# Tips to Solve Physics JEE (Advanced) 2018 Questions – I

### Let Us YOU Solve Physics JEE (Advanced 2018) Questions:

I have the JEE (Advanced) 2018, Paper-1 in front of me. This examination, which was formerly called IIT-JEE, was held on 20 May 2018. The paper contains 54 questions in total, with maximum marks 180, divided equally among Physics, Chemistry and Mathematics. Part I is Physics, having three sections: Section 1 containing six multiple-choice questions (MCQ, total marks 4 x 6 = 24), Section 2 containing eight numerical problems (total marks 3 x 8 = 24), and Section 3 containing four MCQs in two paragraphs (total marks 3 x 4 = 12). Thus, Physics Paper-1 has 18 questions carrying 60 marks in total. Considering the time allotted for the whole paper is three hours, the student ideally should solve Physics part in one hour, that is one minute allotted for one mark scored.

Let us start with Section 1, which contains six MCQs with one or more than one options correct. Obviously, this type of MCQs are more difficult than the other type having only one correct option. This reflects in the marking system: full marks only for choosing all the correct options; less marks for choosing part of the correct options; negative marking for choosing any incorrect option(s). In other words, random guesswork which is a common temptation in questions of this type is heavily penalised. So be warned!

Question 1:  The potential energy of a particle of mass m at a distance r from a fixed point O is given by V(r) = kr² / 2, where k is a positive constant of appropriate dimensions. This particle is moving in a circular orbit of radius R about the point O. If v is the speed of the particle and L is the magnitude of its angular momentum about O, which of the following statements is/are true?

(A) v = √(k/2m). R

(B) v = √(k/m). R

(C) L = √(mk). R²

(D) L = √(mk/2). R²

An alert student must have noted by now there can be only one or two correct options for this MCQ, not more. Both options (A) and (B) give a possible expression for particle speed, v, and therefore, only one of them can be correct (if not both incorrect). Similarly, both (C) and (D) give a possible expression for particle’s angular momentum, L, and hence, at most one of them can be correct. A bit of relief!

Finding v first. What is the acceleration of a particle moving in a circle with constant speed? Subscribe to www.physicsacademyonline.com, and watch the lecture following this path: Chapter Name – Mechanics; Category –  Basic; Topic Name – Circular Motion; Video Name – Derivation of Centripetal Acceleration.

What force can produce such acceleration? Watch the lecture as follows: Chapter Name – Mechanics; Category –  Basic; Topic Name – Newton’s Laws of Motion; Video Name – Newton’s Second Law of Motion.

Potential energy of the particle being given as a function of distance, can you deduce an alternative expression for the (conservative) force? Follow the lecture: Chapter Name – Mechanics; Category –  Basic; Topic Name – Work and Energy; Video Name – Conservative Force from Potential Energy Function, and Energy Diagrams.

You are close to the answer. What is it?

Finding L next. How do you express angular momentum of a particle about any arbitrary origin as a vector product? What is its magnitude? How do you modify it for the simple case of a particle moving in a circle about the centre? Follow the discussion here: Chapter Name – Mechanics; Category –  Basic; Topic Name – Rotational Mechanics; Video Name – Angular Momentum of a Particle, and Its Relation with Torque.

Only when you have completed, confirm your answer from our MP4 video solution below.

Question 2:  Consider a body of mass 1.0 kg at rest at the origin at time t = 0. A force, F = αti +β j, is applied to the body, where α = 1.0 N/s and β = 1.0 N. The torque acting on the body about the origin at time t = 1.0 s is
τ . Which of the following statements is/are true?

(A)  τ  = 1/3 N-m

(B)  The torque vector, τ , is in the direction of the unit vector + k.

(C) The velocity of the body at t = 1.0 s is v = (i + 2j)/2 m/s

(D) The magnitude of displacement of the body at t = 1.0 s 1/6 m

What does an alert mind notice as he reads the question? Unlike the previous question, all the four options are independent. That is, one option being correct does not determine any other option being correct or incorrect. So the number of correct options may be anything from 1 to 4. Don’t feel disheartened. Rather take advantage of the fact that the first two options involve torque vector, the third option involves velocity vector, and the fourth option involves displacement vector. So you will require a calculation based on vector calculus, which will give you all three above quantities – not necessarily in the given order. Can you see a road map here? Things will be easier by the fact the mass of the body, time interval, the two constants in the expression of force are all 1.0 unit each.

How do you express torque of a force about an origin as a vector product? Except force, what is the other quantity involved? Subscribe to  www.physicsacademyonline.com, and watch the lecture following this path: Chapter Name – Mechanics; Category –  Basic; Topic Name – Rotational Mechanics; Video Name – Torque of a Force Acting on a Particle.

So you have to get that first. Noting that the force is time-dependent, how can you derive the body’s acceleration, then velocity, and then displacement as functions of time – all in unit-vector notation? Don’t forget to use the initial conditions at t = 0. Watch the lecture: Chapter Name – Mechanics; Category –  Basic; Topic Name – Motion in One and Two Dimensions; Video Name – Graphical Analysis of Rectilinear Motion. Also helpful, to learn technique of calculation, this lecture: Chapter Name – Mechanics; Category – Advanced; Topic Name – Vectors; Video Name – Advanced-Level Problems on Vectors II.

Now, can you derive torque also as a function of time? The two following lectures are relevant: Chapter Name – Mechanics; Category – Basic; Topic Name – Vectors; Video Names – Vector Product of Two Vectors  & Problems on Vector Product of Vectors.

Put t = 1.0 s in the expressions of v, r and τ. The correct options come out. Only when you have completed, confirm your answer from our MP4 video solution below.

Question 3:  A uniform capillary tube of inner radius r is dipped vertically into a beaker filled with water. The water rises to a height h in the capillary tube above the water surface in the beaker. The surface tension of water is σ . The angle of contact between water and the wall of the tube is θ . Ignore the mass of water in the meniscus. Which of the following statements is/are true?

(A)  For a given material of the capillary tube, h decreases with increase in r

(B) For a given material of the capillary tube, h is independent of σ

(C) If this experiment is performed in a lift going up with a constant acceleration, then h decreases

(D) h is proportional to contact angle θ

This question is rather easy. I am a bit surprised that recalling one single formula correctly leads you to the correct options. The examiner rightly guessed many students do not study Surface Tension, an important fluid property, in detail. If you are among those students, subscribe to www.physicsacademyonline.com, and study this lecture: Chapter Name – Elasticity and Fluid Mechanics; Category –  Basic; Topic Name – Surface Tension; Video Name – Rise or Fall of a Liquid in a Capillary Tube.

There is a ready-made formula waiting for you – sometimes called Jurin’s law – giving the height of water column, h, in terms of various quantities. Picking or leaving options (A), (B) and (D) is a matter of seconds. Option (C) will demand a little more respect.

Whenever they talk of a lift going up or down with constant acceleration or deceleration, how does the situation differ from that on the ground? What other force comes into play in addition to the real forces? Brush up your concept of Pseudo Force, and better still, follow a typical “Lift Problem” here: Chapter Name – Mechanics; Category –  Basic; Topic Name – Newton’s Laws of Motion; Video Name – Pseudo Force with Examples.

Now you can decide how the effective value of free-fall acceleration varies within a lift, and hence, how h varies.

A good question on Physics often increases the appetite for deeper theories and more questions. What happens if the mass of water in the meniscus is not negligible? What happens if the capillary tube is too short? Follow till the end of the first lecture mentioned in this question, and remain prepared for the next year’s Paper.

By the way, don’t forget to confirm your answer from our MP4 video solution below.