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AIEEE Home page 2001 AIEEE PAPER - 1

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t = ax2 + bx
=> Differentiate both side with respect to time (t)
=> dt/dt = 2ax.dx/dt + b.dx/dt
=> 1 = 2axv + bv
=> v.(2ax + b) = 1
=> (2ax + b) = 1/v
=> Again differentiate both side with respect to time (t)
=> 2a.dx/dt = -v-2 . dv/dt
=> 2av = v-2 . acceleration
=> acceleration = -2av3

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Let Car started from point A.
It traveled for distance S with acceleration f. Assume that point is B
AB = S .................... (i)
Using
v2 = u2 + 2as
vb2 = 0 + 2fS
vb2 = 2fS ..................... (ii)

Next the car traveled for time t at contact speed till point C
So, BC = vbt = (2fS)1/2t

Then the car decelerated at f/2 and come to rest till point D
Using
v2 = u2 + 2as
0 = vb2 - 2.f/2.scd
scd = vb2/f = 2S
AB + BC + CD = 3S
=> S + (2fS)1/2t +2S = 15S
=> (2fS)1/2t = 12 S
=> 2fSt2 = 144 S2
=> S = ft2/72

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Acceleration = (change in velocity)/time
change in velocity = Δv
=> Δv = v2-v1
=> Δv2 = v22 + v12 - 2v1v2cos90°
=> Δv2 = v22 + v12 => Δv2 = 52 + 52 = 50
=> Δv = 5
=> Acceleration = Δv/ 10 => Acceleration = 5/10 = 1/ North-West

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First case when it takes time = t1
Horizontol velocity = vx1
Vertical velocity = vy1
Using formula: v=u+at in vertical direction
vy1 = - vy1 + gt1
2vy1 = gt1 .................. (i)
R = vx1t1 .................. (ii)
vy12 + vx12 = v2 = (gt1/2)2 + (R/t1)2 ....................(iii)

Second case when it takes time = t2
2vy2 = gt2 .................. (iv)
R = vx2t2 .................. (v)
vy22 + vx22 = v2 = (gt2/2)2 + (R/t2)2 ....................(vi)
(iii) and (vi) are equal
=> (gt1/2)2 + (R/t1)2 = (gt2/2)2 + (R/t2)2
=> (gt1/2)2 + (R/t1)2 + gR = (gt2/2)2 + (R/t2)2 + gR
=> gt1/2 + R/t1 = gt2/2 + R/t2
=> (g/2)(t1-t2) = R(t1-t2)/(t1t2)
=> t1t2 = 2R/g
=> So, t1t2 is proportional to R

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Centripetal force on the particle having circular motion = mω2r.
Anular velocity of all the paricles on the ring is same.
Force on outer particle = F1 = mω2R1.
Force on outer particle = F2 = mω2R2.
So, F1/F2 = R1/R2

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On the smooth surface force acting on the block in incline direction = F1
F1 = mgsin45°
Time require to travel distance n = t1
Using formula : s = ut + 1/2 at2
d = 1/2 . gsin45° . t12 ...................(i)

On the rough surface force acting on the block in incline direction = F2
F2 = mgsin45° - f , where f=frictional force
f = μkmgcos45° , where μk = coefficient of friction
F2 = mgsin45° - μmgcos45°
Time require to travel distance n = t2
d = 1/2 . (gsin45° - μk gcos45°) . t22 ..........(ii)
cos45° = cos45°
Divising (ii) and (i)
1 = [(1-μk)t12]/[t22]
We know that: t1/t1 = 1
So, 1 = (1-μk)n2
=> μk = 1 - 1/n2

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Let mass of the block = M
Frictional force in incline plane with inclination Φ = μMgcosΦ
Force due to gravity in the incline plane = MgsinΦ
So acceleration in smooth plane = gsinΦ
Retardation in rough plane = μgcosΦ - gsinΦ
The block will come to rest if acceleration = retardation
gsinΦ = μgcosΦ - gsinΦ
=> μ = 2tanΦ

