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SOLUTIONS TO CONCEPTS CHAPTER – 1 1.

a) Linear momentum b) Frequency c) Pressure :

: mv

–1

= [MLT ]

1 0 0 –1 = [M L T ] T

Force [MLT 2 ] –1 –2  = [ML T ] Area [L2 ] 0 0 –1

a) Angular speed  = /t = [M L T ]

 M0L0 T 2   [M0L0T–2] t T –2 2 –2 c) Torque  = F r = [MLT ] [L] = [ML T ] 2 2 2 0 d) Moment of inertia = Mr = [M] [L ] = [ML T ] 2 MLT  [MLT 3I1 ] a) Electric field E = F/q = [IT ] b) Angular acceleration  =

b) Magnetic field B =

F MLT 2   [MT  2I1 ] qv [IT ][LT 1 ]

B  2a MT 2I1 ]  [L]   [MLT 2I2 ] I [I] a) Electric dipole moment P = qI = [IT] × [L] = [LTI] 2 2 b) Magnetic dipole moment M = IA = [I] [L ] [L I] E = h where E = energy and  = frequency. c) Magnetic permeability 0 =

4. 5.

h= 6.

E [ML2 T 2 ]  [ML2 T 1 ] 1  [T ]

Q [ML2 T 2 ]   [L2 T 2K 1 ] mT [M][K ] L  L2 [L] b) Coefficient of linear expansion =  = 1   [K 1 ] L 0 T [L ][R] a) Specific heat capacity = C =

PV [ML1T 2 ][L3 ]   [ML2 T 2K 1(mol) 1 ]  nT [(mol )][K ] Taking force, length and time as fundamental quantity m ( force/acce leration) [F / LT 2 ] F    4 2  [FL 4 T 2 ] a) Density = 2 V Volume [L ] LT 2 –2 b) Pressure = F/A = F/L = [FL ] –2 –1 c) Momentum = mv (Force / acceleration) × Velocity = [F / LT ] × [LT ] = [FT] 1 Force  ( velocity )2 d) Energy = mv 2  2 accelerati on  F   F  =  2   [LT 1 ]2    2   [L2 T  2 ]  [FL]  LT   LT ]  metre g = 10 = 36  105 cm/min2 2 sec The average speed of a snail is 0.02 mile/hr 0.02  1.6  1000 –1 Converting to S.I. units, m/sec [1 mile = 1.6 km = 1600 m] = 0.0089 ms 3600 c) Gas constant = R =

8. 9.

The average speed of leopard = 70 miles/hr In SI units = 70 miles/hour =

70  1.6  1000 = 31 m/s 3600 1.1

Chapter-I 3

10. Height h = 75 cm, Density of mercury = 13600 kg/m , g = 9.8 ms 4

–2

then

Pressure = hfg = 10  10 N/m (approximately) 5

In C.G.S. Units, P = 10 × 10 dyne/cm

11. In S.I. unit 100 watt = 100 Joule/sec 9

In C.G.S. Unit = 10 erg/sec 4

12. 1 micro century = 10 × 100 years = 10

–4

 365  24  60 min

So, 100 min = 10 / 52560 = 1.9 microcentury 13. Surface tension of water = 72 dyne/cm In S.I. Unit, 72 dyne/cm = 0.072 N/m 14. K = kIa b where k = Kinetic energy of rotating body and k = dimensionless constant Dimensions of left side are, 2 –2

K = [ML T ] Dimensions of right side are, Ia = [ML2]a, b = [T–1]b According to principle of homogeneity of dimension, 2 –2 2 –2 –1 b [ML T ] = [ML T ] [T ] Equating the dimension of both sides, 2 = 2a and –2 = –b  a = 1 and b = 2 a

15. Let energy E  M C where M = Mass, C = speed of light a

 E = KM C (K = proportionality constant) Dimension of left side 2 –2

E = [ML T ] Dimension of right side a

–1 b

M = [M] , [C] = [LT ] 2 –2

–1 b

[ML T ] = [M] [LT ]  a = 1; b = 2

So, the relation is E = KMC

2 2 –3 –2

16. Dimensional formulae of R = [ML T I ] 2 3 –1 Dimensional formulae of V = [ML T I ] Dimensional formulae of I = [I] 2 3 –1

