(a) ₉₂U²³⁵+ ₀n¹ ￫ ₃₈Sr⁹⁰+ ₅₄Xe¹⁴³+ 3₀n¹
(b) ₉₂U²³⁵+ ₀n¹ ￫ ₄₃TC¹⁰⁷+ ₄₉In¹²⁴+ 5₀n¹

1 Bq = one disintegration per second
1 curie = 3.7 x 10¹⁰ disintegration per second
                 (Bq)
\(\therefore 1Bq=\frac { 1 }{ 3.7\times { 10 }^{ 10 } } \)Curie
= 2.703 x 10^-11ci

Roentgen

₂₇Co⁶⁰⇾ ₂₈Ni⁶⁰ +_-1e⁰

₈₈Ra²²⁶ ➝₈₆Rn²²²+₂He⁴

₉₀Th²³¹￫₉₁a²³¹ + _-1e⁰(β - decay)

Ga¹⁵⁰⇾X¹⁴⁶+ ₂He⁴
New atomic number - 146

Beta decay.

Beta radiation (i.e. electrons).

Given 
Loss in mass, m = 0.026 g = 2.6 x 10^-5 kg
To find: E= mc²
Solution
E = 2.6 x 10^-5 x (3 X 10⁸)²
Energy, E = 2.3 x 10¹²J

Given
Loss in mass, m = 0.20 g = 2 x 10^-4 kg
To find: E = mc²
Solution
E = 2 x 10^-4 x (3 x 108)²
E = 1.8 x 10¹³J

Given
The energy released from one atom = 2.9 x 10^-8 J
For 350 atoms, E = 2.9 x 10^-8 x 350
To find: mass, m = ?
E = mc²,
\(m=\frac { E }{ { c }^{ 2 } } =\frac { 2.9\times { 10 }^{ 8 }350 }{ { ({ 3\times 10 }^{ 8 }) }^{ 2 } } \)
mass, m=1.13 x 10^-22kg

(a) Given
The energy released E = 8.5 x 10¹⁴ J
To find: Loss in mass, m = ?
Solution
\(m=\frac { E }{ { c }^{ 2 } } =\frac { 8.5\times { 10 }^{ 14 } }{ { ({ 3\times 10 }^{ 8 }) }^{ 2 } } \)
m = 9.4 x 10^-3 kg
(b) Given
Energy released E = 2.9 x 10¹⁷ J
To find: Loss in mass, m =?
Solution
\(m=\frac { E }{ { c }^{ 2 } } =\frac { 2.9\times { 10 }^{ 17 } }{ { ({ 3\times 10 }^{ 8 }) }^{ 2 } } \)
= 3.2 kg

Given
Mass defect = Difference in actual mass and mass of the product
∆m = 3.251 x 10^-28 kg
Velocity of light, c = 3 x 10⁸ ms^-1
To find: Energy, E = ∆mc² = ?
Solution 
E = 3.251 x 10^-28 x (3 x 10⁸)²
E = 2.93 x 10^-11J

1 Bq = 2.703 x 10^-11 curie
∴ 4.3 x 10³ GBq = 4.3 x 103 x 10⁹ x 2.703 x10^-11
= 11.62 x 10¹
= 116.2 curie

The coefficient of linear expansion,
α_L =24 K^-1
The coefficient of cubical expansion,
α_v =3α_L= 3 x 24
= 72 K^-1
The initial temperature, T₁ = 303 K
The initial volume, V₀ = 0.3 m³
The final volume, V₁ = 0.35 m³
Change in volume ΔV = 0.35 - 0.30
= 0.05 m³
To find: The final temperature T₂= ?
ΔV = α_v. V₀ (ΔT)
0.05 = 72 x 0.3 x (T₂ - 303)
0.05 = 21.6 (T₂ - 303)
= 21.6 T₂ - 6544.8
21.6 T₂ = 6544.8 - 0.05
T₂ =\(\frac { 6544.8 }{ 21.6 } \)
Final temperature, T₂ = 303K

Initial temperature, t₁ = 285 k
Final temperature, t₂ = 363 k
Initial length, I₁ = 6.522 m
Change in length, Δl = 0.576 m
To find: Coefficient of linear expansion α_L=?
α_L =\(\frac { \triangle l }{ l\triangle T } =\frac { ({ l }_{ 2 }-{ l }_{ 1 }) }{ { l }_{ 1 }(t_{ 2 }-{ t }_{ 1 }) } \)
=\(\frac { 6.576-6.522 }{ 6.5222(63-285) } \)
=\(\frac { 0.054 }{ 6.522(78) } =\frac { 0.054 }{ 508.7 } \)
α_L =1.06 x 10^-4 K^-1.

