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Electric flux over a given surface is the total number of electric field lines passing through that surface. 

An electric field line is a curve drawn in such a way that the tangent to it at each point represents the direction of the net field at that point. 

Using electric field lines. 

Respiration by animals and plants and burning of fuel.

As the displacement and and centripetal force (electrostatic attraction force between charges ) are perpendicular to each other therefore work done 

W = Fd cos90° = 0

Air contains oxygen and nitrogen as its major constituents of air, These gases retain their properties in air So air is called a mixture.

1. Since the graph is linear till point E. Hooke’s Law is obeyed till point E. 

2. The point E corresponds to the elastic limit because till there the body returns to its original state along EPO. 

3. Elastic - O to E; Plastic - E to B 

4. Up to point E an almost linear increase in strain with stress. After point E, if it is loaded till A, the strain increases faster for some increase in stress. But on unloading the body returns along line AO to a permanent strain (change in shape) of O'. 

5. In the curve C to B, the body undergoes a strain even if it is being unloaded and then fractures. To present fracture the value of stress has to be just less than that of C.

(a) Upto point A, Hooke’s law is obeyed because the graph is straight line from O to A.

(b) Point B, from the graph it is dear that the wire
returns to its original position after being unloaded upto point A only, hence AB is elastic limit.

(c) Elastic region- O to B
Plastic region- B to E

(d) Strain varies directly proportional to load upto
point A and after A, strain increases by greater amount as compared to first case (i.e., O to A) for a given increase in load. Beyond the elastic limit B, the curve does not retrace backwards as the wire is unloaded but returns along dotted line CO’. Point O’ corresponds to strain at zero load, which shows there is a permanent strain in the wire.

\(\frac{\Delta{d}}{d}\,\times100\,=\)\(\frac{\Delta{m}}{m}\,\times100\,+\,\)\(3\frac {\Delta{l}}{l}\,\times100\,\)

= 3% + 3 x 2% = 9%

We have, 

F = qvB sin θ 

B = F/qv sin θ 

[F] = [MLT^-2 ] 

[v] = [LT^-1 ] 

[q] = [AT] 

sin θ is dimensionless 

[B] = \(\frac {|MLT^{-2}|}{|LT^{-1}||AT|}\)

= [ML⁰T ^-2A ^-1 ]

(b) Chlamydomonas

Change in momentum of A = m₁ v₁ – m₁ u₁

Total momentum before collision = m₁ u₁ + m₂ u₂.

Total momentum after collision = m₁ v₁ + m₂ v₂

Final momentum of A = m₁ v₁

Initial momentum of A = m₁ u₁

Initial momentum of B = m₂ u₂

m = 1000 kg 

u = 10 m/s 

v = 0 

t = 5s 

Initial momentum = mu 

= 1000 × 10 = 10000 kg m/s

Final momentum = mV 

= 1000 × 0 = 0 Kg m/s

Velocities are different in both cases due to the difference in masses.

The inertia of an object depends upon its mass. When the mass increases, inertia also increases.

Momentum

Momentum is a characteristic property of moving objects. It is measured as the product of the mass of the body and its velocity.

Momentum = mass × velocity 

Unit of momentum is kg m/s

m = 200g = 200/1000 = 0.2 kg

Initial momentum = mu

= 0.2 × 10 = 2 kg m/s

Final momentum = mv

= 0.2 × 70 = 2 kg m/s

change in momentum = mv – mv = – 2 – 2 = -4 kg m/s

Answer is The cricket ball

I remember how I used to come back to my apartment in New York.

Initially when the bus is at rest our body also follows the same state. But all of a sudden when the bus starts to move the lower part of our body tends to move with the motion of the bus but the upper part rejects this state of motion and continues to be in a state or rest. This results in a sudden jerk backwards when a bus moves. Also a person standing in a bus will be in state of motion, and when the brakes are applied the lower part of our body comes to the state of rest but our upper part is in state of motion. Hence, we tend to fall forward when the bus applies brakes.

From Newton’s third law, we know that every action has an equal and opposite reaction. So when a fireman holds a hose-pipe which is ejecting a large amount of water at high velocity experiences a backward push due to the force of water flowing in the forward direction. Hence making it clear that due to the action of force in the forward direction the force is applied on the pipe in the backward direction, thus making the fireman difficult to hold the hose-pipe.

Mass of the gun M = 4 kg.
Mass of bullet M = 50 g = 50 × 10^-3kg.
Initial velocity of the bullet v = 35ms^-1
Let recoil velocity of the gun be vms^-1
Before firing the bullet both gun and bullet were in rest so total momentum of gun and bullet is zero, After firing
Momentum of bullet = mv,
Momentum of gun = Mv
Total momentum of bullet and gun after firing = mv + Mv
Since there is no external force applied on the system.
So total momentum after firing = total momentum before firing
mv + Mv = 0
Mv = -mv
fi V = \(\frac{mv}{M}= \frac{50\times 10^{-3}\times 35}{4}\) = -0.44 ms^-1
Here negative sign shows that there recoil velocity of the gun is in the direction opposite to the velocity of the bullet.

Mass of rifle m₁ = 4 kg 

Mass of bullet m₂ = 50 g = \(\frac{50}{1000}\) = 0.05 kg 

Initial velocity of rifle u₁ = 0 

Initial velocity of bullet u₂ = 0 

Final velocity of rifle (recoil velocity) v₁ = v 

Final velocity of bullet v₂ = 35 ms^-1 

According to law of conservation of momentum 

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂. 

4(0) + 0.05 (0) = 4v + 0.05 × 35 

4v = – 1.75 

v = – 0.44 m/s

Britain's military success was based on the values taught to school boys in its public schools.

