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JEE Main Test 1
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JEE Main Test 1

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  • Question 1/12
    4 / -1

    Two masses m1 = 5 Kg and m2 = 10 Kg, connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of horizontal surface is 0.15. The minimum weight m that should be put on top of m2 to stop the motion is : , 

    Solutions

    Free Body diagram of the system is 

    From figure equation of motion of the system is 

     

  • Question 2/12
    4 / -1

    A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely proportional to the nth power of R. If the period of rotation of the particle is T, then :

    Solutions

    When particle move in circular motion then central force equals to the centripetal force 

     

  • Question 3/12
    4 / -1

    A telephonic communication service is working at carrier frequency of 10 GHz. Only 10% of it is utilized for transmission. How many telephonic channels can be transmitted simultaneously if each channel requires a bandwidth of 5 kHz?

    Solutions

    Since the carrier frequency is distributed as bandwidth frequency, so 
    10% of 10 GHz = n × 5 kHz, where n is the number of channels 

     

  • Question 4/12
    4 / -1

    A granite rod of 60 cm length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is 2.7 × 103 Kg/m3 and its Young’s modulus is 9.27 × 1010 Pa. What will be the fundamental frequency of the longitudinal vibrations?

    Solutions

     

  • Question 5/12
    4 / -1

    The insoluble alkali metal halide is

    Solutions

    LiF has low solubility in water due to greater force of attraction between lithium ions and fluoride ions in its crystal lattice.

     

  • Question 6/12
    4 / -1

    Which one of the following complexes will most likely absorb visible light? 

    Solutions

    V4+ has electronic config, 4s0 3d1 

    Since it has a d electron which can undergo d-d transition to show colour by absorbing visible light photons.

     

  • Question 7/12
    4 / -1

    Benzene diazonium chloride is usually unstable and decomposes in dry state. It is converted to a specific compound so that it can be kept in room temperature. Identify the compound.

    Solutions

    Benzenediazonium fluoroborate is water insoluble and stable at room temperature.Hence benzene diazonium chloride is stabilized by making its fluoroborate.

     

  • Question 8/12
    4 / -1

    Which of the following sets of quantum numbers represent an impossible arrangement (n, l, m, ms respectively)

    Solutions

    Value of principal quantum number is n. 

    The value of azimuthal quantum number (l) is always an integer between 0 to n-1. 

    The value of magnetic quantum number (m) is an integer between –l to +l. 

    Hence, in arrangement C: 

    value of n = 3 (valid) 

    value of l = 2 (valid as it is between 0 to 2) 

    value of m = -3 (invalid because it is not between -2 to +2)

     

  • Question 9/12
    4 / -1

    The number of real solutions of the equation x2 − 3 |x| + 2 = 0 is

    Solutions

    x2 − 3 |x| + 2 = 0 

    (|x| − 1) (|x| − 2) = 0 

    ⇒ x = ± 1, ± 2.

    Hence there are 4 real solutions.

     

  • Question 10/12
    4 / -1

    The letter of the word ‘EQUATION’ are arranged in all possible ways such that the consonants are in consecutive positions. Find the number of possible arrangements.

    Solutions

    There are 5 vowels and 3 consonants. 

    The consonants can be arranged in 6 ways together. 

    Therefore the number of ways in which the vowels can be arranged is 6 × 5! = 6!

    Number of ways consonants can be arranged is 3!. 

    ∴ n = 3!×6!

    ⇒ n = 4320

     

  • Question 11/12
    4 / -1

    The distance between the line  and the plane = 5 is

    Solutions

    We see that the line and the plane are parallel as =0

    Hence we can pick any point on the line and find it's distance from the plane. We have

    distance = |(2-10+3-5)/3(root3)|

    =

     

  • Question 12/12
    4 / -1

    The number of solutions of the inequality  , 

    where αi ∈ (-π, 2π) for i = 2, 3, ….. , n is:

    Solutions

    Comparing both sides of inequality. 
    sin2αi = 1 
     (i = 2, 3, …..n) 
    no. of solutions = 3 × 3 × 3 × ….. (n – 1) times = 3n – 1.

     

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