Factual information
\(\text { At } \mathrm{t}=0 \text { amplitude } \mathrm{y}=\frac{1}{1+x^{2}}\)
\(\text { At } \mathrm{t}=2 \text { amplitude } \mathrm{y}=\frac{1}{1+(x-2)^{2}}\)
\(\therefore v=\frac{\Delta y}{\Delta t} =\frac{2}{2} =1.0 \mathrm{~ms}^{-1} \)

\(T=2 \pi \sqrt{\frac{\ell}{g}}\)
\(\text { when a bus is moving } g^{\prime} \sqrt{g^{2}+a^{2}}\)
\(\therefore T \propto \frac{1}{\sqrt{g^{2}+a^{2}}}\)

For mono atomic molecule No of degrees of freedom = 3
For diatomic molecule of degrees of freedom = 5
For triatomic molecule No of degree of freedom = 7
\(\text { Total }=\left[3 x_{1}+7\left(\mu_{2}+\mu_{3}\right)\right] N_{A}\)

Work done in an isochoric process is zero.

Moment of inertia of a disc
\(\mathrm{I}_{1}=\frac{M R^{2}}{2}\)
\(\text { Mass of small disc }=\frac{M}{\pi R^{2}} \times \pi \times\left(\frac{R}{2}\right)^{2}\)
\(=\frac{M}{\pi R^{2}} \times \frac{\pi R^{2}}{4}=\frac{M}{4}\)
By the theorem of parallel axis, the moment of inertia of the small disc. About an axis passing through 0 is
\(I_{2} =\frac{1}{2} \times \frac{M}{4}\left(\frac{R}{2}\right)^{2}+\frac{M}{4}\left(\frac{R}{2}\right)^{2} \)
\(=\frac{M}{8} \times \frac{R^{2}}{4}+\frac{M}{4} \times \frac{R^{2}}{4} \)
\(=\frac{M R^{2}}{32}+\frac{M R^{2}}{16}=\frac{M R^{2}+2 M R^{2}}{32} \)
\(I_{2} =\frac{3 M R^{2}}{32} \)
Moment of inertia of the remaining part is I= I₁ - I₂
\(=\frac{M R^{2}}{2}-\frac{3 M R^{2}}{32} \)
\(=\frac{16 M R^{2}-3 M R^{2}}{32}=\frac{13 M R^{2}}{32}\)
\(I =\frac{13 M R^{2}}{32} \)

Initially velocity has maximum value and at maximum height velocity becomes zero. After that the velocity becomes negative

Dimension of Planck's constant is \(\left[\mathrm{ML}^{2} \mathrm{~T}^{-1}\right]\)
Dimension of Gravitational constant is \(\left[\mathrm{M}^{-1} \mathrm{~L}^{+3} \mathrm{~T}^{-1}\right]\)
Dimension of Velocity constant is LT^-1
Dimension of Length is L
\(\therefore \text { Dimension of } \frac{\sqrt{h G}}{C^{\frac{3}{2}}}\)
\(=\frac{\sqrt{\left(\mathrm{ML}^{2} \mathrm{~T}^{-1}\right)\left(\mathrm{M}^{-1} \mathrm{~L}^{3} \mathrm{~T}^{-2}\right)}}{\left(\mathrm{LT}^{-1}\right)^{3 / 2}} \)
\(=\frac{\sqrt{\mathrm{L}^{5} \mathrm{~T}^{-3}}}{\mathrm{~L}^{3 / 2} \mathrm{~T}^{-3 / 2}} \)
\(=\frac{\mathrm{L}^{5 / 2} \mathrm{~T}^{-3 / 2}}{\mathrm{~L}^{3 / 2} \mathrm{~T}^{-3 / 2}} \)
\(=\mathrm{L}^{5 / 2-3 / 2} \mathrm{~T}^{3 / 2+3 / 2}=\mathrm{L}^{1} \mathrm{~T}^{0}=\mathrm{L}\)
Dimension of length = L

Using Graham's law of diffusion
\(\frac { { R }_{ 1 } }{ { R }_{ 2 } } =\sqrt { \frac { { M }_{ 2 } }{ { M }_{ 1 } } } \)
Squaring both side, we get.
\({ \left( \frac { { R }_{ 1 } }{ { R }_{ 2 } } \right) }^{ 2 }={ \left( \sqrt { \frac { { M }_{ 2 } }{ { M }_{ 1 } } } \right) }^{ 2 }\)
\({ \left( \frac { { R }_{ 1 } }{ { R }_{ 2 } } \right) }^{ 2 }=\frac { { M }_{ 2 } }{ { M }_{ 1 } } \)
\({ M }_{ 2 }={ M }_{ 1 }{ \left( \frac { { R }_{ 1 } }{ { R }_{ 2 } } \right) }^{ 2 }\)
\(={ \left( \frac { { 28.7 } }{ { 7.2 } } \right) }^{ 2 }\)
\(=\frac { 823.69 }{ 51.84 } =15.88\)
=16
The gas is identified as oxygen.

