JEE Main: Units, Dimensions And Error Analysis! Quiz

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  • 1/10 Questions
    Suppose refractive index is given as, where P and Q are constants and v is the speed of light and  is the wavelength, then dimensions of Q are the same as that of - ProProfs

    Suppose refractive index is given as, where P and Q are constants and v is the speed of light and  is the wavelength, then dimensions of Q are the same as that of

    • Wavelength
    • Volume
    • Force
    • Acceleration
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About This Quiz

This JEE Main quiz focuses on Units, Dimensions, and Error Analysis, assessing knowledge of physical constants, wave equations, and dimensions in various physical contexts. It is designed for students preparing for engineering entrance exams, enhancing their understanding of fundamental physics concepts.

JEE Main: Units, Dimensions And Error Analysis! Quiz - Quiz

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  • 2. 
    In the Equation   the dimensional formula for will be - ProProfs

    In the Equation   the dimensional formula for will be

    Correct Answer
    A.
    Explanation
    The trigonometrical ratios are the dimensionless quantities so for the equation   the term  will be dimensionless 

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  • 3. 
    The banking of roads is explained with the formula  here v is the speed of the object, r is the radius of the curve formed and g is the gravitational acceleration. Choose the correct statement about this relation. - ProProfs

    The banking of roads is explained with the formula  here v is the speed of the object, r is the radius of the curve formed and g is the gravitational acceleration. Choose the correct statement about this relation.

    • Above expression may be dimensionally correct but numerically it is wrong

    • Above expression is dimensionally as well as numerically correct

    • Above expression is dimensionally as well as numerically wrong

    • Above expression is dimensionally incorrect but numerically it is correct

    Correct Answer
    A. Above expression is dimensionally as well as numerically correct
    Explanation
    The formula is tan dimensionally correct as both the sides are dimensionless. L.H.S. i.e. is a dimensionless trigonometrical ratio and  is also dimensionless as the dimension of v2 and rg are same. Moreover this formula is proved with the help of various theorems and principals so it is also numerically correct. 

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  • 4. 
    A physical Quantity V depends on time as V = V0 (1+ e), where is a constant and t is time, then which of the following is true:   - ProProfs

    A physical Quantity V depends on time as V = V0 (1+ e), where is a constant and t is time, then which of the following is true:  

    • α is a dimensionless constant

    • Has dimensions of T2

    • α has dimensions of V

    • Has dimensions of LT-2

    Correct Answer
    A. Has dimensions of T2
    Explanation
    In the expression V = V0 (1+ ), the term has to be a dimensionless term as it is in the power of an exponential function so for being the dimensionless the dimensions of should be equal to the inverse of the dimensions of T-2 i.e. equal to the dimensions of T2. 

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  • 5. 
    The Equation of stationary wave is then which of the following is false: - ProProfs

    The Equation of stationary wave is then which of the following is false:

    Correct Answer
    A.
    Explanation
    All the options must be checked to find out the wrong statement. First option is correct as  and are dimensionless quantities. In second option x has the same units as that of   i. e. meter. In third option has dimension [L-1T] which is the inverse of v [LT-1]. Now only fourth option is left which is incorrect as has the dimension [T-1], whereas has the dimension [L].
     

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  • 6. 
    What will be the dimensions a b of according to the relation where E =  is Energy, x is displacement and t is time? - ProProfs

    What will be the dimensions a b of according to the relation where E =  is Energy, x is displacement and t is time?

    Correct Answer
    A.
    Explanation
    In the equation E =  the dimension of r.h.s. should be equal to the dimension of l.h.s. both sides must have the dimensions equal to the [ML2T-2]. Now as ‘a’ is added to t2 so the dimensions of ‘a’ will be equal to the dimensions of [T2]. For the dimension of ‘b’ [ML2T-2] = so from here b = [M-1L-3T4], now a b = [T2]  [M-1L-3T4] = [M-1L-3T6].
     

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  • 7. 

    What will be the dimensions of the mass, if C (velocity of light), g (acceleration due to gravity), and P (atmospheric pressure) are considered as the fundamental Quantities in the MKS system?

    • C/g

    • Pg

    • PCg

    • C/P

    Correct Answer
    A. C/P
    Explanation
    Dimensional formula of velocity of light (C) = [LT-1]
    Dimensional formula of acceleration due to gravity (g) = [LT-2]
    Dimensional formula of atmospheric pressure (P) = [ml-1T-2],
    Now let M = CagbPc; [M] = [LT-1] a [LT-2] b [ml-1T-2] c,
    [M1] = [La+b-c] [T-a-2b-2c] [M c];
    a+b-c = 0; -a-2b-2c = 0; c = 1; On solving the equations we get, a = 0, b = 1, c = 1,
    so, M = Pg.
     

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  • 8. 

    For any spring of spring constant k, the frequency of vibrations ‘f’ is given by the formula f = Xmakb, where X is a dimensionless constant. The value of ‘a’ and ‘b’ are, respectively:

    Correct Answer
    A.
    Explanation
    f = Xmakb dimension of f = [T-1], Dimensions of m = [M], Dimensions of spring constant ‘k’ = [MT-2] and X is a dimensionless constant.[T-1] = [M]a [MT-2] b; [T-1] = [M]a+b[T]-2b, now comparing the powers on both sides,a+1 = 0; -2b = -1 so b=1/2 and a= -1/2.

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  • 9. 

    The product of permittivity of free space and the permeability of free space has the dimensions equal to the:

    Correct Answer
    A.
    Explanation
    Dimension of permittivity of free space ( ) = [M-1L-3T4A2]Dimension of permeability of free space ( ) = [MLT-2A-2]Product of and = [M-1L-3T4A2] [MLT-2A-2]                                    = [L-2T2] 

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  • 10. 

    In Searle’s experiment to find Young’s modulus, the diameter of the wire is measured as d=0.050 cm, and the length of the wire is L = 100 cm. When a weight of 10.0 kg is placed, the extension in the wire was found to be 0.200 cm. The maximum permissible error (percentage) in Young’s modulus is:

    • 6.5

    • 4.3

    • 5.8

    • 4.9

    Correct Answer
    A. 4.9
    Explanation
    In Searle’s Experiment, Young’s Modulus Y =  where F is force, A is area, L is the original length and is the change in length due to force F. Here F = Mg and A = 4πd2 so Y = so the relative error in Y; 

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  • Current Version
  • Mar 19, 2023
    Quiz Edited by
    ProProfs Editorial Team
  • Jul 31, 2014
    Quiz Created by
    Tanmay Shankar
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