Us YOU Solve Physics JEE (Advanced 2018) Questions:
I wish to discuss a few more questions from JEE (Advanced) 2018, Physics Paper-2. Let’s pick another problem from Section 2. Remember only correct numerical value will fetch full 3 marks; no marks otherwise.
Question 11: A steel wire of diameter 0.5 mm and Young’s modulus 2 x 1011 N/m2 carries a load of mass M. The length of the wire with the load is 1.0 m. A vernier scale with 10 divisions is attached to the end of this wire. Next to the steel wire is a reference wire to which a main scale, of least count 1.0 mm, is attached. 10 divisions of the vernier scale correspond to 9 divisions of the main scale. Initially, the zero of vernier scale coincides with the zero of main scale. If the load on the steel wire is increased by 1.2 kg, the vernier scale division which coincides with a main scale division is _____________. Take g = 10 m/s² and π = 3.2.
For a wire of given length, diameter and Young’s modulus, what is its extension under a given load? Subscribe to www.PhysicsAcademyOnline.com, and follow this lecture: Chapter Name – Elasticity and Fluid Mechanics; Category – Basic; Topic Name – Elasticity; Video Name – Hooke’s Law and Young’s Modulus.
You get your result in the unit of length. But the question is somewhat different: which vernier scale division coincides with a main scale division. For that, you must know the working principle of a vernier scale, which is as follows:
Least count of a vernier = 1 main scale division – 1 vernier scale division
Extension of wire = least count of vernier x coinciding vernier scale division
Once you have arrived at the final answer, confirm it from the video solution below.
We move to Section 3 of Paper 2. This section contains four MCQs. Each question has two matching lists: List-I and List-II. Four options are given representing matching of elements from List-I and List-II. Only one of these four options corresponds to a correct matching. You score 3 full marks by choosing the correct option, zero marks by choosing no options, and – 1 mark in all other cases.
Question 16: A planet of mass M has two natural satellites of masses m1 and m2 . The radii of their circular orbits are R1 and R2 respectively. Ignore the gravitational force between the satellites. Define v1 , L1 , K1 and T1 to be, respectively, the orbital speed, angular momentum, kinetic energy and time period of revolution of satellite 1. Similarly, v2 , L2 , K2 and T2 denote the corresponding quantities of satellite 2. Given m1/m2 = 2 and R1/R2 = 1/4, match the ratios in List-I to the numbers in List-II.
(A) P → 4 ; Q → 2 ; R → 1 ; S → 3
(B) P → 3 ; Q → 2 ; R → 4 ; S → 1
(C) P → 2 ; Q → 3 ; R → 1 ; S → 4
(D) P → 2 ; Q → 3 ; R → 4 ; S → 1
Although the MCQ is presented in a novel format, your task is quite clear. For the two satellites, the ratios of their masses and orbital radii are known. You calculate the ratios from List-I, match against the numbers in List-II, and choose the correct option.
What is the orbital speed of a satellite revolving round a planet? Watch the lecture: Chapter Name – Mechanics; Category – Basic; Topic Name – Gravitation; Video Name – Natural and Artificial Satellites of Planets.
The ratio v1/v2 comes out in a jiffy. What is the angular momentum of a satellite (particle) in circular motion about centre? If you need to brush up, it’s here: Chapter Name – Mechanics; Category – Basic; Topic Name – Rotational Mechanics; Video Name – Angular Momentum of a Particle and Its Relation to Torque.
The ratio L1/L2 follows. What is the kinetic energy of a satellite? The simple formula for K.E. is enough. For an alternative expression, you may watch: Chapter Name – Mechanics; Category – Basic; Topic Name – Gravitation; Video Name – Mechanical Energy of Satellite-Earth System.
The ratio K1/K2 is found. What is the time period of a satellite? Once again, the simple formula from Circular Motion will do. However, you are always welcome to see some typical problems from: Chapter Name – Mechanics; Category – Basic; Topic Name – Gravitation; Video Name – Problems on Orbital Motion of Satellites.
The ratio T1/T2 is out. Now, that was the lengthy way of solving the question. Usually, for an MCQ of this type, there is a shortcut that saves precious time. Returning to the first step of calculation, the moment you find v1/v2, you know which number it matches with. And luckily, of the four options (A) – (D) given, only one shows that match! Pick that option confidently, and don’t bother to calculate the other ratios. Is not that really sweet?
Try both methods for practice. Return later to compare with the video solution provided.
Question 18: In the List-I below, four different paths of a particle are given as functions of time. In these functions, α and β are unequal positive constants of appropriate dimensions. In each case, the force acting on the particle is either zero or conservative. In List-II, five physical quantities of the particle are mentioned: p is the linear momentum, L is the angular momentum about the origin, K is the kinetic energy, U is the potential energy and E is the total energy. Match each path in List-I with those quantities in List-II which are conserved for that path.
|P.||r(t) = αti + βtj||1.||p|
|Q.||r(t) = α cosωt i + βsinωt j||2.||L|
|R.||r(t) = α(cosωt i + sinωt j)||3.||K|
|S.||r(t) = αti + (βt2/2)j||4.||U|
(A) P → 1, 2, 3, 4, 5 ; Q → 2, 5 ; R → 2, 3, 4, 5 ; S → 5
(B) P → 1, 2, 3, 4, 5 ; Q → 3, 5 ; R → 2, 3, 4, 5 ; S → 2, 5
(C) P → 2, 3, 4 ; Q → 5 ; R → 1, 2, 4 ; S → 2, 5
(D) P → 1, 2, 3, 5 ; Q → 2, 5 ; R → 2, 3, 4, 5 ; S → 2, 5
At first glance this MCQ looks challenging, since each motion under List-I may correspond to one or more numbers under List-II. Moreover, no number is exclusive to any particular motion. But then, the “shortcut technique” discussed in the previous question will come in handy to make the task easier. No need to examine all four motions. Investigate any two of them, check if the given quantities remain conserved, and hopefully, the correct option will come out.
We suggest motions P and S, because in each case the radius vector, r, looks a simple function of time. How do you find the corresponding velocity vector, v, and acceleration vector, a, of the particle? For a typical problem on vector differentiation, see: Chapter Name – Mechanics; Category – Advanced; Topic Name – Vectors; Video Name – Advanced-Level Problems on Vectors II.
Once you find v, you can tell whether linear momentum vector, p, kinetic energy, K, and angular momentum vector, L, remain conserved or not. For the basics on linear momentum, watch: Chapter Name – Mechanics; Category – Basic; Topic Name – Newton’s Laws of Motion; Video Name – Newton’s Second Law of Motion. For the basics on angular momentum, watch: Chapter Name – Mechanics; Category – Basic; Topic Name – Rotational Mechanics; Video Name – Angular Momentum of a Particle and Its Relation with Torque.
The last lecture suggests an alternative way to check if L is conserved. Find a, then force vector, F, and then torque vector, τ , about the origin. You are close!
If a particle moves under conservative force, what can you say about its total (mechanical) energy? Go through this lecture: Chapter Name – Mechanics; Category – Basic; Topic Name – Work and Energy; Video Name – Principle of Conservation of Mechanical Energy and Some Applications.
Now you can easily tell if the total energy, E, and potential energy, U, remain conserved for motions P and S.
Students are advised to try other shortcuts to arrive at the correct choice. For instance, examining motion S alone could be enough! At the end, have a look at the video solution given.