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Suppose velocity of the bullet is 2y at the time of penetrating the wall
So velocity of the bullet at distance 3cm will be = y
Using, v2 = u2 + 2as
=> u2 = 4u2 + 6a
=> a = -u2/2
Let the bullet further penetrate a distance = x
Using, v2 = u2 + 2as
0 = u2 - 2.u2/2.x
=> x = 1

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Let the velocity of the parachutist after 50 m = y
Using, v2 = u2 + 2as
y2 = 2gs = 2 × 9.81 × 50
Suppose he bailed out at height = h
Using, v2 = u2 + 2as
9 = y2 - 2 × 2 × (h-50)
=> 2 × 9.81 × 50 - 9 = 4(h-50)
=> 4(h-50) = 981 - 9 = 972
=> h = 243 + 50 = 293

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F = -kx
=> ma = -kx
=> 0.3a = -15x
=> a = 50x
=> a = - 50 × 0.2
=> a = - 10 ms-2

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Let the velocity of the body at the time of collision = v
1/2 Mv2 = 1/2 kL2
=> M2v2 = MkL2
=> Mv = momemtum = L

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Initial momemtum = final momentum
Initial momemtum = mV = mVi
Final velocity of mass1 m in i direction = Vx
Final velocity of mass1 m in j direction = Vy Final momemtum of mass1 m = mVxi + mVyj
Final momemtum of mass2 m = mv/j
mvi = mVxi + mVyj + mv/j => mvxi + mVyj = mvi - mV/j
=> mvxi + mVyj = mvi - mV/j
=> Vx = V
=> Vy = -V/
=> Vfinal2 = 1+1/3 = 4/3
=> Vfinal = 2 V /

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Initial energy = Final energy
Initial energy = mgh = 100mg
Final energy = 1/2mv2 + mgh = 1/2mv2 + 20mg
So, 100mg = 1/2mv2 + 20mg
=> v2 = 80g × 2 = 1600
=> v = 40 m/s

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If the incline surface is not moving then
acceleration of the block in inclined plane direction = gsinα

Suppose acceleration of the surface is "a" to keep the block stationary
Due to acceleration of the surface, acceleration block = acosα
In order to keep the block stationary :
acosα = gsinα
a = gtanα

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Mass of semicircular disc = M
Suppose there is a circular disc of mass 2M, then
Moment of intertia of circular disc = 1/2(2M)R2
Moment of intertia of circular disc = 1/2(2M)R2 = MR2
=> So, Moment of intertia of semi-circular disc = (1/2)MR2

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The only force that acts on the body A is gravitational force.
There is no force acting perpendicular to the gravitational force.
So the center of the mass of B and C will not shift.

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Let distance of P from C is = y
P must be the center of mass of the T shaped object.
Assume mass per unit length = 1
So, [Mab × (2l-y)] + [M(2l-y) × (2l-y)/2] = [My × y/2[
=> l × (2l-y) + (2l-y) × (2l-y)/2 = y × y/2
=> 4l2 - 2ly + (2l-y)2 = y2
=> 8l2 -2ly - 4ly = 0
=> 8l2 -6ly = 0
=> y = 4l/3

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Acceleration due to gravity at a depth 'd' from earth surface is :
gd = g(1-   d  )
 R





Acceleration due to gravity at height 'h' from earth surface is :
h is very much smaller than R
gh = g(1-   2h  )
 R





gh = gd
By solving it , d=2h

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Work Done =  GMm  =  6.67 × 10-11 × 100 × 0.01  = 6.67 × 10-10
R 0.1

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Capillary action
The height to which the liquid can be lifted is given by:
h =  2γcosθ
ρgr




where,
γ: liquid-air surface tension(T)(T=energy/area)
θ: contact angle
ρ: density of liquid
g: acceleration due to gravity
r: is radius of tube

When entire arrangement is freely falling then relative acceleration = 0
So, Water will rise to the full length of capillary tube i.e. 20cm

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