Initial temperature, t₁ = 0°C
Final temperature, t₂ = 25°C
Coefficient of linear expansion of metal,
α₁ =18 x 10^-6 C^-1
Measured length, l₁ = 50 cm
To find: True (or) Actual length l₂ = ?
\(\frac { \triangle l }{ l\triangle T } =\frac { ({ l }_{ 2 }-{ l }_{ 1 }) }{ { l }_{ 1 }(t_{ 2 }-{ t }_{ 1 }) } \)
l₂ = I₁ (1 + α (t₂ - t₁)
∴ l₂ = 50 x (1 + 18 x 10 ^-6 x (25 - 0)
= 50 x (1 + 450 x 10^-6)
= 50 x (1 + 0.00045)
∴ l₂ = 50.225 cm
The true length of the object at 0⁰C =50.225 cm.

Length of AI sheet, L = 30 m
Coefficient of linear expansion, Δl = 23 x 10^-6 K^-1
Change in temperature ΔT = 300 k
To find: Change in length ΔL =?
ΔL = L α_L ΔL
= 30 x 23 x 10^-6 x 300
= 2.07 x 10^-1
ΔL = 0.207 m

F to (kelvin) K =(F--32) x \(\frac{5}{9}\) + 273
=(95 - 32) x \(\frac{5}{9}\) + 273
=\(\frac { 315 }{ 9 } \) + 273
=35+273 =308 K

Coefficient of linear expansion α_L=\(\frac { \triangle L }{ { L } } \times \frac { 1 }{ \triangle T } \)
Length of a steel piece L =2m
Initial temperature (T_i) = 200 k
Final temperature (T_f) = 250 k
ΔT=T_f - T_i
Increase in temperature, ΔT = 250 -200
=50 k
Increase in length, ΔL = 0.1 m
To find: Coefficient of increase expansion α_L?
α_L =\(\frac { 0.1 }{ 2 } \times \frac { 1 }{ 50 }\)
α_L =0.001 k-1
Coefficient cubical expansion
= 3 x coefficient linear expansion

Object distance, u = - 50 cm
Since the image distance is lesser than object distance and a virtual image is formed. It is a concave lens,
Image distance, v = - 10 cm
To find:
Focal length of the lens, f=?
\(\frac { 1 }{ f } =\frac { 1 }{ v } -\frac { 1 }{ u } \)
\(\frac { 1 }{ f } =\frac { -1 }{ 10 } -\left( \frac { 1 }{ -50 } \right) \)
\(\frac { 1 }{ f } =\frac { -1 }{ 10 } +\frac { 1 }{ 50 } \)
\(\frac { 1 }{ f } =\frac { -50+10 }{ 500 } \)
\(\frac { 1 }{ f } =\frac { -40 }{ 500 } \)
f=\(\frac { -500 }{ 40 } \)
∴ f =-12.5 cm
It is a diverging. ∵ it is concave lens.

Object distance, u =25 cm
virtual image distance, v =-10 cm
To find: Magnification, m = ?
m = \(\frac{\text { Distance of the image }}{\text { Distance of the object }}=\frac{v}{u}\)
= \(\frac { -10 }{ 25 } \) = -0.4
Magnification,
m = -0.4

It is corrected by using convex lens of suitable focal length.

(a) Vision with hypermetropia
(b) Corrected vision using a convex lens
Near point of hypermetropia
eye, v = -1 m = -100 cm
Object distance, u= -25 cm
To find: Power of the lens, P = ?
\(\frac { 1 }{ u } -\frac { 1 }{ v } =\frac { 1 }{ f } \)
Near point of hypermetropia eye
\(\frac { 1 }{ f } =\frac { 1 }{ -100 } -\frac { 1 }{ (-25) } \)
=\(-\frac { 1 }{ 100 } +\frac { 1 }{ 25 } \)
\(\frac { 1 }{ f } =\frac { (-1+4) }{ 100 } \)
=\(\\ \frac { 3 }{ 100 } \) cm
∴ f=\(\frac { 100 }{ 3 } \) cm
Power of the lens, P=\(\frac { 1 }{ f } \) =3D

