Self Studies

Units and Measurements Test - 72

Result Self Studies

Units and Measurements Test - 72
  • Score

    -

    out of -
  • Rank

    -

    out of -
TIME Taken - -
Self Studies

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Self Studies Self Studies
Weekly Quiz Competition
  • Question 1
    1 / -0
    The accuracy in the measurement of the diametre of a hydrogen atom as $${ 1.06\times 10 }^{ -10 }$$ m is
    Solution

  • Question 2
    1 / -0
    The diameter of wire measured with a varnier calliper with keast count 0.1 mm is Scm, then the percentage error in measurement is
  • Question 3
    1 / -0
    Of the following quantities , which one has dimensions different from the remaining three?
    Solution
    Energy density $$=\dfrac{Energy}{Volume}=\dfrac{ML^2T^{-2}}{L^3}=ML^{-1}T^{-2}$$

    Force per unit area $$=\dfrac{Force}{Area}=\dfrac{MLT^{-2}}{L^2}=ML^{-1}T^{-2}$$

    Product of the charge per unit volume and voltage $$=\dfrac{Work}{Charge}\times \dfrac{Charge}{Volume}=\dfrac{Work}{Volume}$$$$=\dfrac{ML^2T^{-2}}{L^3}=ML^{-1}T^{-2}$$

    Angular momentum per unit mass $$=\dfrac{M^1L^2T^{-1}}{M}=L^2T^{-1}$$

    So correct answer is option (D).
  • Question 4
    1 / -0
    If error in measurement of radius of a sphere is $$1\%$$, what will be the error in measurement volume?
    Solution
    $$Volume=\dfrac { 4πR^3 }{ 3 } $$

     $$\dfrac { ΔV }{ V } \times 100=3\dfrac { ∆R }{ R } \times 100$$

     $$=3\times \dfrac { 1 }{ 100 } \times 100$$

     $$=3$$

    ∴Percentage error in volume is $$3$$%
  • Question 5
    1 / -0
    Which of the following measurements is most accurate?
    Solution

  • Question 6
    1 / -0
    The dimensions of latent heat are - 
    Solution

    $$Latent\, heat (L)=\dfrac{Heat}{Mass}$$  . . . . . . (1)

    The dimensional formula of mass $$=[M^1L^0T^0]$$. . . . (2)

    Also, the dimensions of heat $$=$$ dimensions of energy $$=$$ dimensions of work

    Since, $$work = force\times displacement$$ . . . . (3)

    And, the dimensional formula of,

    Displacement $$= [M^0L^1T^0] $$. . . (4)

    Force $$= m \times a = [M^1L^1T^{-2}]$$ . . . (5)

    On substituting equation (4) and (5) in equation (3) we get,

    Work $$=[M^1L^1T^{-2}]\times [L^{1}]$$

    Therefore, the dimensions of work or heat $$=[M^1L^2T^{-2}]$$ . . . . (6)

