Q. No. 11 - 20 carry 3 Marks eachEach question, when worked out will result in one integer from 0 to 9 (both inclusive).Q.During Searleandrsquo;s experiment, zero of the Vernier scale lies between 3.20 x 10-2 m and 3.25 x 10-2 m of themain scale. The 20th division of the Vernier scale exactly coincides with one of the main scale divisions.When an additional load of 2 kg is applied to the wire, the zero of the Vernier scale still lies between 3.20 x10-2 m and 3.25 x10-2 m of the main scale but now the 45th division of Vernier scale coincides with one ofthe main scale divisions. The length of the thin metallic wire is 2 m and its cross-sectional area is 8 x 10-7m2. The least count of the Vernier scale is 1.0 x 10-5 m. The maximum percentage error in the Youngandrsquo;smodulus of the wire isCorrect answer is '4'. Can you explain this answer?

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Q. No. 11 - 20 carry 3 Marks each

Each question, when worked out will result in one integer from 0 to 9 (both inclusive).

During Searle’s experiment, zero of the Vernier scale lies between 3.20 x 10^-2 m and 3.25 x 10^-2 m of the
main scale. The 20th division of the Vernier scale exactly coincides with one of the main scale divisions.
When an additional load of 2 kg is applied to the wire, the zero of the Vernier scale still lies between 3.20 x
10^-2 m and 3.25 x10^-2 m of the main scale but now the 45th division of Vernier scale coincides with one of
the main scale divisions. The length of the thin metallic wire is 2 m and its cross-sectional area is 8 x 10^-7
m². The least count of the Vernier scale is 1.0 x 10^-5 m. The maximum percentage error in the Young’s
modulus of the wire is

Correct answer is '4'. Can you explain this answer?

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Q. No. 11 - 20 carry 3 Marks eachEach question, when worked out will r...

change in lengths = 25(LC)
and the maximum permissible error in elongation is one LC

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Q. No. 11 - 20 carry 3 Marks eachEach question, when worked out will result in one integer from 0 to 9 (both inclusive).Q.During Searle’s experiment, zero of the Vernier scale lies between 3.20 x 10-2 m and 3.25 x 10-2 m of themain scale. The 20th division of the Vernier scale exactly coincides with one of the main scale divisions.When an additional load of 2 kg is applied to the wire, the zero of the Vernier scale still lies between 3.20 x10-2 m and 3.25 x10-2 m of the main scale but now the 45th division of Vernier scale coincides with one ofthe main scale divisions. The length of the thin metallic wire is 2 m and its cross-sectional area is 8 x 10-7m2. The least count of the Vernier scale is 1.0 x 10-5 m. The maximum percentage error in the Young’smodulus of the wire isCorrect answer is '4'. Can you explain this answer?

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Q. No. 11 - 20 carry 3 Marks eachEach question, when worked out will result in one integer from 0 to 9 (both inclusive).Q.During Searle’s experiment, zero of the Vernier scale lies between 3.20 x 10-2 m and 3.25 x 10-2 m of themain scale. The 20th division of the Vernier scale exactly coincides with one of the main scale divisions.When an additional load of 2 kg is applied to the wire, the zero of the Vernier scale still lies between 3.20 x10-2 m and 3.25 x10-2 m of the main scale but now the 45th division of Vernier scale coincides with one ofthe main scale divisions. The length of the thin metallic wire is 2 m and its cross-sectional area is 8 x 10-7m2. The least count of the Vernier scale is 1.0 x 10-5 m. The maximum percentage error in the Young’smodulus of the wire isCorrect answer is '4'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Q. No. 11 - 20 carry 3 Marks eachEach question, when worked out will result in one integer from 0 to 9 (both inclusive).Q.During Searle’s experiment, zero of the Vernier scale lies between 3.20 x 10-2 m and 3.25 x 10-2 m of themain scale. The 20th division of the Vernier scale exactly coincides with one of the main scale divisions.When an additional load of 2 kg is applied to the wire, the zero of the Vernier scale still lies between 3.20 x10-2 m and 3.25 x10-2 m of the main scale but now the 45th division of Vernier scale coincides with one ofthe main scale divisions. The length of the thin metallic wire is 2 m and its cross-sectional area is 8 x 10-7m2. The least count of the Vernier scale is 1.0 x 10-5 m. The maximum percentage error in the Young’smodulus of the wire isCorrect answer is '4'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Q. No. 11 - 20 carry 3 Marks eachEach question, when worked out will result in one integer from 0 to 9 (both inclusive).Q.During Searle’s experiment, zero of the Vernier scale lies between 3.20 x 10-2 m and 3.25 x 10-2 m of themain scale. The 20th division of the Vernier scale exactly coincides with one of the main scale divisions.When an additional load of 2 kg is applied to the wire, the zero of the Vernier scale still lies between 3.20 x10-2 m and 3.25 x10-2 m of the main scale but now the 45th division of Vernier scale coincides with one ofthe main scale divisions. The length of the thin metallic wire is 2 m and its cross-sectional area is 8 x 10-7m2. The least count of the Vernier scale is 1.0 x 10-5 m. The maximum percentage error in the Young’smodulus of the wire isCorrect answer is '4'. Can you explain this answer?.