1. Eton was the most famous of these schools.

2. The English boarding school was the institution that trained English boys for careers in the military, the civil service & the church.

3. Men like Thomas Arnold, head master of the famous Rugby School and founder of the modern public school system, saw team support like cricket and rugby not just as outdoor play, but as an organised way of teaching English boys the discipline.

4. It taught the English boys, the importance of codes of honour and the leadership qualities that helped then build and run the British Empire.

Correct answer is Amateur

(i) When the object is placed in front of the mirror :

(a) between its pole and focus

(b) between the focus and center of curvature 

(ii) In case 

(a) the image is virtual and erect.

(b) the image is real and, inverted.

The key rules for taxation in the tax system of Kautilya are as follows: 

• Timely Tax collection : According to Kautilya, agricultural tax should be collected only when the crop has ripened. The King should not impose taxes at inopportune time to collect wealth. . 
• Appropriate and justified Taxation : The king should not realize taxes at his will. Kautilya was of the opinion that the King should not recover the tax arbitrarily, but rather it should be recovered affectionately and only once. He should not levy heavy taxes on infertile land. 
• Taxation in accordance with capability : Taxes should be levied on all persons, according to their capability. 
• Importance of Financial Discipline : Kautilya gave a lot of emphasis on financial discipline. Appointment of revenue workers should be done very thoughtfully, and it should be ensured that whatever is recovered, its entire amount should be deposited in the state treasury.

Multiples of 4 greater than 25 = 28, 32, 36.

(i) Multiples of 3 – 3, 6, 9, 12, 15

(ii) Multiples of 6 – 6, 12, 18, 24, 30

(iii) Multiples of 12 – 12, 24, 36, 48, 60

(i) Multiples of 4 = 8, 12, 16, 20

(ii) Multiples of 7 = 14, 21, 28, 35

(iii) Multiples of 14 = 28, 42, 56, 70

(iv) Multiples of 19 = 38, 57, 76, 95

Multiples of 7 are 14, 21, 28.

(i) Factors of 7 = 1, 7

(ii) Factors of 9= 1,3,9

(iii) Factors of 16 = 1, 2, 4, 8, 16

(iv) Factors of 25= 1, 5,25

(v) Factors of 48 = 1,2, 3, 4, 6, 8, 12, 16, 24, 48

(vi) Factors of 63 = 1, 3, 7, 9, 21, 63

Four multiples of number 3 is 3, 6, 9, 12.

(i) 8 and 12

Factors of 8 = 1,2, 4, 8
Factors of 12 = 1,2, 3, 4, 6, 12
Therefore, common factors of 8 and 12 = 1, 2, 4

(ii) 10 and 20

Factors of 10 = 1,2, 5, 10
Factors of 20 = 1, 2, 4, 5, 10, 20
Therefore, common factors of 10 and 20 = 1, 2, 5, 10

(iii) 7 and 16

Factors of 7 = 1 , 7
Factors of 16 = 1,2, 4, 8, 16
Therefore, common factors of 7 and 16 = 1

(iv) 18 and 32

Factors of 18 =1,2, 3, 6, 9, 18
Factors of 32 = 1,2, 4, 8, 16, 32
Therefore, common factors of 18 and 32 = 1, 2.

Factors of 20 = 1, 2, 4, 5, 10, 20

Similarly factors of 30 = 1, 2, 3, 5, 6, 10, 15, 30

Highest common factor = 10

Therefore, greatest measurement of a container that can measure total milk of both container exactly is 10 litre.

Factors of 27 = 1, 3, 9, 27

Factors of 30 = 1,2, 3, 5, 6, 10, 15, 30

Therefore common factors of 27 and 30 = 1 and 3.

Factors of 21 = 1, 3, 7, 21

Factors of 28 = 1, 2, 4, 7, 14, 28

Therefore highest common factors of 21 and 28 = 7

Factors of 15 = 1, 3, 5, 15

Factors of 27 = 1, 3, 9, 27

Factors of 36 = 1, 2, 3, 4, 6, 12, 18, 36

Common factors of 15, 27 and 36 = 1, 3

Therefore, highest common factor of 15, 27 and 36 = 3

Factors of 12 = 1, 2, 3, 4, 6, 12

Factors of 18 = 1, 2, 3, 6, 9, 18

Factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24

Common factors of 12, 18, 24 = 1, 2, 3, 6

Therefore highest common factor of 12, 18 and 24 = 6

64 = 1 × 64
64 = 2 × 32
64 = 4 × 16
64 = 8 × 8
64 = 16 × 4
64 = 32 × 2
64 = 64 × 1

Therefore, factors of 64 are – 1, 2, 4, 8, 16, 32 and 64

Highest factor of 64 is = 64

Factors of 20 – 1, 2, 3, 4, 5, 10, 20

Factors of 40 – 1, 2, 3, 4, 5, 8, 10, 20, 40

Factors of 60 – 1, 2, 3,4, 5, 6, 10,12, 15,20, 30,60

Therefore common factors of 20,40 and 60 – 1, 2 4, 5, 10, 20.

Highest factor of 20, 40 and 60 = 20.

Multiple of 8 = 8, 16, 24, 32, 40, 48, 56 ......

Multiple of 6 = 6, 12, 18, 24, 30, 36, 42, 48, 54, .....

Multiple of 4 = 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, ....

Common multiples of 8,6 and 4= 24, 48, 72…

Least common multiples of 8, 6 and 4 = 24.

Multiples of 5 – 5, 10, 15, 20, 25, 30, 35.

Therefore, four odd multiples = 5,15, 25, 35

(B) The Peking Gazetta (China)

Answer is (C) 1690

Answer is (B) 1780

(A) 1938, 1939