According to Bernoulli's theorem, the sum of pressure energy, kinetic energy, and potentialenergy per unit mass of an incompressible, nonviscous fluid in a streamlined flow remains a constant. Mathematically,
\(\frac{P}{\rho}+\frac{1}{2}v^{2}+gh\) = constant
This is known as Bernoulli's equation.
Proof:

Let us consider a flow of liquid through a pipe AB as shown in Figure. Let V be the volume of the liquid when it enters A in a time t which is equal to the volume of the liquid leaving B in the same time. Let a_A, v_A and P_A be the area of cross section of the tube, velocity of the liquid and pressure exerted by the liquid at A respectively.
Let the force exerted by the liquid at A is
F_A= P_Aa_A
Distance travelled by the liquid in time t is
d = v_At
Therefore, the work done is
W = F_Ad = P_Aa_Av_A t
But a_Av_At = a_Ad = V, volume of the liquid entering at A.
Thus, the work done is the pressure energy (at A), W = F_Ad = P_AV
Pressure energy per unit volume at
A = \(\frac{Pressure \ energy}{volume}=\frac{P_{A} V}{V}=P_{A}\)
Pressure energy per unit mass at
A = \(\frac{Pressure \ energy}{volume}=\frac{P_{A} V}{m}=\frac{P_{A}}{\frac{m}{V}}=\frac{P_{A}}{\rho}\)
Since m is the mass of the liquid entering at A in a given time, therefore, pressure energy of the liquid at A is
\(E_{PA}=P_{A}V=P_{A}V\times (\frac{m}{m})=m \frac{P_{A}}{\rho}\)
Potential energy of the liquid at A, P_EA = mg h_A,
Due to the flow of liquid, the kinetic energy of the liquid at A,
\(KE_{A}=\frac{1}{2}m V_{A}^{2}\)
Therefore, the total energy due to the flow of liquid at A, E_A= EP_A+ KE_A + PE_A
\(E_{A}=m \frac{P_{A}}{\rho}+\frac{1}{2}m V^{2}_{A}+mg \ h_{A}\)
Similarly, let a_B, v_B, and P_B be the area of cross section of the tube, velocity of the liquid, and pressure exerted by the liquid at B. Calculating the total energy at EB, we get
\(EB=m \frac{P_{B}}{\rho}+\frac{1}{2}mv^{2}_{B}+mg h_{B}\)
From the law of conservation of energy,
EA = EB
\(m \frac{P_{A}}{\rho}+\frac{1}{2} mv^{2}_{A}+mgh_{A}=m\frac{P_{B}}{\rho}+\frac{1}{2}mv^{2}_{B}+mgh_{B}\)
\(\frac{P_{A}}{\rho}+\frac{1}{2}V^{2}_{A}+gh_{A}=\frac{P_{B}}{\rho}+\frac{1}{2}V^{2}_{B}+gh_{B}\) = constant
Thus, the above equation can be written as
\(\frac{P}{\rho g}+\frac{1}{2}\frac{v^{2}}{g}+h\) = constant
The above equation is the consequence of the conservation of energy which is true until there is no loss of energy due to friction. But in practice, some energy is lost due to friction. This arises due to the fact that in a fluid flow, the layers flowing with different velocities exert frictional forces on each other. This loss of energy is generally converted into heat energy. Therefore, Bernoulli's relation is strictly valid for fluids with zero viscosity or non-viscous liquids. Notice that when the liquid flows through a horizontal pipe, then \(\mathrm{h}=0 \Rightarrow \frac{P}{\rho g}+\frac{1}{2} \frac{v^{2}}{g}=\) constant.

