33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020
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Units and MeasurementsTelegram @unacademyplusdiscounts7−− −∴ R = [ 2 2ML T / IT] 2 3 2= [ ML T I ][I]22. (b) : h 2 −2EI= λc × I= [ ML T ][ L]−1 2[ LT ] × [ ML ]hI = −1[ T ] = frequency.23. (a) : Gravitational constant G2force × (distance)=mass × mass−2 2[MLT ][L ]\ Dimensions of G = = [M–1L 3 T –2 ][M][M]24. (c) : Dimensions of Planck’s constant2 −2Energy [ML T ] 2 −1h = = = [ML T ]Frequency −1[T ]Dimensions of angular momentum L= Moment of inertia (I) × Angular velocity (w)= [ML 2 ][T –1 ] = [ML 2 T –1 ]25. (b) : Dimensions of force = [MLT –2 ]Dimensions of impulse = [MLT –1 ].26. (b) : Impulse = Force × Time.Therefore dimensional formula of impulse= Dimensional formula of force × Dimensionalformula of time = [MLT –2 ][T] = [MLT –1 ] anddimensional formula of linear momentum[p] = [MLT –1 ].27. (a)28. (c) : Units of RC = ohm × ohm –1 × second= second. Therefore dimensions of RC = time.29. (b) : Pressure = ForceAreaTherefore dimensions of pressure = ForceArea = ML–1 T –2 .30. (d) : Dimensions of energy E = [ML 2 T –2 ]Dimensions of volume v = [L 3 ]Dimensions of force F = [MLT –2 ]Dimensions of area A = [L 2 ]Dimensions of voltage V = [ML 2 T –3 A –1 ]Dimensions of charge q = [AT]Dimensions of angular momentum L = [ML 2 T –1 ]\ Dimensions of E 2 − 2v = [ML T ]= −1 −2[ML T ]3[L ]Dimensions of F − 2A = [MLT ]= −1 −2[ML T ]2[L ]Dimensions of Vq 2 −3 −1v = [ML T A ][AT]= [ML3–1 T –2 ][L ]Dimensions of angular momentum is [ML 2 T –1 ] whileother three has dimensions [ML –1 T –2 ]31. (c) : Magnetic flux, φ= = ⎛ ⎝ ⎜ F ⎞BA⎠⎟ AIl= [ −2 ][ 2MLT L ] 2 −2 −1= [ ML T A ].[ A][L]32. (a) : Permeability of free spaceπµ 0 = 2 × force × distancecurrent × current × length−2[MLT ][L]Dimensional formula of µ 0 == [MLT –2 A –2 ][A][A][L]33. (d) : Dimensions of force F = [MLT –2 ]Dimensions of velocity gradient ∆ −1v∆Z = [LT ]= [T –1 ][L]Dimensions of area A = [L 2 ]∆Given F = −ηAv ∆ZDimensional formula for coefficient of viscocity− 2F [MLT ] −1 −1η= = = [ ML T ]⎛ ∆v⎞ 2 −1( A)[L ][T ]⎝⎜∆ ⎠⎟Z34. (c) : Induced emf | ε | =L dIdtwhere L is the self inductance and dI is the rate of changedtof current.\ Dimensional formula of2 −3 −1| ε | [ML T A ] 2 −2 −2L = = = [ ML T A ]dI−1[AT ]dt35. (a) : Torque (t) = Force × distanceDimensional formula for (t) = [MLT –2 ][L]= [ML 2 T –2 ]Charge36. (a) : Capacitance C =Potentialdifference[AT]Dimensions of C == [M –1 L –2 T 4 A 2 ]2 −3 −1[ML T A ]Potential differenceResistance R =Current= [ML 2 T −3 A −1] = [ML 2 T –3 A –2 ][A]Dimensional formula of CR= [M –1 L –2 T 4 A 2 ][ML 2 T –3 A –2 ] = [T]As the (CR) has dimensions of time and so is called timeconstant of CR circuit.37. (d) : Angular momentum L= Moment of inertia I × Angular velocity w.\ Dimensional formula L = [ML 2 ][T –1 ]= [ML 2 T –1 ]2e38. (d) : Dimensions of2 3 2F d4πε = × = −[ ] [ ML T ]0Dimensions of G = [M –1 L 3 T –2 ],Dimensions of c = [LT –1 ]pe q rl ∝ ⎛ 2 ⎞G c⎝ ⎜ ⎟4πε 0 ⎠
