Q: A temperature difference can generate e.m.f. in some materials. Let S be the e.m.f. produced per unit temperature difference between the ends of a wire, the electrical conductivity and the thermal con

aligned & S = emf per unit temperature difference \\ & = Electrical conductivity \\ & k = Thermal conductivity \\ & [ S ]= [ ML ^2 ~T ^ -3 I ^ -1 ~K ^ -1 ] \\ & [ ]= [ M ^ -1 ~L ^ -3 ~T ^3 I ^2 ] \\ & [ K ]= [ M ^1 ~L ^1 ~T ^ -3 ~K ^ -1 ] \\ & [ Z ]= S ^2 ~K = [ M ^1 ~L ^1 ~T ^ -3 ~K ^ -2 ] [ M ^1 ~L ^1 ~T ^ -3 ~K ^ -1 ] aligned [ Z ]= [ K ^ -1 ]

Q: Young’s modulus of elasticity Y is expressed in terms of three derived quantities, namely, the gravitational constant G , Planck’s constant h and the speed of light c , as Y = c &#94

Dimensions of Young's modulus: M 1 L – 1 T – 2 Dimensions of velocity of light: M 0 L 1 T – 1 Dimensions of Planck's constant: M 1 L 2 T – 1 Dimensions of Gravitational constant: M – 1 L 3 T – 2 Now, as given in the question Y = c α h β G γ ⇒ M 1 L – 1 T – 2 = M 0 L 1 T – 1 α M 1 L 2 T – 1 β M – 1 L 3 T – 2 γ Compari

Q: In a particular system of units, a physical quantity can be expressed in terms of the electric charge e , electron mass m e . Planck's constant h , and Coulomb's constant k = 1 4 π ε

Given here: B = e α m e β h γ k δ The dimensions of e = Current × time = I T and m e = M , Planck's constant h = Energy Frequency = M L 2 T - 2 T - 1 = M L 2 T - 1 and k = 1 ε 0 = M - 1 L - 3 T 4 I 2 - 1 = M L 3 T - 4 I - 2 Using Principle of homogeneity, M 1 T - 2 I - 1 = I T α M β M L 2 T - 1 γ M L 3 T - 4 I - 2 δ Comparing both sides, we get β + γ + δ = 1

Q: A physical quantity S → is defined as S → = E → × B → μ 0 . where E → is electric field, B → is magnetic field a

Dimension of Force Length × Time = MLT - 2 LT = MT - 3 Dimensions of Power Area = Force × Velocity Area = MLT - 2 × LT - 1 L 2 = MT - 3 , Dimension of Electric field E , E = MA - 1 LT - 3 Dimensions of Magnetic field B ; B = MA - 1 T - 2 Dimension of magnetic permeability μ 0 ; μ 0 = MA - 2 T - 2 L Therefore, dimensions of S , S = E B μ 0 = MA - 1 LT - 3 MA - 1 T - 2 MA - 2 T - 2 L = MT - 3

Q: Sometimes it is convenient to construct a system of units so that all quantities can be expressed in terms of only one physical quantity. In one such system, dimensions of different quantities are giv

momentum Force = X p - r ⇒ m v m a = X p - r ⇒ v a = X p - r but given that v a = X β - p So, p - r = β - p ⇒ 2 p = β + r . . . 1 s p e e d 2 a c c e l e r a t i o n = p o s i t i o n ⇒ X 2 β X p = X α ⇒ 2 β - p = α ⇒ p + α = 2 β . . . 2 and f o r c e l i n e a r m o m e n t u m = a c c e l e r a t i o n s p e e d ⇒ X r X q = X p X β &#86

Question 1

In a new system of units, the units of mass, length, time and current are 5 kg, 5 m, 5 s and 5 A, respectively. If _0 and _0 are the permeability and permittivity of free space, respectively, then in

Accepted Answer

The dimensional formula for permeability of free space _0 is [M L T^ -2 A^ -2 ]. The dimensional formula for permittivity of free space _0 is [M^ -1 L^ -3 T^4 A^2]. The dimensional formula for _0 _0 is: ( [M L T^ -2 A^ -2 ] [M^ -1 L^ -3 T^4 A^2] )^ 1/2 = ( [M^2 L^4 T^ -6 A^ -4 ] )^ 1/2 = [M L^2 T^ -3 A^ -2 ] Let the magnitude of 1 SI unit in the new system be n. Using the principle of dimensional homogeneity: n_1 u_1 = n_2 u_2 1 [M_1 L_1^2 T_1^ -3 A_1^ -2 ] = n [M_2 L_2^2 T_2^ -3 A_2^ -2 ] Given

Question 2

A temperature difference can generate e.m.f. in some materials. Let S be the e.m.f. produced per unit temperature difference between the ends of a wire, the electrical conductivity and the thermal con

