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Units & Dimensions Test - 4

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Units & Dimensions Test - 4
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Weekly Quiz Competition
  • Question 1
    1 / -0

    Which of the following is not the name of a physical quantity ?

    Solution

    Kilogram represent unit of physical quantity and not the physical quantity.

  • Question 2
    1 / -0

    Light year is the unit of

    Solution

    The light-year is a unit of length used to express astronomical distances and measures about 9.46 trillion kilometres or 5.88 trillion miles. As defined by the International Astronomical Union, a light-year is the distance that light travels in vacuum in one Julian year. 

  • Question 3
    1 / -0

    PARSEC is a unit of

    Solution

    The parsec (symbol: pc) is a unit of length used to measure large distances to astronomical objects outside the Solar System. A parsec is defined as the distance at which one astronomical unit subtends an angle of one arcsecond, which corresponds to 648000π astronomical units.

  • Question 4
    1 / -0

    Which of the following system of units is NOT based on the unit of mass, length and time alone

    Solution

    The SI system of units is a modern system and hence involves all the quantities that can't be derived using all the other quantities of the set. While the rest systems are old and local methods and hence are not scientifically accurate and explainable and thus only have three basic quantities while SI have 7.

  • Question 5
    1 / -0

    In the S.I. system the unit of energy is-

    Solution

    Joule is the SI unit of energy

  • Question 6
    1 / -0

    Unit of pressure in S.I. system is-

    Solution

    The SI unit of pressure is Pascal (represented as Pa) which is equal to one newton per square metre (N/m-2 or kg m-1s-2). Interestingly, this name was given in 1971. Before that pressure in SI was measured in newtons per square metre.

  • Question 7
    1 / -0

    The mutual inductance has unit of-

    Solution

    The unit of inductance in the SI system is the henry (H), named after American scientist Joseph Henry, which is the amount of inductance which generates a voltage of one volt when the current is changing at a rate of one ampere per second.

  • Question 8
    1 / -0

    In SI unit the angular acceleration has unit of-

    Solution

    Angular acceleration. Angular acceleration is the rate of change of angular velocity. In three dimensions, it is a pseudovector. In SI units, it is measured in radians per second squared (rad/s2), and is usually denoted by the Greek letter alpha (α).

  • Question 9
    1 / -0

    The SI unit of the universal gravitational constant G is

    Solution

    We know that,       g = G. m1 m2 / r2
    Where we know g has dimensions of acceleration
    Thus [G] = [ g.r2 /m1 m2]
    = N m2kg-2

  • Question 10
    1 / -0

    Which of the following statement is wrong ?

    Solution

    Unit of surface tension is Newton per metre

  • Question 11
    1 / -0

    What are the dimensions of length in force × displacement/time

    Solution

    [F] = MLT-2
    [displacement/time]  = LT-1
    Thus we get [F x displacement/time] = ML2T-3
    Thus the answer is 2

  • Question 12
    1 / -0

    The angular frequency is measured in rad s-1. Its dimension in length are :

    Solution

    Unit of angular frequency is rad/sec which can be said as angle/time. As angle is dimensionless and time has dimension T, we get the dimension of angular frequency as T-1

  • Question 13
    1 / -0

    The dimensional formula of coefficient of viscosity is

    Solution

    Coefficient of viscosity (η)= Fr/Av   

     F= tangential Force, Area, r= distance between the layers, v= velocity.

    Dimensional Formula of Force = M1L1T-2.
    Dimensional Formula of Area= M0L2T0.
    Dimensional Formula of distance= M0L1T0.
    Dimensional Formula of velocity= M0L1T-1.

    Putting these values in above equation we get,

    [η]= [M1L1T-2][M0L1T0] / [M0L2T0] [M0L1T-1] = [M1L-1T-1]

  • Question 14
    1 / -0

    A pair of physical quantities having the same dimensional formula is :

    Solution

    The dimensions of angular momentum are M L2T−1
    That of torque is  M L2T−2
    Also dimension of energy is  M L2T−2
    Where as same of force is  M LT−2
    And of power is  M L2T−3
    Thus we get torque and energy have the same dimensional formulas.

  • Question 15
    1 / -0

    Dimensions of pressure are the same as that of

    Solution

    [P] = [F/A] = MLT-2 / L2  = ML-1T-2
    [F/V] = MLT-2 / L3  = ML-2T-2
    [E/V] = ML2T-2 / L3  = ML-1T-2

  • Question 16
    1 / -0

     If momentum (P), area (A) and time (T) are taken to be fundamental quantities, then energy has the dimensional formula 

    Solution

    [E] = [F][d]
    = [P/T][A]½ 
    [E] = P1A½T-1

  • Question 17
    1 / -0

    If radian correction is not considered in specific heat measurement. The measured value of specific heat will be

    Solution

    Correct Answer :- b

    Explanation : If radian correction is not considered in specific heat measurement. The measured value of specific heat will be more than its actual value.