Newton's inverse square Law:
Newton considered the orbits of the planets as circular. For circular orbit of radius r,' the centripetal acceleration towards the center is
\(a=\frac { { V }^{ 2 } }{ r } \) ...(1)

Here v is the velocity and r, the distance of the planet from the center of the orbit.
The velocity in terms of known quantities r and T, is
v = \(\frac { 2\pi r }{ T } \)    ...(2)
Here T is the time period of revolution of the planet. Substituting this value of v in equation we get,
a = \(\frac { \left( \frac { 2\pi }{ T } \right) ^{ 2 } }{ r } =-\frac { 4\pi ^{ 2 }t }{ { T }^{ 2 } } \)   ...(3)
Substituting the value of 'a' from (3) in Newton's second law, F = ma, where 'm' is the mass of the planet
F = \(\frac { 4\pi mr }{ { T }^{ 2 } } \)   ...(4)
From Kepler's third law
 \(\frac { r^{ 3 } }{ { T }^{ 2 } } \) = k(constant)   ...(5)
\(\frac { r }{ { T }^{ 2 } } =\frac { k }{ { r }^{ 2 } } \)   ....(6)
By substituting equation (6) in the force expression, we can arrive at the law of gravitation
\(F=\frac { 4\pi ^{ 2 }mk }{ { r }^{ 2 } } \) ...(7)
Here negative sign implies that the force is attractive arid it acts towards the center. In equation (7), mass of the planet 'm'. comes explicitly. But Newton strongly felt that according to his third law, if Earth is attracted by the Sun, then the Sun must also be attracted by the Earth with the same magnitude of force. So he felt that the Sun's mass (M) should also occur explicitly in the expression for force. From this insight, he equated the constant 4π²k to GM which turned out to be the law of gravitation
F = \(-\frac { GMm }{ { r }^{ 2 } } \)
Again the negative sign in the above equation implies that the gravitational force is attractive.

At any instant t, the projectile has velocity components along both x-axis arid j-axis. The resultant of these two components gives the velocity of the projectile at that instant t, as shown in the figure

The velocity component at any t along horizontal (x-axis) is v_x = u_x+a_xt
Since, u_x = u, a_x = 0, we get
v_x= u
The component of velocity along vertical direction (y-axis) is v_y = u_y + a_yt
Since, u_y = 0, a_y = g, we get
v_y = gt
Hence the velocity of the particle at any instant is
\(\vec { v } =u\hat { i } +gt\hat { j } \)
The speed of the particle at any instant t is given by
\(\therefore v=\sqrt { { v }_{ x }^{ 2 }+{ v }_{ y }^{ 2 } } \)
\(v=\sqrt { { u }^{ 2 }+{ g }^{ 2 }{ t }^{ 2 } } \)

From law of conservation of momentum, along x-axis,
\(mu_1+0=mv_1\cos\theta_1+mv_2\cos\theta_2\)
\(u_1=v_1\cos\theta_1+v_2\cos\theta_2\)   ....(1)
along y-axis,  \(0=v_1\sin\theta_1-v_2\sin\theta_2\)   ....(2)
From energy of conservation,
\({1\over2}{mu}_{1}^{2}={1\over2}{mv}_{1}^{2}+{1\over 2}{mv}_{2}^{2}\)
\({u}_{1}^{2}={v}_{1}^{2}+{v}_{2}^{2}\)   ...(3)
By using equation (1) and (2) in equation (3) we get,
\(2v_1 v_2\cos(\theta_1+\theta_2)=0\)
\(\cos(\theta_1+\theta_2)=\cos{\pi \over 2}\)
\(\theta_1+\theta_2={\pi \over 2}\)
\(\therefore\) This shows that the particle move perpendicular to each other after collision.   

mass, m=1200 kg
v=54 km/h=\(54\times \frac { 5 }{ 18 } \) m/sec=15 m/s
Angular momentum, L =r m v= 300 \(\times\) 1200 \(\times\)15
=54,00,000
=5.4 x 10⁶ kg m²/s

