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    Units and Measurement: Fill Ups

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    Harshit Gautam

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    Units and Measurement: Fill Ups

    Uploaded by

    Harshit Gautam
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    Units and Measurement: Fill Ups

    Uploaded by

    Harshit Gautam

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    © All Rights Reserved
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    of 11

    Units and Measurement

    Fill Ups

    Q.1. Planck’s constant has dimension ________. (1985 - 2 Marks)

    Ans. ML 2 T –1

    Solution.

    Q.2. In the formula X = 3YZ2, X and Z have dimensions of capacitance and


    magnetic induction respectively. The dimensions of Y in MKSQ sytem are
    _________, ________. (1988 - 2 Marks)

    Ans. M –3 L –2 T 4 Q 4

    Solution.

    Q.3. Th e equation of state for real gas is given by The


    dimensions of the constant a is _________.

    Ans. ML5 T –2

    Solution.
    Subjective Questions

    Q.1. Give the MKS units for each of the following quantities. (1980)

    (i) Young’s modulus

    (ii) Magnetic Induction

    (iii) Power of a lens

    Ans. (i) N/m2;

    (ii) Tesla;

    (iii) Dioptre;

    Solution. The M.K.S. unit of Young’s modulus is Nm–2.

    The M.K.S. unit of magnetic induction is Tesla.

    The M.K.S. unit of power of lens is Dioptre.

    Q.2. A gas bubble, from an explosion under water, oscillates with a period T
    proportional to padbEc. Where ‘P’ is the static pressure, ‘d’ is the density of
    water and ‘E’ is the total energy of the explosion. Find the values of a, b and c.

    Ans. a = – 5/6, b = 1/2, c = 1/3

    Solution. Given that T ∝ PadbEc

    ⇒ [M0L0T1] = [ML–1T–2]a [ML–3]b [ML2T–2]c

    ∴ [M0L0T1] = [Ma + b + c L–a –3b + 2c T–2a – 2c]

    ∴ a + b + c = 0, – a – 3b + 2c = 0

    – 2a – 2c = 1

    On solving, we get

    a = – 5/6, b = 1/2, c = 1/3


    Q.3. Write the dimensions of the following in terms of mass, time, length and
    charge (1982 - 2 Marks)

    (i) magnetic flux

    (ii) rigidity modulus

    Ans. (i) [M1L2T–1Q–1] (ii) [ML–1T–2]

    Solution. Magnetic Flux = [M1L2T–1Q–1]

    Modulus of Rigidity = [ML–1T–2]

    Q.4. Match the ph ysical quan tities given in column I with dimensions
    expressed in terms of mass (M), length (L), time (T), and charge (Q) given in
    column II and write the correct answer against the matched quantity in a
    tabular form in your answer book.

    Column I Column II

    Angular momentum ML2T–2

    Latent heat ML2Q–2

    Torque ML2T–1

    Capacitance ML3T–1Q–2

    In ductan ce M–1 L–2 T2 Q2

    Resistivity L2T–2

    Solution.. Angular Momentum [ML2T –1] ;

    Latent heat [L2T –2] ;

    Torque [ML2T–2]

    Capacitance [M–1L–2T2Q2] ;
    Inductance [ML2Q–2];

    Resistivity [ML3T–1Q–2]

    Q.5. Column -I gives three physical quant ities. Select the appropriate units for
    the choices given in Column-II. Some of the physical quantities may have more
    than one choice correct :

    Column I Column II

    Capacitance (i) ohm-second

    In ductan ce (ii) coulomb2–joule–1

    Magnetic Induction (iii) coulomb (volt)–1

    (iv) newton (amp-metre)–1

    (v) volt-second (ampere) –1

    Ans. Capacitance coulomb-volt–1, coulomb2-joule–1

    In ductance ohm-sec, volt-second (ampere) –1

    Magnetic Induction newton (ampere-metre)–1

    (b) Refer to solution of Q. 3, type D

    Q.6. If nth division of main scale coincides with (n+1) th divisions of vernier
    scale. Given one main scale division is equal to ‘a’ units. Find the least count of
    the vernier. (2003 - 2 Marks)

    Ans. a/n + 1 units

    Solution. (n + 1) divisions of vernier scale = n divisions of main scale.

    ∴ One vernier division = a/n + 1 main scale division


    Q.7. A screw gauge having 100 equal divisions and a pitch of length 1 mm is
    used to measure the diameter of a wire of length 5.6 cm. The main scale reading
    is 1 mm and 47thcircular division coincides with the main scale. Find the curved
    surface area of wire in cm 2to appropriate significant figure.

    (use π = 22/7)

    Ans. 2.6 cm2

    Solution.

    Diameter = MSR + CSR × (least count)

    = 1 mm + 47 × (0.01) mm = 1.47 mm Surface Area = πDl

    = 2.6 cm2 (Rounding off to two significant figures)

    Q.8. In Searle’s exper iment, which is used to find Young’s Modulus of


    elasticity, the diameter of experimental wire is D = 0.05 cm (measured by a
    scale of least count 0.001 cm) and length is L = 110 cm (measured by a scale of
    least count 0.1 cm). A weight of 50 N causes an extension of X = 0.125 cm
    (measured by a micrometer of least count 0.001cm). Find maximum possible
    error in the values of Young’s modulus.