Concave lens should be used to correct myopic eye.
Object distance u = Infinity
Far position of the defective eye v = -90 cm
To find: Power of the lens, P = ?
Nature of the lens =?
\(\frac { 1 }{ f } =\frac { 1 }{ v } -\frac { 1 }{ u } \)
\(\frac { 1 }{ f } =\frac { 1 }{ -90 } +\frac { 1 }{ \infty } \)
\(\frac { 1 }{ f } =\frac { 1 }{ -90 } \)
∴ f=-90 cm (or) 0.9 m
Power of the lens, P \(\frac { 1 }{ f } =\frac { 1 }{ 0.9 } \) =1.11 D
Concave lens of 1.11D is used to see distance object.

Power, P = -4D
To find: Focal length of the lens, f = ?
For correct real vision, f = ?
P=\(\frac { 1 }{ f } \) =\(\frac { 1 }{ P } \Rightarrow \frac { 1 }{ -4 } \)
(i) f =-0.25 m
P=+1.5 D
P=\(\frac { 1 }{ f } \)
∴ f=\(\frac { 1 }{ P } \Rightarrow f=\frac { 1 }{ 1.5 } \)
(ii) Focal length of correcting, f = + 0.667 m

Focal length, f = - 20 cm (concave lens)
h₀ = 5 cm
Height of the object, h₀ = 5 cm
Image distance v = -15 cm
\(\frac { 1 }{ f } =\frac { 1 }{ v } -\frac { 1 }{ u } \quad \frac { 1 }{ v } -\frac { 1 }{ f } \)
=\(\frac { 1 }{ u } \)
\(\frac { 1 }{ u } =\frac { 1 }{ -15 } +\frac { 1 }{ (20) } \)
=\(\frac { -20+15 }{ 300 } =\frac { -5 }{ 300 } \)
u = -60 cm
Magnification, m = \(\frac { { h }_{ i } }{ { h }_{ 0 } } =\frac { v }{ u } \)
h_i = \(\frac { v }{ u } \).h₀
Size of the image, h_i =\(\frac { -15 }{ -60 } \) x 5
= \(\frac { 5 }{ 4 } \) = 12.5 cm
Image is diminished & Virtual

Object distance, u = -20 cm
Focal length, f = -10 cm (Concave lens)
To find: Position & size of the image
i.e., v = ? ; h_i =?
\(\frac { 1 }{ f } =\frac { 1 }{ u } -\frac { 1 }{ v } \)
\(\frac { 1 }{ v } =\frac { 1 }{ f } +\frac { 1 }{ u } =-\frac { 1 }{ 10 } -\frac { 1 }{ 20 } \)
\(\frac { 1 }{ v } =\frac { -3 }{ 20 } \)
∴ v = -6.7 cm
m = \(\frac { v }{ u } =\frac { -20/3 }{ -20 } =\frac { 1 }{ 3 } \)
Height of the image
h_i = h₀ x m = 6 x \(\frac { 1 }{ 3 } \) = 2 cm
Image is virtual, at a distance 6.7 cm from lens on the same side as the object and has a height of 2 cm.

Object distance, u =-15 cm
Focal length, f =10 cm
To find: Nature of the image = ?
\(\frac { 1 }{ f } =\frac { 1 }{ u } -\frac { 1 }{ v } \)
\(\frac { 1 }{ v } =\frac { 1 }{ f } +\frac { 1 }{ u } =\frac { 1 }{ 10 } +\frac { 1 }{ -15 } \)
\(\frac { 1 }{ v } =\frac { 3-2 }{ 30 } =\frac { 1 }{ 30 } \)
\(\frac { 1 }{ v } =\frac { 1 }{ 30 } \)
∴ v=30 cm [image distance]
Magnification, m=\(\frac { v }{ u } =\frac { 30 }{ -15 } \) =-2
∴ Image is real, inverted, 30 cm from lens on the opposite side of the object & magnified 2 times.

