Units And Dimensions(12th Grade > Physics ) Questions and answers for exam Preparation

Question 1. Dimensions of coefficient of viscosity are

ML2T−2
ML2T−1
ML−1T−1
MLT

Explanation:-

Answer: Option C. -> ML−1T−1
:
C

F = - η . A \frac{d v}{d x} \Rightarrow [η] = [M L^{- 1} T^{- 1}]

where A is the area, v the velocity and dx the change in displacement.

Question 2. From the equation

t a n θ = \frac{r g}{v^{2}}

, one can obtain the angle of banking

θ

for a cyclist taking a curve (the symbols have their usual meanings). Then say, it is

Both dimensionally and numerically correct
Neither numerically nor dimensionally correct
Dimensionally correct only
Numerically correct only

Explanation:-

Answer: Option C. -> Dimensionally correct only
:
C
Given equation is dimensionally correct because both sides are dimensionless but numerically wrong because the correct equation is

t a n θ = \frac{v^{2}}{r g} .

Question 3. The frequency of vibration f of a mass m suspended from a spring of spring constant K is given by a relation of this type f =C

m^{x} K^{y}

; where C is a dimensionless quantity. The value of x and y are

x = 12, y = 12
x = -12, y = - 12
x = 12, y = - 12
x = - 12, y = 12

Explanation:-

Answer: Option D. -> x = - 12, y = 12
:
D
By putting the dimensions of each quantity both sides we get

[T^{- 1}] = [M]^{x} [M T^{- 2}]^{y}

Now comparing the dimenstions of quantities in both sides we get x+y =0 and 2y =1

∴

x = -

\frac{1}{2}

,y =

\frac{1}{2}

Question 4. Dimensional formula

M L^{- 1} T^{- 2}

does not represent the physical quantity

Young's modulus of elasticity
Stress
Strain
Pressure

Explanation:-

Answer: Option C. -> Strain
:
C
Strain

= \frac{Δ L}{L} \Rightarrow

dimensionless quantity

Question 5. If the speed of light (c), acceleration due to gravity (g) and pressure (p) are taken as the fundamental quantities, then the dimension of gravitational constant is

c2g0p−2
c0g2p−1
cg3p−2
c−1g0p−1

Explanation:-

Answer: Option B. -> c0g2p−1
:
B
Let [G]

\propto c^{x} g^{y} p^{z}

by substituting the following dimensions :
[G] =

[M^{- 1} L^{3} T^{- 2}], [c] = [L T^{- 1}], [g] = [L T^{- 2}]

[p] = [M L^{- 1} T^{- 2}]

compare the dimension of M, L and T onboth sides
we can get x=0,y=2,z=1

∴ [G] \propto c^{0} g^{2} p^{- 1}

Question 6. The physical quantity which has dimensional formula as that of

\frac{E n e r g y}{M a s s X L e n g t h}

is

Force
Power
Pressure
Acceleration

Explanation:-

Answer: Option D. -> Acceleration
:
D

\frac{E n e r g y}{M a s s \times L e n g t h} = \frac{[M L^{2} T^{- 2}]}{[M] [L]} = [L T^{- 2}]

This has the same dimension as acceleration.

Question 7. The dimension of

\frac{1}{\sqrt{ϵ_{0} μ_{0}}}

is that of

Velocity
Time
Capacitance
Distance

Explanation:-

Answer: Option A. -> Velocity
:
A
Speed of light

= \frac{1}{\sqrt{ϵ_{0} μ_{0}}}

Therefore the dimension of the quantity is same as velocity.

Question 8. Dimensional formula

M L^{- 1} T^{- 2}

does not represent the physical quantity

Young's modulus of elasticity
Stress
Strain
Pressure

Explanation:-

Answer: Option C. -> Strain
:
C
Strain

= \frac{Δ L}{L} \Rightarrow

dimensionless quantity

Question 9. The SI unit of universal gas constant (R) is

Watt K−1 mol−1
Newton K−1 mol−1
Joule K−1 mol−1
Erg K−1 mol−1

Explanation:-

Answer: Option C. -> Joule K−1 mol−1
:
C
PV = nRT

\Rightarrow

R =

\frac{P V}{n T}

=

\frac{J o u l e}{m o l e \times K e l v i n} = J K^{- 1} m o l^{- 1}

Question 10. 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 these

Explanation:-

Answer: Option B. -> 2 ×102 cm3
:
B
Volume V = l

\times

b

\times

t
= 12

\times

6

\times

2.45 = 176.4

c m^{3}

V = 176.4

\times 10^{2} c m^{3}

Since, the minimum number of significant figure is one in breadth, hence volume will also contain only one significant figure. Hence, V = 2

\times 10^{2} c m^{3} .