Eamcet Pb Physics Jr Inter Physics 01 01units and Dimensions

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MEASUREMENT AND UNITS & DIMENSIONS PREVIOUS EAMCET BITS

1. When a wave traverses a medium the displacement of a particle located at ‘x’ at a time ‘t ‘ is given by ( )y = asin bt - cx , where a, b, c are constants of the wave, which of the following is a quantity with dimensions ? [EAMCET 2009 E]

1) y/a 2) bt 3) cx 4) b/c Ans: 4

Sol : 1. y Displacement L= = = Noa amplitude L

dimensions.

2. bt = No dimension. Since angle has no dimensions. 3. cx = No dimensions. Since angle has no dimensions. 4. Angle has no dimensions, according to the principle of homogeneity of dimensions.

-1b x Lbt = cx = = = LT = velocityc t T

∴ ⇒

2. The energy [ ]E , angular momentum ( )L and universal gravitational constant ( )G are chosen as fundamental quantities. The dimensions of universal gravitational constant in the dimensional formula of planks constant ( )h is. [EAMCET 2008 E]

1) 0 2) -1 3) 53

4) 1

Ans: 1 Sol : From the given problem a b ch = E L G Where a, b, c are constants From the principle of homogeneity of dimensions Dimensions of ⎡ ⎤⎣ ⎦

2 -1h = ML T

Dimensions of ⎡ ⎤⎣ ⎦2 -2E = ML T

Dimensions of ⎡ ⎤⎣ ⎦2 -1L = ML T

Dimensions of ⎡ ⎤⎣ ⎦-1 3 -2G = M L T

a b c2 -1 2 -2 2 -1 -1 3 -2ML T = ML T ML T M L T⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤∴⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

.................⇒1 a+b-cM = M a + b - c = 0 1 ...................⇒2 2a+2b=3cL = L 2a + 2b + 3c = 0 2 ...............⇒-1 -2a-b-2cT = T - 2a - b - 2c = 1 3 Solving 1, 2 & 3 we get c = 0 3. Some physics constants are given in List – I and their dimensional formula are given in List – II.

Match the following: [EAMCET 2007 E] List – I List – II 1) Planck’s constant i) 1 2ML T− −⎡ ⎤⎣ ⎦

Measurement and Units & $ dimensions

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2) Gravitational constant ii) 1 1ML T− −⎡ ⎤⎣ ⎦

3) Bulk modulus iii) 2 1ML T−⎡ ⎤⎣ ⎦

4) Coefficient of viscosity iv) 1 3 2M L T− −⎡ ⎤⎣ ⎦

The correct answer is 1 2 3 4 1 2 3 4 1) iv iii ii i 2) ii i iii iv 3) iii ii i iv 4) iii iv i ii Ans: 4

Sol : 1. Plancks constant 2 -2

2 -1-1

Energy ML T= = = ML TFrequency T

⎡ ⎤⎣ ⎦

2. Gravitational constant ( )2

-1 3 -2

1 2

FrG = = M L Tm m

⎡ ⎤⎣ ⎦

3. Bulk modulus -1 -2Bulk Stress F/A= = = ML TdvVolume strainv

⎡ ⎤⎣ ⎦

4. Co-efficient of viscosity F dv= from F = adv dxA.dx

⎛ ⎞η⎜ ⎟⎝ ⎠

-2

-1 -1-1

2

MLT= = ML TLTL

L

⎡ ⎤⎣ ⎦⎛ ⎞⎜ ⎟⎝ ⎠

2006 : 4. If C, R, L and I denote capacity, resistance, inductance and electric current respectively, the

quantitates having the same dimensions of time are [EAMCET 2006 E] a) CR b) L/R c) LC d) 2LI 1) a and b only 2) a and c only 3) a and d only 4) a, b and c only Ans: 4 Sol : Dimensional formula of capacity ⎡ ⎤⎣ ⎦

-1 -2 4 2= M L T A

Dimensional formula of Resistance ⎡ ⎤⎣ ⎦2 -3 -2= M L T A

Dimensional formula of Inductance ⎡ ⎤⎣ ⎦2 -2 -2= M L T A

Dimensional formula of Current ⎡ ⎤⎣ ⎦0 0 0= M L T A

a. ( )( )-1 -2 4 2 2 -3 -2CR = M L T A ML T A 0 0 0= M L TA = Time

b. 2 -2 -2

2 -3 -2

L ML T A= = T = TimeR ML T A

c. ( )( )2 -2 -2 -1 -2 4 2LC = ML T A M L T A = T = Time

d. 2 2 -2 -2 2 2 -2LI = ML T A x A = ML T = energy


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5. Names of units some physical quantities are given in List – I and their dimensional formulae are given in List II. Match the correct pairs in the lists [EAMCET 2005 E]

