IIT JAM Physics MCQ Test 1 - Physics MCQ

Force acting on the system are shown in Fig.
For equilibrium in horizontal direction.
T₁ cos∝ = T₂ cosβ .....(1)
For equilibrium in vertical direction,
T₁ sin∝ = T₂ sinβ - mg = 0 .....(2)
From(1)  ....(3)
Put in (2)


T₁(sinα cosβ+cosαsinβ) =  mgcosβ
T₁sin (α+β) = mg cosβ

Put in (3)

As the normal force does no work on the particle, its energy is conserved.
So (L.E. + P.E.)_P1= (K.E. + P.E.)_P2
(K + U)_P1 = (K + U)P2......(1)
Initially the particle in rest at P₁ so kinetic energy (V = 0) will be zero at P₁.
So from equation(1)

mgR cosθ₁ is potential energy at 
P₁ = mgh
h → height = Rcosθ₁

At point P₂ potential energy is mgh = mg Rsinθ₂

 [using 2]

 
 

Using the conservation of energy law in the form
Loss of PE = gain of KE


From Fig.
h = I - I cosθ

u = -5km/s = -5 x 10³m/s

F = -(-5 x 10³) (100) = 5 x 10⁵N


= 5 x 10³log_e100 = (5 x10³)
= 5 x 10³ x 2.303 x 2
= 2.303 x 10⁴ m/s.

If X- axis is taken vertically, Z-axis towards north and Y-axis along east, then the velocity of the bulled is v = 500km/sec. and angular velocity w = w  because the angular velocity vector w of the earth is directed parallel to its axis and is inclined at 30⁰ to the horizontal
Here, 

Hence Coriolis acceleration
= 2w x v

 m/sec² towards west.

By conservation of momentum, P_cannon = -P_ball'

By the property of ellipse 
Major axis 2a = maximum distance + minimum distance

Maximum distance =  maximum altitude + radius of earth = (4100+6400)
Minimum distance from centre of earth = 6400 + 1100
So, 2a = (4100+6400)+(6400+1100) = 18000km

For attractive central force,

So,  ...(1)
 ...(2)
Now substitute equation(2) in euqation of orbit



and from equation(1), seq q= 2au



 

f(r) a -rⁿ
f = -krⁿ
Where k is a constant


= arⁿ⁺¹   where (new constant)
Condition for stable orbit

If the potential energy  function for the central force is of the form  and centrifugal energy 

The condition for the stability in the radial motion is

So, differentiate equation (1) w.r.to r 
  ...(2)


Thus any circular orbit with r = r₀ under a central force is stable

 


hence 

The centre of mass    Q(dm = λdx)
 

The position of the center of mass is marked by a dot in the figure. As figure suggests, we treat this point as if it held a real particle, assigning to it a mass M equal to the system mass of m₁ + m₂ m₃ = 16.4kg and assuming that all external forces are applied at that point. We find the center of mass from equation




The x component of the net external force acting on the center of mass is

Suppose the circular disc of radius a with centre O is made up of 
(i) Circular section of radius b with centre O₁ and
(ii) Remaining portion  of disc with cm at O₂
Taking O as origin, and O₁, O₂ on X-axis, (y = 0, z = 0), figure the position of cm of disc is given by
 ...(i)
If s is surface density of material of the disc.

In this coordinate system, we have the three coordinates as 


These coordinate are related to the cartesian coordinates by the relations

Position vector r is given by

Hence, the velocity is given by 
Differentiating above equation with respect to time, we get acceleration

Two types of central force always produce closed orbits: F(r) = αr (a linear force) and F(r) = α / r² (an inverse-square law).

Retarding force = mg +F.
Retardation a = 
Distance s = 
Downward motion: Force = mg -F
Acceleration a' = 
Distance's =  
Sinces =  S, we have 

If X-axis is taken vertically, Z-axis towards north and Y-axis along east, then the velocity of the bullet is v = 500 km/sec. and angular velocity - UTP2  because the angular velocity vector ww of the earth is directed parallel to its axis and is inclined at 30⁰ to the horizibtal.

Hence coriolis acceleration

Time of journey, t 
Deflection of the bullet due to the coriolis acceleration

Vertical displacement of the bulet due to the gravity


Coriolis force = -2m w x v

=3.6 x 10^-4 newton towwards east.

The body moves in a circular orbit of radius r = 5ms
So, energy of the system will always be equal to v_e at r = 5m
Also, ve will be maximum at r = 5m and dve/dr will be zero at r = 5m



The body moves in a circular orbit of radius r = 5m
So, energy of the system will always be equal to v_e at r = 5m
Also, v_e will be maximum at r = 5m and dv_e/dr will be zero at r = 5m

0 to 2

Let the man was initially at A and moves towards B with uniform velocity u = 20m/s.
Let the man in car meets the man B at point Q, then
AQ= 100 + BQ
⇒ 
⇒ 
⇒ 

The value of t will be real, if 

By the law of conservation of energy 

Total energy at A = total energy at B

Let V be the velocity acquire by the block.


If n' is speed of bullet on emerging out of block, then by law of conservation of momentum, we get 
mn +M x 0 = MV +Mn'

Herem mass of rocket, 
m₁ = 0.1 kg,
mass of fuel, m₂ = 0.02 kg, t = 3s
Initial velocity u = 0
final velocity n = 20m/s
from n = u+at

Thrust on the rocket, 
If s is the distance moved by the rocket in 3 sec, then from

Energy content of the fuel = work done on 
rocket = Thrust x distance = 

Energy/kg = 20/0.02 = 1000 J/kg

Resolving the forces and finding their horizontal and vertical component
mg = R + Fcosθ
Frictional force = F sinθ
By Eq.(ii) we get

Let a boy standing at A throw a ball with a velocity u at an angle q with the horizontal, which just enters window W.
As the boy is at 78.4 m from the building and the ball just enters the window 39.2m above the ground, therefore

And horizontal range,

Dividing (i) by (ii), we get

Substituting in (ii), we get

Consider a satellite of mass m is orbiting around the earth with orbit velocity n. Let h be the height of satellite from the surface of earth and r be the orbital radius of the satellite. Then

Total energy the orbiting satellite (E)


Work done in shifting the satellite from height h to h₁ is 
W = difference in energy of satellite in two orbits 

= 34.92m/s²

With our choice of axes, we find the following velocity components:

The x- component of impulse is equal to the x-component of momentum change, and the same is true for they y-components.

The components of the average net force on the ball are 

The magnitude and directionn of the average force are