Accepted Answer

aligned & S = emf per unit temperature difference \ & = Electrical conductivity \ & k = Thermal conductivity \ & [ S ]= [ ML ^2 ~T ^ -3 I ^ -1 ~K ^ -1 ] \ & [ ]= [ M ^ -1 ~L ^ -3 ~T ^3 I ^2 ] \ & [ K ]= [ M ^1 ~L ^1 ~T ^ -3 ~K ^ -1 ] \ & [ Z ]= S ^2 ~K = [ M ^1 ~L ^1 ~T ^ -3 ~K ^ -2 ] [ M ^1 ~L ^1 ~T ^ -3 ~K ^ -1 ] aligned [ Z ]= [ K ^ -1 ]

Question 3

A dimensionless quantity is constructed in terms of electronic charge e, permittivity of free space _0, Planck's constant h and speed of light c. If the dimensionless quantity is written as e^ _0^ h^

Accepted Answer

For the quantity to be dimensionless aligned & e ^ _0^ h ^ c ^ d = M ^0 ~L ^0 ~T ^0 ~A ^0 \ & ( AT )^ ( M ^ -1 ~L ^ -3 ~T ^4 ~A ^2 )^ ( ML ^2 ~T ^ -1 )^ ( LT ^ -1 )^ = A ^0 M ^0 ~L ^0 ~T ^0 \ & +2 =0, +4 - - =0,- + =0 ~\&~-3 +2 + =0 \ & =-2 , = ~\&~ = aligned Option (1) satisfies the given condition

Question 4

Young&rsquo;s modulus of elasticity Y  is expressed in terms of three derived quantities, namely, the gravitational constant G , Planck&rsquo;s constant h and the speed of light c , as Y = c &#94

Accepted Answer

Dimensions of Young's modulus:  M 1 L &#8211; 1 T &#160; &#8211; 2 Dimensions of velocity of light:  M 0 L 1 T &#8211; 1 Dimensions of Planck's constant:  &#160; M 1 L 2 &#160; T &#8211; 1 Dimensions of Gravitational constant:  M &#8211; 1 L 3 T &#8211; 2 Now, as given in the question Y = c &#945; h &#946; G &#947; &#8658; M 1 L &#8211; 1 T &#160; &#8211; 2 = M 0 L 1 T &#8211; 1 &#945; &#160; M 1 L 2 &#160; T &#8211; 1 &#946; M &#8211; 1 L 3 T &#8211; 2 &#947; Compari

Question 5

In a particular system of units, a physical quantity can be expressed in terms of the electric charge e , electron mass m e . Planck's constant h , and Coulomb's constant k = 1 4 &#960; &#949;

Accepted Answer

Given here:  B = e &#945; m e &#946; h &#947; k &#948; The dimensions of  e = Current &#215; time = I T  and  m e = M , Planck's constant  h = Energy Frequency = M L 2 T - 2 T - 1 = M L 2 T - 1  and  k = 1 &#949; 0 = M - 1 L - 3 T 4 I 2 - 1 = M L 3 T - 4 I - 2 Using Principle of homogeneity, M 1 T - 2 I - 1 = I T &#945; M &#946; M L 2 T - 1 &#947; M L 3 T - 4 I - 2 &#948; Comparing both sides, we get   &#946; + &#947; + &#948; = 1 &#160; &#160; &#160;

Question 6

A physical quantity  S &#8594;  is defined as  S &#8594; = E &#8594; &#215; B &#8594; &#956; 0 . where  E &#8594;  is electric field,  B &#8594;  is magnetic field a

Accepted Answer

Dimension of  Force Length &#215; Time = MLT - 2 LT = MT - 3 Dimensions of Power Area = Force &#215; Velocity Area = MLT - 2 &#215; LT - 1 L 2 = MT - 3 ,  Dimension of Electric field  E , &#160; E = MA - 1 LT - 3 Dimensions of Magnetic field  B ; &#160; B = MA - 1 T - 2 Dimension of magnetic permeability  &#956; 0 ; &#160; &#956; 0 = MA - 2 T - 2 L Therefore, dimensions of  S , S = E B &#956; 0 = MA - 1 LT - 3 MA - 1 T - 2 MA - 2 T - 2 L = MT - 3

Question 7

Sometimes it is convenient to construct a system of units so that all quantities can be expressed in terms of only one physical quantity. In one such system, dimensions of different quantities are giv

Accepted Answer

momentum Force = X p - r &#8658; m v m a = X p - r &#8658; v a = X p - r but given that  v a = X &#946; - p So,   p - r = &#946; - p &#8658; 2 p = &#946; + r &#160; &#160; &#160; . . . 1 s p e e d 2 a c c e l e r a t i o n = p o s i t i o n &#8658; X 2 &#946; X p = X &#945; &#8658; 2 &#946; - p = &#945; &#8658; p + &#945; = 2 &#946; &#160; &#160; &#160; . . . 2 and  f o r c e l i n e a r &#160; m o m e n t u m = a c c e l e r a t i o n s p e e d &#8658; X r X q = X p X &#946; &#86

Units and Dimensions — Previous Year Questions & Solutions