How To Solve NEET 2018 With A Smile – I

  1. Home
  2. /
  3. NEET 2018
  4. /
  5. How To Solve NEET 2018 With A Smile – I

Let Us YOU Solve NEET 2018

Regular visitors to this blog site must know that, presently, we have been guiding students to solve Physics questions from various competitive examinations of repute. In our earlier blogs, we discussed questions from JEE (Advanced) 2018 & JEE (Main) 2017 and 2018, in that order. While JEE (Main) is the gateway to B.E. and B. Tech courses in most Engineering Institutes except IITs in India, JEE (Advanced) is the final hurdle to be crossed by more ambitious students who will study in one of the Indian Institutes of Technology (IITs).

Now, we shall shift our attention to the premier Medical Entrance Examination in India, known as National Eligibility cum Entrance Test (NEET-UG). This test is taken by those students who wish to pursue medical course (MBBS) and dental course (BDS) in government or private medical colleges and dental colleges, respectively. More than 66,000 MBBS and BDS seats are filled through NEET-UG. Up till 2018, NEET-UG has been conducted in pen-and-paper mode by Central Board of Secondary Education (CBSE). But as per the latest report, significant changes are on the cards. From 2019 onwards, NEET-UG will be conducted by the newly formed National Testing Agency (NTA) in online mode. Instead of being held once a year, it will be held twice a year – in February and May – for the academic session beginning in August of that year. There will be no change in syllabus and pattern of questions at the moment.

I have in front of me the NEET 2018 question paper. This examination was held in off-line mode on Sunday, 06 May 2018, at 1000-1300 hours. The paper contains 180 questions in total, with maximum marks 720, divided equally among Physics, Chemistry, Botany and Zoology. In my paper, set-QQ (order of questions varies from set to set), Physics questions come last. Each one is a multiple-choice question (MCQ), with one correct option. For each correct response, the student gets 4 marks. For each incorrect response, one mark is deducted from the total score. The duration of the exam is three hours, which boils down to 45 minutes for 45 questions. However, considering the questions are relatively simple, a well-prepared student can definitely manage one question a minute.

To remind once again, it is the readers of this blog who will themselves solve the questions. We @ PhysicsAcademyOnline.com. believe in the teaching principle: Theory First, Problems Next. We have built a large collection of quality video lectures, on both Physics theories and problems at basic and advanced levels, which can be accessed at the click of a mouse. As you attempt any question, we connect you to the relevant theory/theories through our lectures, so that you can solve the question with supreme confidence. A video solution is provided at the end for checking purpose only. The same mechanism will be followed in the questions which appear next.

Question 137:  Two wires are made of the same material and have the same volume. The first wire has cross-sectional area A, and the second wire has cross-sectional area 3A. If the length of the first wire is increased by ∆ l  on applying a force F, what force is needed to stretch the second wire by the same amount?

(1) 9F                    (2) F                             (3) 4F                             (4) 6F

Given the ratio of cross sections of the two wires, what is the ratio of their lengths? What is the formula for Young’s modulus of the wire material? Subscribe to PhysicsAcademyOnline.com., and watch this lecture:  Chapter Name – Elasticity and Fluid Mechanics; Category –  Basic; Topic Name – Elasticity; Video Name – Hooke’s Law and Young’s Modulus.

For the same extension of the wires, how does the applied force vary with length and cross-sectional area? Answer follows in a jiffy.

 

 

Question 139:  A small sphere of radius r falls from rest in a viscous liquid. As a result, heat is produced due to viscous force. The rate of production of heat when the sphere attains its terminal speed is proportional to:

(1) r3                      (2) r4                          (3) r5                            (4) r2

The rate of production of heat is equal to the power expended in falling against viscous force. What is the formula for power? Study from the lecture: Chapter Name – Mechanics; Category –  Basic; Topic Name – Work and Energy; Video Name – Power.

What is the formula for viscous force on the sphere? What is the formula for terminal speed of the sphere? Both can be found in:   Chapter Name – Elasticity and Fluid Mechanics; Category –  Basic; Topic Name – Fluid Dynamics and Viscosity; Video Name – Motion of a Solid Body in a Viscous Fluid.

Select the correct option.

 

 

 

Leave a Reply

Your email address will not be published. Required fields are marked *