**1.** 11. Find the acceleration of the moon with respect to the
earth from the following data : Distance between the
earth and the moon = 3.85 x 10 5 km and the time taken
by the moon to complete one revolution around the earth
= 27.3 days.

1 None

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SolutionsDistance between Earth and Moon $\\$ $T$ = $27.3$ $days$ =$ 24 $x$ 3600 $x$ (27.3)sec$= $2.36 $x$ 10^6 sec$ $\\$ v = $\frac{2\pi f}{T}$ =$\frac{ 2\times3.14 \times3.85\times10^8}{2.36 \times 10^6} $ = $1025.42 m/sec$ $\\$ $a = \frac{v^2}{r}$ =$\frac{(1025.42)^2}{64 \times 10^5}$ = $0.00273m/sec^2$=$2.73\times10^-3m/sec^2$

**2.** 2. Find the acceleration of a particle placed on the surface
of the earth at the equator due to earth's rotation. The
diameter of earth = 12800 km and it takes 24 hours for
the earth to complete one revolution about its axis.

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SolutionsDiameter of earth = 12800km $\\$ Radius R = 6400km = 64 x 10^5 $\\$ V = $\frac{2 \pi R}{T}$ = $\frac {2\times3.14 \times64\times10^5}{24\times3600}$ m/sec=465.185 $\\$ a = $\frac{V^2}{R}$= $\frac{(46.5185)^2}{64\times10^5}$ = $0.0338m/sec^2$

**3.** 3. A particle moves in a circle of radius 1.0 cm at a speed
given by v = 2.0 t where v is in cm/s and t in seconds.
(a) Find the radial acceleration of the particle at t = 1 s.
(b) Find the tangential acceleration at t = 1 s. (c) Find
the magnitude of the acceleration at t = 1 s.el.

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SolutionsV = 2t, r= 1cm $\\$ a) Radial acceleration at t= 1 sec $\\$ a= $\frac{V^2}{R}$ = $\frac{2^2}{1}$ = $4cm/sec^2$ $\\$ b) Tangential acceleration at t =1 sec $\\$ a= $\frac{dv}{dt}$ = $\frac{d}{dt}$(2t) = $2cm/sec^2$ $\\$ c)Magnitude of acceleration at t=1 sec $\\$ a=$\sqrt4^2+\sqrt2^2$ = $\sqrt20 cm/sec^2$

**4.** 4. A scooter weighing 150 kg together with its rider moving
at 36 km/hr is to take a turn of radius 30 m. What
horizontal force on the scooter is needed to make the
turn possible ?

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SolutionsGiven that m = 150kg. $\\$ v=36km/hr = 10 m/sec, r=30m $\\$ Horizontal force needed is $\frac{mv^2}{r}$ = $\frac{150\times(10)^2}{30} $ = $\frac{150\times100}{30}$ = 500N

**5.** 5. If the horizontal force needed for the turn in the previous
problem is to be supplied by the normal force by the
road, what should be the proper angle of banking ?

5 None

SolutionsIn the diagram $R cos\theta =mg ..(i)$ $\\$ $R sin\theta= \frac{mv^2}{r}$ $\\$ Dividing equation (i) by equation (ii) $Tan\theta = \frac{mv^2}{rmg}=\frac{v^2}{rg}$ $\\$ $v=36km/hr = 10m/sec, r=30m$ $\\$ $Tan\theta =\frac{v^2}{rg} = \frac{100}{30\times10} =(1/3)$ $\\$ $\Rightarrow \theta = tan^-1 (1/3)$

**6.** 6. A park has a radius of 10 m. If a vehicle goes round it
at an average speed of 18 km/hr, what should be the
proper angle of banking ?

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SolutionsRadius of Park = r = 10m $\\$ speed of vehicle = 18km/hr =5m/sec $\\$ $ Angle of banking tan \theta = \frac{v^2}{rg} $ $\\$ $\Rightarrow \theta = tan^-1 \frac {v^2}{rg} = tan^-1 \frac{25}{100} = tan^-1 (1/4) $

**7.** 7. If the road of the previous problem is horizontal (no
banking), what should be the minimum friction
coefficient so that a scooter going at 18 km/hr does not
skid ?

