Question Nahi Banta

 Q. A car is negotiating a curved road of radius R. The road is banked at an angle θ. The coefficient of friction between the tyres of the car and the road is μs. The minimum safe velocity on this road is :( Twisted Neet 2016 | Circular Motion Q 1 ) Don't forget to LIKE , SHARE & SUBSCRIBE my Channel. Please Press BELL icon to receive latest updates. Don't miss Hint based videos / concept related videos/ twisted qus videos available only on this channel , from which you can learn the most solid framework of problem solving/ clarity in application of different concepts/ formation of other problems with a small change in data given or language. Check out text solution( No need to run videos all the time, solutions are available in written format)just a click away on questionnahibanta.blogspot.com  If you have anything to say/discuss ,feel free to contact Email : questionnahibanta@gmail.com ,     If you want to be part of Exam related discussion/t...

Q. A car is negotiating a curved road of radius R. The road is banked at an angle  θ . The coefficient of friction between the types of the car and the road is  μ s . The maximum safe velocity on this road is : Description :   Neet 2016 | Circular Motion Q 1 | It may not even appear as a qus, for many students as this problem and its answer , are now part of lecture notes. But if this is not the case then my advice is that , Don’t pick such qus in paper. As you can see in solution   so many steps are involved from making the FBD , division of forces into components , identifying centripetal force to applying condition of slipping in case of maximum speed.   Don't forget to LIKE , SHARE & SUBSCRIBE my Channel. Please Press BELL icon to receive latest updates. Don't miss Hint based videos / concept related videos/ twisted qus videos available only on this channel , from which you can learn the most solid framework of problem solving/ clarity in applicati...

Q. Two stones of masses  m  and  2m  are whirled in horizontal circles, the heavier one in a radius  r/2  and the lighter one in radius  r . The tangential speed of lighter stone is  n  times that of the value of heavier stone when they experience same centripetal forces. The value of  n  is Description : Only thing needs to be identify, the force which is responsible for centripetal force i.e circular dynamics. Rest is the comparison between the values of this force for the two given masses. Answer is a numerical value so things are a lot easy. Don't forget to LIKE , SHARE & SUBSCRIBE my Channel. Please Press BELL icon to receive latest updates. Don't miss Hint based videos / concept related videos/ twisted qus videos available only on this channel , from which you can learn the most solid framework of problem solving/ clarity in application of different concepts/ formation of other problems with a small change in data give...

Q. A car of mass m is moving on a level circular track of radius R.If μs  represents the static friction between the road and tyres of the car, the maximum speed of the car in circular motion is given by Description :   We have seen similar qus in year 2010 . qus is based on very well known concept of circular motion i.e circular turning on road by Friction only. Although no  numerical value in answer , but still its answer is result based . Students want to know why even such qus came into paper. Answer is simple , Examiner wants to check whether you even know the formula or not. Which implies while doing Revision , give importance to formula. Don't forget to LIKE , SHARE & SUBSCRIBE my Channel. Please Press BELL icon to receive latest updates. Don't miss Hint based videos / concept related videos/ twisted qus videos available only on this channel , from which you can learn the most solid framework of problem solving/ clarity in application of different concepts/...

A car of mass  1000kg  negotiates a banked curve of radius  90m  on a frictionless road. If the banking angle is  45 ∘ , the speed of the car is Description :  Aipmt /Neet 2012 | Circular Motion q 1 | What more can you expect from a Qus Answer is a numerical value , angle given is easy to process , mathematical calculation ( square root ) is easy   All in all no surprise element , Straight formula based. Sometimes Students want to know why even such qus came into paper. Answer is simple , Examiner wants to check whether you even know the formula or not. Which implies while doing Revision , give importance to formula. Don't forget to LIKE , SHARE & SUBSCRIBE my Channel. Please Press BELL icon to receive latest updates. Don't miss Hint based videos / concept related videos/ twisted qus videos available only on this channel , from which you can learn the most solid framework of problem solving/ clarity in application of different concepts/ formatio...

Q. A gramophone record is revolving with an angular velocity  ω . A coin is placed at a distance  r  from the centre of the record. The static coefficient of friction is  μ . The coin will revolve with the record if Description :   Although qus is based on very well known concept of circular motion i.e circular turning on road by Friction only, still inequality relation and no   numerical value in answer ,can make it little bit difficult for some students. Since Static Friction is a self adjusting force while it will provide required centripetal force , it will limit this motion according to its maximum value or limiting value. From this inequality idea will pop up. But if you are not well aware of the concept of circular turning on road by friction only then things may not be easy for you. Don't forget to LIKE , SHARE & SUBSCRIBE my Channel. Please Press BELL icon to receive latest updates. Don't miss Hint based videos / concept related videos/ twisted...

Q. A tube of length L is filled completely with an incompressible liquid of mass M and closed at both the ends. The tube is then rotated in a horizontal plane about one of its ends with a uniform angular velocity  ω.  The force exerted by the liquid at the other end is Description :  Aipmt /Neet 2006 | Circular Motion q 1 | you had seen my Hint Video Fotu  Banao on this and if still not able to solve then you must understand . It may not be an easy qus  for many students as  1) Recognition of centripetal force can be a problem 2)Relating this force with force exerted by liquid 3)Identification of force exerted by liquid like what kind of force, reason of force etc 4)Final & most important thing ; value of radius . As many students can think its equal to length. Concept of center of mass may not strike. I hope you must understand the solution and things will be easier now. Try to learn and next time , do it yourself. Don't forget to LIKE , SHARE ...

