.1x = 0.1x=0.1 m

Potential energy U=12kx2U = frac{1}{2} k x^2U=21​kx2:
U=12⋅100⋅(0.1)2=0.5 JU = frac{1}{2} cdot 100 cdot (0.1)^2 = 0.5 text{ J}U=21​⋅100⋅(0.1)2=0.5 J

Question 8: A 10 kg object is lifted to a height of 5 m above the ground. Calculate the work done against gravity.
Solution: Given:

• Mass, m=10m = 10m=10 kg
• Height, h=5h = 5h=5 m
• Acceleration due to gravity, g=9.8g = 9.8g=9.8 m/s2^22

Work done W=mghW = mghW=mgh:
W=10⋅9.8⋅5=490 JW = 10 cdot 9.8 cdot 5 = 490 text{ J}W=10⋅9.8⋅5=490 J

Question 9: A ball is thrown horizontally from a height of 20 m with a velocity of 10 m/s. Calculate the time it takes to reach the ground.
Solution: Given:

• Initial height, h=20h = 20h=20 m
• Initial horizontal velocity, u=10u = 10u=10 m/s
• Acceleration due to gravity, g=9.8g = 9.8g=9.8 m/s2^22

Time ttt to reach the ground can be found using h=ut+12gt2h = ut + frac{1}{2}gt^2h=ut+21​gt2:
20=0⋅t+12⋅9.8⋅t220 = 0 cdot t + frac{1}{2} cdot 9.8 cdot t^220=0⋅t+21​⋅9.8⋅t2
Solving for ttt:
t=2⋅209.8≈2.02 st = sqrt{frac{2 cdot 20}{9.8}} approx 2.02 text{ s}t=9.82⋅20​​≈2.02 s

Question 10: An object of mass 4 kg is pushed with a force of 20 N. Calculate its acceleration.
Solution: Given:

• Force, F=20F = 20F=20 N
• Mass, m=4m = 4m=4 kg

Acceleration aaa can be found using a=Fma = frac{F}{m}a=mF​:
a=204=5 m/s2a = frac{20}{4} = 5 text{ m/s}^2a=420​=5 m/s2

Also Read: CBSE Class 11th Physics Notes