7 None

SolutionsThe road is horizontal (no banking) $\\$ $\frac{mv^2}{R} =\mu N$ $\\$ $\frac{mv^2}{R}=\mu mg \hspace{1cm} v=5m/sec, R =10m$ $\\$ $\Rightarrow \frac{25}{10} = \mu g \Rightarrow \mu = \frac{25}{100} = 0.25 $ $\\$

**8.** 8. A circular road of radius 50 m has the angle of banking
equal to 30°. At what speed should a vehicle go on this
road so that the friction is not used ?

8 None

Solutions$Angle\hspace{1cm} of\hspace{0.5cm} banking = \theta = 30 ^{\circ}$ $\\$ Radius =r = 50m $\\$ $ tan\theta = \frac{v^2}{rg} \Rightarrow tan 30^{\circ} = \frac{v^2}{rg}$ $\\$ $\Rightarrow \frac{1}{\sqrt{3}} = \frac{v^2}{rg} \Rightarrow = v^2 = \frac{rg}{\sqrt{3}} = \frac{50\times10}{\sqrt{3}}$ $\\$ $\Rightarrow v = \sqrt\frac{500}{\sqrt{3}} = 17m/sec.$ $\\$

**9.** 9. In the Bohr model of hydrogen atom, the electron is
treated as a particle going in a circle with the centre at
the proton. The proton itself is assumed to be fixed in
an inertial frame. The centripetal force is provided by
the Coulomb attraction. In the ground state, the electron
goes round the proton in a circle of radius
5.3 x 10 -11 m. Find the speed of the electron in
the ground state. Mass of the electron = 9.1 x 10^ -3 kg
and charge of the electron = 1.6 x 10 -19 C.

1 None

9 None

SolutionsElectron revolves around the proton in a circle having proton at the centre. $\\$ Centripetal force is provided by coulomb attraction. $\\$ $r=5.3 \rightarrow t 10^-11m \hspace{1cm} m=mass\hspace{0.5cm}of \hspace{0.5cm}electron=9.1\times10^-3kg $ $\\$ $charge\hspace{0.5cm}of\hspace{0.5cm} electron = 1.6\times10^-19c$ $\\$ $ \frac{mv^2}{r} = k\frac{q^2}{r^2} \Rightarrow v^2 = \frac{kq^2}{rm} =\frac {9\times10^9\times1.6\times1.6\times10^-38}{5.3\times10^-11\times9.1\times10^-31}=\frac{23.04}{48.23} \times 10^{13}$ $\\$ $v^2= 0.477\times10^13=4.7\times10^{12}$ $\\$ $\Rightarrow v = \sqrt{4.7\times10^{12}}= 2.2\times10^6 m/sec$ .

**10.** 10. A stone is fastened to one end of a string and is whirled
in a vertical circle of radius R. Find the minimum speed
the stone can have at the highest point of the circle.

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SolutionsAt the highest point of a vertical circle $\\$ $\frac{mv^2}{R}= mg$ $\\$ $ \Rightarrow v^2 = Rg \Rightarrow v = \sqrt{Rg} $

**11.** 11. A ceiling fan has a diameter (of the circle through the
outer edges of the three blades) of 120 cm and rpm 1500
at full speed. Consider a particle of mass 1 g sticking at
the outer end of a blade. How much force does it
experience when the fan runs at full speed ? Who exerts
this force on the particle ? How much force does the
particle exert on the blade along its surface ?

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SolutionsDiameter of the ceiling fan= 120cm $\\$ $\therefore Radius = r =60cm =0/6m $ $\\$ Mass of particle on the outer end of the blade is 10. $\\$ $ n = 1500 rev/min =25 rev/sec $ $\\$ $\omega = 2\pi n = 2 \pi \times25 = 157.14$ $\\$ $Force\hspace{0.25cm}of\hspace{0.25cm}the\hspace{0.25cm}particle\hspace{0.25cm}on\hspace{0.25cm} the\hspace{0.25cm} blade = Mr\omega^2=(0.001)\times0.6\times(157.14) = 14.8N $ $\\$ The fan runs at full speed in a circular path. This exerts the force on the particle (inertia). $\\$The particle also exerts a force of 14.8N on the blade along the surface.

