12th Grade > Physics
UNITS AND DIMENSIONS MCQs
Total Questions : 40
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Answer: Option A. -> Pressure and stress
:
A
Pressure =ForceArea=ML−1T−2
Stress =RestoringforceArea=ML−1T−2
:
A
Pressure =ForceArea=ML−1T−2
Stress =RestoringforceArea=ML−1T−2
Answer: Option A. -> V=πPγ48ηl
:
A
Formula for viscosity η=πPγ48Vl⇒ V = πPγ48ηl
:
A
Formula for viscosity η=πPγ48Vl⇒ V = πPγ48ηl
Answer: Option C. -> a = LT−2, b = L, c = T
:
C
From the principle of dimensional homogeneity [v] = [αt]⇒[α]=[LT−2]. similarly [b] = [L]and [c] = [T]
:
C
From the principle of dimensional homogeneity [v] = [αt]⇒[α]=[LT−2]. similarly [b] = [L]and [c] = [T]
Answer: Option B. -> Fv3A−2
:
B
L ∝vxAyFz⇒ L = kvxAyFz
Putting the dimenstions in the above relation
[ML2T−1]=k[LT−1]x[LT−2]y[MLT−2]z
⇒[ML2T−1]=k[MzLx+y+zT−x−2y−2z]
Comparing the pwers of M,LandT
z=1 ...(i)
x+y+z =2...(ii)
-x-2y-2z =-1 ...(iii)
On solving (i),(ii), and (iii) x=3, y=-2 ,z=1
So dimension of L in terms of v,Aandf
[L] = [Fv3A−2]
:
B
L ∝vxAyFz⇒ L = kvxAyFz
Putting the dimenstions in the above relation
[ML2T−1]=k[LT−1]x[LT−2]y[MLT−2]z
⇒[ML2T−1]=k[MzLx+y+zT−x−2y−2z]
Comparing the pwers of M,LandT
z=1 ...(i)
x+y+z =2...(ii)
-x-2y-2z =-1 ...(iii)
On solving (i),(ii), and (iii) x=3, y=-2 ,z=1
So dimension of L in terms of v,Aandf
[L] = [Fv3A−2]
Answer: Option B. -> x=1,y=-1,z=1
:
B
By substituting the dimension of given quantities
[ML−1T−2]x[MT−3]y[LT−1]z=[MLT]0
By comparing the power of M,L,Tin both sides x + y = 0 .....(i)
-x + z = 0 ......(ii)
-2x - 3y - z = 0 ......(iii)
The only values of x,y,zsatisfying ((i),(ii)and (iii)corresponds to (b).
:
B
By substituting the dimension of given quantities
[ML−1T−2]x[MT−3]y[LT−1]z=[MLT]0
By comparing the power of M,L,Tin both sides x + y = 0 .....(i)
-x + z = 0 ......(ii)
-2x - 3y - z = 0 ......(iii)
The only values of x,y,zsatisfying ((i),(ii)and (iii)corresponds to (b).
Answer: Option A. -> AB
:
A
Quantities having different dimensions can only be divided or multiplied but they cannot be added or subtracted.
:
A
Quantities having different dimensions can only be divided or multiplied but they cannot be added or subtracted.
Answer: Option B. -> Dyne ×cm4
:
B
Units of 'a'and ′PV2 are same from the equation.
Units of P = Dyne/cm2.
Units of V2=cm6.
Units of a = Units of PV2 = Dyne ×cm4
:
B
Units of 'a'and ′PV2 are same from the equation.
Units of P = Dyne/cm2.
Units of V2=cm6.
Units of a = Units of PV2 = Dyne ×cm4
Answer: Option A. -> Pressure
:
A
Y=(FA)(Changeinlengthlength)
=(pressuredimension)∗(Nodimension)
:
A
Y=(FA)(Changeinlengthlength)
=(pressuredimension)∗(Nodimension)
Answer: Option A. -> sec
:
A
[LR] is a time constant so its unit is same as the unit of time which is seconds.
:
A
[LR] is a time constant so its unit is same as the unit of time which is seconds.
Answer: Option D. -> Angular momentum
:
D
[h] =[Angular~momentum]=[ML2T−1] (Else use the Bohr's 2nd postulate)
:
D
[h] =[Angular~momentum]=[ML2T−1] (Else use the Bohr's 2nd postulate)