(i) The centripetal acceleration is given by \(a=\frac { { v }^{ 2 } }{ r } \) This expression explicitly depends on Moon's speed which is non trivial. We can work with the formula
ω²R_m = a_m 
(ii) a_m is centripetal acceleration of the Moon due to Earth's gravity.
ω is angular velocity
(iii) R_m the distance between Earth and the Moon, which is 60 times the radius of the Earth.
R_m = 60R = 60 \(\times\)6.4 \(\times\)10⁶ = 384 \(\times\)10⁶ m
(iv) As we know the angular velocity ω = \(\frac { 2\pi }{ T } \)and
T = 27.3 days = 27.3 \(\times\) 24 \(\times\) 60 \(\times\) 60 second
=  2.358 \(\times\) 10⁶ sec
(v) By substituting these values in the formula for acceleration 
a_m = \(\frac { (4\pi ^{ 2 })(384\times { 10 }^{ 6 }) }{ (2.358\times { 10 }^{ 6 }) } \)  = 0.00272 ms^-2
= 2.72 x 10^-3ms^-2
The centripetal acceleration of Moon towards the Earth is 0.00272 m s^-2

Use of screw gauge in measuring radius of a thin wire in the range of 10^-5 m
The wire whose diameter is to be determined should be clamped between the jaws of the screw gauge. The reading on pitch scale (P.S.R) is noted. Then the reading of the reading of the head scale coinciding with the pitch scale is noted (A.S.C.) The zero correction is applied to head scale incidence. (C.H.S.S.).
The total reading is given by
T.R = P.S.R + (C.H.S.C. X L.C)
The procedure is repeated for at least six different positions of the wire. The mean of the reading taken gives the diameter of the wire. Half of this gives radius of the wire 'r' in the range of 10^-5 m.
Use of vernier caliper in measuring smaller distances in the range of 10^-4 m
The sphere is kept between the two jaws. The main scale reading (MSR) is noted (i.e), the main scale division immediately before the zero of the vernier scale. Then the vernier scale division which coincides with some main scale division (VSD) is noted. Zero correction made with this VSD gives VSR. Multiply this VSR by least count and add with MSR. This will give the diameter of the sphere. Observations for different positions of the sphere is hence forth recorded. The mean of the readings taken gives the diameter of the sphere. Half of this gives the diameter of the sphere. Half of this gives radius in the range of 10^-4 am
(ii) Write a note on triangulation method and radar method to measure larger distances.
Triangulation method for the height of an accessible object
Let AB = h be the height of the tree or tower to be measured. Let C be the point of observation at distance x from B. Place a range finder at C and measure the angle of elevation, ∠ACB = θ as shown in Figure

From right angled triangle ABC,
\(\tan \theta=\frac{A B}{B C}=\frac{h}{x}\)
(or) height h = x tan θ
Knowing the distance x, the height h can be determined.
Radar Method: In Radar method radio waves are sent from transmitters which, after reflection from the planet, are detected by the receiver. By measuring, the time interval (r) between the instants the radio waves are sent and received, the distance of the planet can be determined as to get the actual distance of the object. This method can also be used to determine the height, at which an aeroplane flies from the ground.
\(Speed =\frac{\text { Distance travelled }}{\text { Time taken }} \) (Speed is explained in unit 2 )
Distance (d)= Speed of radio waves x Time taken
\(d=\frac{v \times t}{2}\)
where v is the speed of the radio wave. As the time taken (r) is for the distance covered during the forward and backward path of the radio waves, it is divided by 2 to get the actual distance of the object. This method can also be used to determine the height, at which an aeroplane flies from the ground.

As the temperature increases, the length of the prong of the tuning fork increases. This increases the wavelength of the stationary waves set up in the tunning fork, As frequency,  \(v\ \alpha\ {1\over \lambda}\)so frequency of the tunning fork decreases.

For a closed organ Pipe

\(\ell=\frac{\lambda}{4} \)
\(\therefore \lambda=4 \ell\)
For a open organ pipe \(L=\frac{\lambda}{2} \quad \therefore \lambda=2 L\)
Fundamental frequency of a closed pipe
\(f_{c} =250 \mathrm{~Hz}  \)
\(f_{0} =\frac{V}{\lambda}  \)
\(=\frac{V}{2 L}\)
Fundamental frequency of open organ pipe
\(f_{o} =2\left(\frac{V}{4 L}\right)  \)
\(=2 \times f_{c}  \)
\(=2 \times 250=500 \mathrm{~Hz}\)
∴ Frequency of open organ pipe =500 Hz.

