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Hi, There is no doubt that CBSE board is one of the best board of india and it also covers lots of portion of IIT-JEE. As for as selction in IIT-JEE is concerned a lot of hard work is required throughout the year. So you can choose CBSE board All the best.
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x y + 3 y2 - x + 4 y - 7 = 0 ..(1) 2 x y + y2 - 2 x - 2 y + 1 = 0 ..(2) Multiplying (1) by 2 and substracting from eq.2 We get 5 y2 + 10 y - 15 = 0 y2 + 2 y- 3 = 0 (y+3) (y-1) = 0 y = -3 , 1 when y = -3, x = 2 Note: y ยน 1 as y=1 will make x infinite or reduce the equation to an identity. Hence the solution of equation is x = 2, y = -3
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Inertial frame of reference Hence, within the inertial frame, an object or body accelerates only when a physical force is applied, and (following Newton's first law of motion), in the absence of a net force, a body at rest will remain at rest and a body in motion will continue to move uniformly?ie. in a straight line and at constant speed. Non-inertial reference frame non inertial frame of reference is one in which a body violates Newton's Laws of Motion, mainly the First Law. In such a frame, despite no real force acting on a body at rest, it might move; or one that was already moving come at rest or change it's direction of motion. For comparison see an inertial frame. thanx for writing! do not hesitate for inquiring!bye.
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Dear Faisal, In my openion you should be very cool on the final day. No study should be on final day. Because IIT exam is not like board exams so that if you will read before the exam that you may expect in exam. Your brain should be cool and it should actively work during the exam. Everybody works hard whole the year. But during exam you should have capability of thinking and problem solving. you can revise formula in which you are not comfortable. If any doubt pl. ask me. All the best.
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The whole kinetic energy of the second mass is converted to compress the spring. Therefore the maximum compression of the spring 1/2 M V2 = 1/2 K X2 0.1 x 1 = 10 K X2 x = 0.1 m Now since the mass returns with the velocity 0.6 m/s. Therefore the loss in K.E = 1/2 M V2 - 1/2 M v2 This is equal to 1/2 K A2 where A is amplitude 1/2 M (V2 - v2) = 1/2 K A2 0.1 [1 - (0.6)2] = 10 A2 A = 0.1 + 0.8 = 0.08 m
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Hi, Better to have a new form because they are very strict in following the instruction. If suppose they have rejected your form than you will loose one year. So beter to purchase new form. All the best for your exam Neeraj
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Dear Akshay, You have got very good percentage in class Xth. It means you are a good student and you are having a potential to go in IIT. But you have to be very sincere regarding this. First of all you should join a class room coaching. You will get an environment over there. and you will feel comptetion wih other students. You can also visit goiit.com regularly. We are there to help you. So all the best and start working hard for IIT. Bye Neeraj
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Hi, You can have Brilliant Tutorials correspondence course and you can visit www.goiit.com regularly. In Brilliant they are having materials for Physics, Chemistry and Maths. They are conducting regularly exams. Goiit is also going to provide help in clearing your doubts. We are always there to help you. Following books you can refer IIT Maths by TMH Publication IIT Chemistry by P.Bahadur IIT Physics by Arihant publication
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Hi surya, There is no time criteria for the preparation of IIT JEE. You should put efforts as much as you can. First of all you should not leave any topic until you are not comfortable with it. And you can judge yourself. You should make daily next day time table. Don't make timetable of the whole week. You always remeber that you have to work hard. All the best Neeraj