Object size, h_o = 5 cm
Object size, h₁ =5 cm
Object distance, u=-30 cm
Focal length of convex lens, f=20 cm
To find : Size of image h_i = ?
Nature of image = ?
Position of image = ?
\(\frac { 1 }{ f } =\frac { 1 }{ v } -\frac { 1 }{ u } \)
\(\frac { 1 }{ v } =\frac { 1 }{ u } +\frac { 1 }{ f } \)
=\(\frac { -1 }{ 30 } +\frac { 1 }{ 20 } =\frac { -2+3 }{ 60 } \)
(or) \(\frac { 1 }{ v } =\frac { 1 }{ 60 } \) ⇒ = 60 cm
Magnification =\(=\frac{\text { height of the image size }}{\text { height of the object }}\)
=\(\frac { { h }_{ 2 } }{ { h }_{ 1 } } \)
m=\(\frac { { h }_{ 1 } }{ { h }_{ 0 } } \Rightarrow \frac { v }{ u } =\frac { 60 }{ -30 } \)=-2
∴ h₂=h₁ x (-2) =5 x (-2) =-10 cm
(i) The image is real, inverted and magnified.
(ii) Size of the image, h₂ = -10 cm

μ_a sin i = μ_dia sin r         ............(1)
(Reactive index of air, μ_a = 1)
(Angle of incidence, i = 45°)
in (1)
(1) ⇒ (sin 45⁰) =2.42 (sin r)
sin r=\(\left( \frac { sin{ 45 }^{ 0 } }{ 2.42 } \right) \)
r=sin^-1\(\frac { \frac { 1 }{ \sqrt { 2 } } }{ 2.42 } \)
r=sin^-1\(\left[ \frac { 1 }{ \sqrt { 2 } \times 2.42 } \right] \)
Angle of refraction,
r=16.977 =17⁰.

μ₁ sin i = μ₂ sin r
μ_a sin i = μ_w sin r
μ_a = 1
μ_w = 1.33
r = 25'
To find : Angle of incidence, i = ?
sin i =\(\frac { { \mu }_{ w } }{ { \mu }_{ a } } \)
sin r =\(\frac { 1.33\times sin25^{ 0 } }{ 1 } \)
i=sin^-1\(\left( \frac { 1.33\times sin25^{ 0 } }{ 1 } \right) \)
Angle of incidence,
i=34.2⁰.

Velocity of light in air, c = 3 x 10⁸ ms^-1
Velocity of light in glass, v = 2 x 10⁸ ms^-1
To find: Refractive index of glass,
μ =\(\frac { c }{ v } \)=?
μ =\(\frac { 3\times { 10 }^{ 8 } }{ 2\times { 10 }^{ 8 } } \)=1.5
μ =1.5 (No unit)

Refractive index of diamond with respect to kerosene
=\(=\frac{\text { Refractive index of diamond }}{\text { Refractive index of kerosene }}\)
=\(\frac { 3.42 }{ 1.44 } \)=1.68

Velocity of light through air, ca = 3 x 10⁸ ms^-1
To find : Speed of light through kerosene,
c_k =?
Refractive index of kerosene, μ_k= 1.44
μ_k=\(\frac { { c }_{ a } }{ { c }_{ k } } \) ⇒ 1.44
μ_k= \(\frac { 3\times 10^{ 8 } }{ { c }_{ k } } \)
=2.083 x 10⁸ ms^-1
Speed of light through kerosene,
c_k = 2 083 x 10⁸ ms^-1

Diameter of the objective lens, D = 5 m
Wavelength of light, λ = 6000Å or
6000 x 10^-10 m
To find: Resolving power of telescope =?
Resolving power of telescope =\(\frac { D }{ 1.22\lambda } \)
=\(\frac { 5 }{ 1.22\times 6000\times 10^{ -10 } } \)
=\(\frac { 5 }{ 7.32\times { 10 }^{ -7 } } \)
Resolving power of telescope = 0.683 x 10⁷.

Given
Temperature, T 10°C
Time, t = 0.274 s
Distance, D = ?
Velocity, v = (v_o + 0.61T) ms^-1
v = [331.4 + 0.61 x 10]
= 331.4 + 6.1
v = 337.5 ms^-1
Distance, D = v x t
D = 337.5 x 0.274
D = 92.48 m
Distance due to reflecting surface = \(\frac{distance}{2}\)
= 46.2 m

Given
Air temperature = 23°C
Velocity of sound in air, Vs = (331.4 + 0.61T) ms^-1
= 331.4 + 0.61 x 23
= 331.4 +13.8
V_s = 345.2 ms^-1

