Units & Dimensions Test

Result

Units & Dimensions Test - 6

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Question 1
1 / -0

The pairs of physical quantities that have the same dimensions are

A
Reynolds number and coefficient of friction

B
Latent heat and gravitational potential

C
Curie and frequency of light wave

D
Planck's constant and torque.

Solution

Reynold's number and coefficient of friction are dimensionless quantities. Curie is the number of atoms decaying per unit time and frequency is the number of oscillations per unit time. Latent heat and gravitational potential both have the same dimension corresponding to energy per unit mass.

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Question 2
1 / -0

The respective number of significant figures for the numbers 23.023, 0.0003 and 2.1 × 10^–3 are

A
5, 1, 2

B
4, 4, 2

C
5, 1, 5

D
5, 5, 2

Solution

(i) All the non-zero digits are significant.
(ii) All the zeros between two non-zero digits are significant, no matter where the decimal point is, if at all.
(iii) If the number is less than 1, the zero(s) on the right of decimal point but to the left of the first non-zero digit are not significant.
(iv) The power of 10 is irrelevant to the determination of significant figures. According to the above rules, 23.023 has 5 significant figures. 0.0003 has 1 significant figures. 2.1 × 10^–3 has 2 significant figures.

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Question 3
1 / -0

Let [ε₀] denote the dimensional formula of the permittivity of the vaccum, and [μ₀] that of the permeability of the vacuum. If M = mass, L = length, T = time and I = electric current :
[JEE-1998' 2/200]

A
[ε₀] = M^-1 L^-3 T² I

B
[μ₀] = MLT^-2 I^-2

C
[ε₀] = M^-1 L^-3 T⁴ I²

D
[μ₀] = M L² T^-1 I

Solution

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Question 4
1 / -0

A quantity X is given by ε₀L . where ε₀ is the permittivity of free space, L is length, ΔV is potential difference and Δt is time interval. The dimensional formula for X is the same as that of
[JEE Sc.2000'3/105]

A
Resistance

B
Charge

C
Voltage

D
Current

Solution

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Question 5
1 / -0

Pressure depends on distance as, P = , where α, β are constants, z is distance, k is Boltzmann's constant and q is temperature. The dimension of β are
[JEE-2004s '3/84]

A
M⁰ L⁰ T⁰

B
M^-1 L^-1 T^-1

C
M⁰ L² T⁰

D
M^-1 L¹ T²

Solution

Unit of K is joules per kelvin or the dimensional formula of Kis [ML²T^-2θ^-1].
αz/kθ should be dimensionless,
So, dimensional formula of α = [ML²T^-2]/[L] = MLT^-2
Dimensional formula of P = [ML^-1T^-2]
rArr β = [α/P]
= [MLT^-2]/[ML^-1T^-2]
= M⁰L²T⁰

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