Q. A particle moves along a circle of radius  (20/π)  m with constant tangential acceleration. If the velocity of the particle is 80 m/s at the end of the second revolution after motion has begun, the tangential acceleration is Description :  This problem is using 1) one revolution distance in radian 2) relation between angular acceleration & tangential acceleration 3) relation between angular velocity & linear velocity 4) equation of motion in case of constant circular acceleration Since various relations are used along with equation of motion, its an average level qus Don't forget to LIKE , SHARE & SUBSCRIBE my Channel. Please Press BELL icon to receive latest updates. Don't miss Hint based videos / concept related videos/ twisted qus videos available only on this channel , from which you can learn the most solid framework of problem solving/ clarity in application of different concepts/ formation of other problems with a small change in data give...

Two bodies of mass 4 kg and 6 kg are tied to the ends of a massless string. The string passes over a pulley which is frictionless (see figure). The acceleration of the system in terms of acceleration due to gravity (g) is : Neet 2020 NLM & Frctn q 1|| In my previously uploaded two related videos on this. In Direct approach video i told you about the formulas you can apply to get direct answer of two types of qus and in the hint video Fotu Banao , i discussed how to make Fotu of this qus, definitely the first step towards problem solving .From the FBD(Fotu) we can write down equation of Newton's Second law of motion for both the blocks & can get our answer. Don't forget to LIKE , SHARE & SUBSCRIBE my Channel. Please Press BELL icon to receive latest updates. Don't miss Hint based videos / concept related videos/ twisted qus videos available only on this channel , from which you can learn the most solid framework of problem solving/ clarity in application of...

Neet 2019 NLM & Frctn q 1 If you haven't seen my hint video . Funda Kya for this qus , then you should watch it first then try to solve by yourself . This qus is a mix of Vectors & Newton's law. As i told you earlier ,in this qus we need to apply Triangle law of vector addition & since in this qus all vectors are added in such a way that there resultant will be zero. According to Newton's second law of motion , we knew that Net force zero means either mass of particle is zero or acceleration is zero. Clearly mass cannot be zero then only acceleration could be zero and if acceleration is zero then Velocity will remains constant. Don't forget to LIKE , SHARE & SUBSCRIBE my Channel. Please Press BELL icon to receive latest updates. Don't miss Hint based videos / concept related videos/ twisted qus videos available only on this channel , from which you can learn the most solid framework of problem solving/ clarity in application of different concepts...

A block of mass m is placed on a smooth inclined wedge ABC of inclination θ as shown in the figure. The wedge is given an acceleration 'a' towards the right. The relation between a and θ for the block to remain stationary on the wedge is :- Neet 2018 NLM & Friction q 2  I told earlier during discussion of qus of year 2004 that you should remember results of such kind of qus(as we had similar qus in year 2004)because this type of qus acts as standard qus . They may come in some different form or they can be a part of a long qus. Either way , you will be get benefited as here in this case . qus is asking about relation between acceleration and angle of inclination (In 2004 normal reaction force was asked) After watching first hint video Fotu Banao you can understand that gravitational pull down the incline is going to be balanced by component of Pseudo along the incline upwards. By equating them , we will get the desired answer. Don't forget to LIKE , SHARE & SUBSCRI...

Which one of the following statements is incorrect ?  (1) Rolling friction is smaller than sliding friction  (2) Limiting value of static friction is directly proportional to normal reactions  (3) Frictional force opposes the relative motion  (4) Coefficient of sliding friction has dimensions of length Neet 2018 NLM & Frctn q 1 its a theoretical qus where we need to check every option one by one. Generally people deal casually with theoretical qus in exam since no calculation is involved but let me tell you . Chances of negative are more in such qus because of this casual attitude Don't forget to LIKE , SHARE & SUBSCRIBE my Channel. Please Press BELL icon to receive latest updates. Don't miss Hint based videos / concept related videos/ twisted qus videos available only on this channel , from which you can learn the most solid framework of problem solving/ clarity in application of different concepts/ formation of other problems with a small change in data...

Two blocks A and B of masses 3 m and m respectively are connected by a massless and inextensible string. The whole system is suspended by a massless spring as shown in figure. The magnitudes of acceleration of A and B immediately after the string is cut, are respectively :- Neet 2017 NLM & Frctn q 1, i hope you already watched previous hint videos fotu banao , funda kya & firangi ko follow and you must understand , its alengthy qus where too many steps involved from making FBD , writing equation of newton's second law for string ,spring & blocks in equilibrium to disturbing the equilibrium by cutting. the block connected to string will fall freely since tension becomes zero whereas block connected to spring will accelerate upwards because spring force will become greater than its weight. Don't forget to LIKE , SHARE & SUBSCRIBE my Channel. Please Press BELL icon to receive latest updates. Don't miss Hint based videos / concept related videos/ twisted qus...