2 –3 –2

[ML T I ] = [ML T I ] [I]  V = IR a b

17. Frequency f = KL F M M = Mass/unit length, L = length, F = tension (force) –1 Dimension of f = [T ] Dimension of right side, a a b –2 b c –1 c L = [L ], F = [MLT ] , M = [ML ] –1

–2 b

–1 c

[T ] = K[L] [MLT ] [ML ] 0 0 –1

M L T = KM

b+c

a+b–c

–2b

Equating the dimensions of both sides, b+c=0

…(1)

–c + a + b = 0 –2b = –1

…(2) …(3)

Solving the equations we get, a = –1, b = 1/2 and c = –1/2 –1 1/2

 So, frequency f = KL F M

–1/2

K 1/ 2 1/ 2 K F F M   L L M 1.2

Chapter-I 18. a) h =

2SCos rg

LHS = [L]

MLT 2  [MT 2 ] L

Surface tension = S = F/I =

Density =  = M/V = [ML–3T0] Radius = r = [L], g = [LT–2] RHS =

2Scos  [MT 2 ]   [M0L1T0 ]  [L]  3 0 rg [ML T ][L][LT 2 ]

LHS = RHS So, the relation is correct b) v =

p where v = velocity 

LHS = Dimension of v = [LT–1] –1 –2

Dimension of p = F/A = [ML T ] Dimension of  = m/V = [ML–3] RHS =

p [ML1T 2 ]   [L2 T 2 ]1/ 2 = [LT 1 ]  [ML3 ]

So, the relation is correct. c) V = (pr4t) / (8l) LHS = Dimension of V = [L3] –1 –2

Dimension of p = [ML T ], r = [L ], t = [T] Coefficient of viscosity = [ML–1T–1] RHS =

pr 4 t [ML1T 2 ][L4 ][T]  8 l [ML1T 1 ][L]

So, the relation is correct. d) v =

1 (mgl / I) 2 –1

LHS = dimension of v = [T ] RHS =

(mgl / I) =

[M][LT 2 ][L] 2

[ML ]

–1

= [T ]

LHS = RHS So, the relation is correct. 19. Dimension of the left side =  Dimension of the right side = So, the dimension of

dx 2

(a  x )



L 2

(L  L )

1 1 a  –1 sin   = [L ] a x

1 a 

 (a  x ) ≠ a sin  x  2

So, the equation is dimensionally incorrect. 1.3

= [L ]

Chapter-I 20. Important Dimensions and Units : Physical quantity Force (F) Work (W) Power (P) Gravitational constant (G) Angular velocity () Angular momentum (L) Moment of inertia (I) Torque () Young’s modulus (Y) Surface Tension (S) Coefficient of viscosity () Pressure (p) Intensity of wave (I) Specific heat capacity (c) Stefan’s constant () Thermal conductivity (k) Current density (j) Electrical conductivity () Electric dipole moment (p) Electric field (E) Electrical potential (V) Electric flux () Capacitance (C) Permittivity () Permeability () Magnetic dipole moment (M) Magnetic flux () Magnetic field (B) Inductance (L) Resistance (R)

Dimension [M1L1T 2 ]

SI unit newton

1 2 2

[M L T ]

joule

1 2 3

[M L T ]

watt

1 3 2

[M L T ]

N-m /kg

[T 1]

radian/s

1 2 1

[M L T ]

kg-m /s

[M1L2 ]

kg-m2

1 2 2

[M L T ]

N-m

[M1L1T 2 ]

N/m2

[M1T 2 ]

N/m

[M1L1T 1]

N-s/m2

[M1L1T 2 ]

N/m (Pascal)

[M1T 3 ]

watt/m

[L2T 2K 1] 1 3

J/kg-K

4

[M T K ]

watt/m2-k4

[M1L1T 3K 1]

watt/m-K

1 2

[I L ]

ampere/m2

[I2T3M1L3 ]



[L1I1T1]

–1

m 

C-m

1 1 1 3

[M L I T ]

V/m

1 2 1 3

[M L I T ]

volt

1 3 1 3

[M T I L ] 2 4

1 2

2 4

1 3

volt/m

[I T M L ]

farad (F)

[I T M L ]

C /N-m

[M1L1I2T 3 ]

Newton/A2

[I1L2 ]

N-m/T

[M1L2I1T 2 ]

Weber (Wb)

1 1 2

[M I T ]

tesla

1 2 2 2

[M L I T ]

henry

1 2 2 3

[M L I T ]

ohm ()

**** 1.4

SOLUTIONS TO CONCEPTS CHAPTER – 2 1.