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Following explation includes the rotation of complex number and multiplication of complex number. Kindly go throuh it if you have any question you can ask me Multiplying a complex number by i. In our goal toward finding a geometric interpretation of complex multiplication, let's consider next multiplying an arbitrary complex number z = x + yi by i. z i = (x + yi) i = ?y + xi.  Let's interpret this statement geometrically. The point z in C is located x units to the right of the imaginary axis and y units above the real axis. The point z i is located y units to the left, and x units above. What has happened is that multiplying by i has rotated to point z 90ยฐ counterclockwise around the origin to the point z i. Stated more briefly, multiplication by i gives a 90ยฐ counterclockwise rotation about 0. You can analyze what multiplication by ? i does in the same way. You'll find that multiplication by ? i gives a 90ยฐ clockwise rotation about 0. When we don't specify counterclockwise or clockwise when referring to rotations or angles, we'll follow the standard convention that counterclockwise is intended. Then we can say that multiplication by ? i gives a ?90ยฐ rotation about 0, or if you prefer, a 270ยฐ rotation about 0. A geometric interpretation of multiplication. To completely justify what we're about to see, trigonometry is needed, and that is done in an optional section. For now, we'll see the results without the justification. We've seen two special cases of multiplication, one by reals which leads to scaling, the other by i which leads to rotation. The general case is a combination of scaling and rotation.  Let z and w be points in the complex plane C. Draw the lines from 0 to z, and 0 to w. The lengths of these lines are the absolute values | z| and | w|, respectively. We already know the length of the line from 0 to zw is going to be the absolute value | zw| which equals | z| | w|. (In the diagram, | z| is about 1.6, and | w| is about 2.1, so | zw| should be about 3.4. Note that the unit circle is shaded in.) What we don't know is the direction of the line from 0 to zw. The answer is that "angles add". We'll determine the direction of the line from 0 to z by a certain angle, called the argument of z, sometimes denoted arg( z). This is the angle whose vertex is 0, the first side is the positive real axis, and the second side is the line from 0 to z. The other point w has angle arg( w). Then the product zw will have an angle which is the sum of the angles arg( z) + arg( w). (In the diagram, arg( z) is about 20ยฐ, and arg( w) is about 45ยฐ, so arg( zw) should be about 65ยฐ.) In summary, we have two equations which determine where zw is located in C: |zw| = |z| |w| arg(zw) = arg(z) + arg(w)
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Problem : y3 - y2 - 2 y + 1 = 0 Sol 1 y = (1/3) + 7 2/3 / 3. [(-1 + 3 i  3) / 2 ] 1/3 + (1/3) [(-1 + 3 i 3) / 2 ] 1/3 Sol 2 y = (1/3) - (7/2) 2/3 (1+i 3) / [(-1 + 3 i 3) ] 1/3 - (1/ 6) ( 1- i 3) [ (7 / 2 ) [(-1 + 3 i 3) ] 1/3 Sol 3 y = (1/3) - (7/2) 2/3 (1- i 3) / [(-1 + 3 i 3) ] 1/3 - (1/ 6) ( 1 + i 3) [ ( 7 / 2) [( -1 + 3 i 3) ] 1/3
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(1-x 2) + (1-y 2) = a (x-y) Put x sin A y = sin B cos A + cos B = a (sin A - sin B) 2 cos (A+B)/2 . cos (A-B)/2 = 2 a cos (A+B)/2 . sin (A-B)/2 cos (A-B)/2 = a sin (A-B)/2 ........(1) [ cos (A+B)/2 ยน 0] For if cos (A+B)/2 = 0 then (A+B)/2 =  /2 A + B = or A = - B sin A = Sin B x = y which does not satisfy given equation From eq. 1 cot (A-B)/2 = a A - B = 2 cot-1 a or sin-1 x - sin-1y - 2 cot-1 a Differentiating above eq. w.r.t.x 1/ (1-x2) - 1 / (1-y2) dy / dx = 0 dy/dx = (1-y2) / (1-x2)
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Que: (sin 5x - 2 sin 3x + sin x) / (cos5x - cosx) Sol: Numerator : (sin 5x - sin 3x) + ( sin x - sin 3x) = [ 2 cos (5x+3x) / 2 . sin (5x-3x)/2 ] + [2 cos (x+3x)/2 . sin (x-3x)/2 ] = [ 2 cos 4x . sin x] + [ 2 cos 2x . sin (-x)] = [ 2 cos 4x . sin x] - [ 2 cos 2x . sin x] = 2 sinx [ cos 4x - cos 2x] = 2 sin x [ 2 sin (4x+2x)/2 . sin (2x - 4x) /2] = 2 sin x [ 2 sin3x . sin (-x)] = - 4 sin x sin 3x sin x Denominator (cos5x - cosx) = 2 sin (5x+x)/2 . sin (x-5x)/2 = 2 sin 3x . sin (-2x) = - 2 sin 3x . sin 2x = - 2 sin 3x . (2.sinx .cosx) = - 4 sin3x . sinx. cosx Hence ( - 4 sin x sin 3x sin x) / ( - 4 sin3x . sinx. cosx ) = sin x / cos x = tan x
