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in gauss theorem the flux is due to the charge inside the enclosing body. here it is ZERO.
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i think u must mention the co eff. of restitution.
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i have an easy ans. the bottommost pt. should be at rest , hence, the force is 0 there. F * 2R = Ia/r & I = 1/2MR^2 + MR^2 = 3/2MR^2. => a = 4f/3m
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is semi conductor in bits syllabus although it is not mentioned but some were saying that it was there in exam.
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vel. of light in any med. = vel. in vacuum/refractive index of medium. when light ray comes back then the index is same again.
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pot. inside a shell remains constant. upto R1 the only charge is q1 &pot.is KQI/R1 but the pot. due to q2, at r2 , will be KQ2/R2. HOPE U UNDERSTOOD.
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WHAT IS UR DOUBT ABT. THE SOL.
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Pot. diff. will be independent of the outer charge qb. Va =KQ1/R1 +KQ2/R2 Vb=K(Q1+Q2)/R2 Va - Vb = KQ1 (1/R1 -1/R2)
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pls tell me if semi conductors is in bitsat syllabus,it is not mentioned in brochure.
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select 5 nos.from the given nos. 0,1,2,3,4,5 & then apply the above two cases.
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Make 2 cases: 1) take 1,2,3,4,5 it gives 120 ways. 2)take 0,1,2,4,5 4 * 24 = 96 total- 216
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NH4Cl is correct but in the aieee exam it was given that N2 is correct at fiitjee site.
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Cl2 reacts with excess of NH3 to give ?
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kanavdm, it will be difficult to get pec or dce but u have a god scope in ur state nit.
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I got hyperbola but ellipse is also correct b'cos we have to use mode of the distance therefore both the ans are correct.ihope all of u got it.
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always apply normality eqn to such problems no. of millimoles of Mn2+ = (NV)KMnO4 =(.102)(3)(21.5) &AMT. CAN BE FOUND by multiplying with m.wt.
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