All questions of Measurement for Grade 9 Exam
Can you explain the answer of this question below:The error in two readings of A are 0.01 and -0.03, then mean absolute error is ?- A:0.002
- B:-0.004
- C:0.02
- D:-0.02
The answer is c.
The error in two readings of A are 0.01 and -0.03, then mean absolute error is ?
A:
0.002
B:
-0.004
C:
0.02
D:
-0.02
|
Bhargavi Bajaj answered |
Mean absolute error = (|0.01|+|-0.03|)/2
= 0.02
= 0.02
Which is the most accurate atomic clock?- a)Carbon-12 clock
- b)Carbon atom clock
- c)caesium atom clock
- d)Stop clock
Correct answer is option 'C'. Can you explain this answer?
Which is the most accurate atomic clock?
a)
Carbon-12 clock
b)
Carbon atom clock
c)
caesium atom clock
d)
Stop clock
|
Nandini Patel answered |
According to the most accurate caesium atomic clock in the world, 1 second is the time that elapses during 9,192,631,770 cycles of the radiation produced by the transition between two levels of the caesium 133 atom.
Can you explain the answer of this question below:The method used to measure the distance of a planet from the earth is- A:Parallax method
- B:Archimedes’ method
- C:Newton’s method
- D:Galileo’s method
The answer is a.
The method used to measure the distance of a planet from the earth is
A:
Parallax method
B:
Archimedes’ method
C:
Newton’s method
D:
Galileo’s method
|
Yash Ghoshal answered |
The astronomers mainly use parallax method to find distances to objects like other planets. To calculate the distance to a star or to a planet, astronomers observe it from different places along Earth's orbit around the Sun.
The most suitable instrument for measuring the size of an atom is.- a)Electron microscope
- b)Vernier calliper
- c)Optical Microscope
- d)Screw gauge
Correct answer is 'A'. Can you explain this answer?
The most suitable instrument for measuring the size of an atom is.
a)
Electron microscope
b)
Vernier calliper
c)
Optical Microscope
d)
Screw gauge
|
Mira Sharma answered |
Only electron microscope has resolution of 0.5-0.6 Angstrom which is less than the size of an atom.
Can you explain the answer of this question below: Which of the following is caused by careless handling?
- A:
Systematic error
- B:
Gross error
- C:
Random error
- D:
None of the mentioned
The answer is b.
Which of the following is caused by careless handling?
Systematic error
Gross error
Random error
None of the mentioned
|
|
Nandini Iyer answered |
Gross errors are mostly due to lack of knowledge, judgment and care on the part of experiment. That is Gross error is caused by careless handling.
Which of the following is caused by careless handling?- a)Systematic error
- b)Gross error
- c)Random error
- d)None of the mentioned
Correct answer is option 'B'. Can you explain this answer?
Which of the following is caused by careless handling?
a)
Systematic error
b)
Gross error
c)
Random error
d)
None of the mentioned
|
Trisha Vashisht answered |
Gross errors are mostly due to lack of knowledge, judgment and care on the part of experiment. That is Gross error is caused by careless handling.
The error in two readings of A are 0.01 and -0.03, then mean absolute error is ?- a)0.002
- b)-0.004
- c)0.02
- d)-0.02
Correct answer is 'C'. Can you explain this answer?
The error in two readings of A are 0.01 and -0.03, then mean absolute error is ?
a)
0.002
b)
-0.004
c)
0.02
d)
-0.02
|
Anagha Sharma answered |
Mean absolute error = (|0.01|+|-0.03|)/2
= 0.02
= 0.02
Error in the measurement of radius of a sphere is 1%. The error in the calculated value of its volume is- a)1%
- b)3%
- c)5%
- d)7%
Correct answer is option 'B'. Can you explain this answer?
Error in the measurement of radius of a sphere is 1%. The error in the calculated value of its volume is
a)
1%
b)
3%
c)
5%
d)
7%
|
Raghav Bansal answered |
∵ V = (4/3)πr3
% error is volume = 3 × error in radius = 3 × 1 = 3%
% error is volume = 3 × error in radius = 3 × 1 = 3%
Angstrom is the unit of- a)time
- b)mass
- c)light
- d)length
Correct answer is option 'D'. Can you explain this answer?
Angstrom is the unit of
a)
time
b)
mass
c)
light
d)
length
|
Vishal Giri answered |
Ya ri8 angstrom this unit used to measure length. 1 A = 10^ - 10 m
1 fermi is equal to:- a)10-10 m
- b)1010 m
- c)1015 m
- d)10-15 m
Correct answer is option 'D'. Can you explain this answer?