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Let me explain you the laws of floatation When a body floats in a liquid then equilibrium is governed by two forces (i) The wt. of a body acts vertically downwards through centre of gravity of the body (ii) The resultant upward thrust exerted on the body by the liquid i.e. force of buoyancy acts vertically upward through the centre of gravity of displaced liquid i.e. centre of buoyancy When a body floats freely: (i) The wt of the body is equal to the wt. of the displaced liquid (ii) The centre of gravity of the body and centre of gravity of the displaced liquid are in the same vertical line The only essential condition for a body to float in a liquid is that the weight of the liquid displaced by the body should be equal to the weight of the body itself Since air is pumped in side the water bottle, hence body will float until the wt. of the liquid displaced by the body is equal to the wt. of the body itself otherwise it will start submerge.
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Moment of inertia of continuous mass system is I =  r2 dm dm = mass of the element r = is the distance of element dm from its perpendicular line I am giving you two examples of explaining M.I (Moment of Inertia) [1] Moment of Inertia of rod Consider a uniform rod of mass m and length l. Suppose M.I is to be calculated about the bisector. Take origin at middle point of the rod. Consider the element of the rod between a distance x and x+dx from origin. As rod is uniform Mass per unit length of rod = m / l Mass of element = (m / l ) . dx The perpendicular distance of the element from the axis is x. The M.I of this element about axis passing through middle point of rod and perpendicular to it. dI = (m / l) . dx . x2 x will changed from - l/2 to + l/2 I = (x2 dx = m l2 ) / 12 (integration limit ? l/2 to + l/2) [2] Moment of inertia of ring Moment of inertia of a circular ring about its axis (the line perpendicular to the plane of the ring through its center) Suppose the radius of ring is R and its mass is m. As all the elements of the rings are at the same perpendicular distance R from the axis the moment of inertia of the ring is I = r2 dm = R2 dm = M R2
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Dear Tanmay, I am not able to get you. Kindly explain your problem.
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(i) work done by the friction will be zero (ii) If friction is static it will affect the acceleration of center of mass which is also a translational acceleration I am giving you explanation for (ii) question When a sphere is rolled on a horizontal table it slows down and eventually stops. The forces acting on the sphere are (a) weight mg (b) friction at the contact and (c) the normal force As the center of the sphere decelerates the friction should be opposite to its velocity. But friction will have a clockwise torque that should increase the angular velocity of the sphere. There must be an anticlockwise torque that causes the decrease in the angular velocity. Infact when sphere rolls on the table both the sphere and surface deform near the contact. The contact is not a single point as we normally assume rather there is an area of contact. The front part pushes the table a bit more strongly than the back part. As a result the normal force does not pass through the center. It is shifted towards the right. This force then has an anticlockwise torque and net torque causes an angular deceleration
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( a , c ) is the correct answer There is a exchange of linear velocities. However the sphere and ball cannot exert torques on each other, as their surfaces are frictionless and the angular velocities of the spheres do not change. Hence sphere will stop moving but continue to rotate with an angular velocity w and ball will move with a speed v without rotating. Therefore the answer is (a, c)
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Hi, There are many methods for solving above problem. You can choose whatever is convenient to you
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We can consider a function f(x) = sin ( x) Since f(x) is an odd function having a period 2 Hence f(2) = sin (2 ) so f(4) = sin (2 . 2 ) Since function is periodic with period 2 Hence f(4) = f(2)
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