    On substituting equation (2) and (6) in equation (1) we get,

    Or, $$L = \dfrac{[M^1L^2T^{-2}] }{ [M^1L^0T^0]}=[M^0L^2T^{-2}]$$

    Therefore, latent heat is dimensionally represented as $$[M^0L^2T^{-2}]$$

  • Question 7
    1 / -0
    If pressure $$P$$, area $$A$$ and length $$L$$ are taken to be fundamental quantities in a system of units, then energy has dimensional formula.
    Solution
    We know that
    $$P = \dfrac {P}{A} = \dfrac {MLT^{-2}}{L^{2}}$$
    $$= ML^{-1} T^{-2}$$
    $$A = L^{2}$$
    $$L = L$$
    Let $$E = P^{x} A^{y} L^{z}$$
    $$E = ML^{2}T^{-2}$$
    $$= (ML^{-1}T^{-2})^{x} (L^{2})^{y} (L)^{2}$$
    $$= M^{x} L^{-x + 2y + z} T^{-2x}$$
    $$x = 1, -2x = -2$$
    $$x = 1$$
    $$-x + 2y + z = 2$$
    $$z = 0, 2y = 2 + 1$$
    $$y = \dfrac {3}{2}$$
    $$\therefore E = P^{1} A^{3/2} L^{0}$$.
  • Question 8
    1 / -0
    A uniform rod undergoes a longitudinal strain of $$2$$. find percentage change in its volume if poisson's ratio of the material is $$0.50$$
    Solution
    We know that realtion between moduler of elasticity $$\left( \varepsilon \right),$$ possion's ratio $$\left( u\right)$$ and balls modules elasticity $$\left(h\right).$$
    Given by,
    $$\Rightarrow \varepsilon  =3h\left( 1-2u\right)$$
    here given $$u=0.3$$
    $$h\rightarrow \alpha$$
    and we also know $$\rightarrow h=\dfrac{\dfrac{-\triangle p}{\triangle v}}{v}$$
    If $$h$$ tends to infinite then $$\triangle v$$ tends to zero.
    $$\%$$ change in volume will be zero.
    Hence, solve.

  • Question 9
    1 / -0
    The error in the measurement of length of a simple pendulum is $$0.1\%$$ and error in the time period is $$2 \%$$. The possible maximum error in the quantity having dimensional formula $$LT^{-2}$$ is
    Solution
    Given,
    Error $$\%$$ in time $$=2\%=\dfrac{2}{100}=\dfrac{1}{50}=2\%$$
    Error $$\%$$ in length $$=0.1\%$$
    Find $$\%$$ error in $$\dfrac{1}{12}$$
    $$\therefore$$ as $$\dfrac{\triangle A}{A}=\left( \dfrac{\triangle 2}{2}\right) +\left( \dfrac{\triangle Q}{Q}\right)$$    if $$A=\dfrac{t}{Q}$$
    Thus, 
    $$\Rightarrow \%$$ error $$=\left( \dfrac{\triangle L}{L}\right) +2\left( \dfrac{\triangle T}{T}\right)$$
                       $$=\left( 2\%\right) +2\left( .1\%\right)$$
                       $$=2\%+.2\%$$
                       $$=2.2\%$$
    Hence, the answer is $$2.2\%.$$

  • Question 10
    1 / -0
    Which of the following is a dimensional constant?
    Solution
    Hint:- Check the dimensional formula of all given quantities

    Explanation :-
    A) Refractive index:- It is the ratio of velocity of light in a  vacuum to its velocity in specified medium.
    $$\mu=\dfrac{c}{v}$$
    $$[\mu]=\dfrac{[c]}{[v]}=\dfrac{[M^0L^0T^{-1}]}{[M^0L^1T^{-1}]}$$
    Hence, refractive index is dimensionless quantity.
    B) Passions ratio:- The ratio of the proportional decreases in a lateral measurement to the proportional increases in length in a sample of the material that is elastically stretched.
    Hence, it is dimensionless quantity.
    C) Stress:- It is the ratio of force to the area.
    $$s=\dfrac{F}{A}$$
    $$[]=\dfrac{[MLT^{-2}]}{[L^2]}=[ML^{-1}T^{-2}]$$
    It is not a dimensionless quantity. But it is not a constant value.
    D) Gravitational constant:- It is a universal constant, denoted by $$G$$.
    $$G=6.674\times 10^{-11}m^3/kgs^2$$
    Dimensional value, $$[G]=[M^{-1}L^3T^{-2}]$$

    The correct option is D.

Self Studies
User
Question Analysis
  • Correct -

  • Wrong -

  • Skipped -

My Perfomance
  • Score

    -

    out of -
  • Rank

    -

    out of -
Re-Attempt Weekly Quiz Competition
Self Studies Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Self Studies
Click on Allow to receive notifications
Allow Notification
Self Studies
Self Studies Self Studies
To enable notifications follow this 2 steps:
  • First Click on Secure Icon Self Studies
  • Second click on the toggle icon
Allow Notification
Get latest Exam Updates & FREE Study Material Alerts!
Self Studies ×
Open Now