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Question No. 40 and 41 are based on the following paragraph.Consider a block of conducting material of resistivity ‘ρ’ shown in the figure. Current ‘I’ enters at ‘A’ and leaves from ‘D’. We apply superposition principle to find voltage ‘ΔV’ developed between ‘B’ and ‘C’. The calculation is done in the following steps:(i) Take current ‘I’ entering from ‘A’ and assume it to spread over a hemispherical surface in the block.(ii) Calculate field E(r) at distance ‘r’ from A by using Ohm’s law E =ρj, where j is the current per unit area at ‘r’.(iii) From the ‘r ’ dependence of E(r), obtain the potential V(r) at r.(iv) Repeat (i), (ii) and (iii) for current ‘I’ leaving ‘D’ and superpose results for ‘A’ and ‘D’.Q.40.ΔV measured between B and C isa)b)c)d)Correct answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Question No. 40 and 41 are based on the following paragraph.Consider a block of conducting material of resistivity ‘ρ’ shown in the figure. Current ‘I’ enters at ‘A’ and leaves from ‘D’. We apply superposition principle to find voltage ‘ΔV’ developed between ‘B’ and ‘C’. The calculation is done in the following steps:(i) Take current ‘I’ entering from ‘A’ and assume it to spread over a hemispherical surface in the block.(ii) Calculate field E(r) at distance ‘r’ from A by using Ohm’s law E =ρj, where j is the current per unit area at ‘r’.(iii) From the ‘r ’ dependence of E(r), obtain the potential V(r) at r.(iv) Repeat (i), (ii) and (iii) for current ‘I’ leaving ‘D’ and superpose results for ‘A’ and ‘D’.Q.40.ΔV measured between B and C isa)b)c)d)Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Question No. 40 and 41 are based on the following paragraph.Consider a block of conducting material of resistivity ‘ρ’ shown in the figure. Current ‘I’ enters at ‘A’ and leaves from ‘D’. We apply superposition principle to find voltage ‘ΔV’ developed between ‘B’ and ‘C’. The calculation is done in the following steps:(i) Take current ‘I’ entering from ‘A’ and assume it to spread over a hemispherical surface in the block.(ii) Calculate field E(r) at distance ‘r’ from A by using Ohm’s law E =ρj, where j is the current per unit area at ‘r’.(iii) From the ‘r ’ dependence of E(r), obtain the potential V(r) at r.(iv) Repeat (i), (ii) and (iii) for current ‘I’ leaving ‘D’ and superpose results for ‘A’ and ‘D’.Q.40.ΔV measured between B and C isa)b)c)d)Correct answer is option 'A'. Can you explain this answer?.

Solutions for Question No. 40 and 41 are based on the following paragraph.Consider a block of conducting material of resistivity ‘ρ’ shown in the figure. Current ‘I’ enters at ‘A’ and leaves from ‘D’. We apply superposition principle to find voltage ‘ΔV’ developed between ‘B’ and ‘C’. The calculation is done in the following steps:(i) Take current ‘I’ entering from ‘A’ and assume it to spread over a hemispherical surface in the block.(ii) Calculate field E(r) at distance ‘r’ from A by using Ohm’s law E =ρj, where j is the current per unit area at ‘r’.(iii) From the ‘r ’ dependence of E(r), obtain the potential V(r) at r.(iv) Repeat (i), (ii) and (iii) for current ‘I’ leaving ‘D’ and superpose results for ‘A’ and ‘D’.Q.40.ΔV measured between B and C isa)b)c)d)Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.

