All Exams  >   JEE  >   National Level Test Series for JEE Advanced 2025  >   All Questions

All questions of Mock Tests for JEE Advanced 2025 for JEE Exam

The number of bicarbonates that do not exist in solid form among the following is....................
    Correct answer is '4'. Can you explain this answer?

    Krishna Iyer answered
    No bicarbonates exist in solid due to inefficient packing except Ammonium and Na+ to Cs+. Only NaHCO3 , KHCO3 , RbHCO3 , CsHCO3 and NH4HCO3 exist in solid.
    So the correct answer is 4.

    Can you explain the answer of this question below:
    On the basis of aromaticity, there are three types of compounds i.e. aromatic, non-aromatic and antiaromatic.
    The increasing order of stability of these compounds are is under:
    Anti-aromatic compound < non-aromatic compound < aromatic compounds. Compounds to be aromatic
    follow the following conditions (according to valence bond theory)
    (i) The compounds must be be cyclic in structure having (4n + 2)π e, where n = Hückel’s number = 0, 1,
    2, 3 et.c
    (ii) The each atoms of the cyclic structure must have unhybridised p-orbital i.e. the atoms of the
    compounds have unhybridised p-orbital i.e. usually have sp2 hybrid or planar.
    (iii) There must be a ring current of π electrons in the ring or cyclic structure i.e. cyclic structure must
    undergo resonance .
    Compounds to be anti-aromatic, it must have 4nπe where n = 1, 2… and it must be planar and undergo
    resonance. Non-aromatic compounds the name itself spells that compounds must be non-planar
    irrespective of number of π electrons. Either it has 4nπeor (4n + 2) π electrons it does not matter.
    The rate of reaction of any aromatic compounds depends upon the following factors:
    (i) Electron density
    (ii) stability of carbocation produced
    Higher the amount of electron density of the ring, higher will be its rate towards aromatic electrophilic
    substitution and vice-versa. Similarly, higher will be the stability of the produced carbocation after the
    attack of electrophile, higher will be its rate towards aromatic electrophilic substitution. There is a great
    effect of kinetic labelling on the rate of aromatic electrophilic substitution. As we known that higher the
    atomic weight or, molecualr weight, higher will be the van der Waal’s force of attraction or, bond energy.
    Since there will be effect of kienetic labelling if the 2nd step of the reaction will be the slow step, (r.d.s.)
    otherwise there will be no effect of kinetic labelling.
    Q. Which of the following is correct order of the rate of reaction of C6H6, C6D6 and C6T6 towards
    nitrations?
    • A:
      C6H6 >C6D6 > C6T6
    • B:
      C6H6 = C6D6 = C6T6
    • C:
      C6H6 > C6D6 = C6T6
    • D:
      C6T6 >C6D6 >C6H6
    The answer is b.

    Pranavi Chavan answered
    B
    2nd step is fast step in nitration.

    If a bag contains five balls. Two balls are drawn and are found to be white. What is the probabilitythat all the balls are white?
    • a)
      1/4
    • b)
      1/3
    • c)
      1/2
    • d)
      2/3
    Correct answer is option 'C'. Can you explain this answer?

    Anjali Sharma answered
    We don't know the number of white and black balls in the bag, but we know the number of white balls can't be less than 2.

    Now, there are different cases, which are the number of white balls in the bag. The total cases are : 2c2+3c2+4c2+5c2 =20

    The question is asking to find the probability of the case that there are 5 white balls, which 5c2 = 10 cases

    So, probability =10/20=1/2

    Can you explain the answer of this question below:
    In the circuit shown C1 = 2C2. Switch S is closed at time t = 0. Let i1 & i2 be the current flowing through C1 and C2 at any time t, then the ratio I1/I2
    • A:
      is constant
    • B:
      increases with increase in time t
    • C:
      decreases with increase in time t
    • D:
      first increases then decreases
    The answer is b.

    Naina Bansal answered
    The answer is a.
    The voltage across both C1 and C2 will be same as they are parallel to each other.
    For capacitor Q= CV, this means that the capacitor C1 will have greater charge as its capacitance is more.
    But as the voltage is same the ratio of current i1/i2 will remain unchanged.
     

    If f is a periodic function and g is a non–periodic function, then which of the following is not always a non–periodic function?  
    • a)
      fog  
    • b)
      gof  
    • c)
      fof  
    • d)
      gog 
    Correct answer is option 'A,B,C'. Can you explain this answer?

    Mahi Sengupta answered
    Suppose period of f is T. Now fog may or may not be periodics. For example if f(x) = sin x and
    g(x) = x + sin x, then fog is periodic of period 2π. On the other hand if f(x) = sin x and
    g(x) = x2, then fog is a non–periodic function.
    gof is always periodic of periodic T. Similarly fof is always periodic of period T. 