**12.** 12. A mosquito is sitting on an L.P. record disc rotating on
a turn table at 33$\frac{1}{3}$ per minute. The distance 3
of the mosquito from the centre of the turn table is
10 cm. Show that the friction coefficient between the
record and the mosquito is greater than It 2/81. Take
g =$10 m/s^2.$

12 None

SolutionsA pendulum is suspended from the ceiling of a car taking a turn $\\$ $r=10m, v=36km/hr=10m/sec\hspace{0.1cm} g=10m/sec^2$ $\\$ From the figure Tsin$\theta={mv^2}{r} ..(i)$ $\\$ T$cos\theta=mg$ ..(ii) $\\$ $\Rightarrow \frac{sin\theta}{cos\theta}=\frac{mv^2}{mg}\Rightarrow tan\theta=\frac{v^2}{rg} \Rightarrow \theta=ta{n}^-1\frac{v^2}{rg}$ $\\$ $=tan^-1\frac{100}{10\times10}=tan^-1(1) \Rightarrow \theta = 45^{\circ} $

**13.** 13. A simple pendulum is suspended from the ceiling of a
car taking a turn of radius 10 m at a speed of 36 km/h.
Find the angle made by the string of the pendulum with
the vertical if this angle does not change during the turn.
Take g= $10 m/sec^2$.

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Solutions**14.** 14. The bob of a simple pendulum of length 1 m has mass
100 g and a speed of 1.4 m/s at the lowest point in its
path. Find the tension in the string at this instant.

14 None

SolutionsBob has a velocity of 1.4m/sec,when the string makes an angle of 0.2 radian. $\\$ m=100g=0.1kg r=1m, v=1.4m/sec $\\$ From the diagram, $\\$ T - mgcos$\theta$=$\frac{mv^2}{R}$ $\\$ $\Rightarrow T = \frac{mv^2}{R} +mgcos\theta$ $\\$ $\Rightarrow T = \frac{0.1\times{1.4}^2}{1} + 0.1\times9.8\times[1-\frac{{\theta}^2}{2}]$ $\\$ $\Rightarrow = 0.196 + 9.8 \times [1-\frac{({2})^2}{2}] (\therefore cos \theta = 1-\frac{{\theta}^2}{2} for small \theta) $ $\\$ $\Rightarrow = 0.196 + (0.98) \times (0.98)= 0.916 + 0.964= 1.156N \approx 1.16N$ $\\$

**15.** 15. Suppose the bob of the previous problem has a speed of
1.4 m/s when the string makes an angle of 0.20 radian
with the vertical. Find the tension at this instant. You
can use cos$\theta$ = 1- $\theta^2 /2\hspace{0.2cm} and\hspace{0.2cm} sin\theta$ = 0 for small$\hspace{0.1cm}$$ \theta$. $\\$

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Solutions**16.** 16. Suppose the amplitude of a simple pendulum having a
bob of mass m is $\theta_{\circ}$ . Find the tension in the string when
the bob is at its extreme position.

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SolutionsAt the extreme position, velocity of the pendulum is zero.$\\$ So there is no centrifugal force. $\\$ So T=mgcos$\theta_{\circ}$

**17.** 17. A person stands on a spring balance at the equator.
(a) By what fraction is the balance reading less than his
true weight ? (b) If the speed of earth's rotation is
increased by such an amount that the balance reading
is half the true weight, what will be the length of the
day in this case ?