As shown in the figure,   The angle between A and B = 110° – 20° = 90°   | A | = 3 and | B | = 4m

 B

 R

y 

 A

A 2  B 2  2AB cos  = 5 m   Let  be the angle between R and A Resultant R =

 4 sin 90  –1  = tan 1  = tan (4/3) = 53°  3  4 cos 90   Resultant vector makes angle (53° + 20°) = 73° with x-axis. 2.

  Angle between A and B is  = 60° – 30° =30°   | A | and | B | = 10 unit

 B



60° A



102  10 2  2.10.10.cos30 = 19.3    be the angle between R and A 10 sin30  1  –1  –1 1   = tan    tan   = tan (0.26795) = 15°  10  10 cos30  2  3  

30°

 Resultant makes 15° + 30° = 45° angle with x-axis.  3.

 x component of A = 100 cos 45° = 100 / 2 unit  x component of B = 100 cos 135° = 100 / 2  x component of C = 100 cos 315° = 100 / 2 Resultant x component = 100 / 2 – 100 / 2 + 100 / 2 = 100 / 2  y component of A = 100 sin 45° = 100 / 2 unit  y component of B = 100 sin 135° = 100 / 2  y component of C = 100 sin 315° = – 100 / 2 Resultant y component = 100 / 2 + 100 / 2 – 100 / 2 = 100 / 2 Resultant = 100 y component Tan  = =1 x component –1

  = tan (1) = 45° The resultant is 100 unit at 45° with x-axis. 4.

      a  4i  3j , b  3i  4 j  a) | a | 42  3 2 = 5  b) | b | 9  16 = 5     c) | a  b || 7 i  7 j | 7 2   d) a  b  ( 3  4)iˆ  ( 4  3)ˆj  ˆi  ˆj   | a  b | 12  ( 1)2  2 . 2.1

45°

315°

135°

Chapter-2 5.

x component of OA = 2cos30° =

A 2m 30°

x component of BC = 1.5 cos 120° = –0.75

1.5m 60° x 90° D B 1m

x component of DE = 1 cos 270° = 0 y component of OA = 2 sin 30° = 1

y component of BC = 1.5 sin 120° = 1.3 y component of DE = 1 sin 270° = –1 Rx = x component of resultant =

3  0.75  0 = 0.98 m

Ry = resultant y component = 1 + 1.3 – 1 = 1.3 m So, R = Resultant = 1.6 m If it makes and angle  with positive x-axis y component Tan  = = 1.32 x component –1

  = tan

1.32

 6.

  | a | = 3m | b | = 4

a) If R = 1 unit 

3 2  42  2.3.4. cos  = 1

 = 180°

3 2  4 2  2.3.4. cos  = 5

 = 90°

3 2  4 2  2.3.4. cos  = 7

 = 0° Angle between them is 0°.  7.

AD  2 î  0.5Ĵ  4K̂ = 6iˆ  0.5 ˆj AD =

0.5 km

AE 2  DE2 = 6.02 KM

Tan  = DE / AE = 1/12

 2m

–1

0.5 km E

= tan (1/12) The displacement of the car is 6.02 km along the distance tan–1 (1/12) with positive x-axis. 8.

In ABC, tan = x/2 and in DCE, tan = (2 – x)/4 tan  = (x/2) = (2 – x)/4 = 4x  4 – 2x = 4x  6x = 4  x = 2/3 ft a) In ABC, AC =

AB 2  BC2 =

2 10 ft 3

b) In CDE, DE = 1 – (2/3) = 4/3 ft CD = 4 ft. So, CE =

CD2  DE2 =

F BC = 2 ft AF = 2 ft DE = 2x 2–x E G D

4 10 ft 3

AG2  GE 2 = 2 2 ft.  Here the displacement vector r  7 î  4 ĵ  3k̂

c) In AGE, AE = 9.

a) magnitude of displacement =

74 ft

b) the components of the displacement vector are 7 ft, 4 ft and 3 ft. 2.2

r Y