List –I List – II a) Pa s (e) 2 2 1L T K− −⎡ ⎤⎣ ⎦

b) Nm K–1 (f) 3 1MLT K− −⎡ ⎤⎣ ⎦ c) 1 1J kg K− − (g) 1 1ML T− −⎡ ⎤⎣ ⎦

d) 1 1Wm K− − (h) 2 2 1ML T K− −⎡ ⎤⎣ ⎦

a b c d a b c d 1) h g e f 2) g f h e 3) g e h f 4) g h e f Ans: 4 Sol : a. Dimensional formula of Pa – s = [ ]⎡ ⎤⎣ ⎦

-1 -2ML T T

= ⎡ ⎤⎣ ⎦-1 -1ML T

b. Dimensional formula of [ ]⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦-1 -2 -1Nm - k = MLT L K

= ⎡ ⎤⎣ ⎦2 -2 -1ML T K

c. Dimensional formula of -1 -1J kg k = ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦ ⎣ ⎦2 -2 -1 -1ML T M K

= ⎡ ⎤⎣ ⎦2 -2 -1L T K

d. Dimensional formula of ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦ ⎣ ⎦-1 -1 2 -3 -1 -1Wm k = ML T L K

= ⎡ ⎤⎣ ⎦-3 -1MLT K

6. The position of a particle at time ‘t’ is given by the equation x(t) = ( )At0V 1 eA

− V0 = constant and

A > 0. Dimension of V0 and A respectively are [EAMCET 2004 E] 1) 1M LT and T−° ° 2) 1 2M LT and LT− −° 3) 1M LT and T−° 4) 1 1M LT and T− −° Ans: 4 Sol : Since exponential functions has no dimensions ⎡ ⎤∴ ⇒ ⎣ ⎦

0 0 0 -1AT = M L T A = T

According to the principle of homogeneity of dimensions

⎡ ⎤⇒ ⎣ ⎦-10

0Vx = V = xA = LTA


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7. In planetary motion the areal velocity of position vector of a planet depends on angular velocity (ω) and the distance of the planet from sun (r). If so the correct relation for areal velocity is

[EAMCET 2003 E]

1) dA rdt

∝ ω 2) 2dA rdt

∝ ω 3) 2dA rdt

∝ ω 4) dA rdt

∝ ω

Ans : 3

Sol : In the problem it is given that Areal velocity dAdt

depends on angular frequency ( ω) and distance

of the planet from the sun (r)

∴ ωa bdA α rdt

Where a and b are dimensions of ω and r .

⇒ ωa bd A = K rd t

K is proportionality constant writing dimensional formulas on both sides [ ]⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦a b2 -1 -1L T = T L

From the principle of homogeneity of dimensions ⇒2 bL = L b = 2

a⇒-1 -aT = T = 1

8. The Vander Waal’s equation for a gas is ( )2

aP V b nRTV

⎛ ⎞+ − =⎜ ⎟⎝ ⎠

where P, V, R, T and n

represent the pressure volume, universal gas constant, absolute temperature and number of moles of a gas respectively, ‘a’ and ‘b’ are constant. The ratio b/a will have the following dimensional formula : [EAMCET 2002 E]

1) 1 2 2M L T− − 2) 1 1 1M L T− − − 3) 2 2ML T 4) 2MLT− Ans: 1 Sol : According to the principle of homogeneity of dimensions physical quantities having same

dimensional formula can be added or subtracted.

∴ ⎡ ⎤ ⎡ ⎤⇒ ⎣ ⎦ ⎣ ⎦22 -1 -2 3

2

aP = a = Pv = ML T Lv

5⎡ ⎤⎣ ⎦-2= ML T

⎡ ⎤⇒ ⎣ ⎦3

3 -1 -2 +25 -2

b Lv = b = L = = M L Ta ML T

9. In C.G.S system the magnitude of the force is 100 dynes. In another system where the fundamental physical quantities are kilogram, meter and minute, the magnitude of the force is

[EAMCET 2001 E] 1) 0.0.36 2) 0.36 3) 3.6 4) 36 Ans: 4 Sol : From 1 1 2 2n u n u=

2 21 1 1 1 1 2 2 2n M L T n M L T− −⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦

2

1 1 12 1

2 2 2

M L Tn nM L T

−⎡ ⎤ ⎡ ⎤ ⎡ ⎤

= ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦


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2

2g cm sn 100kg m min

−⎡ ⎤ ⎡ ⎤ ⎡ ⎤= ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦

2g cm s100

1000g 100cm 60s

−⎡ ⎤ ⎡ ⎤ ⎡ ⎤= ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦

2n 3.6= 10. The fundamental , physical quantities that have same dimensions in the formulae of torque and

angular momentum are [EAMCET 2000 E] 1) mass, time 2) time, length 3) mass, length 4) time, mole Ans: 3 Sol : Dimensional formula of Torque ⎡ ⎤⎣ ⎦