  • Question 18
    1 / -0

    The specific resistance has the unit of-

    Solution

    We know that specific resistance, ⍴ = R A / L
    Where R is net resistance, A is area and
    L is length
    Hence [⍴] = ohm x m.m / m
    = ohm.m

  • Question 19
    1 / -0

    One watt-hour is equivalent to

    Solution

    Power = energy / time.
    energy = power × time.
    energy = watt × hour
    = 1watt × 1hour
    = 1 × (60×60)
    = 3600
    = 3.6 × 103 joule

  • Question 20
    1 / -0

    The density of mercury is 13600 kg m-3. Its value of CGS system will be :

    Solution

    We know that density of Hg = 13600 kg/m.m.m
    = 13600 x 1000 g / 100cm x 100cm x 100cm
    = 13.6 g/cm.cm.cm

  • Question 21
    1 / -0

    What is the unit for measuring the amplitude of a sound?

  • Question 22
    1 / -0

    The unit of intensity of magnetisation is-

    Solution

    Intensity of magnetisation : It represents the extent to which a specimen is magnetised when placed in a magnetising field. Or in other words the intensity of magnetisation of a magnetic material is defined as the magnetic moment per unit volume of the material. 

    It indicates how the sample affected by magnetic field when placed in it.

    M = Magnetic moment/volume = μM / V

    unit SI : M= Amp.metre-1

  • Question 23
    1 / -0

    The M.K.S. units of coefficient of viscosity is-

    Solution

    We know that coefficient of viscosity (η)= Fr/Av where F = tangential force, r = distance between the layers , v = velocity and A is the area of the surface.
     Thus we get [η] = MLT-2.L / L2. (L/T)
    = M / LT
    Thus its unit is kg / m sec

  • Question 24
    1 / -0

    For 10(at+3) , the dimension of a is-

    Solution

    As 'at' and 3 are added in the equation, we get at and 3 have same dimensions i.e.1
    Thus a has dimensions same as 1/t.

  • Question 25
    1 / -0

    The pressure of 106 dyne/cm2 is equivalent to

    Solution

    We know that 105 dyne = 1N
    And 104cm2  = 1 m2
    Thus we get 10 dyne / cm2 = N / m2
    Hence 106 dyne / cm2 =105 N / m2

  • Question 26
    1 / -0

    The sum of the numbers 436.32, 227.2 and 0.301 in appropriate significant figures is

    Solution

    The result of an addition or subtraction in the number having different precisions should be reported to the same number of decimal places as present in the number having the least number of decimal places.

    The final result should, therefore, be rounded off to one decimal place, i.e. 663.8

  • Question 27
    1 / -0

     The SI unit of length is the meter. Suppose we adopt a new unit of length which equals to x meters. The area 1m2 expressed in terms of the new unit has a magnitude-

    Solution

    We have  1 unit = x meters
    So  1 unit2 = x2 meter2
    Hence, we get  1 meter2 = 1/x2 unit2

  • Question 28
    1 / -0

     r = 2 g/cm3 convert it into MKS system -

    Solution

    multiply the mass / volume value by 1000

  • Question 29
    1 / -0

    Given that v is the speed, r is radius and g is acceleration due to gravity. Which of the following is dimension less

    Solution

    v2r/g= (L1T-1)2L1/LT-2=L2
    v2/rg= (LT-1)2 L1(L1 T-2) ​=M0L0T0
    v2g/r​=(LT-1)2 LT-2/L​=L2T-4
    v2rg=(LT-1)2(L1) (LT-1) =L4T-3
    So, option D is correct.

  • Question 30
    1 / -0

     The value of G = 6.67 × 10_11 N m2 (kg)_2. Its numerical value in CGS system will be :

    Solution


    G = 6.67 x 10-11 Nm2/kg2

    so, when expressed in VGS units we shall convert N to dynes, m to cm and kg to g, thus

    G = 6.67 x 10-11 x [(105dynes x (102)2cm2 ) / (103)2 g2]

    =  6.67 x 10-11 x [ (105 x 104 ) / 106 ] 

    Thus,

    in CGS system

    G = 6.67 x 10-8 dyne.cm2/g2

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