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Solutionsa)Net force on the spring balance. $\\$ R=mg -m$\omega^2$r $\\$ So the fraction less than the true weight (3mg) is $\\$ =$\frac{mg-(mg-m\omega^2r)}{mg}= \frac{\omega^2}{g}={[{\frac{2\pi}{24\times3600}}^2]}\times \frac{6400\times10^3}{10}=3.5\times 10^-3$ $\\$ b) When the balance reading is half the true weight, $\\$ $\frac{mg-(mg-m\omega^2r)}{mg}=1/2 $ $\\$ $\omega^2r = g/2 \Rightarrow \omega= \sqrt{\frac{g}{2r}}=\sqrt{\frac{10}{2\times6400\times10^3}}rad/sec$ $\\$ $\therefore\hspace{0.2cm}Duration\hspace{0.2cm} of\hspace{0.2cm} the day is$ $\\$ $ T = \frac{2\pi}{\omega} = 2\pi \times \sqrt{\frac{2\times6400\times10^3}{9.8}}sec=2\pi\sqrt{\frac{64\times10^6}{49}}sec= \frac{2\pi\times8000}{7\times3600}hr=2hr $ $\\$

**18.** 18. A turn of radius 20 m is banked for the vehicles going
at a speed of 36 km/h. If the coefficient of static friction
between the road and the tyre is 0.4, what are the
possible speeds of a vehicle so that it neither slips down
nor skids up ?

1 None

18 None

SolutionsGiven v=36km/hr =10m/s r=20m, $ \mu=0.4$ $\\$ The road is banked with an angle , $\\$ $tan^-1[\frac{v^2}{rg}]=tan^-1[\frac{100}{20\times10}]=tan^-1 \frac{1}{2} or tan\theta=0.5$ $\\$ When the car travels at max speed such that it slips upwards, $ \mu R1$ $\\$ acts downwards as shown in Fig.1 $\\$ So $R1 - mgcos\theta -\frac{mv1^2}{r}sin\theta=0 ..(i) $ $\\$ And $ \mu R1 + mgsin\theta - \frac{mv1^2}{r}cos\theta = 0 ..(ii) $ $\\$ Solving the equations we get, $\\$ V1= $\sqrt{rg\frac{tan\theta- \mu }{1+\mu tan \theta}} =\sqrt{20\times10\times\frac{0.1}{1.2}} = 4.082 m/s = 14.7 km/hr $ $\\$ So the possible speeds lie between 14.7 km/hr and 54km/hr

**19.** 19. A motorcycle has to move with a constant speed on an
overbridge which is in the form of a circular arc of radius
R and has a total length L. Suppose the motorcycle
starts from the highest point. (a) What can its maximum
velocity be for which the contact with the road is not
broken at the highest point ? (b) If the motorcycle goes
at speed 1/42 times the maximum found in part (a),
where will it lose the contact with the road ? (c) What
maximum uniform speed can it maintain on the bridge
if it does not lose contact anywhere on the bridge ?

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SolutionsSo the possible speeds are between 15.7 km/hr and 54 km/hr $\\$ R= radius of the bridge $\\$ L=total length of the overbridge $\\$ a) At the highest point $\\$ mg=$\frac{mv^2}{R} \Rightarrow v^2 = Rg \Rightarrow v = \sqrt{Rg} $ $\\$ b) Given,$ v =\frac{1}{\sqrt{2} }\sqrt{Rg} $ $\\$ suppose it loses contact at . So, at B, mgcos$\theta = \frac{mv^2}{R} $ $\\$ $\Rightarrow v^2=Rg_2cos\theta $ $\\$ $[\sqrt{\frac{Rv}{2}}]^2=Rgcos\theta \Rightarrow\frac{Rg}{2}=Rgcos\theta\Rightarrow cos\theta =\frac{1}{2} \Rightarrow60^{\circ}=\frac{\pi}{3} $ $\\$ $\theta=\frac{l}{r}\rightarrow l=r\theta=\frac{\pi R}{3}$ $\\$ So it will lose contact at distance $\frac{\pi R}{3}$ from the highest point $\\$ c) Let the uniform speed on the bridge be v. $\\$ The chance of losing contact is maximum at the end of the bridge for which a= $\frac{L}{2R}$ $\\$ So, $\frac{mv^2}{R}=mgcos\alpha \Rightarrow v=\sqrt{gRcos[\frac{l}{2R}] }$ $\\$

**20.** 20. A car goes on a horizontal circular road of radius R, the
speed increasing at a constant rate —dv/dt = a. The friction
coefficient between the road and the tyre is $\mu$. Find the
speed at which the car will skid.