Given 
Frequency g car's distance, n = 800 Hz
Velocity of source, Vs = 120ms^-1
Velocity of sound, v = 340 ms^-1
To find: Frequency emitted by the car, n' = ?
Solution
Apparent frequency, n' =\(\left( \frac { v }{ v-{ v }_{ s } } \right) \times n\)
\(=\frac { 340 }{ 340-120 } \times 800\)
\(=\frac { 340 }{ 220 } \times 80\)
Apparent frequency, n' = 1236 Hz

Given
Speed of the train, V_s =30 rns^-1
Frequency of the train, n =600Hz
Speed of sound, v =340 ms^-1
To find: Apparent frequency n =?
Solution
\(n'=\left( \frac { v }{ v-{ v }_{ s } } \right) \times 600\)
\(=\frac { 340 }{ 310 } \times 600\)
Apparent frequency n' = 658 Hz

Given
Frequency of the source, n = 1000 Hz
Speed of observer, v_L = 2 v
To find: Observer approaches the source
apparent frequency n' =?
Solution 
\(n'=\left( \frac { v+{ v }_{ L } }{ v } \right) n\)
\(=\left( \frac { v+2v }{ v } \right) \times 1000\)
\(=\frac { 3b }{ v } \times 1000\)
Apparent frequency n' = 3000 Hz

Mass of body 1, m₁ = 50 kg
Mass of body 2, m₂ = 10kg
Distance, R = 10m
Universal gravitation
constant, G = 6.67 x 10^-11 Nm² kg^-2
To find :Force of gravitation, F =\(\frac { G{ m }_{ 1 }{ m }_{ 2 } }{ { R }^{ 2 } } \)
F=\(\frac { 6.67\times 10^{ -11 }\times 50\times 10 }{ 10^{ 2 } } \)
Force, F = 33.35 x 10^-11 N

Given:
Weight = 50 kg
To find: Acceleration, a = ? (downward)
Reaction, R = 400 N
R = m (g - a)
400 = 50 (10 - a)
400 = 500 - 50 a
500 = 500 - 400
50 a = 100
a=\(\frac{100}{50}\)
Downward acceleration, a = 20 ms^-1

Force, F = 50 N
Distance, d = 5 cm
To find: Momentum of force= F x d
Momentum of force, 50 x 5 x 10^-2
= 250 x 10^-2
= 2.5 Nm

Time, t = 10 s
Mass of body, m = 10 kg
Initial velocity of the body, u₁ = 0
Distance, d = 500 cm
= 500 x 10^-2 m
To find : Force, F = ?
F=m₁\(\frac { ({ u }_{ 1 }-{ v }_{ 1 }) }{ t } \)

v₁=\(\frac { distance }{ time\quad taken } \)
F=v₁=\(\frac { 500\times 10^{ -2 } }{ 10 } \)
Force, F=0.5 N

Acceleration due to gravity of a planet,
g_p =\(\frac { 1 }{ 2 } \) g_e
Radius of the planet R_p =\(\frac { 1 }{ 2 } \) R_e
To find : Mass of the planet M_p = ?
g_p =\(\frac { 1 }{ 2 } \)
g_e =\(\frac { { GM }_{ p } }{ \left( \frac { 1 }{ 2 } .{ R }_{ e } \right) ^{ 2 } } =\frac { 1 }{ 2 } \)
g_e =\(\frac { { GM }_{ p } }{ \left( \frac { 1 }{ 2 } .{ R }_{ e } \right) ^{ 2 } } \)            ......(1)
g_e =\(\frac { { GM }_{ e } }{ { R }_{ e }^{ 2 } } \)        ........(2)

\(\frac { 1 }{ 2 } =\frac { 4{ M }_{ p } }{ { M }_{ e } } \)
⇒ M_p =\(\frac { 1 }{ 8 } \).M_e
Mass of the planet, M_p =\(\frac { { M }_{ e } }{ 8 } \).

