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 Multiple Choice QuestionsMultiple Choice Questions

1.

The velocity v of a particle at time t is given by straight v equals at plus fraction numerator straight b over denominator straight t plus straight c end fraction comma where a, b and c are constants, The dimensions of a, b and c are respectively:

  • open square brackets LT to the power of negative 2 end exponent close square brackets comma space open square brackets straight L close square brackets space and space open square brackets straight T close square brackets
  • open square brackets straight L squared close square brackets space open square brackets straight T close square brackets space and space open square brackets LT squared close square brackets
  • open square brackets LT squared close square brackets comma space open square brackets LT close square brackets space and space open square brackets straight L close square brackets
  • open square brackets LT squared close square brackets comma space open square brackets LT close square brackets space and space open square brackets straight L close square brackets

A.

open square brackets LT to the power of negative 2 end exponent close square brackets comma space open square brackets straight L close square brackets space and space open square brackets straight T close square brackets

According to principle of homogeneity of dimensions, the dimensions of all the terms in a physical expression should be same.
   The given expression is 
                      straight v equals at plus fraction numerator straight b over denominator straight t plus straight c end fraction
From principle of homogeneity
     open square brackets straight a close square brackets space open square brackets straight t close square brackets space equals space open square brackets straight v close square brackets
space space space space space open square brackets straight a close square brackets space equals space fraction numerator open square brackets straight v close square brackets over denominator open square brackets straight t close square brackets end fraction space equals space fraction numerator open square brackets LT to the power of negative 1 end exponent close square brackets over denominator open square brackets straight T close square brackets end fraction space equals space open square brackets LT to the power of negative 2 end exponent close square brackets
Similarly comma space space open square brackets straight c close square brackets space equals space open square brackets straight t close square brackets space equals space open square brackets straight T close square brackets
Further comma space space space space space fraction numerator open square brackets straight b close square brackets over denominator open square brackets straight t plus straight c close square brackets end fraction space equals open square brackets straight v close square brackets

or space space space space space space space space space space space space open square brackets straight b close square brackets space equals space open square brackets straight v close square brackets space open square brackets straight t plus straight c close square brackets
or space space space space space space space space space space space space open square brackets straight b close square brackets space equals space open square brackets LT to the power of negative 1 end exponent close square brackets space open square brackets straight T close square brackets space equals space open square brackets straight L close square brackets

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2.

A force F is given by F =at + bt2 ,where, t istime. What are the dimensions of a and b ?

  • M L T -1 and M L T0

  •  M L T -3 and M L2 T4

  • M L T -4 and M L T1

  • M L T -3 and M L T -4


D.

M L T -3 and M L T -4

Force F = at + bt2

From the principle of homogeneity

Dimension of at = dimension of F

Dimensionofa=Ft=MLT-2T=MLT-3Dimensionofb=dimensionofBb=Ft2=MLT-2T2=MLT-4


3.

The dimensional formula for Young's modulus is

  • [ML-1T-2]

  • [M0LT-2]

  • [MLT-2]

  • [ML2T-2]


A.

[ML-1T-2]

The formula for Young's modulus is

Y =stressstrain=F/AΔl/l

Y=ML-1T-2M0L0T0=ML-1T-2


4.

The length, breadth and thickness of ablock are given by l = 12 cm, b = 6 cm andt = 2.45 cm. The volume of the block according to the idea of significant figures should be

  • 1 × 102 cm3

  • 2 × 102 cm3

  • 1.76 × 102 cm3

  • None of these


B.

2 × 102 cm3

Using relation for volume

Given:- length of a block = 12 cm

breadth of a block = 6 cm

thickness of a block = 2.45cm

V =length × breadth × thickness

=12 × 6 × 2.45 = 176.4

=1.764 ×102 cm3

The minimum number of significant figures is 1in thickness, hence the volume will contain only onesignificant figure.

Therefore V= 2 ×102 cm3


5.

In a system of units if force (F), acceleration (A),and time (T) are taken as fundamental units,then the dimensional formula of energy is:

  • FA2T

  • FAT2

  • F2AT

  • FAT


B.

FAT2

Let energy E = Fx Ay T3

Writing the dimensions on both sides

ML2T-2=MLT-2xLT-2yT2ML2T-2=MxLx+yT-2x-2y+zOncomparingbothsidesx=1x+y=1y=2-x=2=2-1=1and-2x-2y+z=-2z=-2×1-2×1+zDimensionalformulaofenergy=FAT2


6.

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-1I-1]


D.

[ML2T-1I-1]

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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7.

If C and R denote capacity and resistance, the dimensions of C R are :

  • M0L0T

  • ML0T

  • M0L0T2

  • not expressible in terms of M, L and T


A.

M0L0T

RC is the time constant of the circuit, so its dimensions are [M0L0T].


8.

The length, breadth and thickness of a block are given by l = 12 cm, b = 6 cm and t = 2.45 cm. The volume of the block according to the idea of significant figures should be:

  • 1 × 102 cm3

  • 2 × 102 cm3

  • 1.763 × 102 cm3

  • none of the above


B.

2 × 102 cm3

Using the relation for volume
V = length × breadth × thickness
= 12 × 6 × 2.45
= 176.4 cm
= 1.764 × 102 cm3

The minimum number of significant figures is 1 in breadth, hence, the volume will contain only one significant figure. Therefore,

V = 2× 102 cm3


9.

The potential energy of a particle varies with distance x from a fixed origin asV=Axx+B;where A and B are constants. The dimensions of AB are

  • ML5/2T-2

  • ML2T-2

  • M3/2L3/2T-2

  • ML7/2T-2


D.

ML7/2T-2

Given,v=Axx+B....(i)Dimensionsofv=dimensionsofpotentialenergy=ML2T-2FromEq.(i),DimensionsofB=dimensionsofx=M0LT0DimensionsofA=dimensionsofv×dimensionofx+Bdimensionsofx=ML2T-2M0LT0M0L1/2T0=ML5/2T-2Hence,dimensionsofAB=ML5/2T-2M0LT0=ML7/2T-2


10.

The SI unit of gravitational potential is

  • J

  • J-kg-1

  • J-kg

  • J-kg2


B.

J-kg-1

Gravitationalpotential=workmassHence,SIunitofgravitationalpotential=unitofworkunitofmass=Jkg=J-kg-1