\(\eta =1-\frac{T_{2}}{T_{1}}  \)
\(\eta =\frac{45}{100}=0.45  \)
\(T_{1} =327^{\circ} \mathrm{C}  \)
\(=327+273=600 \mathrm{~K}  \)
\(\therefore 0.45 =1-\frac{T_{2}}{600}   \)
\(\therefore \frac{T_{2}}{600} =1-0.45=0.55  \)
\(\therefore T_{2} =600 \times 0.55=330 \mathrm{~K}  \)
\(\eta =60 \%  \)
\(= \frac{60}{100}=0.6\)
\(0.6 =1-\frac{T_{2}}{T_{1}}  \)
\(0.6 =1-\frac{330}{T_{1}}  \)
\(\frac{330}{T_1} =1-0.6=0.4\)
\({T_{1}} =\frac{330}{0.4}=\frac{3300}{4}=825 \mathrm{~K}  \)
= 825 - 273 = 552 ^oC
Intake temperature = 552 ^oC

Hooke's law of elasticity states that within the elastic limit, the strain produced in a body is directly proportional to the stress applied.
strain ∝ stress

Let the particle of mass m cross a horizontal distance x in time t.

Angular momentum \(\overrightarrow{L}=\int{\overrightarrow{\tau}}{dt}\)
But \(\overrightarrow{\tau}=\overrightarrow{r}\times \overrightarrow{F}\)
\(\overrightarrow{r}=x\hat { i } +y\hat { j} \) and \(\hat { F }=-mg \hat { j } \)
\(\therefore \overrightarrow { \tau }=(x\hat { i }+y\hat { j } ) \times (-mg \hat { j } )\)
\(\overrightarrow { \tau } =-mgx(\hat { i } \times \hat { j } )=-mgx\hat { k } \)
\(\overrightarrow { L } = -mg \int{(xdt)}\hat {k} =-gv\ cos\theta (\int t dt)\hat { k } \)
Let initial time t = 0 and final time t = t_f
\(\overrightarrow {L}=-mg \cos \theta \left( ^{ t }\int _{ 0 }^{ f }{ tdt } \right) \hat k = -\frac{1}{2}mgv\ cos \theta\ t^{2}_{f} \hat{k}\)
Negative sign indicates, \(\overrightarrow {L}\) point inwards. 

We know that when the object falls towards the Earth, it experiences acceleration due to gravity g = 9.8 m s^-2 downward. We can choose the coordinate system as shown in the figure. The acceleration is along the negative y direction.
\(\vec { a } =g(-\hat { j } )-g\hat { j } \)

K.E. = \(\frac { { p }^{ 2 } }{ 2m } \) so p=\(\sqrt { 2mk } \)
Increase in K.E. = 300% of k = 3k
Final K.E., k^' = k+ 3k= 4k
Final momentum, p^' = \(\sqrt { 2m{ k }^{ \prime } } =\sqrt { 2m\times 4k } =2\sqrt { 2mk } \)=2p
% Increase in momentum = \(\frac { { p }^{ \prime }-p }{ p } \times 100\)=100%

To find the force, we need to find the acceleration experienced by the particle.
Th e acceleration is given by a=\(\frac { { d }^{ 2 }y }{ { dt }^{ 2 } } \) or a=\(\frac { dv }{ dt } \)
Here v = velocity of the particle in y direction
v = \(\frac { dy }{ dt } \) = u - gt
The momentum of the particle = mv = m(u - gt)
a = \(\frac { dv }{ dt } \) = -g
The force acting on the object is given by F = ma = -mg
The negative sign implies that the force is acting on the negative y direction. This is exactly the force that acts on the object in projectile motion.

3

Using the equation,
\(|\vec A-\vec B|=\sqrt{5^2+7^2-2\times 5\times 7\cos60^o}=\sqrt{25+49-35}=\sqrt{39}\) units
The angle that \(\vec A-\vec B\) makes with the vector \(\vec A\) is given by
\(\tan\alpha_2=\frac{7\sin60^o}{5-7\cos60^o}=\frac{7\sqrt{3}}{10-7}=\frac{7}{\sqrt{3}}\)= 4.041
\(\alpha_2=tan^{-1}(4.041)=76^o\)

(a) A frame of reference which is at rest or which is moving with a uniform velocity along a straight line is called an inertial frame of reference.
(b) In the inertial frame of reference Newton's laws of motion holds good.
Example: The lift at rest, lift moving (up or down) with constant velocity, car moving with constant velocity on a straight road.