20 None

SolutionsSince the motion is non uniform, the acceleration has both radial and tangential components. $\\$ $a_r=\frac{V^2}{r} $ $\\$ $a_t=\frac{dv}{dt}=a$ $\\$ Resultant magnitude = $\sqrt{[\frac{v^2}{r}]^2+a^2}$ $\\$ Now $\mu N= \sqrt{[\frac{v^2}{r}]^2 + a^2} \Rightarrow \mu mg = m \sqrt{[\frac{v^2}{r}]^2+a^2}\Rightarrow\mu^2g^2=\frac{V^4}{r2}+a^2$ $\\$ $\Rightarrow v^4=(\mu^2g^2-a^2)r^2\Rightarrow v=[(\mu^2g^2-a^2)r^2]^{1/4}$

**21.** 21. A block of mass m is kept on a horizontal ruler. The
friction coefficient between the ruler and the block is g.
The ruler is fixed at one end and the block is at a
distance L from the fixed end. The ruler is rotated about
the fixed end in the horizontal plane through the fixed
end. (a) What can the maximum angular speed be for
which the block does not slip ? (b) If the angular speed
of the ruler is uniformly increased from zero at an
angular acceleration a, at what angular speed will the
block slip ?

21 None

Solutions**22.** 21. A block of mass m is kept on a horizontal ruler. The
friction coefficient between the ruler and the block is g.
The ruler is fixed at one end and the block is at a
distance L from the fixed end. The ruler is rotated about
the fixed end in the horizontal plane through the fixed
end. (a) What can the maximum angular speed be for
which the block does not slip ? (b) If the angular speed
of the ruler is uniformly increased from zero at an
angular acceleration a, at what angular speed will the
block slip ?

22 None

Solutionsa) When the ruler makes uniform circular motion in the horizontal plane $\\$ $\mu=mg = m\omega_1^2L$ $\\$ $\omega_1=\sqrt{\frac{\mu g}{L}}$ $\\$ b) When the ruler makes uniformly accelerated circular motion $\\$ $\mu mg=\sqrt{({m\omega_2}^2L)^2 + (mL\alpha)^2}\Rightarrow \omega_2^4 + \alpha^2=\frac{\mu^2g^2}{L^2}\Rightarrow \omega_2=[(\frac{\mu g}{L})^2-\alpha^2]^{1/4} $

**23.** 22. A track consists of two circular parts ABC and CDE of
equal radius 100 m and joined smoothly as shown in
figure (7-E1). Each part subtends a right angle at its
centre. A cycle weighing 100 kg together with the rider
travels at a constant speed of 18 km/h on the track.
(a) Find the normal contact force by the road on the
cycle when it is at B and at D. (b) Find the force of
friction exerted by the track on the tyres when the cycle
is at B, C and D. (c) Find the normal force between the
road and the cycle just before and just after the cycle
crosses C. (d) What should be the minimum friction
coefficient between the road and the tyre, which will ensure that the cyclist can move with constant speed ?
Take g= 10 $\frac{m}{s^2}$

23 None

SolutionsRadius of the curves = 100m $\\$ Weight= 100Kg $\\$ Velocity = 18km/hr=5m/sec $\\$ a) at B mg$\frac{mv^2}{R}=N \Rightarrow N=(100\times10)-\frac{100\times25}{100}=1000-25=975N$ $\\$ At d, N= mg+$\frac{mv^2}{R}=1000+25=1025N$ $\\$ b)At B & D the cycle has no tendency to slide. So at B & D,frictional force is zero. $\\$ At 'C', mgsin$\theta = F \Rightarrow F = 1000\times \frac{1}{\sqrt2}=707N $ $\\$ c) (i) Before 'C' mgcos$\theta-N=\frac{mv^2}{R} \Rightarrow N= mgcos\theta-\frac{mv^2}{R}=707-25=683N $ $\\$ (ii) N-mgcos$\theta=\frac{mv^2}{R}+mgcos\theta=25+707=732N $ $\\$ d) To find out the minimum desired coeff. of friction,we have to consider a point just before C.( Where N is minimum) $\\$ Now, $\mu N= mgsin\theta \Rightarrow \mu\times682 = 707$ $\\$ So,$ \mu=1.037$ $\\$

**24.** 23. In a children's park a heavy rod is pivoted at the centre
and is made to rotate about the pivot so that the rod
always remains horizontal. Two kids hold the rod near
the ends and thus rotate with the rod (figure 7-E2). Let
the mass of each kid be 15 kg, the distance between the
points of the rod where the two kids hold it be 3.0 m
and suppose that the rod rotates at the rate of 20
revolutions per minute. Find the force of friction exerted
by the rod on one of the kids.

