Please find the notes in the form of question and answer for the Physics Subject for the IIT JEE Examination which is very helpful in preparation as the answers have been explained in detail.

Question 1
A particle of mass M at rest is acted by a force F for a time t
Find the distance travelled by the particle in time t
Find the velocity of the particle at time t
Find the kinetic energy
Find the Workdone by the force

Detailed Solution
Given
Mass of the particle M
Force-F
Time -t
Also intial velocity u-0

As per Newton first law
F-ma
So a-F/M

Also as per equation of rectilinear motion
S-ut+(1/2)at2

S-(1/2)(P t2/M)

Also
V-u+at
So v-Pt/M

Also Kinetic Energy-(1/2)Mv2
KE-(1/2)(P2 t2/M)

Now Workdone -Fs
W-F* (1/2)(P t2/M)
W -(1/2)(P2 t2/M)

Question 2:
4 point charges each +Q is placed on the circumference of circle of radius R in such a way that they form a square.
Find the potential at the center
Find the electric field at the center

Detail Solution:

Since they are placed such a way they form square that means diagonal should be along the diameter

So V-(1/4пε0)[Q/R+Q/R+Q/R+Q/R]
-(Q/Rпε0)
Electric Field at the center will cancel out for the charges diagonically opposite charges ..So it will be zero at the center

Question 3
Consider the following statement
A particle executing uniform circular motion has
P. tangential velocity
Q Radial Velocity
R .Radial acceleration
S Tangential acceleration

Of these statements
a. P and R are correct
b. P and S are correct
c. Q and S are correct
d. Q and R are correct

Detailed Solutions
A is correct

Explanation:
In uniform circular motion,magnitude of velocity is constant only direction is changing,so tangential acceleration is not present only radial acceleration

Also there is no radial velocity otherwise it would not move in uniform circle

Question 4:
A particle moving with velocity V collides with another particle of the same mass which is at rest .The velocity of the center of mass after the collision is
a. 2V
b V
c. V/2
d None of these

Detailed Solutions
C is correct
Explanation
Velocity of center before collision
Vcm- (mV+m(0))/(m+m)
-V/2

Velocity of CM will not change during the collision as no external force acts on the system

Question 5:
Suppose two bodies are there
Body A whose Mass is m and velocity is v.
Body B whose Mass is M and velocity is V

It is given that M>m and KE of both the bodies are same

Which one of the following is correct
a. mv > MV
b. MV> mv
c. mv-MV
d. Mv-mV

Detailed Solutions:
B is correct

Explanation

KE of A-(1/2)mv2
KE of B-(1/2)MV2

Now m v2- MV2

(V/v)-(m/M)1/2
Now as M > m so (m/M) < 1

So V < v or (V/v) < 1
Now
(mv/MV)-(V/v)

Or MV > mv

Question 6

What is true of Inelastic collision
a. Momentum is conserved but energy is not
b. Momentum is not conserved but energy is conserved
c. Neither momentum nor energy is conserved
d. Both the momentum and energy is conserved

Ans a

Question 7.

A particle moves in the plane xy with velocity given by , where and are the unit vectors of the x and y axis and a and b are constants. At the initial moment of time the particle was located at point x-y-0. Find
(a) The equation of particle’s trajectory y(x)
(b) The curvature radius of trajectory as a function of x.

Detail Solutions
. (a) As given in the question velocity vector of particle is

From this we have x and y components of velocity i.e.
and (1)
From equation 1 we can calculate equation describing motion of particle along x and y axis. Thus integrating for x
or x-at (2)
Now from equation 2 we have
dy - bx dt - bat dt
Integrating it we get
Or (3)
From equations 2 and 3 we get
(4)
This is the required equation of particle’s trajectory.
(b) Radius of curvature of trajectory y(x) is given as
(5)
Differentiating path given in equation 4 for its first and second derivatives we find
and (6)
Using equation 5 and 6 we find radius of curvature of trajectory as follows:

Question 8.

Careful radioactivity measurements on a sample of indicate that there carbon disintegrations per second. The half life of is 5568 years . Calculate the size of the sample in grams.

Detailed Solution

The activity of a radioactive sample is given by
A-λN disintegration/second (1)
Where λ, is the disintegration constant and is given by

And N is the number of atoms in the sample and is given by

Where, - Avogadro’s number
- Mass of the sample
- Atomic mass
Combining all these formulas equation 1 becomes,

On solving for we get

Numerically, Or, Answer.

Question 9
A gas is contained in a vertical, frictionless piston cylinder device. The piston has a mass of 20 kg with a cross-sectional area of 20 cm2 and is pulled with a force of 100N. If the atmospheric pressure is 100 kPa, determine the pressure inside.

Detailed Solutions

Total forces on the piston
Weight of the piston acting downward-200N
Force due to Atmospheric pressure downward -100*103*20*10-4 N-200N
Let P be the Pressure of the gas
Then Force acting due to pressure of gas on the piston upward-PA
Force with the piston is being pulled upward-100N

Total Upward force-Total Downward force
PA + 100-200 +200
20*10-4*P-300
P-150kPa

Question 10

The heat capacity of a substance is found to be varies as temperature
C-C0(1+aT).
A sample of mass m of that substance is heated from tempertaure T1 to T2.How much heat is required

Detailed Solutions:
dQ-mCdT
Now integrating with upper and lower limit as T1 and T2
Q-∫mC0(1+aT)dT
Q-mC0[T+aT2/2]
Q- mC0[2T2-2T1-a(T12-T22)]/2

Q- mC0(T2-T1)(2+a(T2+T1))/2