Ratio of mass of two planets = M₁ : M₂ = 1:2
Ratio of radius of the planets = R₁ : R₂ = 1:2
To find : Acceleration due two planets
= g₁ : g₂ =?
g = \(\frac { GM }{ { R }^{ 2 } } \) gravitational
Constituent G = 6.674 x 10^-11 Nm² kg^-2
g₁ = \(\frac { { GM }_{ 1 } }{ { R }_{ 1 }^{ 2 } } ;{ g }_{ 2 }=\frac { { GM }_{ 2 } }{ { R }_{ 2 }^{ 2 } } \)
\(\frac { { g }_{ 1 } }{ { g }_{ 2 } } =\frac { { M }_{ 1 }{ R }_{ 2 }^{ 2 } }{ { M }_{ 2 }{ R }_{ 1 }^{ 2 } } =\frac { 1 }{ 2 } \times \frac { 1 }{ 2 } \)
=\(\frac { 1 }{ 4 } \) = 1:4 (or) 1:2
g₁ : g₂ = 1:2

Mass of the car m = 2000 kg
Speed of the car v = 20 ms^-1
Time t = 0.05 s
To find : Impulse = ?
Impulse = F x t
Force =m x a
a =\(\frac { v }{ t } =\frac { 20 }{ 0.05 } \) =400 ms^-2
F = ma ⇒ 2000 x 400
= 8 x 10⁵ N
Impulse = F x t ⇒ 8 x 10⁵ x 0.5
4 x 10⁴ N s
The magnitude of impulse exerted by the wall on the car = 4 x 10⁴ Ns

Mass of first body, m₁ = 2 kg
Initial velocity of first body, u₁ = 40 ms^-1
Initial velocity of second body, u₂ = 0
Final velocity of first body,
v₁ z = Final velocity of the second body, v₂
i.e., v₁ = v₂= 20 ms^-1
To find : Mass of second body, m² = ?
According to law of conservation of momentum,
m₁u₁ + m₂u₂ = m₁v₁+ m₂v₂
m₁u₁ + m₂u₂ = v₁m₂ + m₂v₂
2 x 40 + m₂ x 0 =20 x 2 + 20 m₂
80 =40 + 20m₂
20 m₂ =40 ⇒ m₂
=\(\frac{40}{20}\) =2 kg
Mass of second body, m₂ = 2 kg

Given
Power of toaster P = 120w
Time, t = 20 min = 20 x 60
\(=\frac { 20 }{ 60 } h\)

E = 40 w h (or) 144 000 ws

Length of the wire I = 1 m
Area A = πr²
Diameter of the wire, d = 0.4mm
radius of the wire, r\(\frac { d }{ 2 } =0.2\times { 10 }^{ -3 }m\)
\(=3.14\times (0.2\times { 10 }^{ -3 })\)
Resistance of the wire R = 20Ω
Resisting Power =?
\(\rho =\frac { RA }{ l } \)
\(A={ \pi r }^{ 2 }\)
= 3.14 x 0.2 x 10^-3 x 0.2 x 10^-3
= 0.1256 x 10^-6m²
ρ = 20 x 0.1256 x 10^-6
ρ = 2.512 x 10^-6Ωm

Given 
For resistance in parallel, the effective resistance R = ?
\(\frac { I }{ { R }_{ p } } =\frac { 1 }{ { R }_{ 1 } } +\frac { 1 }{ { R }_{ 2 } } +\frac { 1 }{ { R }_{ 3 } } \)
\(=\frac { 1 }{ 20 } +\frac { 1 }{ 100 } +\frac { 1 }{ 50 } \)
\(\frac { 1 }{ { R }_{ p } } =\frac { 5+1+2 }{ 100 } =\frac { 8 }{ 100 } \)
\({ R }_{ p }=\frac { 100 }{ 8 } =12.5\Omega \)
\({ R }_{ p }=12.5\Omega \)
i) Total current flowing 
through the circuit \({ I }_{ p }=\frac { { V }_{ p } }{ { R }_{ p } } \)
\({ I }_{ p }=\frac { 125 }{ 12.5 } \)
\({ I }_{ p }=10A\)
(ii) In a parallel circuit, the potential drop across each resistor is same.
∵ V₁ = V₂ = V₃ = 125 V
(iii) The current through each resistor according to ohm's law V = IR
\({ I }_{ 1 }=\frac { { V }_{ 1 } }{ { R }_{ 1 } } =\frac { 125 }{ 20 } =6.25A\)
\({ I }_{ 2 }=\frac { { V }_{ 2 } }{ { R }_{ 2 } } =\frac { 125 }{ 100 } =1.25A\)
\({ I }_{ 3 }=\frac { { V }_{ 3 } }{ { R }_{ 3 } } =\frac { 125 }{ 50 } =2.50A\)
(iv) Total current flowing the circuit
I_p= I₁+ I₂+I₃
I=6.25 + 1.25 + 2.50 = 10A
(v) The electric power of each resistor
P₁=V₁I₁= 125 x 6.25 = 781.25 W
P₂ = V₂I₂ = 125 x 1.25 = 156.25 W
P₃ = V₃I₃ = 125 x 50 = 312.50 W