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Solutionsd=3m$\Rightarrow R=1.5m $ $\\$ R=distance from the centre to one of the kids $\\$ N=20 rev per min=20/60=1/3 rev per sec $\\$ $\omega=2\pi r=2\pi /3$ $\\$ $m=15kg$ $\\$ $\therefore Frictional force F=mr\omega^2=15\times(1.5)\times\frac{(2\pi)^2}{9}=5 \times (0.5) \times 4\pi^2=10\pi^2 $ $\\$ $\therefore$ Frictional force for one of the kids is $10\pi^2 $ $\\$

**25.** 24. A hemispherical bowl of radius R is rotated about its
axis of symmetry which is kept vertical. A small block
is kept in the bowl at a position where the radius makes
an angle 0 with the vertical. The block rotates with the
bowl without any slipping. The friction coefficient
between the block and the bowl surface is u. Find the
range of the angular speed for which the block will not
slip.

25 None

SolutionsIf the bowl rotates at maximum angular speed, the block tends to slip upwards. So, the frictional force acts downwards. Here, r$=Rsin\theta $ $\\$ From FBD-1 $\\$ R$_1-mgcos\theta-m\omega_1^2(Rsin\theta) sin\theta=0$ ..(i) [because r=$Rsin\theta$] $\\$ and $\mu R_1mgSin\theta-m\omega_1^2(Rsin\theta]$ $\\$ and $\mu R_1mgsin\theta-m\omega_1^2(Rsin\theta)cos\theta=0 $ ..(ii) $\\$ Substituting the value of $R_1$ from Eq (i) in Eq(ii), it can be found out that $\\$ $\omega1=[\frac{g(sin\theta+\mu cos\theta}{Rsin\theta(cos\theta-\mu sin\theta)}]^{1/2} $ $\\$ Again, for minimum speed, the frictional force$ \mu R_2 $acts upward from FBD-2, it can be proved that $\\$ $\omega_2=[\frac{g(sin\theta-\mu cos\theta)}{Rsin\theta(cos\theta + \mu sin\theta)}]^{1/2} $ $\\$ $\therefore$ the range of speed is between $\omega_1$ and$\omega_2$ $\\$

**26.** 25. A particle is projected with a speed u at an angle 0 with
the horizontal. Consider a small part of its path near
the highest position and take it approximately to be a
circular arc. What is the radius of this circle ? This
radius is called the radius of curvature of the curve at
the point.

26 None

SolutionsParticle is projected with speed 'u' at an angle $\theta$. At the highest point, the vertical component of velocity is '0'. $\\$ So, at the point, velocity =ucos$\theta$ $\\$ centripetal force = m$u^2cos^2(\frac{\theta}{r}) $ $\\$ At the highest pt. $\\$ mg=$\frac{mv^2}{r} \Rightarrow r=\frac{u^2cos^2\theta}{g} $ $\\$

**27.** 26. What is the radius of curvature of the parabola traced
out by the projectile in the previous problem at a point
where the particle velocity makes an angle 0/2 with the
horizontal ?

27 None

SolutionsLet 'u' be the velocity at the pt where it makes an angle $\theta/2$ with the horizontal. The horizontal component remains unchanged $\\$ So, v cos$\theta/2=\omega cos\theta \Rightarrow v =\frac{ucos\theta}{cos\frac{\theta}{2}} $ ...(ii) $\\$ From the figure $\\$ mg$ cos(\theta/2)=\frac{mv^2}{r} \Rightarrow r = \frac{V^2}{gcos(\theta/2)} $ $\\$ putting the value of 'v' from equn(i) $\\$ $r=\frac{u^2cos^2\theta}{gcos^3(\theta/2)} $ $\\$