Here you can find the meaning of Question No. 40 and 41 are based on the following paragraph.Consider a block of conducting material of resistivity ‘ρ’ shown in the figure. Current ‘I’ enters at ‘A’ and leaves from ‘D’. We apply superposition principle to find voltage ‘ΔV’ developed between ‘B’ and ‘C’. The calculation is done in the following steps:(i) Take current ‘I’ entering from ‘A’ and assume it to spread over a hemispherical surface in the block.(ii) Calculate field E(r) at distance ‘r’ from A by using Ohm’s law E =ρj, where j is the current per unit area at ‘r’.(iii) From the ‘r ’ dependence of E(r), obtain the potential V(r) at r.(iv) Repeat (i), (ii) and (iii) for current ‘I’ leaving ‘D’ and superpose results for ‘A’ and ‘D’.Q.40.ΔV measured between B and C isa)b)c)d)Correct answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of Question No. 40 and 41 are based on the following paragraph.Consider a block of conducting material of resistivity ‘ρ’ shown in the figure. Current ‘I’ enters at ‘A’ and leaves from ‘D’. We apply superposition principle to find voltage ‘ΔV’ developed between ‘B’ and ‘C’. The calculation is done in the following steps:(i) Take current ‘I’ entering from ‘A’ and assume it to spread over a hemispherical surface in the block.(ii) Calculate field E(r) at distance ‘r’ from A by using Ohm’s law E =ρj, where j is the current per unit area at ‘r’.(iii) From the ‘r ’ dependence of E(r), obtain the potential V(r) at r.(iv) Repeat (i), (ii) and (iii) for current ‘I’ leaving ‘D’ and superpose results for ‘A’ and ‘D’.Q.40.ΔV measured between B and C isa)b)c)d)Correct answer is option 'A'. Can you explain this answer?, a detailed solution for Question No. 40 and 41 are based on the following paragraph.Consider a block of conducting material of resistivity ‘ρ’ shown in the figure. Current ‘I’ enters at ‘A’ and leaves from ‘D’. We apply superposition principle to find voltage ‘ΔV’ developed between ‘B’ and ‘C’. The calculation is done in the following steps:(i) Take current ‘I’ entering from ‘A’ and assume it to spread over a hemispherical surface in the block.(ii) Calculate field E(r) at distance ‘r’ from A by using Ohm’s law E =ρj, where j is the current per unit area at ‘r’.(iii) From the ‘r ’ dependence of E(r), obtain the potential V(r) at r.(iv) Repeat (i), (ii) and (iii) for current ‘I’ leaving ‘D’ and superpose results for ‘A’ and ‘D’.Q.40.ΔV measured between B and C isa)b)c)d)Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of Question No. 40 and 41 are based on the following paragraph.Consider a block of conducting material of resistivity ‘ρ’ shown in the figure. Current ‘I’ enters at ‘A’ and leaves from ‘D’. We apply superposition principle to find voltage ‘ΔV’ developed between ‘B’ and ‘C’. The calculation is done in the following steps:(i) Take current ‘I’ entering from ‘A’ and assume it to spread over a hemispherical surface in the block.(ii) Calculate field E(r) at distance ‘r’ from A by using Ohm’s law E =ρj, where j is the current per unit area at ‘r’.(iii) From the ‘r ’ dependence of E(r), obtain the potential V(r) at r.(iv) Repeat (i), (ii) and (iii) for current ‘I’ leaving ‘D’ and superpose results for ‘A’ and ‘D’.Q.40.ΔV measured between B and C isa)b)c)d)Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice Question No. 40 and 41 are based on the following paragraph.Consider a block of conducting material of resistivity ‘ρ’ shown in the figure. Current ‘I’ enters at ‘A’ and leaves from ‘D’. We apply superposition principle to find voltage ‘ΔV’ developed between ‘B’ and ‘C’. The calculation is done in the following steps:(i) Take current ‘I’ entering from ‘A’ and assume it to spread over a hemispherical surface in the block.(ii) Calculate field E(r) at distance ‘r’ from A by using Ohm’s law E =ρj, where j is the current per unit area at ‘r’.(iii) From the ‘r ’ dependence of E(r), obtain the potential V(r) at r.(iv) Repeat (i), (ii) and (iii) for current ‘I’ leaving ‘D’ and superpose results for ‘A’ and ‘D’.Q.40.ΔV measured between B and C isa)b)c)d)Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice JEE tests.