Given

V= 125 V, R₁ = 20\(\Omega \); R₂=30\(\Omega \);R₃50\(\Omega \)
To find:
(i) R_s =?
(ii) I_s =?
(iii) I_s = ?
(iv) V₁= ?, V₂ = ?, V₃₌ ?
(i) Resistance in series, R_s = R₁+R₂+R₃
R_s = 20 + 30 + 50
= 1000
R_s = 100 \(\Omega \)
(ii) Total current, I_s =\(\frac { V }{ { R }_{ s } } \)
\(\\ I=\frac { 125 }{ 100 } =1.25A\)
I=1.25A
(iii) Since resistors are in series, the current through each resistor is same.
I_s = I₁= I₂=I₃ =1.25A
Voltage drop across R₁ is V₁ = I₁ R₁
1.25 x 20
= 25V
V₁ = 25V
Voltage drop across R₂ is V₂ = I₂R₂
Voltage drop across V₂ 1.25 x 30
V₂ = 37.5 V
Voltage drop across R₃is V₃ = I₃R₃
V₃ = 1.25 x 50
V₃ = 62.5 V
Total Voltage, V₁ + V₂ + V₃
25 + 37.5 + 62.5
= 125V

Given :
Potential V = 110 V
Current passing through bulb, I = 0.9 A
Time, t = 12h / day
For 30 days, time t = 12 x 30
= 360h
To find: Energy consumed, E = ?
Solution
E = VIt
= 0.9 x 110 x 12 x 30
= 35.64 kwh
Energy, E = 35.64 kwh

Given:
Current passing through heater I= 20 A
To find: Resistance of heating element R = ?
Solution
\(V=IR\Rightarrow \frac { V }{ I } =\frac { 220 }{ 20 } =11\Omega \)
Resistance, R = 11Ω

Electrical energy,
E = 1.2 kwh = 1.2 x 10³ wh
Time, t =30 min\(\frac{1}{2}\) hour
To find: Current drawn by the heater, I = ?
Solution
Energy, E = Vlt \(\therefore I=\frac { E }{ Vt } \)
\(I=\frac { 1.2\times { 10 }^{ 3 }\times 2 }{ 120 } =20A\)
Current, I = 20A

Given :
Potential difference, V = 110 V
Charge, q = 5 C
To find: Work done, W = ?
Solution
\(V=\frac { W }{ q } \therefore q\times V=W\)
W= 5 x 110 = 550 J
Work, W = 550J

Given:
Amount work done W = 220 J
Quantity of charge q = 20 C
To find : Potential difference between
two points, V = ?
Solution :
\(V=\frac { W }{ q } =\frac { 220 }{ 20 } \)
=11V
Voltage, V = 11 volt

Current flowing through the wire, I = 6 A
Time taken, t = 2 miniatures = 2 x 60 = 120 S
To find : Amount of charge passing
through wire, q = ?
Solution
\(I=\frac { q }{ t } \Rightarrow =6\times 120=720C\)
Charge, q = 720 C

Given :
Charge passing through bulb, q = 30C
Current drawn by 100 W bulb, I = 680 x 10^-3 A
To find: Time taken to pass
through the bulb t =? t=\(\frac{q}{I}\)
Solution :
\(I=\frac { q }{ t } =\frac { 30 }{ 680\times { 10 }^{ -3 } } \)
=441.173
Time required, t =7.35 min

Given:
Mass of the first body, m₁ = 20 kg
Mass of the second body, m₂ = 15 kg
Initial velocity of first body, u₁ = 40 ms^-1
Initial velocity of second body, u₂ = 0
(Second body is at rest initially)
Final velocity of the first body,
v₁ = Final velocity of the second body, v₂
i.e., v₁ = v₂ = v
According to law of conservation of momentum,
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
m₁u₁ + m₂u₂ =(m₁+m₂)v
20 x 10 + 15 x 0 = (20+15)v
800 = 35 v
Velocity, v=\(\frac{800}{35}\)
=22.85 ms^-1