1 fermi is equal to:
a)
10-10 m
b)
1010 m
c)
1015 m
d)
10-15 m
|
|
Rohit Shah answered |
One fermi or femtometre is equal 10^-15 m. This unit is used to measure nuclear dimensions.
Can you explain the answer of this question below: Approximate ratio of 1 fermi to 1 angstrom is …………..
- A:
10-5m
- B:
10-15m
- C:
10-9m
- D:
10-6m
The answer is a.
Approximate ratio of 1 fermi to 1 angstrom is …………..
10-5m
10-15m
10-9m
10-6m
|
|
Shreya Gupta answered |
As, 1 fermi = 10-15 m and 1 angstrom = 10^-10 m;
So, approximate ratio = 10^-5 m
The accuracy in the measurement of time for 20 oscillations of pendulum for the following measurements, t1 = 39.6 s, t2 = 39.9 s and t3 = 39.5 s is:
- a)± 0.005 s
- b)± 0.3 s
- c)± 0.5 s
- d)± 0.4 s
Correct answer is option 'A'. Can you explain this answer?
The accuracy in the measurement of time for 20 oscillations of pendulum for the following measurements, t1 = 39.6 s, t2 = 39.9 s and t3 = 39.5 s is:
a)
± 0.005 s
b)
± 0.3 s
c)
± 0.5 s
d)
± 0.4 s
|
Niharika Nair answered |
Given,
t1 = 39.6 s
t2 = 39.9s
t3 = 39.5 s
Least count of measuring instrument = 0.1 s
[As measurements have only one decimal place] Percision in the measurement = Least count of the measuring instrument =0.1 s
Mean value of time for 20 oscillations
t1 = 39.6 s
t2 = 39.9s
t3 = 39.5 s
Least count of measuring instrument = 0.1 s
[As measurements have only one decimal place] Percision in the measurement = Least count of the measuring instrument =0.1 s
Mean value of time for 20 oscillations
Absolute errors in the measurements
Δt1 = t - t1 = 39.7 - 39.6 = 0.1s
Δt2 = t - t2 = 39.7 - 39.9 = -0.2 s
Δt3 = t - t3 = 39.7 - 39.5 = 0.2 s
Δt1 = t - t1 = 39.7 - 39.6 = 0.1s
Δt2 = t - t2 = 39.7 - 39.9 = -0.2 s
Δt3 = t - t3 = 39.7 - 39.5 = 0.2 s
Mean absolute error
(rounding-off upto decimal place)∴ Accuracy of measurement = ±0.2 s
The most suitable instrument for measuring the size of an atom is.- a)Electron microscope
- b)Vernier calliper
- c)Optical Microscope
- d)Screw gauge
Correct answer is option 'A'. Can you explain this answer?
The most suitable instrument for measuring the size of an atom is.
a)
Electron microscope
b)
Vernier calliper
c)
Optical Microscope
d)
Screw gauge
|
Rashi Sharma answered |
Only electron microscope has resolution of 0.5-0.6 Angstrom which is less than the size of an atom.
Which of the following is caused by careless handling?- a)Systematic error
- b)Gross error
- c)Random error
- d)None of the mentioned
Correct answer is 'B'. Can you explain this answer?
Which of the following is caused by careless handling?
a)
Systematic error
b)
Gross error
c)
Random error
d)
None of the mentioned
|
.mie. answered |
Ys the answr is B ... as systematic errors arises due to faulty instruments.... random errors are those errors, which occur irregularly and hence are random. These can arise due to random and unpredictable fluctuations in experimental conditions For example, when the same person repeats the same observation, it is very likely that he may get different readings every time... Gross Errors: This category basically takes into account human oversight and other mistakes while reading, recording and the readings ...(^_^)
What is the percentage error in the physical quantity A if it is related to four other physical quantities a, b, c and d as 
The percentage error in measurement of a, b, c and d are 1%, 3%, 2% and 2% respectively.
Correct answer is '14'. Can you explain this answer?
What is the percentage error in the physical quantity A if it is related to four other physical quantities a, b, c and d as The percentage error in measurement of a, b, c and d are 1%, 3%, 2% and 2% respectively.
The percentage error in measurement of a, b, c and d are 1%, 3%, 2% and 2% respectively.
|
Soumya Mukherjee answered |
Percentage error in A
A student measures the distance covered and time taken by a body under free fall which was intially at rest. He uses this data for estimating g, the acceleration due to gravity. If the maximum percentage error in measurement of the distance and the time are e1 and e2 respectively, then the maximum possible percentage error in the estimation of g is- a)e2 - e1
- b)e1 + 2e2
- c)e1 + e2
- d)e1 - 2e2
Correct answer is option 'B'. Can you explain this answer?