Telegram @unacademyplusdiscounts8 NEET-AIPMT Chapterwise Topicwise Solutions Physics\ [L 1 ] = [ML 3 T –2 ] p [M –1 L 3 T –2 ] q [LT –1 ] r[] b −1On comparing both sides and solving, we get= [ LT ] or, [b] = [L].[ T]1 1p= , q=and r = –2⎛ a ⎞2 244. (c) : Equation P + b⎝⎜1∴ ∝ ⎡ 12 /V ⎠⎟ = θ a .2 Since is addedV2V2Ge ⎤al ⎢ ⎥to the pressure, therefore dimensions of and pressure22c ⎣⎢4πε0⎦⎥(P) will be the same.V39. (a) : According to question,a aAnd dimensions of = = [ML –1 T –2 ]l ∝ h p c q G r2V [ L3 ]2l = k h p c q G r ...(i) or a = [ML 5 T –2 ].Writting dimensions of physical quantities on both sides, 45. (b) : Relative density, refractive index and[M 0 LT 0 ] = [ML 2 T –1 ] p [LT –1 ] q [M –1 L 3 T –2 ] rApplying the principle of homogeneity of dimensions, wegetPoisson’s ratio all the three are ratios, therefore they aredimensionless constants.46. (d) : Dimensions of P = [ML –1 T –2 ]p – r = 0...(ii) Dimensions of r = [L]2p + q + 3r = 1...(iii) Dimensions of v = [LT –1 ]–p – q – 2r = 0 ...(iv) Dimensions of l = [L]Solving eqns. (ii), (iii) and (iv), we get1 3\ Dimensions of h = [ ][ 2 2P r − x ]p= r = , q =−[ 4vl]2 2−1 −2 2From eqn. (i), we get l=K hG[ML T ][L ]== [ML –1 T –1 ]−132 /[LT ][L]c240. (c) : [v c ] = [h x r y r z −αt] (given) ... (i) 47. (b) : Given : p=p0eWriting the dimensions of various quantities in at 2 is a dimensionless ∴ α = = = − 2[ T ]2 2eqn. (i), we gett [ T ][M 0 LT –1 ] = [ML –1 T –1 ] x [ML –3 T 0 ] y [M 0 LT 0 ] z48. (c) : Let k = P x S y c z ...(i)= [M x + y L –x –3y + z T –x ]k is a dimensionlessApplying the principle of homogeneity of dimensions, we Dimensions of k = [M 0 L 0 T 0 ]getDimensions of P = Forcex + y = 0; –x – 3y + z = 1; – x = –1Area = [MLT −2] = [ML –1 T –2 ]2[L ]On solving, we get x = 1, y = –1, z = –1Energy41. (d) : Let mass m ∝ F a V b T cDimensions of S =Area time = [ML 2 T −2] = [MT –3 ]×2or m = kF a V b T c [L ][T] ...(i) Dimensions of c = [LT –1 ]where k is a dimensionless constant and a, b and c are the Substituting these dimensions in eqn (i), we getexponents.[M 0 L 0 T 0 ] = [ML –1 T –2 ] x [MT –3 ] y [LT –1 ] z .Writing dimensions on both sides, we getApplying the principle of homogeneity of dimensions, we[ML 0 T 0 ] = [MLT –2 ] a [LT –1 ] b [T] cget[ML 0 T 0 ] = [M a L a + b T –2a – b + c ]x + y = 0.....(ii)Applying the principle of homogeneity of dimensions, we –x + z = 0 .....(iii)get–2x – 3y – z = 0 .....(iv)a = 1...(ii) Solving (ii), (iii) and (iv), we geta + b = 0...(iii) x = 1, y = –1, z = 1–2a – b + c = 0 ...(iv)Solving eqns. (ii), (iii) and (iv), we get49. (d) : f = am x k y .....(i)a = 1, b = –1, c = 1Dimensions of frequency f = [M 0 L 0 T –1 ]From eqn. (i), [m] = [FV –1 T]Dimensions of constant a = [M 0 L 0 T 0 ]Dimensions of mass m = [M]42. (c) : As n 1 u 1 = n 2 u 2Dimensions of spring constant k = [MT –2 ]g 100g4 = n403 2 ⇒ n3 2 =Putting these value in equation (i), we getcm ( 10 cm)[M 0 L 0 T –1 ] = [M] x [MT –2 ] yb43. (b) : v at Applying principle of homogeneity of dimensions, we gett cx + y = 0As c is added to t, \ [c] = [T]1 1–2y = –1 or y= , x=−[at] = [LT –1 [ LT− 12 2] −2] or, [] a = =[ LT ]x km[T]50. (c) : Units of b = =2t svvv2
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8 NEET-AIPMT Chapterwise Topicwise Solutions Physics
\ [L 1 ] = [ML 3 T –2 ] p [M –1 L 3 T –2 ] q [LT –1 ] r
[] b −1
On comparing both sides and solving, we get
= [ LT ] or, [b] = [L].
