1:52:21 The two hands will coincide at the time (120/11) minutes past 2 o' clock i.e. 2 : (120/11) Hour and minute hand of a clock are perpendicular to each other 44 times in a day.

Sir the accn of lift may be changing.

Lift is going from 11th floor to first floor.... First reading in weighing machine decrease, then increase and then normalise... WHY ? Comment your explanation.... Like and share !!

Thanks for teaching this difficult chapter with so much simple way..... Praying God for your well being... ❤

Q1 ans is - g/3 Q2 ans is - T=30N , acceleration=5m/s^2

Q1 ans is = g/3 Q2 ans is= T 15N a=5m/s^2

Option 'C'

Ans Ra = 100N and Rb =50√3N

Sir ans is - Ra= -50/√3 N Rb= 100/√3 N

Option - C

Amazing sir!

Thank you sir!

Amazing video!

totally new approach to Newton's first law....thank you verymuch sir...!

enjoyed the class ...❤

so many views ,lo like kitna kam....mein to kardia....😊 beautiful explanation...

jai ho....sor aapka

so beautiful explanation

so many views ...so less likes....kaun kaun is message ko padhne ke bàad like dala😅

nice sir.....

thanks a lot .....

thank you sir...

ans is C

we are so grateful...beautiful effort

C

Option C

Option C

Option C

C

Option A

Ans 15

❤

Thank you sir 🙏

Thaaaank you sir..

Thank You Sir..

Funda clear... ❤

Is it 30 kg sir 😊

Answer to last qn is 20kg

Thank you sir...

Nice sir

🙏

Sir please make one shot lecture of vector and projectile motion with numerical ❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤

Thank you for the hard work and dedication you put into preparing your lessons Sir..🙏🏻

With that they can ask time period of pendulum and geff concept can be used ❤❤😊

Let the inclined plane be at an angle'θ' from the ground Let us consider the FBD of a liquid particle on the surface of the liquid. Let mass of the particle be m Then the forces acting on the particle to keep it in dynamic equilibrium are: its weight(mg) vertically downwards, pseudo force(ma) and a force normal to the surface of the liquid(Let N) mg can be represented as the sum of its components in the direction normal to the inclined plane and parallel to the inclined plane. They will be mg cosθ and mg sinθ respectively. Then accn will be mg sinθ/m = g sinθ Therefore ma = mg sinθ The pseudo force cancels the component of the weight of the particle in the dirn parallel to the inclined plane, both being mg sin In order to maintain dynamic eqm, N must be equal and opposite to the component of weight in the dirn normal to the inclined plane. As because N is perpendicular to the surface of the liquid, correct position will be that of option B

Let the accelerated fluid subtend an angle 'A' to the horizontal surface of the liquid, Assuming a cylinder of length L and cross sectional area A the forces exerted on it are, ma, pghA1, pghA2, mgsinx From FBD, ma+pghA1=pghA2+mgsinx =pgh1=pgh2 =h1=h2 As tanA= (h2-h1)/L =tanA=0 As A is between 0 and 90 A=0° Therefore the accelerated fluid subtends 0 degree with the surface of the liquid Hence, position B is correct

My favourite teacher on You tube.... Kass mey bhi aapse off line padh paati😢

Thanks sir...

Acha laga sir video dekh k ❤🫶🏻

Thank You Sir 🙏