IIT JAM Physics MCQ Test 2 - Physics MCQ

The situationn is shown in figure. Let n' be the velocity of the block with the embedded bullet after impact.
Applying the law of conservation of momentum, i.e.,

Momentum before impact = Momentum after impact
mn = (M + m)n'......(1)
Applying the law of conservation of mechanical energy, i.e., K.E. of the combined mass = P.E. at the highest point 
 .....(2)
From figure,
h = L -L cosq = L(1-cosq).
From equations (2) and (3), we get
 ....(4)
Substituting the value of n' from equation (1), we get

or 


 

Law of conservation of momentum gives
mu₁ = mv₁cosq₁ + mv₂cosq₂ .......(1)
and 0 mv₁sinq₁ - mv₂sinq₁ ....(2)

Energy conservation gives
 ....(3)
Squaring and adding (1) and (2) and comparing with (3) we get θ₁ + θ₂ = 90^o

As the wall is smooth, the vertical component of the impulse it receives is zero. The total time of flight is given by

The time of flight t₁ from G to A = 
The time of flight l₂ from A to G = 
where e = coefficient of restitution.


Solving we get e = 

Weight of the portion BC of the chain lying on the table, W = mg/2(downwards)
Thrust force 


F₁ = lv₂ (where, λ = m/l, is mass per unit length of chain)

 (downwards)
Net force eerted by the  chain on the table is 

So, from Newton's thrid law the force exerted by the table on the chain will be 3/2 mg (vertically upwards).

where
m = mass
r = radius
clearly 
⇒ I₁>I₃>I₂

When the rod rotates through an angle q, the centre of gravity falls through a distance h from DBG'C,

 ..(1)
The decrease in P.E. is equal to the kinetic energy of rotation 
 ...(2)
From equation (1) and (2), we get 

 ...(i)

or  ...(ii)
Frin eqs, (1) and (2), we get

We divide the sphere into thin disks of thickness dx. The radius r or the disk shown is 

Its volume is 
dV = pr²   dx = p(R²-x²)dx
and its mass is 
dm = r dV = pr (R²-x²) dx

We will find the moment of inertia of a sphere about an axis through its center 
The mass element is a disk with thickness dx.
The moment of inertia of a disk of radius r and mass dm is 

Integrating this expression from x = 0 to x = R gives the moment of inertia of the light hemisphere. From symmetry, the total I for the entire sphere is just twice of this: 

Carrying out the integration, we obtain

The mass M of the sphere of the volume V = 4pR³/3 is 

By comparing the expression for I and M, we find

We choose a thin cylindrical shell of radius r, thickness dr, and length L as a volume element. All parts of this element are at very nearly the same distance from the axis. The volume of the element is very nearly equal to that of a flat sheet with thickness dr, length L, and width 2pr(the circumference of the shell).
Then
dm = p dV = p (width x length x thickness)
= r(2prL dr) .....(1)

The moment of inertia is given by
 [use equation(1)]


If we express the moment of inertia in terms of the total mass M of the body. 
Then
 Q m = rV ......(3)
and the volume of the cylindrical shell is 
 [Volume of the cylinder V = pR²h] ...(4)
From equation (3) and (4) we get 
 ...(5)
using equation (5) in equation (2), we get

Let radius & height of solid cylinder is 2r and 2h. Assume that it is made up by small disc having radius 2r. Now we take a disc of width of dx and at x distance from zz' axis volume of thin disc is = π(2r)²dx
so, mass dm = 
where    is mass per unit volume
so moment of inertia about the diameter AB is 