[ T]
1 1
p= , q=
and r = –2
⎛ a ⎞
2 2
44. (c) : Equation P + b
⎝
⎜
1
∴ ∝ ⎡ 12 /
V ⎠
⎟ = θ a .
2 Since is added
V
2
V
2
Ge ⎤
a
l ⎢ ⎥
to the pressure, therefore dimensions of and pressure
2
2
c ⎣⎢
4πε0
⎦⎥
(P) will be the same.
V
39. (a) : According to question,
a a
And dimensions of = = [ML –1 T –2 ]
l ∝ h p c q G r
2
V [ L
3 ]
2
l = k h p c q G r ...(i) or a = [ML 5 T –2 ].
Writting dimensions of physical quantities on both sides, 45. (b) : Relative density, refractive index and
[M 0 LT 0 ] = [ML 2 T –1 ] p [LT –1 ] q [M –1 L 3 T –2 ] r
Applying the principle of homogeneity of dimensions, we
get
Poisson’s ratio all the three are ratios, therefore they are
dimensionless constants.
46. (d) : Dimensions of P = [ML –1 T –2 ]
p – r = 0
...(ii) Dimensions of r = [L]
2p + q + 3r = 1
...(iii) Dimensions of v = [LT –1 ]
–p – q – 2r = 0 ...(iv) Dimensions of l = [L]
Solving eqns. (ii), (iii) and (iv), we get
1 3
\ Dimensions of h = [ ][ 2 2
P r − x ]
p= r = , q =−
[ 4vl]
2 2
−1 −2 2
From eqn. (i), we get l=
K hG
[ML T ][L ]
=
= [ML –1 T –1 ]
−1
32 /
[LT ][L]
c
2
40. (c) : [v c ] = [h x r y r z −αt
] (given) ... (i) 47. (b) : Given : p=
p0e
Writing the dimensions of various quantities in at 2 is a dimensionless ∴ α = = = − 2
[ T ]
2 2
eqn. (i), we get
t [ T ]
[M 0 LT –1 ] = [ML –1 T –1 ] x [ML –3 T 0 ] y [M 0 LT 0 ] z
48. (c) : Let k = P x S y c z ...(i)
= [M x + y L –x –3y + z T –x ]
k is a dimensionless
Applying the principle of homogeneity of dimensions, we Dimensions of k = [M 0 L 0 T 0 ]
get
Dimensions of P = Force
x + y = 0; –x – 3y + z = 1; – x = –1
Area = [MLT −2
] = [ML –1 T –2 ]
2
[L ]
On solving, we get x = 1, y = –1, z = –1
Energy
41. (d) : Let mass m ∝ F a V b T c
Dimensions of S =
Area time = [ML 2 T −2
] = [MT –3 ]
×
2
or m = kF a V b T c [L ][T]
...(i) Dimensions of c = [LT –1 ]
where k is a dimensionless constant and a, b and c are the Substituting these dimensions in eqn (i), we get
exponents.
[M 0 L 0 T 0 ] = [ML –1 T –2 ] x [MT –3 ] y [LT –1 ] z .
Writing dimensions on both sides, we get
Applying the principle of homogeneity of dimensions, we
[ML 0 T 0 ] = [MLT –2 ] a [LT –1 ] b [T] c
get
[ML 0 T 0 ] = [M a L a + b T –2a – b + c ]
x + y = 0
.....(ii)
Applying the principle of homogeneity of dimensions, we –x + z = 0 .....(iii)
get
–2x – 3y – z = 0 .....(iv)
a = 1
...(ii) Solving (ii), (iii) and (iv), we get
a + b = 0
...(iii) x = 1, y = –1, z = 1
–2a – b + c = 0 ...(iv)
Solving eqns. (ii), (iii) and (iv), we get
49. (d) : f = am x k y .....(i)
a = 1, b = –1, c = 1
Dimensions of frequency f = [M 0 L 0 T –1 ]
From eqn. (i), [m] = [FV –1 T]
Dimensions of constant a = [M 0 L 0 T 0 ]
Dimensions of mass m = [M]
42. (c) : As n 1 u 1 = n 2 u 2
Dimensions of spring constant k = [MT –2 ]
g 100g
4 = n
40
3 2 ⇒ n
3 2 =
Putting these value in equation (i), we get
cm ( 10 cm)
[M 0 L 0 T –1 ] = [M] x [MT –2 ] y
b
43. (b) : v at Applying principle of homogeneity of dimensions, we get
t c
x + y = 0
As c is added to t, \ [c] = [T]
1 1
–2y = –1 or y= , x=−
[at] = [LT –1 [ LT
− 1
2 2
] −2
] or, [] a = =[ LT ]
x km
[T]
50. (c) : Units of b = =
2
t s
vvv
2
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