Explain the processes of nuclear fission and nuclear fusion by using the plot of binding energy per nucleon ( B E / A ) versus the mass number A .

The plot of binding energy per nucleon versus mass number is given below: We use the above plot to understand the processes of nuclear fission and fusion as follows: Nuclear fission: When we move from the heavy nuclei region to the middle region of the plot, we find that there will be an increase in the overall binding energy and hence results in the release of energy. This indicates that energy can be released when a heavy nucleus ( A ~ 240 ) breaks into two roughly equal fragments. This process is called nuclear fission. Nuclear fusion: As seen from the plot, when we move from lighter nuclei to heavier nuclei, we again find that there will be gain in the overall binding energy and hence release of energy takes place. This indicates that energy can be released when two or lighter nuclei fuse together to form a heavy nucleus. This process is called nuclear fusion.

Derive an expression for the electric field at any point on the equatorial line of an electric dipole.

Consider an electric dipole consisting of charges - q and + q separated by a distance 2 a . Let E A be the electric field at point Q due to charge - q and E B be the electric field at point Q due to charge + q where Q is a point on the equatorial line of the dipole whose electric field is to be measured. Let x be the distance between the centre of the dipole and point Q . Now, resultant electric field at point Q would be, E Q = E A + E B E A = 1 4 π ε 0 q A Q 2 ⇒ 1 4 π ε 0 q x 2 + a 2 ( ∵ According to Pythagoras, A Q = x 2 + a 2 2 ) E B = 1 4 π ε 0 q B Q 2 ⇒ 1 4 π ε 0 q x 2 + a 2 ( ∵ According to Pythagoras, B P = x 2 + a 2 2 ) Now, according to parallelogram law of vector addition, the resultant electric field would become, E Q = E A 2 + E B 2 + 2 E A E B cos 2 θ Now, since E A = E B we can say, E A = E B = E ' E Q = E ' 2 + E ' 2 + 2 E ' E ' cos 2 θ ⇒ 2 E ' 2 ( 1 + cos 2 θ ) ⇒ E ' 2 ( 1 + cos 2 θ ) ⇒ E ' 2 2 cos 2 θ E Q = 2 E ' cos θ - - - ( i ) Now, from △ A O Q , cos θ = a x 2 + a 2 Also, we know that, E ' = 1 4 π ε 0 q x 2 + a 2 Putting both these values in equation ( i ) , we get, E Q = 2 4 π ε 0 q x 2 + a 2 a x 2 + a 2 E Q = p 4 π ε 0 x 2 + a 2 3 2 Since we know that in dipoles, x > > > 2 a , then a 2 can be neglected. ∴ E = p 4 π ε 0 x 3 2

Photodiode is a device

which is always operated in reverse bias

which is always operated in forward bias

in which photo current is independent of intensity of incident radiation

which may be operated in forward or reverse bias.

A carbon resistor of 47 ± 4.7 kΩ is to be marked with rings of different colours for its identification. The colour code sequence will be:

Yellow - Violet - Orange - Silver

Violet - Yellow - Orange - Silver

MEDIUM

12th Rajasthan Board

IMPORTANT

Earn 100

Draw a labelled ray diagram of a compound microscope showing image formation at near point of normal eye.

Important Questions on Solved Physics Paper-2

EASY

12th Rajasthan Board

IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 11 - Dual Nature of Radiation and Matter>Exercise>Q 8

Two monochromatic beams A and B of equal intensity

I

, hit a screen. The number of photons hitting the screen by beam A is twice that by beam B. Then what inference can you make about their frequencies?

HARD

12th Rajasthan Board

IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 11 - Dual Nature of Radiation and Matter>Exercise>Q 9

Explain the effect of potential difference on photoelectric current.

MEDIUM

12th Rajasthan Board

IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 11 - Nuclei>Exercise>Q 14

Calculate the binding energy of an

α -

particle. Given mass of proton=

1.0073 u

, mass of neutron=

1.0087 u

. and mass of alpha particle=

4.0015 u

MEDIUM

12th Rajasthan Board

IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 11 - Solved Physics Paper-2>Exercise>Q 33

A nucleus contains no electrons, but it can emit them. How?

HARD

12th Rajasthan Board

IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 11 - Solved Physics Paper-2>Exercise>Q 34

Two parallel wires separated by

2 m

carry currents of

3 A

each in the opposite directions. Find the magnetic induction for the points between the wires at a distance

50 cm

from each wire.

EASY

12th Rajasthan Board

IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 11 - Solved Physics Paper-2>Exercise-Choice>Q 35

Name the rule which relates the line integral of magnetic field and current enclosed.

MEDIUM

12th Rajasthan Board

IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 11 - Solved Physics Paper-2>Exercise>Q 36

Answer the following questions:
You have learnt that plane and convex mirrors produce virtual images of objects. Can they produce real images under some circumstances? Explain.

MEDIUM

12th Rajasthan Board

IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 11 - Solved Physics Paper-2>Exercise-Choice>Q 37

(i) A giant refracting telescope at an observatory has an objective lens of focal length $15 m$ . If an eyepiece of focal length $1.0 c m$ is used, what is the angular magnification of the telescope?

(ii) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is $3.48 \times 10^{6} m$ , and the radius of lunar orbit is $3.8 \times 10^{8} m$ .

Draw a labelled ray diagram of a compound microscope showing image formation at near point of normal eye.

Important Questions on Solved Physics Paper-2

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