IIT JEE , AIEEE Forum - Articles Category , Community shelf - Few selected problems in physics with solutions Part 2

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A plano convex lens of focal length 40 cms is silvered at plane surface. Find the focal length of arrangement after silvering.

Let the refractive index of the lens be <Mu>

The focal length of any resulting optic device can be found by letting parallel rays from infinity (parallel to the principal axis) fall on the device and finding where the image is formed.

When the plane surface isn't silvered
When parallel rays fall on the curved surface,
u = -<alpha>
Then by the refraction formula
<Mu>/v - 1/ -<alpha> = (<Mu> -1)/R
v =<Mu>R/(<Mu> -1)
This serves as object for refraction at the plane surface.
v' = v / <Mu> = R/(<Mu>-1)
Hence focal length of this plano convex lens = R/(<Mu>-1) = 40 cm

Once the plane surface is silvered
Now, when the parallel rays from infinity fall on the unsilvered surface,
u = - <alpha>
Then by the refraction formula
<Mu>/v - 1/ -<alpha> = (<Mu>-1)/R
v =<Mu>R/(<Mu>-1)
This serves as the object for reflection from the silvered plane surface. For the reflection at this surface, the image is formed at the same distance but on the opposite side.
v' = -<Mu>R/(<Mu> -1)
This serves as the object for the next refraction at the unsilvered curved surface.
1/v'' - <Mu>/v' = (1-<Mu>)/R
v'' = -R/2(<Mu>-1)
Hence, focal length of this new optical device = R/2(<Mu>-1) = 20 cm.

The velocity of a particle moving along X axis is v =(40 -10t)m/s, where,t is in sec. At the time t=0,the x co-ordinate of particle was zero. Find the time when the particle is at a distance of 60m from origin.

v = 40 - 10t
₀^x dx = ₀^t (40 - 10t) dt
x = 40t - 5t²
Now at t = 4 s, the particle comes to rest. That is it turns around towards the -ve X axis at t = 4 s.
Now, position of the particle at t = 4 s is
x(4 s) = 80 m.
Hence, the particle once crosses the x = +60m point before turning, and then again when its retuning back. Moreover, when it crosses the origin to go into the negative X axis, it reaches the point x = -60 m at some instant and then continues to move forward.
Hence there are three time instants when the particle is at a distance of 60 m from the origin, twice at +60m and once at -60m.
For x = +60 m,
40t - 5t² = 60
t = 2 s, 6 s
For x = -60 m,
40t - 5t² = -60
t = 4 + 2

7 s (discarding the -ve root)

Hence the time instants when the particle is at a distance of 60 m from the origin are
t = 2 s, 6 s, 4 + 2

7 s

Find the force of interaction between the two dipoles placed parallel to each other on the same line.

Consider any two dipoles of dipole moments p₁ and p₂ placed parallel to each other along a straight line. Let the distance between them be d, and the length of the dipoles be 2l₁ and 2l₂.

Net force on the second dipole due to the first is the sum of force experienced by the positive and negative charges of the second dipole due to the electric field of the first dipole.

F = F₊ + F_-
= (1/4<pi>

)[2p₁q₂/(d+l₁)³ + 2p₁(-q₂)/(d-l₁)³]e_p2
where e_p2is the unit vector along p₂.

F =-(1/4<pi> )(2p₁q₂)[6d²l₂ + 2l₂³/(d²-l₁²)³]e_p2
If the distance between the two dipoles d, is much greater than the dimensions of the dipoles, then
d²>>l₁²
F =-(1/4<pi>

)(6p₁p₂/d⁴) e_p2The force is directed away from the dipole moment direction. Hence its an attractive force.

A car goes out of control and slides off a steep embarkment of height h at angle <theta> to the horizontal.If lands in a ditch at a distance R from the base. Find the speed at which the car leaves the slope.

The car flies off tangentially from the outer circle. Let its speed be v at the moment it flies off the circular path. Then at that moment its at a height h above the ground and the base of the embarkment is at a distance hcosec<theta> along the incline.

When the car flies off the embarkment, it flies as a projectile projected horizontally from a height h. The quantity R is measured from the base of the embarkment and not from the point vertically below the point of projection.

Range of the car =

R² - h²cot² <theta>Then,
v

(2h/g) =

(R² - h²cot² <theta>)
v =

g(R² - h²cot² <theta>)/2h

Two particles 1 & 2 move with constant velocity v₁ & v₂ . At their initial moment their radius vector are equal to r₁ & r_{2 .}How must these four be interrelated for the particles to collide?

As the particles move with constant velocities their position vectors are
r₁(t) = r₁ + v₁t
r₂(t) = r₂ + v₂t
At the instant of collision, say at t = T,
r₁(T) = r₂(T)
r₁ + v₁T = r₂ + v₂T
r₁ - r₂ = (v₂ - v₁)T
Hence r₁ - r₂ and (v₂ - v₁) are parallel vectors. As they are parallel vectors, unit vectors along them must be same.
Hence,
r₁ - r₂/|r₁ - r₂| = (v₂ - v₁)/|(v₂ - v₁)|

Calculate the electric field at any axial point of two dipoles placed parallel to each other.

Let the dipoles be placed at a distance d from each other, both the dipole moments being oriented from left to right. Consider an axial point at a distance L from the right dipole.Then the electric field at this axial point is the vector sum of the electric fields of these two individual dipoles.

E = E₁+ E₂
= (1/4<pi>

)[2p₁/(d+L)³ + 2p₂/L³]
=(1/2<pi>

)[p₁/(d+L)³ + p₂/L³]

A simple pendulum of length L has a bob of mass m,with a charge q on it. A vertical sheet of charge , with charge

per unit area, passes through the point of suspension of the pendulum. At equilibrium,the string makes an angle <theta> with the vertical. It's time period of oscillation is T in this position

(A) tan<theta> = q /2

mg
(B)tan<theta> = q

/

mg
(C) T<2<pi>

(L/g)
(D) T>2<pi>

(L/g)

The answer is A and D

The forces acting on the bob are
The tension T at an angle <theta> with the vertical
mg downwards
q

/2

horizontally, away from the sheet
For equilibrium,
Tcos<theta> = mg
Tsin<theta> = q

/2

Dividing these two eauations,
tan<theta> = q

/2

mg ---------------(1)

Now suppose the bob is deflected by an arbitrary small angle <alpha> from the equilibrium position, away from the sheet.
Then, the equation of motion is
mgsin(<theta>+<alpha>) - (q

/2

)cos(<theta>+<alpha>) = m(dv/dt)
Expanding and applying cos

1 and sin<alpha>

<alpha>
mgsin<theta> - (q

/2

)cos<theta> + [mgcos<theta> - (q

/2

)sin<theta>]<alpha> = m(dv/dt)
Now from the relationship proved in (1), the first two terms cancel out.
[mgcos<theta> - (q

/2

)sin<theta>]<alpha> = m(d(L

)/dt)
Now,

= |d(<theta>+<alpha>)/dt| = - d<alpha>/dt (As <alpha> is an decreasing function of time). The above equation can then be written as
d²<alpha>/dt² + [(g/L)cos<theta> - (q

/2

mL)sin<theta>]<alpha> = 0
Hence the time period is
T = 2<pi>

[L/(gcos<theta> - (q

/2

m)sin<theta>)]
The denominator is definitely less than g. Hence The time period is greater than 2<pi>

(L/g)

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