\({ \left( \frac { { T }_{ 2 } }{ { T }_{ 1 } } \right) }^{ 2 }={ \left( \frac { { R }_{ 2 } }{ { R }_{ 1 } } \right) }^{ 3 }(or)\ { \left( \frac { { T }_{ 1 } }{ 5 } \right) }^{ 2 }={ \left( \frac { { 4 }R_{ 1 } }{ { R }_{ 1 } } \right) }^{ 3 }\)
\({ T }_{ 2 }=\sqrt { 64\times 25 } \)
= 40h

(i) This is a thermodynamic process in which the thermodynamic system returns to its initial state after undergoing a series of changes.
(ii) Since the system comes back to the initial state, the change in the internal energy is  zero.
(iii) In cyclic process, heat can flow in to system and heat flow out of the system.

Consider one mole of a gas.
p - pressure; v - volume; T - temperature;
M - molecular mass of the gas; density p.
Accumulate to kinetic theory, the pressure excited by the gas is
v = \(\frac{1}{3}\)pc² = p =\(\frac{1}{3}\)\(\frac{m}{3}\).c2 (or) pv = \(\frac{1}{3}\)Mc²....(1)
pv = \(\frac{2}{3}\).\(\frac{1}{3}\)Mc²
\(\frac{1}{2}\)Mc². is the ave. K.E.E of one mole of the gas.
\(\therefore\) pv =\(\frac{2}{3}\) .E 
The ideal gas eqn for one mole of a gas is
pv = RT \(\therefore\) RT  = \(\frac{2}{3}\) E   (or) E=\(\frac{3}{2}\).RT
The above equation gives the mean K.E of one mole of gas.
\(\therefore\)The mean K. E per molecule is proportional to the absolute temperature of the gas.

Periodic motion is any motion which repeats itself in a fixed time internal.
Example:
Hands in pendulum clock, swing of a cradle, the revolution of the Earth around the Sun, waxing and waning of Moon, etc.
Non periodic motion is any motion which does not repeat itself after a regular interval of time.
Example: Occurrence of Earth quake, eruption of volcano, etc.

a. Since k₁ and k₂ are parallel, k_u = k₁ + k₂ Similarly, k₃ and k₄ are parallel, therefore, k_d = k₃ + k₄ But k_u and k_d are in series,
therefore, \({ k }_{ eq }=\frac { { k }_{ u }{ k }_{ d } }{ { k }_{ u }+{ k }_{ d } } \)
If all the spring constants are equal then, k₁ = k₂ = k₃ = k₄ = k
Which means, k_u = 2k and k_d = 2k
Hence, \({ k }_{ eq }=\frac { { 4k }^{ 2 } }{ 4k } =k\)
b. Since k₁ and k₂ are parallel, k_A = k₁ + k₂ Similarly, k₄ and k₅ are parallel,
therefore, k_B = k₄ + k₅
But k_A, k₃, k_B, and k₆ are in series,
therefore, \(\frac { 1 }{ { k }_{ eq } } =\frac { 1 }{ { K }_{ A } } +\frac { 1 }{ { K }_{ 3 } } +\frac { 1 }{ { K }_{ B } } +\frac { 1 }{ { K }_{ 6 } } \)
If all the spring constants are equal
then, k₁ = k₂ = k₃ = k₄ = k₅ = k₆ = k
which means, k_A = 2k and k_B = 2k
\(\frac { 1 }{ { k }_{ eq } } =\frac { 1 }{ { 2K } } +\frac { 1 }{ { K } } +\frac { 1 }{ { 2K } } +\frac { 1 }{ { K } } =\frac { 3 }{ { K } } \)
\({ k }_{ eq }=\frac { k }{ 3 } \)

In this, air from nearly regions get concentrated in a small space, so I decreases considerably. Since I.\(\omega\) = constant so co increases so high.

Linear momentum remains conserved.
Resultant initial momentum   \(p=\sqrt{{p}_{1}^{2}+{p}_{2}^{2}+2{p}_{1}{p}_{2}\cos \ \theta}\)
\({\{ 2m\left( {v \over 2} \right) \}}^{2}=\{mv\}^2+2\{ mv\}\{ mv \}\ \cos\theta\)
\(1 = 1 + 1 +2 \times 1 \times 1 \cos\theta\)
\(\cos \theta=-{1\over2}\)
\(\theta=120°\)

Radius of gold nucleus = 41.3\(\times\)10^-15 m.
Volume (v) \(= {{4}\over{3}}\pi r^3\)
\(={{4}\over{3}}\times3.14\times(41.3\times{10}^{-15})^3\)
v = 2.95\(\times\)10^-40 m.