12th Grade > Physics
UNITS AND DIMENSIONS MCQs
Total Questions : 40
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Answer: Option C. -> ML−1T−1
:
C
F=−η.Advdx⇒[η]=[ML−1T−1] where A is the area, v the velocity and dx the change in displacement.
:
C
F=−η.Advdx⇒[η]=[ML−1T−1] where A is the area, v the velocity and dx the change in displacement.
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 tanθ=v2rg.
:
C
Given equation is dimensionally correct because both sides are dimensionless but numerically wrong because the correct equation is tanθ=v2rg.
Answer: Option D. -> x = - 12, y = 12
:
D
By putting the dimensions of each quantity both sides we get [T−1]=[M]x[MT−2]y
Now comparing the dimenstions of quantities in both sides we get x+y =0 and 2y =1
∴ x = - 12,y = 12
:
D
By putting the dimensions of each quantity both sides we get [T−1]=[M]x[MT−2]y
Now comparing the dimenstions of quantities in both sides we get x+y =0 and 2y =1
∴ x = - 12,y = 12
Answer: Option C. -> Strain
:
C
Strain =ΔLL⇒ dimensionless quantity
:
C
Strain =ΔLL⇒ dimensionless quantity
Answer: Option B. -> c0g2p−1
:
B
Let [G] ∝cxgypz
by substituting the following dimensions :
[G] = [M−1L3T−2],[c]=[LT−1],[g]=[LT−2]
[p]=[ML−1T−2]
compare the dimension of M, L and T onboth sides
we can get x=0,y=2,z=1
∴[G]∝c0g2p−1
:
B
Let [G] ∝cxgypz
by substituting the following dimensions :
[G] = [M−1L3T−2],[c]=[LT−1],[g]=[LT−2]
[p]=[ML−1T−2]
compare the dimension of M, L and T onboth sides
we can get x=0,y=2,z=1
∴[G]∝c0g2p−1
Answer: Option D. -> Acceleration
:
D
EnergyMass×Length=[ML2T−2][M][L]=[LT−2]
This has the same dimension as acceleration.
:
D
EnergyMass×Length=[ML2T−2][M][L]=[LT−2]
This has the same dimension as acceleration.
Answer: Option A. -> Velocity
:
A
Speed of light =1√ϵ0μ0
Therefore the dimension of the quantity is same as velocity.
:
A
Speed of light =1√ϵ0μ0
Therefore the dimension of the quantity is same as velocity.
Answer: Option C. -> Strain
:
C
Strain =ΔLL⇒ dimensionless quantity
:
C
Strain =ΔLL⇒ dimensionless quantity
Answer: Option C. -> Joule K−1 mol−1
:
C
PV = nRT ⇒ R = PVnT = Joulemole×Kelvin=JK−1mol−1
:
C
PV = nRT ⇒ R = PVnT = Joulemole×Kelvin=JK−1mol−1
Answer: Option B. -> 2 ×102 cm3
:
B
Volume V = l × b × t
= 12 × 6 × 2.45 = 176.4 cm3
V = 176.4 ×102cm3
Since, the minimum number of significant figure is one in breadth, hence volume will also contain only one significant figure. Hence, V = 2 ×102cm3.
:
B
Volume V = l × b × t
= 12 × 6 × 2.45 = 176.4 cm3
V = 176.4 ×102cm3
Since, the minimum number of significant figure is one in breadth, hence volume will also contain only one significant figure. Hence, V = 2 ×102cm3.