By theorem of "parallel axes"  moment of inertia w.r.t. axis xz is 


So, Total moment of inertia,

The following statements are reference to X-rays: 
1. They produce heat when absorbed by matter : X-ray beam attenuation. As the x-ray beam passes through tissue, photons get absorbed so there is less energy; this is known as attenuation. It turns out that higher energy photons travel through tissue more easily than low-energy photons (i.e. the higher energy photons are less likely to interact with matter). 
2. They are generated when fast-moving electrons strike a metal target : An electromagnetic wave called x-ray is produced when fast moving electrons impact on high atomic weight substances. However, when these fast moving electrons strike on a metal target, a photoelectric effect occurs. 
3. They can penetrate through a thin sheet of aluminium : X-Rays will Pass Straight through Aluminium of Almost ANY Thickness as it's Density is so Low !! But the Crystalline Structure of Diamond makes it Totally Opaque to X-Rays !! X-Rays do not "see"! But yes, they can pass easily through a single sheet of aluminum foil.

 We know that in elastic collision both kinetic energy and momentum remains conserved.

According to principle of Conservation of momentum under mutual attractive force. The momentum of centre of mass remains constant, since the system is initially at rest, hence is momentum is zero.

Let the positron more along the +ve x-direction and the neutrino along +y direction with momenta 

Weight of the portion BC of the chain lying on the table W = mg/2 (downwards)
Thrust force 

 is mass per unit length of chain 


Net force exerted by the chain on the table is 

So, from Newton's third law the force exerted by the table on the chain will be 
3/2 mg (vertically upwards)

(b) Final momentum =  (m₁+m₂) V = 0.7V =

Now applying the law of conservation of momentum

where v is the velocity of the combined particle in the laboratory system. Note that v is identical to V, i.e.,   
Therefore 
Thus 

Since,  is exact differential work done by this force is independent of path. So the force is conservative
Torque about origin is zero, so angular momentum of particle is constant about origin.
Equilibrium position of particle is given by
F(n) = 0


 
 so potential energy is minimum at x = 3 ^1/3
So, x = 3 ^1/3 is point of stable equilibrium.
Hence, the particle will move towards x = 1

If a force or a vector quantity is conservative the 


And if  is conservative then a scalar potential v exist such that 

⇒ v = -xyz

Stream function ψ is given by

Equation of streamline is given by

where u and v are velocity components of fluid velocity 

So, velocity of fluid

The flow is irrotational if 


Since,  the velocity of fluid  can be expressed as gradient of scaler potential 

Here the effective height of orifice cannot be taken as (H-h) due to the presence of two liquids. Using Berrnoulli,s theorem just inside and just outside the orifice, we have


where V is the velocity of efflux


...(i)
2. Let be the time taken by the liquid to hit the ground. Then
....(ii)
The horizontal range x = Vt
 [use equation (1) and (2)]

...(iii)
3. Q 
Differentiate x with respect to h and equate it to zero for maximum h.



Value of h put in equation(3)

The period of oscillation of Foucalt's pendulum at the place of latitude is given as 

at l = 45⁰

The apparent weight is given by 

Taking downward direction positive, the force in vertically upward direction is 
F = mg + ma₀



Therefore the weight of the astronaut in the aboratory is 60kg.

For two slits,

For three slits,

The initial momentum of the system (gun + bullet) = 0
Final momentum of the system = m₁n₁ + m₂n₂
where m₁ and m₂ are the masses of the gun and bullet respectively and n₁ and n₂ are their velocities finally.
According to principle of conservation of linear momentum.
Initial momentum = Final momentum



The gun will move in opposite direction to that of the bullet with velocity 5 m/s.

Linear momentum is conserved during the collision. We can write, for the linear momentum components in the x and y directions,

in which M = m_A + m_B. Dividing equation (2) by equation (1) yields

Thus

From equation (2) we then have

It is given that the block slides a distance of 0.06m and comes to the rest for the first time.
μ = 0.1
By the energy conservation

To complete the vertical circle 

l₁/l_{2 = 5}

Let 

We know that velocity equation is givven by 




The maximum velocity of the rodect is 
v = 4000 ln 5 - 600

= 5837.75 m/sec

Weight of the portion BC of the chein 

Thrust force 


Net force exerted by the chain on the table is 

The force enerfed by the table on the chain will be 5/4 mg.