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An electron is projected horizontally with a kinetic energy 10 kev. A magnetic field of strength 1.0 X 10-7 T exists in the vertically upward direction. Calculate the sideways deflection of the electron in traveling through 1m. Make appropriate approximations. Ans: 1.5 cm approx.
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An electron is projected horizontally with a kinetic energy 10 kev. A magnetic field of strength 1.0 X 10-7 T exists in the vertically upward direction. Calculate the sideways deflection of the electron in traveling through 1m. Make appropriate approximations. Ans: 1.5 cm approx.
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At the topmost point, the total downward force is Mg + N. This downward force is radially inward at the topmost point. So this force itself serves as the centripetal force at the topmost point. So, Mg + N = Mv2/R. Now, N i.e normal contact force is self adjustible. So, for minimising v , N assumes its minimum value i.e. 0. So, Mg = Mv2/R You can also solve it by using the concept of centrifugal (pseudo) force. All bodies going in a circle experiences a centrifugal (pseudo) force mv2/r in the radially outward direction. So, at the topmost point this force acts upward and balances the downward force (Mg+N). So, Mv2/R = (Mg+N). The rest is same.
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An electron is projected horizontally with a kinetic energy 10 kev. A magnetic field of strength 1.0 X 10-7 T exists in the vertically upward direction. Calculate the sideways deflection of the electron in traveling through 1m. Make appropriate approximations. Ans: 1.5 cm approx.
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In the left triangle the (3+7)=10 ohm is in parallel with the 10 ohm. Thus for that part Req = 5 ohm. Now (5+10) ohm =15 ohm is in parallel with 30 ohm. Thus Total Req = 10 ohm.
Since per volt resistance R/V= 200, so V/R = i = 1/200. So, total voltage drop = i.(total resistance) = 1/200 . (80X200 +32000) = 80+160 = 240 volts.
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mv2/L = mg does not yield any answer because it does not contain the h term in it. Also you must include the contact force by the ring in this equation. i.e mv2/L = mg - N (in the vertically extreme position.) Now from energy calculations we have found that for minimum height, v needs to be minimum. For this purpose, N will maximise and become equal to mg. So, mv2/L = mg - mg = 0.
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In electricity when we define potential, potential energy etc. we generally define it for a positive charge. Keeping in mind this point, let us take two points A and B, such that A is at higher potential while B is at lower potential. This means that if a positive charge moves towards A, it will face repulsion. So, in the same case, a negative charge will face attraction which means that A is actually the lower potential point for a negative charge. Thus the point which serves as higher potential for a +ve charge serves as lower potential for a negative charge, and vice versa. So, the negative charge which actually moves from higher to lower potential (defined for a negative charge), conventionally moves from lower to higher potential.
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A shorter process:
Loop law: 24i + 18 - 30 = 0 or i = 0.5
From Current div. rule: [50X0.5/(50+R)] R = 18 So, R = 128.57 =130approx
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Current always flows from a higher potential to a lower potential. So, the normal course of current flow is: However, while charging, it is necessary to flow the current in the opposite direction through the battery so that the chemical reactions opposite to those in discharging (normal current flow) takes place. In order to do so, we have to provide a potential difference higher than that between the plates of the emf as shown in the following diagram: Emf Charger When the internal resistance is drawn and Loop law is applied, we get the following diagram and equation: emf , int. resist Charger(V) E + ir -V = 0 The rest is same as Raman sir's
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The second one is false because in the question it is asked about the kinetic energy of one single molecule. Moreover there is no requirement of being monoatomic, the gas just has to be ideal. The third one is correct.
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It is not given in the question that the c.s.area of the pipe is uniform.
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From Bernoulli's equation: P 1 + 1/2  v 12 + gh 1 = P 2 + 1/2 v 22 + gh 2 As v1 = v2 So , the equation reduces to : P 1 + gh 1 = P 2 + gh 2 But since h1 is not equal to h2 So, P1 not equal to P2 Hence false. II) No single gas molecule can have 3/2.PV. The energy of one gas molecule (ideal gas irespective of whether it is monoatomic or diatomic) is 3PV/2NA. Hence false.
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Thanks rphy for uploading those stuffs. Can you re-upload the files in zip format, coz doc files are too huge to be uploaded thru slow connections. If you zip it, it may reduce to less than half.
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Thanks studyid, thanks everybody!
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