Units-And-Dimensions-Quiz-7

Units and Dimensions Quiz-7

Dear Readers,

As per analysis for previous years, it has been observed that students preparing for NEET find Physics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Biology due to their medical background. 

Furthermore, sections such as Physics are dominantly based on theories, laws, numerical in comparison to a section of Biology which is more of fact-based, life sciences, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced. 

Q1. “Pascal-Second” has dimension of

•  Force
•  Energy
•  Pressure
•  Coefficient of Viscoscity

Solution
(d) NSm^-2=Nm^-2×S= Pascal-second

Q2.In the equation X=3YZ², X and Z have dimensions of capacitance and magnetic induction respectively. In MKSQ system, the dimensional formula of Y is

•  [M^-3L^-2T^-2Q^-4]
•  [ML^-2]
•  [M^-3L^-2Q⁴T⁸]
•  [M^-3L^-2Q⁴T⁴]

Solution
(d) [capacitance X]=[M^-1L^-1T²Q²] [Magnetic induction Z] = [MT^-1 Q^-1] [Z²]=[M² T^-2 Q^-2] Given, X=3YZ²or Y=X/(3Z² ) or [Y]=([X])/[Z]² ∴ [Y] =([M^-1 L^-2 T² Q²])/([M²T^-2Q^-2])=[M^-3L^-2 T⁴ Q⁴]

Q3. The dimensional formula for impulse is same as the dimensional formula for

•  Momentum
•  Force
•  Rate of change of Momentum
•  Torque

Solution
(a) Momentum = mv=[MLT^-1] Impulse = Force × Time = [MLT^-2)]×[T]=[MLT^-1]

Q4. If E,M,L and G denote energy, mass, angular momentum and gravitational constant respectively, then the quantity (EL²/M⁵G²) has the dimensions of

•  Angle
•  Length
•  Mass
•  Time

Solution
(a) [E]=[ML²T^-2 ] [M]=[M] [L]=[ML²T^-1 ] [G]=[M^-1 L³T^-2 ] [(EL²)/(M⁵G²)]=([ML²T^-2 ] [ML²T^-1]²)/([M]⁵[M^-1L³T^-2]²) =([ML²T^-2][M² L⁴T^-2])/([M⁵ ][M^-2 L⁶T^-4])=([M³L⁶T^-4])/([M〗³L⁶T^-4]) =[m⁰L⁰T⁰]= Angle

Q5.Unit of electric flux is

•  Vm
•  Nm /C^-1
•  Vm^-1
•  C Nm^-1

Solution
 (a) If E is the intensity of electric field over a small area element dS and θ is angle between E and outdrawn normal to area element. Therefore, electric flux through this element is dϕ_E=(dS)(E cos⁡θ ) =E dS cos⁡θ=E.dS Hence,ϕ_E=E .S = V/d .S ∴Unit of ϕ_E=(volt ×metre²)/metre =volt-metre

Q6. Dimensions of frequency are

•  M⁰L^-1T⁰
•  M⁰L⁰T^-1
• M⁰L⁰T¹
•  M¹T^-2

Solution
(b) Frequency = 1/T=[M⁰L⁰T^-1]

Q7. Students I, II and III perform an experiment for measuring the acceleration due to gravity (g) using a simple pendulum. They use different lengths of the pendulum and/or record time for different number of oscillations. The observations are shown in the table Least count for length =0.1 cm Least count for time =0.1 s

If E_I,E_IIand E_IIIare the percentage errors in g,i.e.,(∆g/g×100) for students I, II and III, respectively

•  E_I=0
•  E_Iis minimum
•  E_I=E_II
•  E_IIis maximum

Solution
(b) % error in g=∆g/g×100=(∆l/l)×100+2(∆T/T)×100 E_I=0.1/64×100+2(0.1/128)×100=0.3125% E_II=0.1/64×100+2(0.1/64)×100=0.4687% E_III=0.1/20×100+2(0.1/36)×100=1.055%

Q8.The unit of the coefficient of viscosity in S.I. system is

•  m/kg-s
•  Tm-s/kg²
•  kg/m-s²
•  kg/m-s

Solution
d) [η]=ML^-1T^-1so its unit will be kg/m-sec

Q9.The physical quantity angular momentum has the same dimensions as that of

•  Work
•  Force
•  Momentum
•  Planck's Constant

Solution
(d) Angular momentum, [J]=[Iω]=[ML²T^-1] Planck’s constant, [h]=[E]/[v] =[ML²T^-1]

Q10. The physical quantity having the dimensions [M^-1L^-3A²] is

•  Resistance
•  Resistivity
•  Electrical Conductivity
• Electro Motive Force

Solution
c) Resistivity, ρ=m/(ne^2 τ) ∴[ρ]=[M]/[L^-3][AT][T^2 ] =[ML³A^-2T^-3] So, electrical conductivity σ=1/ρ ⇒ [σ]=1/[ρ] =[M^-1L^-3A²T³]