During impulse on first body:
Mass of the first body,
m₁ = 20 x 10^-3 kg
Time, t₁ = 4 s
Initial velocity, u₁= 0
Final velocity, v₁ =100 m/s
During impulse on second body:
Mass of the second body,
m₂ =10kg
Time, t₂ = 2 minutes
= 2 x 60 s = 120 s
To find: Final velocity, v₂ =?
Initial velocity, u₂ = 0
(i) Impulse on the first body,
F.t₁ =m₁(v₁-u₁)             [∵ u₁=0]
F.t₁ =m₁v₁
F.t₁ =20 x 10^-3 x 100
F.t₁ =2 x 10^-3 x 10³ =2 Js         .............(1)
(ii) Impulse on the second body
F.t₂ =m₂(v₂-u₂)        [∵ u₂=0]
F.t₂ =m₂v₂
Ft₂ =10 x v₂                    ...............(2)
Dividing al and CD, we get

Velocity of the second body, v₂ =6 m/s.

Given:
Ratio of two masses, 2 : 3
i.e = m₁ : m₂ = 2: 3
Mass of first body, m₁ = 2 kg
Mass of second body, m₂ = 3 kg
Acceleration of the first body, a₁ = 6 ms^-2
To find: Acceleration of the second body, a₂ =?
F₁ = F₂
\(\frac { { m }_{ 1 } }{ { m }_{ 2 } } =\frac { { a }_{ 2 } }{ { a }_{ 1 } } \)
\(\frac { 2 }{ 3 } =\frac { { a }_{ 2 } }{ 6 } \)
a₂ =\(\frac { 6\times 2 }{ 3 } \) = 4 ms^-2
Acceleration of the second body,
a₂ = 4 ms^-1

Mass = 100g = 0.1 kg;
Initial speed u = 20 ms^-1
Final velocity v = 0
Change in momentum = ?
mv - mu = 0.1 (0 - 20)
Change in momentum = -2 kg ms^-1.

Given:
Mass of a body, m = 20 kg
To find: Apparent weight, R= ?
(i) Whether lift is descending with the same acceleration of 2m/s²
(ii) When the lift is descending with the same acceleration of 2m/s²
(iii) When the lift is descending with a constant velocity 2ms^-1.
(i) While ascending,
R = m (g+ a)
R = 20 [10 + 2] = 240N
= 24 kg weight
(ii) While descending,
R = m (g- a)
= 20 [10 - 2]
= 20 x 8 = 160N = 160 N
(iii) At constant velocity,
a = 0,
R = mg = 20 x 10 = 200N
= 20 kg weight.

Given:
Mass of the bullet, m = 50 x 10^-3 kg
Initial speed of the bullet, u = 300 m/s^-1
Final speed, v = 0
time, t = 1s
To find: (i) Impulse (of a force), J =?
(ii) Force, F = ?
(i) Impulse, J = Change in momentum
= m (v - u)
= 50 X 10^-3 [0 - 300]
= -15 Ns
(ii) Impulse = F x t
-15 = F x 1s
F = \(\frac { -15 }{ 1 } \) = -15 N

Given:
Mass, m =72 kg
Acceleration due to gravity, g =10 ms^-2
Constant acceleration, a =3 ms²
To find: Apparent weight, R = ?
(i) While descending with constant velocity.
(ii) While ascending with constant acceleration
(i) While descending with constant velocity,
a = 0, R = mg
R = 72 x 10 = 720 N
(ii) While ascending with constant acceleration
a = 3 ms^-2
R = m (g + a)
= 72 (10 + 3)
Apparent weight,
R = 72 x 13 = 936 N

Gram molecular of CO₂ = 44g
Number of molecules present in 44g of CO₂
= 6.023 X 10²³
∴ Number of molecules present in 11g of CO₂
= \(\frac{6.023\times{10}^{23}}{44}\)x 11
= 1.53 x 10²³ molecules.

Number of moles = \(\frac{Mass}{Atomic \ mass}\)
∴ Mass = Number of moles x Atomic mass
= 0.5 x 16
= 8g.

Number of moles = \(\frac{Mass}{Molecular \ mass}\)
= \(\frac{2}{40}\) = 0.05

Number of moles = \(\frac{Mass}{Atomic \ mass}\)
= \(\frac{81}{27}\) = 3 moles

Atomic masses of C= 12, O= 16
Gram molecular mass of CO₂ = 1(C) + 2(O) = 1(12) + 2(16) = 12 + 32 = 44
Gram molecular mass of CO₂ = 44g.