2 -2= ML T

Dimensional formula of angular momentum = mvr ⎡ ⎤⎣ ⎦2 -1= ML T

∴Mass and length have same dimensions 11. If pressure P, velocity V and time T are taken as fundamental physical quantities, the dimensional

formula of the force is [EAMCET 2000 E] 1) 2 2PV T 2) 1 2 2P V T− − 3) 2PVT 4) 1 2P VT− Ans: 1 Sol : From the given problem a b cFαP v T Where a, b, c are dimensions a b cF = K P v T K is dimensionless proportionality constant Dimensional formula of Force ⎡ ⎤⎣ ⎦

-2= MLT

Dimensional formula of Pressure ⎡ ⎤⎣ ⎦-1 -2= ML T

Dimensional formula of Velocity ⎡ ⎤⎣ ⎦-1= LT

Dimensional formula of Time [ ]= T

∴ [ ]⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦ ⎣ ⎦a b c-2 -1 -2 -1MLT = ML T LT T

[ ] [ ]⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦-a+b -2a-b+c-2 aMLT = M L T

From the principle of homogeneity of dimensions ⇒ 1 aM = M a = 1 ⇒ ∴1 -a+bL = L - a + b = 1 ⇒ ∴-2 -2a-b+cT = T - 2a - b + c = -2 Solving a = 1, b = 2, c = 2.


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MEDICAL 12. If energy E , velocity V and time T are taken as fundamental quantitates, the dimensional formula

for surface tension is [EAMCET 2009 M] 1) 1 2 2E V T− 2) 2 1 2E V T− 3) 2 2 1E V T− − 4) 2 2 1E V T− − − Ans 1

Sol. From the problem surface tension (s) a b cE V T∝ a b cS KE V T= where a, b, c are dimensions, K is a proportionality constant

∴ ( ) ( ) ( )a b c0 2 2 2 1ML T ML T LT T− − −=

equating the powers ∴ a = 1, b = - 2 and c = – 2 ∴ 1 2 2s E V T− −∝ 13. If power (p)surface tension (T) and plancks constant (h) are arranged so that the dimensions of

time in their dimension formulae are in ascending order, then which of the following is correct [EAMCET 2008 M] 1) P, T, h 2) P, h, T 3) T, P, h 4) T, h, P Ans : 1

Sol : Dimensional formula of power ⎡ ⎤⎣ ⎦2 -3Work= = ML T

time

Dimensional formula of surface tension ⎡ ⎤⎣ ⎦-2Force= = MT

Length

Dimensional formula of plancks constant ⎡ ⎤⎣ ⎦2 -1Energy= = ML T

frequency

Ascending order of time is -3, -2, -1 14. If force (F), work (W) and velocity (V) are taken as fundamental quantities then the dimensional

formula of time (T) is [EAMCET 2007 M] 1) 1 1 1W F v 2) 1 1 -1W F v 3) -1 -1 -1W F v 4) 1 -1 -1W F v Ans : 4 Sol : From the given problem a b cTαF w v Where a, b, c are dimensions ∴ a b cT = K F w v K is a dimensionless constant

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦ ⎣ ⎦a b c-2 2 -2 -1T = MLT ML T LT

[ ] [ ] [ ]a+b a+2b+c -2a-2b-cT = M L T Comparing the powers we get a + b = 0 ……….……………..…..1 a + 2b+c = 0 ….…………………. 2 -2a -2b - c = 1……………………. 3 Solving equations 1, 2 & 3 we get a = -1, b = 1, c = -1 ∴ -1 1 -1T = F w v


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15. Some physics constants are given in List – I and their dimensional formula are given in List – II. Match the following: [EAMCET 2006 M]

List – I List – II 1) Planck’s constant i) 1 2ML T− −⎡ ⎤⎣ ⎦

2) Gravitational constant ii) 1 1ML T− −⎡ ⎤⎣ ⎦

3) Bulk modulus iii) 2 1ML T−⎡ ⎤⎣ ⎦

4) Coefficient of viscosity iv) 1 3 2M L T− −⎡ ⎤⎣ ⎦

The correct answer is 1 2 3 4 1 2 3 4 1) iv iii ii i 2) ii i iii iv 3) iii ii i iv 4) iii iv i ii Ans: 4