A student measures the distance covered and time taken by a body under free fall which was intially at rest. He uses this data for estimating g, the acceleration due to gravity. If the maximum percentage error in measurement of the distance and the time are e1 and e2 respectively, then the maximum possible percentage error in the estimation of g is
a)
e2 - e1
b)
e1 + 2e2
c)
e1 + e2
d)
e1 - 2e2
|
|
Priya Deshpande answered |
From the releation
h= 1/2gt2 ⇒ g = 2h/t2
h= 1/2gt2 ⇒ g = 2h/t2
Taking natural logarithms on both sides, we get
ln (g) = ln (h) - 2ln (t) + ln (2)
Differentiating
for maximum permissible errors , we add both the errors . so
At a certain place, the acceleration due to gravity is determined using a simple pendulum. The length of the pendulum is 25.0 cm and a stopwatch with least count is 1 s measures the time taken by the pendulum for 40 oscilations to be 50 s. the accuracy in the measurement of g is- a)4.40 %
- b)3.40 %
- c)2.40 %
- d)5.40 %
Correct answer is option 'A'. Can you explain this answer?
At a certain place, the acceleration due to gravity is determined using a simple pendulum. The length of the pendulum is 25.0 cm and a stopwatch with least count is 1 s measures the time taken by the pendulum for 40 oscilations to be 50 s. the accuracy in the measurement of g is
a)
4.40 %
b)
3.40 %
c)
2.40 %
d)
5.40 %
|
|
Tanishq Unni answered |
Can you explain the answer of this question below: Approximate ratio of 1 fermi to 1 angstrom is …
- A:
10-5m
- B:
10-15m
- C:
10-9m
- D:
10-6m
The answer is a.
Approximate ratio of 1 fermi to 1 angstrom is …
10-5m
10-15m
10-9m
10-6m
|
|
Bhargavi Bajaj answered |
To find the approximate ratio of 1 fermi to 1 angstrom, we need to convert both units to a common unit of length.
1 fermi (fm) is a unit of length in the metric system, and it is equal to 10^-15 meters (m).
1 angstrom (Å) is also a unit of length in the metric system, and it is equal to 10^-10 meters (m).
To find the ratio, we can divide 1 fermi by 1 angstrom:
Ratio = 1 fm / 1 Å
Since both units are in meters, we can convert the units before dividing:
Ratio = (1 fm / 10^-15 m) / (1 Å / 10^-10 m)
Simplifying the expression:
Ratio = (1 fm * 10^-10 m) / (1 Å * 10^-15 m)
Ratio = 10^-10 / 10^-15
Using the rules of exponents, we can subtract the exponents:
Ratio = 10^5
Therefore, the approximate ratio of 1 fermi to 1 angstrom is 10^5.
1 fermi (fm) is a unit of length in the metric system, and it is equal to 10^-15 meters (m).
1 angstrom (Å) is also a unit of length in the metric system, and it is equal to 10^-10 meters (m).
To find the ratio, we can divide 1 fermi by 1 angstrom:
Ratio = 1 fm / 1 Å
Since both units are in meters, we can convert the units before dividing:
Ratio = (1 fm / 10^-15 m) / (1 Å / 10^-10 m)
Simplifying the expression:
Ratio = (1 fm * 10^-10 m) / (1 Å * 10^-15 m)
Ratio = 10^-10 / 10^-15
Using the rules of exponents, we can subtract the exponents:
Ratio = 10^5
Therefore, the approximate ratio of 1 fermi to 1 angstrom is 10^5.
What is the number of significant figures in 0.310×103 ?- a)6
- b)2
- c)4
- d)3
Correct answer is option 'D'. Can you explain this answer?
What is the number of significant figures in 0.310×103 ?
a)
6
b)
2
c)
4
d)
3
|
|
Suresh Kumar answered |
10^ powers are not considered as significant figures....
& in the case of decimal points....the zero present at the beginning & in between the numbers is considered as significant digit....
Hence;in this case the significant figures are 3...
& in the case of decimal points....the zero present at the beginning & in between the numbers is considered as significant digit....
Hence;in this case the significant figures are 3...
Chapter doubts & questions for Measurement - Physics 2025 is part of Grade 9 exam preparation. The chapters have been prepared according to the Grade 9 exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for Grade 9 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
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