Unit 1: Physical Quantities and Measurement
NOTE :
Correction and Comments about Numericals and Short Question Answers
- Short Question 1.8
Problem: Wrong answer given in notes.
Correct Answer: It is the measurement or value that an instrument shows when it should be showing zero.
- Short Question 1.9
Problem: Wrong answer given in notes.
Correct Answer: By knowing the zero error, necessary correction can be made to find correct measurement. Otherwise our reading will be incorrect.
- Numerical 1.7
Problem : Wrong values of pitch of the screw guage(0.1mm) and number of division(100).
Correction: The vale of pitch of the screw should be 0.5mm and number of divisions should be 50 according to the question. There is no change in answer, it remains the same.
Unit 2: Kinematics
Problem: Wrong answer given in notes.
Correct Answer: It is the measurement or value that an instrument shows when it should be showing zero.
Problem: Wrong answer given in notes.
Correct Answer: By knowing the zero error, necessary correction can be made to find correct measurement. Otherwise our reading will be incorrect.
Problem : Wrong values of pitch of the screw guage(0.1mm) and number of division(100).
Correction: The vale of pitch of the screw should be 0.5mm and number of divisions should be 50 according to the question. There is no change in answer, it remains the same.
NOTE :
Correction and Comments about Numericals and Short Question Answers
- Numerical 2.9
problam : The answer of numerical 2.9 of 2nd part where the distance is asked to find is, not coming correct.so please check this.your answer is 266.53m but through calculator is 296.14m.plz solve this query.
Answer : Actually, the value of 266m comes when you do not round off the value of acceleration (means a= -0.3331666875 m/s*s) and all other velocities. So your answer is also correct because you round off the value of acceleration. In exams I would say that write the value of distance 296m because this is the value you computed after rounding off and it is correct also.
- Numerical 2.8
Problem : Incomplete Numerical and missing page number 14.
Remaining Part:
0 = 6vi – 180
vi = 30m/s
Now calculating height reached by the ball
using formula 2as = vf2 – vi2
putting values in the above formula
s = (vf2 – vi2)/2a
s = (0 – 30*30)/2*9.8
s = 45m approx
Hence the height reached by the ball is 45 meters and the initial velocity is 30 m/s
Unit 3: Dynamics
- Numerical 2.9
problam : The answer of numerical 2.9 of 2nd part where the distance is asked to find is, not coming correct.so please check this.your answer is 266.53m but through calculator is 296.14m.plz solve this query.
Answer : Actually, the value of 266m comes when you do not round off the value of acceleration (means a= -0.3331666875 m/s*s) and all other velocities. So your answer is also correct because you round off the value of acceleration. In exams I would say that write the value of distance 296m because this is the value you computed after rounding off and it is correct also. - Numerical 2.8
Problem : Incomplete Numerical and missing page number 14.
Remaining Part:
0 = 6vi – 180
vi = 30m/s
Now calculating height reached by the ball
using formula 2as = vf2 – vi2
putting values in the above formula
s = (vf2 – vi2)/2a
s = (0 – 30*30)/2*9.8
s = 45m approx
Hence the height reached by the ball is 45 meters and the initial velocity is 30 m/s
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