Sol : 1. Plancks constant 2 -2

2 -1-1

Energy ML T= = = ML TFrequency T

⎡ ⎤⎣ ⎦

2. Gravitational constant ( )2

-1 3 -2

1 2

FrG = = M L Tm m

⎡ ⎤⎣ ⎦

3. Bulk modulus -1 -2Bulk Stress F/A= = = ML TdvVolume strainv

⎡ ⎤⎣ ⎦

4. Co-efficient of viscosity F dv= from F = adv dxA.dx

⎛ ⎞η⎜ ⎟⎝ ⎠

-2

-1 -1-1

2

MLT= = ML TLTL

L

⎡ ⎤⎣ ⎦⎛ ⎞⎜ ⎟⎝ ⎠

16. According to Bermoulli’s theorem 2P V gh

d 2+ + = constant. The dimensional formula of the

constant is (P = Pressure, d= density, h= height, V = velocity, g = acceleration due to gravity) [EAMCET 2005 M] 1) 0 0 0M L T⎡ ⎤⎣ ⎦ 2) 0 0M LT⎡ ⎤⎣ ⎦ 3 0 2 2M L T−⎡ ⎤⎣ ⎦ 4) 0 2 4M L T−⎡ ⎤⎣ ⎦

Ans: 3

Sol : According to Bernoulli’s theorem 2p v+ + gh = constant

d 2

According to the principle of homogeneity of dimensions only same quantities can be added or subtracted.

∴ Dimensional formula of p =d

Dimensional formula of 2v

2

= Dimensional formula of gh = Dimensional formula of constant


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= 0 2 -2M L T 17. The correct order in which the dimension of ‘ Length ‘ increases in the following physical

quantities is [EAMCET 2004 M] a) permittivity b) resistance c) magnetic permeability d) stress 1) a, b, c, d 2) d, c,b, a 3) a, d, c, b 4) c, b, d, a Ans: 3

Sol : a) From coulombs law .πε

1 22

0

Q Q1F =4 r

( ) ( )2⇒ ε

π1 2

0 -2 2

IT x ITQ Q= =F4 r MLT xL

⎡ ⎤⎣ ⎦-1 -3 4 2= M L T I

b) From ohms law ⇒ =v w wv = iR R = =i Q i it.i

2 -2

2

ML T=I T

⎡ ⎤⎣ ⎦2 -3 -2= ML T I

c) From coulombs inverse square law

⇒2

0 1 202

1 2

μ m m Fd 4πF = μ =4π d m m

Where 1 2m ,m = pole strength = [ ]I L

⎡ ⎤⎣ ⎦-2 2

-2 -20 2 2

MLT x Lμ = = MLT II L

d) Stress ⎡ ⎤⎣ ⎦-2

-1 -22

Force MLT= = = ML Tarea L

∴ As < 3 < -1 < 1 < 2 ∴ A, D, C, B 18. The dimensional equation for magnetic flux is [EAMCET 2003 M] 1) 2 2 1ML T I− − 2) 2 2 2ML T I−− 3) 2 2 1ML T I− −− 4) 2 2 2ML T I− −− Ans: 1

Sol : Magnetic flux φ = BA but F = mB or FB =m

∴ φ =-2 2FA Force x Area MLT x L= =

m PoleStrength I L

⎡ ⎤⎣ ⎦2 -2 -1= ML T I

19. The dimensional formula of coefficient of kinematic viscosity is [EAMCET 2002 M] 1) 0 1 1M L T− − 2) 0 2 1M L T− 3) 2 1ML T− 4) 1 1ML T− − Ans: 2 Sol : Coefficient of kinematic viscosity = 0 2 1M L T−


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20. The dimensional formula of the product of two physical quantities P and Q is 2 2ML T− . The dimensional formula of P/Q is MT–2. P and Q respectively are [EAMCET 2001 M]

1) Force, Velocity 2) Momentum , Displacement 3) Force , Displacement 4)Work, Velocity Ans: 3 Sol : Check all the options which satisfies 2 -2PQ = ML T and -2P / Q = ML 1. Let force = P Velocity = Q ∴ -2 -1 2 -3PQ = MLT xLT = ML T 2. PQ = Momentum x displacement

⎡ ⎤⎣ ⎦-1 2 -1= MLT x L = ML T

3. PQ = force x displacement = ⎡ ⎤⎣ ⎦-2 2 -2MLT xL = ML T

⎡ ⎤⎣ ⎦-2

-2P MLT= = M TQ L

∴ As both the conditions are satisfied P = Force Q = displacement

21. The physical quantity which has dimensional formula as that of Energymass length×

is [EAMCET 2000M]

1) Force 2) Power 3) Pressure 4) Acceleration Ans: 4 Sol :

2 -2

-2Energy ML T= = L T =mass x length M xL

acceleration

22. The dimensional formula of magnetic induction is [EAMCET 2000 M] 1) 1 1MT A− − 2) 2 1MT A− − 3) 1MLA− 4) 2MT A Ans: 2

Sol : From ⇒-2

-2 -1F Force MLTF = mB B = = = = MT Am pole strength AL

Eamcet Pb Physics Jr Inter Physics 01 01units and Dimensions

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