Solutions for Q. No. 11 - 20 carry 3 Marks eachEach question, when worked out will result in one integer from 0 to 9 (both inclusive).Q.During Searle’s experiment, zero of the Vernier scale lies between 3.20 x 10-2 m and 3.25 x 10-2 m of themain scale. The 20th division of the Vernier scale exactly coincides with one of the main scale divisions.When an additional load of 2 kg is applied to the wire, the zero of the Vernier scale still lies between 3.20 x10-2 m and 3.25 x10-2 m of the main scale but now the 45th division of Vernier scale coincides with one ofthe main scale divisions. The length of the thin metallic wire is 2 m and its cross-sectional area is 8 x 10-7m2. The least count of the Vernier scale is 1.0 x 10-5 m. The maximum percentage error in the Young’smodulus of the wire isCorrect answer is '4'. 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 Q. No. 11 - 20 carry 3 Marks eachEach question, when worked out will result in one integer from 0 to 9 (both inclusive).Q.During Searle’s experiment, zero of the Vernier scale lies between 3.20 x 10-2 m and 3.25 x 10-2 m of themain scale. The 20th division of the Vernier scale exactly coincides with one of the main scale divisions.When an additional load of 2 kg is applied to the wire, the zero of the Vernier scale still lies between 3.20 x10-2 m and 3.25 x10-2 m of the main scale but now the 45th division of Vernier scale coincides with one ofthe main scale divisions. The length of the thin metallic wire is 2 m and its cross-sectional area is 8 x 10-7m2. The least count of the Vernier scale is 1.0 x 10-5 m. The maximum percentage error in the Young’smodulus of the wire isCorrect answer is '4'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of Q. No. 11 - 20 carry 3 Marks eachEach question, when worked out will result in one integer from 0 to 9 (both inclusive).Q.During Searle’s experiment, zero of the Vernier scale lies between 3.20 x 10-2 m and 3.25 x 10-2 m of themain scale. The 20th division of the Vernier scale exactly coincides with one of the main scale divisions.When an additional load of 2 kg is applied to the wire, the zero of the Vernier scale still lies between 3.20 x10-2 m and 3.25 x10-2 m of the main scale but now the 45th division of Vernier scale coincides with one ofthe main scale divisions. The length of the thin metallic wire is 2 m and its cross-sectional area is 8 x 10-7m2. The least count of the Vernier scale is 1.0 x 10-5 m. The maximum percentage error in the Young’smodulus of the wire isCorrect answer is '4'. Can you explain this answer?, a detailed solution for Q. No. 11 - 20 carry 3 Marks eachEach question, when worked out will result in one integer from 0 to 9 (both inclusive).Q.During Searle’s experiment, zero of the Vernier scale lies between 3.20 x 10-2 m and 3.25 x 10-2 m of themain scale. The 20th division of the Vernier scale exactly coincides with one of the main scale divisions.When an additional load of 2 kg is applied to the wire, the zero of the Vernier scale still lies between 3.20 x10-2 m and 3.25 x10-2 m of the main scale but now the 45th division of Vernier scale coincides with one ofthe main scale divisions. The length of the thin metallic wire is 2 m and its cross-sectional area is 8 x 10-7m2. The least count of the Vernier scale is 1.0 x 10-5 m. The maximum percentage error in the Young’smodulus of the wire isCorrect answer is '4'. Can you explain this answer? has been provided alongside types of Q. No. 11 - 20 carry 3 Marks eachEach question, when worked out will result in one integer from 0 to 9 (both inclusive).Q.During Searle’s experiment, zero of the Vernier scale lies between 3.20 x 10-2 m and 3.25 x 10-2 m of themain scale. The 20th division of the Vernier scale exactly coincides with one of the main scale divisions.When an additional load of 2 kg is applied to the wire, the zero of the Vernier scale still lies between 3.20 x10-2 m and 3.25 x10-2 m of the main scale but now the 45th division of Vernier scale coincides with one ofthe main scale divisions. The length of the thin metallic wire is 2 m and its cross-sectional area is 8 x 10-7m2. The least count of the Vernier scale is 1.0 x 10-5 m. The maximum percentage error in the Young’smodulus of the wire isCorrect answer is '4'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice Q. No. 11 - 20 carry 3 Marks eachEach question, when worked out will result in one integer from 0 to 9 (both inclusive).Q.During Searle’s experiment, zero of the Vernier scale lies between 3.20 x 10-2 m and 3.25 x 10-2 m of themain scale. The 20th division of the Vernier scale exactly coincides with one of the main scale divisions.When an additional load of 2 kg is applied to the wire, the zero of the Vernier scale still lies between 3.20 x10-2 m and 3.25 x10-2 m of the main scale but now the 45th division of Vernier scale coincides with one ofthe main scale divisions. The length of the thin metallic wire is 2 m and its cross-sectional area is 8 x 10-7m2. The least count of the Vernier scale is 1.0 x 10-5 m. The maximum percentage error in the Young’smodulus of the wire isCorrect answer is '4'. Can you explain this answer? tests, examples and also practice JEE tests.

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