    Screw gauge and meter scale are free from error. (2004 - 2 Marks)

    Ans. 1.09 × 1010 Nm -2

    Solution.
    KEY CONCEPT : Maximum error in Y is given by

    It is given that W = 50 N; D = 0.05 cm = 0.05 × 10 –2m;

    X = 0.125 cm = 0.125 × 10–2m;

    L = 110 cm = 110 × 10 –2m

    ∴ Maximum possible error in the value of

    Y = DY = 0.0489 × 2.24 × 1011

    = 1.09× 1010 N/m2

    Q.9. The side of a cube is measured by vernier callipers (10 divisions of a


    vernier scale coincide with 9 divisions of main scale, where 1 division of main
    scale is 1 mm). The main scale reads 10 mm and first division of vernier scale
    coincides with the main scale. Mass of the cube is 2.736 g. Find the density of
    the cube in appropriate significant figures.

    Ans. 2.66 gm/cm3


    Match the Following

    DIRECTIONS (Q. No. 1) : Each question contains statements given in two


    columns, which have to be matched. The statements in Column-I are labelled A, B,
    C and D, while the statements in Column-II are labelled p, q, r and s. Any given
    statement in Column-I can have correct matching with ONE OR MORE statement(s)
    in Column-II. The appropriate bubbles corresponding to the answers to these
    questions have to be darkened as illustrated in the following example :

    If the correct matches are A-p, s and t; B-q and r; C-p and q; and D-s then the correct
    darkening of bubbles will look like the given.

    Q.1. Some physical quantities are given in Column I and some possible SI units
    in which these quantities may be expressed are given in Column II. Match the
    physical quantities in Column I with the units in Column II and indicate your
    answer by darkening appropriate bubbles in the 4 × 4 matrix given in the
    ORS. (2007)

    Column I Column II

    (A) GMeMs , G – universal gravitational constant,


    (p) (volt) (coulomb)(metre)
    Me – mass of the earth, Ms – mass of the Sun

    (B) 3RT/M , R – universal gas constant,


    (q) (kilogram) (metre)3(second)–2
    T – absolute temperature, M – molar mass

    (r) (metre)2 (second)–2


    F – Force, q – charge, B – magnetic field

    Me – mass of the earth, Re – radius of the (s) (farad) (volt)2 (kg)–1


    earth
    Ans. (A) → p, q ; (B) → r, s ; (C) → r, s ; (D) → r,s

    Solution. A : p → q

    Reason : Unit of GMeMs = Fr2 = Nm2

    = kg m3s–2

    Also (volt) (coulomb) (metre) = (joule) (metre)

    = (N - m) (m) = Nm2 = kg m3s–2

    B:r→s

    C:r→s

    ∴ Unit of v2 is m2s–2 which is further equal to FV2 kg–1.

    D:r→s

    DIRECTIONS (Q. No. 2) : Following question has matching lists. The codes for the
    lists have choices (a), (b), (c) and (d) out of which ONLY ONE is correct.
    Q.2. Match List I with List II and select the correct answer using the codes
    given below the lists: (JEE Adv. 2013)

    List I List II

    P. Boltzmann constant 1. [ML2T-1]

    Q. Coefficient of
    2. [ML–1T–1]
    viscosity

    R. Planck constant 3. [MLT–3K–1]

    S. Thermal
    4. [ML2T–2K–1]
    conductivity

    Codes:
    P Q R S

    (a) 3 1 2 4

    (b) 3 2 1 4

    (c) 4 2 1 3

    (d) 4 1 2 3

    Ans. (c)
    Solution.

    (c) is the correct option.


    Integer Value Correct Type

    Q.1. To find the distance d over which a signal can be seen clearly in foggy
    conditions, a railways-engineer uses dimensions and assumes that the distance
    depends on the mass density ρ of the fog, intensity (power/area) S of the light
    from the signal and its frequency f. The engineer finds that d is proportional to
    S1/n. The value of n is (JEE Adv. 2014)

    Ans. (3)

    Solution.

    Q.2. During Searle’s experiment, zero of the Vernier scale lies between 3.20 ×
    10–2 m and 3.25 × 10–2 m of the main scale. The 20th division of the Vernier
    scale exactly coincides with one of the main scale divisions. When an additional
    load of 2 kg is applied to the wire, the zero of the Vernier scale still lies between
    3.20 × 10–2 m and 3.25 × 10–2 m of the main scale but now the 45th division of
    Vernier scale coincides with one of the main scale divisions. The length of the
    thin metallic wire is 2 m and its cross-sectional area is 8 × 10–7m2. The least
    count of the Vernier scale is 1.0 × 10–5 m. The maximum percentage error in the
    Young’s modulus of the wire is (JEE Adv. 2014)

    Ans. (4)

    Solution.

    Here F, a and L are accurately known.


    Q.3. The energy of a system as a function of time t is given as E(t) = A 2 exp(–αt,)
    where a = 0.2 s–1. The measurement of A has an error of 1.25%. If the error in
    the measurement of time is 1.50%, the percentage error in the value of E(t) at t
    = 5 s is (JEE Adv. 2015)

    Ans. (4)

    Solution. E = A2 e–0.2t

    ∴ loge E = 2 loge A –0.2t

    On differentiating we get

    As errors always add up therefore

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