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1 A.ABIDHA BEGUM .,M.Sc.,B.Ed.,M.Phil.,PGDCA., PG ASST., (PHYSCIS), P.M.G.HR.SEC.SCHOOL,KARUR http://www.trbtnpsc.com/2016/03/10th-12th-public-exam-answer-keys.html
Public Examination March-16
12th Standard - Physics PART – I I
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36. a = 10 cm = 10 x 10
-2 m
I = 10 A ; 𝜇0= 4 x 10-7
H/m
B=𝜇0𝐼
2𝜋𝑎 =
4𝜋×10−7× 10
2𝜋×10×10−2 = 2 x 10
-5 N
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2 A.ABIDHA BEGUM .,M.Sc.,B.Ed.,M.Phil.,PGDCA., PG ASST., (PHYSCIS), P.M.G.HR.SEC.SCHOOL,KARUR http://www.trbtnpsc.com/2016/03/10th-12th-public-exam-answer-keys.html
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3 A.ABIDHA BEGUM .,M.Sc.,B.Ed.,M.Phil.,PGDCA., PG ASST., (PHYSCIS), P.M.G.HR.SEC.SCHOOL,KARUR http://www.trbtnpsc.com/2016/03/10th-12th-public-exam-answer-keys.html
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4 A.ABIDHA BEGUM .,M.Sc.,B.Ed.,M.Phil.,PGDCA., PG ASST., (PHYSCIS), P.M.G.HR.SEC.SCHOOL,KARUR http://www.trbtnpsc.com/2016/03/10th-12th-public-exam-answer-keys.html
PART- III
51.
OR
q1 =+1𝜇 C q2 =+3 𝜇 C
q3=-5 𝜇 C r =60 cm = 60 x 10
-2 m
𝑈 =1
4𝜋𝜀0 𝑞1𝑞2
𝑟12+
𝑞1𝑞3
𝑟13+
𝑞2𝑞3
𝑟23
=9 x109 1×10−6×3×10−6
6×10−1+
1×10−6×(−5)×10−6
6×10−1+
3×10−6×(−5)×10−6
6×10−1
=−51
2× 10−2
U =-25.55J = 0.255J
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5 A.ABIDHA BEGUM .,M.Sc.,B.Ed.,M.Phil.,PGDCA., PG ASST., (PHYSCIS), P.M.G.HR.SEC.SCHOOL,KARUR http://www.trbtnpsc.com/2016/03/10th-12th-public-exam-answer-keys.html
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6 A.ABIDHA BEGUM .,M.Sc.,B.Ed.,M.Phil.,PGDCA., PG ASST., (PHYSCIS), P.M.G.HR.SEC.SCHOOL,KARUR http://www.trbtnpsc.com/2016/03/10th-12th-public-exam-answer-keys.html
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7 A.ABIDHA BEGUM .,M.Sc.,B.Ed.,M.Phil.,PGDCA., PG ASST., (PHYSCIS), P.M.G.HR.SEC.SCHOOL,KARUR http://www.trbtnpsc.com/2016/03/10th-12th-public-exam-answer-keys.html
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8 A.ABIDHA BEGUM .,M.Sc.,B.Ed.,M.Phil.,PGDCA., PG ASST., (PHYSCIS), P.M.G.HR.SEC.SCHOOL,KARUR http://www.trbtnpsc.com/2016/03/10th-12th-public-exam-answer-keys.html
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9 A.ABIDHA BEGUM .,M.Sc.,B.Ed.,M.Phil.,PGDCA., PG ASST., (PHYSCIS), P.M.G.HR.SEC.SCHOOL,KARUR http://www.trbtnpsc.com/2016/03/10th-12th-public-exam-answer-keys.html
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62.
Part - IV
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