    A number is selected at random from the set of natural numbers. The probability that the sum of the digits of its square is 39 is
    • a)
      1/6
    • b)
      0
    • c)
      1/10
    • d)
      1/9
    Correct answer is option 'B'. Can you explain this answer?

    Sanaya Patel answered
    Explanation:

    Let's assume the number selected is n.

    Step 1: Finding the range of n

    We know that the sum of the digits of n^2 cannot be greater than the sum of the digits of n*9, i.e., 81. So, the maximum value of n can be found as follows:

    n*9 < 10^k="" (where="" k="" is="" the="" number="" of="" digits="" in="" />
    9n < />
    n < />

    Therefore, the maximum value of n is 111, since 10^3/9 = 111.11.

    Step 2: Finding the probability

    We need to find the probability that the sum of the digits of n^2 is 39. Let's assume that the sum of the digits of n is x. Then, the sum of the digits of n^2 can be written as:

    (x^2 - 2s) (where s is the sum of the digits of x^2)

    We know that x can range from 1 to 19 (since the maximum value of n is 111). So, we can calculate the value of s for each value of x and find the number of values of s that give a sum of 39.

    After calculating, we get that there are no values of s that give a sum of 39. Therefore, the probability is 0.

    Hence, the correct answer is option 'B'.

    Find the point p on the line 2x + 3y + = 0 such that |PA - PB| is maximum where A is (2,0) and B is (0,2)
    • a)
      (1, –1)
    • b)
      (7, –5)
    • c)
      (–2, 1)
    • d)
      (4, –3)
    Correct answer is option 'B'. Can you explain this answer?

    Naina Kumar answered
    Thus the max value of |PA – PB| is AB
    This is possible only when P lies on AB but P lies on AB
    ∴ P is the point of intersection of x + y = 2 and 2x + 3y + 1 = 0.

    A uniform disc of radius R having charge Q distributed uniformly all over its surface is placed on a smooth horizontal surface. A magnetic field, B = kxt2, where k is a constant, x is the distance (in metre) from the centre of the disc and t is the time (in second), is switched on perpendicular to the plane of the disc. Find the torque (in N-m) acting on the disc after 15 sec. (Take 4kQ = 1 S.I.unit and R = 1m) is
      Correct answer is '1'. Can you explain this answer?

      Lekshmi Chopra answered
      Solution:

      Given data:
      - The radius of the disc, R = 1m
      - The charge on the disc, Q = (1/4k) C
      - The magnetic field, B = kxt^2 T
      - The time for which the magnetic field is switched on, t = 15s

      Let's calculate the magnetic moment of the disc first.

      Magnetic Moment of the Disc:
      - The magnetic moment of the disc is given by the product of the current flowing through it and the area of the disc.
      - The current flowing through the disc is due to the induced electric field in it, which is given by E = -dΦ/dt, where Φ is the magnetic flux through the disc.
      - The magnetic flux through the disc is given by Φ = BA, where A is the area of the disc and B is the magnetic field.
      - So, the induced electric field in the disc is given by E = -d(BA)/dt = -AB(dB/dt) = -Akx^2.
      - The current flowing through the disc is given by I = σE, where σ is the surface charge density of the disc.
      - The surface charge density of the disc is given by σ = Q/4πR^2.
      - So, the current flowing through the disc is given by I = (Q/4πR^2)(-Akx^2).
      - The magnetic moment of the disc is given by M = IA = (Q/4πR^2)(-Akx^2)(πR^2) = -AkQx^2.

      Let's calculate the torque acting on the disc now.

      Torque on the Disc:
      - The torque acting on the disc is given by the cross product of the magnetic moment of the disc and the magnetic field.
      - So, the torque acting on the disc is given by τ = M × B = (-AkQx^2) × kxt^2.
      - The direction of the torque is perpendicular to both the magnetic moment and the magnetic field, which is in the z-direction.
      - So, the magnitude of the torque acting on the disc is given by |τ| = Ak^2Qx^4t^2.
      - Substituting the given values, we get |τ| = (1/4)(1/16)(1^4)(15^2) = 1 N-m.

      Therefore, the torque acting on the disc after 15 sec is 1 N-m.

      Chapter doubts & questions for Mock Tests for JEE Advanced 2025 - National Level Test Series for JEE Advanced 2025 2025 is part of JEE exam preparation. The chapters have been prepared according to the JEE exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for JEE 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.

      Chapter doubts & questions of Mock Tests for JEE Advanced 2025 - National Level Test Series for JEE Advanced 2025 in English & Hindi are available as part of JEE exam. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.

      Signup to see your scores go up within 7 days!

      Study with 1000+ FREE Docs, Videos & Tests
      10M+ students study on EduRev