Unit 1-5 (Test) PDF

Unit – 1 to 5 (Test) By – C.M. Prakash 1) nks vkos”k ds xksys ,d nwljs ls d nwjh ij F cy yxkrs gSA a) 1 J ;fn vkos”kksa dks nksxquk rFkk muds chp dh nwjh Hkh nksxquh dj b) 1 C/1 J nh tk, rks cy gksxk &...

Unit – 1 to 5 (Test) By – C.M. Prakash 1) nks vkos”k ds xksys ,d nwljs ls d nwjh ij F cy yxkrs gSA a) 1 J ;fn vkos”kksa dks nksxquk rFkk muds chp dh nwjh Hkh nksxquh dj b) 1 C/1 J nh tk, rks cy gksxk & c) 1 J/1 C a) F d) 1 JC b) F/2 1V is equal to – c) F/4 a) 1 J d) 4F b) 1 C/1 J Two charged sphere are separated by a distance d c) 1 J/1 C exert a force F on each other. If charges are doubled d) 1 JC and distance between them is also doubled then the 6) fujis{k fo|qr”khyrk dk eku gS & force is – a) 8.85 X 10-12 C2N-1m-2 a) F b) 8.85 X 1012 C2N-1m-2 b) F/2 c) 885 X 10-12 C2N-1m-2 c) F/4 d) 8.85 X 10-15 C2N-1m-2 d) 4F The value of absolute permittivity is – 2) fo|qr {ks= E rFkk foHko V ds chp laca/k gS & a) 8.85 X 10-12 C2N-1m-2 a) E = -dV/dx b) 8.85 X 1012 C2N-1m-2 b) E = dV/dx c) 885 X 10-12 C2N-1m-2 c) V = dE/dx d) 8.85 X 10-15 C2N-1m-2 d) V = dE/dv 7) la/kkfj= esa lafpr mTkkZ ds fy, dkSu lk O;atd lgh ugha gS The relation between electric field E and potential V a) ½ CV2 is – b) Q2/2C a) E = -dV/dx c) ½ QV b) E = dV/dx d) QV2 c) V = dE/dx Which one is the not correct expression for energy stored d) V = dE/dv in a capacitor 3) ,d lekarj IysV la/kkfj= ds IysVksa ds chp ,d ijkoS|qr Mkyk a) ½ CV2 tk, rks mldh /kkfjrk dk eku & b) Q2/2C a) ?kVrk gS c) ½ QV b) c N2 b) 450 The relation between number of turns of primary coil c) 900 and secondary coil of step up transformer is – d) 600 a) N1 = N2 28) ,d xSYosuksehVj dks ,sehVj esa ifjofrZr fd;k tk ldrk gS & b) N1 > N2 a) Js.kh esa cM+s eku dk “kaV tksM+dj c) N2 < N1 b) lekarj esa NksVs eku dk “kaV tkM+dj d) N1 >> N2 c) Js.kh esa NksVs eku dk izfrjks/k tksM+dj 33) izsjd dq.Myh ds fy, /kkjk rFkk oksYVrk ds e/; dykarj gS & d) lekarj esa cM+s eku dk izfrjks/k tkM+dj a) 00 A galvanometer can be converted into an ammeter b) 450 by – c) 900 a) Introducing a shunt of large value in series d) 600 b) Introducing a shunt of small value in parallel For an inductor coil the phase difference between c) Introducing a resistance of small value in series current and voltage is – d) Introducing a resistance of large value in parallel a) 00 29) izsfjr fo|qr okgd cy dh /kzqork fdlds }kjk ikbZ tkrh gS b) 450 a) cks;ks&lkorZ dk fu;e c) 900 b) Qysfeax dk nkW;k gkFk dk fu;e d) 600 c) ysat dk fu;e 34) fo|qr pqEcdh; rjax ds pky dk O;atd gS & 1 d) ,sfEi;j dk ifjiFkh; fu;e a) √∈0 The polarity of induced emf is found by – 1 a) Biot-Savart’s law b) √ 0 𝜖0 𝜇 b) Flemming’s right hand rule ∈ c) √ 0 c) Lenz’s law 𝜇 0 d) Ampere’s circuital law 𝜇 d) √ 0 30) ysat dk fu;e fdlds laj{k.k ds fu;e dk ifj.kke gS & ∈ 0 a) mtkZ The expression for speed of electromagnetic waves is b) vkos”k – 1 c) fo|qr okgd cy a) √∈0 d) /kkjk 1 Lenz’s law is a consequence of the law of b) √𝜇0 𝜖0 conservation of – ∈ c) √𝜇0 a) Energy 0 b) Charge d) 𝜇0 √∈ c) Emf 0 d) Current 35) fuEu esa ls fdldk rjaxnS/;Z lcls de gS & 31) vuksU; izsjdRo dk SI ek=d gS & a) lq{e rjaxsa a) farad b) ijkcSaxuh fdj.ksa b) ampere c) xkek fdj.ksa c) henry d) n`”; izdk”k d) tesla Which of the following having minimum wavelength SI unit of mutual inductance is – – a) farad a) Microwaves b) ampere b) UV rays c) henry c) Gamma rays d) tesla d) Visible light 32) fdlh mPpk;h VªkalQkWeZj ds izkFkfed rFkk f}rh; dq.Myh ds Qsjksa ds la[;k ds e/; laca/k gS & a) N1 = N2 4 All the best