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Units and Measurement

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Question
CBSEENPH11020782
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Dimensions of resistance in an electrical circuit, in terms of the dimension of mass M, of length L, of time T and of current I, would be:

  • [ML2T-3I-1]

  • [ML2T-2]

  • [ML2T-1I-1]

  • [ML2T-3I-2]

Solution

D.

[ML2T-3I-2]

Resistance 
straight R space equals space fraction numerator potential space difference over denominator current end fraction space equals space straight V over straight i space equals space straight W over qi left parenthesis therefore space potnetial space difference space is space equal space to space work space done space per space unit space charge right parenthesis equals space fraction numerator left square bracket Dimensions space of space work right square bracket over denominator left square bracket Dimensions space of space charge right square bracket left square bracket Dimensions space of space current right square bracket end fraction space equals space fraction numerator left square bracket ML squared straight T to the power of negative 2 end exponent right square bracket over denominator left square bracket IT right square bracket left square bracket straight I right square bracket end fraction space equals space left square bracket ML squared straight T to the power of negative 3 end exponent straight I to the power of negative 2 end exponent right square bracket

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Question
CBSEENPH11020525
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If energy (E), velocity (v) and time (T) are chosen  as the fundamental quantities, the dimensional formula of surface tension will be

  • [Ev-2T-1]

  • [Ev-1T-2]

  • [Ev-2T-2]

  • [E-2v-1T-3]

Solution

C.

[Ev-2T-2]

We space know space that surface space tension space left parenthesis straight S right parenthesis space equals space fraction numerator Force space left square bracket straight F right square bracket over denominator Length space left square bracket straight L right square bracket end fraction so comma space left square bracket straight S right square bracket space equals space fraction numerator MLT to the power of negative 2 end exponent over denominator left square bracket straight L right square bracket end fraction space equals space left square bracket ML to the power of 0 straight T to the power of negative 2 end exponent right square bracket Energy space left parenthesis straight E right parenthesis space equals space force space straight x space displacement rightwards double arrow space left square bracket straight E right square bracket space equals space left square bracket ML squared straight T squared right square bracket velocity space left parenthesis straight v right parenthesis space equals displacement over time left square bracket straight v right square bracket space equals space left square bracket LT to the power of negative 1 end exponent right square bracket straight S proportional to space straight E to the power of straight a straight v to the power of straight b straight T to the power of straight c Where space straight a comma space straight b comma space straight c space are space constants From space the space principle space of space homogeneity left square bracket LHS right square bracket space equals left square bracket RHS right square bracket rightwards double arrow space left square bracket ML to the power of 0 straight T to the power of negative 2 end exponent right square bracket space equals space left square bracket ML squared straight T to the power of negative 2 end exponent right square bracket to the power of straight a space left square bracket LT to the power of negative 1 end exponent right square bracket to the power of straight b left square bracket straight T right square bracket to the power of straight c rightwards double arrow space left square bracket ML to the power of 0 straight T to the power of negative 2 end exponent right square bracket space equals space left square bracket straight M to the power of straight a straight L to the power of 2 straight a plus straight b end exponent space straight T to the power of negative 2 straight a minus straight b plus straight c end exponent right square bracket Equating space the space power space on space both space sides comma space we space get straight a equals space 1 comma space 2 straight a space plus straight b space equals 0 comma space straight b equals negative 2 minus 2 straight a minus straight b plus straight c space equals negative 2 straight c equals space left parenthesis 2 straight a plus straight b right parenthesis minus 2 space equals space 0 minus 2 space equals negative 2 left square bracket straight S right square bracket space equals space left square bracket Ev to the power of negative 2 end exponent straight T squared right square bracket space equals space left square bracket Ev to the power of negative 2 end exponent straight T squared right square bracket

Question
CBSEENPH11020562
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If force (F), velocity (v) and time (T) are taken as fundamental units, then the dimensions of mass are

  • [FvT-1]

  • [FvT-2]

  • [Fv-1T-1]

  • [Fv-1T]

Solution

D.

[Fv-1T]

We know that,
F = ma
That is,
F = mv/t
straight m space equals space Ft over straight v
So, left square bracket straight M right square bracket space equals space fraction numerator left square bracket straight F right square bracket left square bracket straight T right square bracket over denominator left square bracket straight V right square bracket end fraction space equals space left square bracket space Fv to the power of negative 1 end exponent straight T right square bracket

Question
CBSEENPH11020717
Wired Faculty App

If the dimensions of a physical quantity are given by Ma Lb Tc, the physical quantity will be 

  • pressure if a = 1, b = -1, c= 2

  • velocity if a = 1, b = 1, c = -2

  • acceleration if a = 1, b = 1, c =-2

  • force if a = 0, b = -1, c =- 2

Solution

A.

pressure if a = 1, b = -1, c= 2

Dimensions of velocity = [ M0L1T-1]
Here, a = 0, b =1, c = -1
Dimension of acceleration = [ M0L1T-2]
Dimension of force = [ M1L1T-2]
Dimension of pressure = [M1L-1T-2
The physical quantity is pressure.

Question
CBSEENPH11020741
Wired Faculty App

If the error in the measurement of radius of sphere is 2%, then the error in the determination of volume of the sphere will be

  • 4%

  • 6%

  • 8%

  • 2%

Solution

B.

6%

Volume space of space straight a space sphere space equals space 4 over 3 straight pi space left parenthesis radius right parenthesis cubed or space straight V space equals space 4 over 3 πR cubed Taking space logarithium space on space both space sides comma space we space have space log space straight V space equals space log space 4 over 3 space πR cubed Taking space logarithium space on space both space sides comma space we space have log space straight V space equals space log space 4 over 3 straight pi space plus 3 space log space straight R Differentiating comma space we space get comma fraction numerator increment straight V over denominator straight V end fraction space equals space 0 space plus fraction numerator 3 increment straight R over denominator straight R end fraction Accordingly comma space fraction numerator increment straight R over denominator straight R end fraction space equals 2 percent sign Thus comma space fraction numerator increment straight V over denominator straight V end fraction space equals space 3 space straight x space 2 percent sign space equals space 6 percent sign 

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