12th Grade > Physics
UNITS AND DIMENSIONS MCQs
Total Questions : 40
| Page 4 of 4 pages
Answer: Option A. -> −32,12,1
:
A
By substituting the dimension of each quantity we get T = [ML−1T−2]a[L−3M]b[MT−2]c=Ma+b+cL−a−3bT−2a−2c
∴a+b+c=0,−a−3b=0 and −2a−2c=1
Solving we get, a=−32,b=12 and c=1
:
A
By substituting the dimension of each quantity we get T = [ML−1T−2]a[L−3M]b[MT−2]c=Ma+b+cL−a−3bT−2a−2c
∴a+b+c=0,−a−3b=0 and −2a−2c=1
Solving we get, a=−32,b=12 and c=1
Answer: Option C. -> Increased 16 times
:
C
Energy = force × distance, so if both are increased by 4 times then energy will increase by 16 times.
:
C
Energy = force × distance, so if both are increased by 4 times then energy will increase by 16 times.
Answer: Option B. -> M−1L3T−2
:
B
F=Gm1m2d2⇒ G =Fd2m1m2
∴[G]=[MLT−2][L2][M2]=[M−1L3T−2]
:
B
F=Gm1m2d2⇒ G =Fd2m1m2
∴[G]=[MLT−2][L2][M2]=[M−1L3T−2]
Answer: Option D. -> AT2
:
D
Acceleration =distancetime2⇒ A = LT−2⇒ L= AT2
:
D
Acceleration =distancetime2⇒ A = LT−2⇒ L= AT2
Answer: Option C. -> v2∝ g λ
:
C
Letvx=kgyλzρδ
Now by substituting the dimensions of each quantities and equating the powers of M,L and T, we get δ = 0 and x = 2,y = 1,z = 1.
:
C
Letvx=kgyλzρδ
Now by substituting the dimensions of each quantities and equating the powers of M,L and T, we get δ = 0 and x = 2,y = 1,z = 1.
Answer: Option A. -> m/sec
:
A
Quantities of similar dimensions can be added or subtracted so unit of awill be same as that of velocity.
:
A
Quantities of similar dimensions can be added or subtracted so unit of awill be same as that of velocity.
Answer: Option C. -> c12G−12h12
:
C
Letm∝cxGyhz
by substituting the following dimentions :
[c] = LT−1;[G] = [M−1L3T−2]and [h] = [ML2T−1]
Now comparing both sides we will get
x = 12;y=−12,z=+12
So m∝c12G−12h12
:
C
Letm∝cxGyhz
by substituting the following dimentions :
[c] = LT−1;[G] = [M−1L3T−2]and [h] = [ML2T−1]
Now comparing both sides we will get
x = 12;y=−12,z=+12
So m∝c12G−12h12
Answer: Option D. -> ms−2
:
D
ct2 must have dimensions of L⇒c must have dimensions of LT2 i.e.,LT−2
:
D
ct2 must have dimensions of L⇒c must have dimensions of LT2 i.e.,LT−2
Answer: Option D. -> C2m−2N−1
:
D
F=14πϵq1q2r2⇒ϵ=14πq1q2Fr2=C2m−2N−1
:
D
F=14πϵq1q2r2⇒ϵ=14πq1q2Fr2=C2m−2N−1
Answer: Option D. -> Angular momentum per unit mass
:
D
Energy per unit volume =[ML2T−2][L3]=[ML−1T−2]
Force per unit area =[MLT−2][L2]=[ML−1T−2]
Product of voltage and charge per unit volume
=V×QVolume=VltVolume=Power×TimeVolume
⇒[ML2T−3][T][L3]=[ML−1T−2]
Angular momentum per unit mass =[ML2T−1][M]=[L2T−1]
So angular momentum per unit mass has different dimension.
:
D
Energy per unit volume =[ML2T−2][L3]=[ML−1T−2]
Force per unit area =[MLT−2][L2]=[ML−1T−2]
Product of voltage and charge per unit volume
=V×QVolume=VltVolume=Power×TimeVolume
⇒[ML2T−3][T][L3]=[ML−1T−2]
Angular momentum per unit mass =[ML2T−1][M]=[L2T−1]
So angular momentum per unit mass has different dimension.
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