Live Session - Units & Measurement -II : 22 May 2020Contact Number: 9667591930 / 8527521718Live Session - Units & Measurement -II : 22 May 2020Contact Number: 9667591930 / 8527521718The period of oscillation of a simple pendulum is given by \(T = 2\pi \sqrt{\frac{L}{g}}\) where \(L\) is about \(100~\text{cm}\) and is known to have \(1~\text{mm}\) accuracy. The period is about \(2~\text{s}\). The time of \(100\) oscillations is measured by a stopwatch of least count \(0.1~\text{s}\). The percentage error in \(g\) is:
1. \(0.1\%\)
2. \(1\%\)
3. \(0.2\%\)
4. \(0.8\%\)
The random error in the arithmetic mean of 100 observations is x; then random error in the arithmetic mean of 400 observations would be
1. 4x
2.
3. 2x
4.
A body travels uniformly a distance of (13.8 0.2) m in a time (4.0 ± 0.3) sec. The velocity of the body within error limits is:
1. (3.45 ± 0.2) ms-1
2. (3.45 ± 0.3) ms-1
3. (3.45 ± 0.4) ms-1
4. (3.45 ± 0.5) ms-1
Accuracy of measurement is determined by
(1) Absolute error
(2) Percentage error
(3) Both
(4) None of these
In the context of the accuracy of measurement and significant figures in expressing the results of the experiment, which of the following is/are correct?
(1) Out of the two measurements 50.14 cm and 0.00025 Amperes, the first one has greater accuracy.
(2) If one travels 478 km by rail and 397 m by road, the total distance traveled is 478 km.
1. Only (1) is correct
2. Only (2) is correct
3. Both are correct
4. None of them is correct.
The relative density of material of a body is found by weighing it first in air and then in water. If the weight in air is (5.00 ± 0.05) Newton and weight in water is (4.00 ± 0.05) Newton. Then the relative density along with the maximum permissible percentage error is
1. 5.0 ± 11%
2. 5.0 ± 1%
3. 5.0 ± 6%
4. 1.25 ± 5%
If there is a positive error of \(50\%\) in the measurement of the velocity of a body, then the error in the measurement of kinetic energy is:
1. \(25\%\)
2. \(50\%\)
3. \(100\%\)
4. \(125\%\)
A student measures the distance traversed in free fall of a body, initially at rest in a given time. He uses this data to estimate g, the acceleration due to gravity. If the maximum percentage errors in measurement of the distance and the time are and respectively, the percentage error in the estimation of g is
1.
2.
3.
4.
If a screw gauge has a pitch of 1.5 mm and 300 divisions on circular scale, which of the following reading can be made from this screw gauge?
(1) 1.05 mm
(2) 3.01 cm
(3) 0.030000 m
(4) 2.1185 dm
Two full turns of the circular scale of gauge cover a distance of \(1\) mm on scale. The total number of divisions on the circular scale is \(50.\) Further, it is found that screw gauge has a zero error of \(-0.03\) mm. While measuring the diameter of a thin wire a student notes the main scale reading of \(3\) mm and the number of circular scale division in line, with the main scale as \(35.\) The diameter of the wire is:
1. \(3.32\) mm
2. \(3.73\) mm
3. \(3.67\) mm
4. \(3.38\) mm
Consider a screw gauge without any zero error. What will be the final reading corresponding to the final state as shown?
It is given that the circular head translates \(P\) MSD in \({N}\) rotations. (\(1\) MSD \(=\) \(1~\text{mm}\).)
1. \( \left(\frac{{P}}{{N}}\right)\left(2+\frac{45}{100}\right) \text{mm} \)
2. \( \left(\frac{{N}}{{P}}\right)\left(2+\frac{45}{{N}}\right) \text{mm} \)
3. \(P\left(\frac{2}{{N}}+\frac{45}{100}\right) \text{mm} \)
4. \( \left(2+\frac{45}{100} \times \frac{{P}}{{N}}\right) \text{mm}\)
A screw gauge has some zero error but its value is unknown. We have two identical rods. When the first rod is inserted in the screw, the state of the instrument is shown by diagram (I). When both the rods are inserted together in series then the state is shown by the diagram (II). What is the zero error of the instrument? \(1~\text{msd}= 100~\text{csd}=1~\text{mm}\)
| 1. | \(-0.16~\text{mm}\) | 2. | \(+0.16~\text{mm}\) |
| 3. | \(+0.14~\text{mm}\) | 4. | \(-0.14~\text{mm}\) |
One cm on the main scale of vernier callipers is divided into ten equal parts. If 20 divisions of vernier scale coincide with 8 small divisions of the main scale. What will be the least count of callipers?
1. 0.06 cm
2. 0.6 cm
3. 0.5 cm
4. 0.7 cm
Find the zero correction in the given figure.
1. \(0.4\) mm
2. \(0.5\) mm
3. \(-0.5\) mm
4. \(-0.4\) mm
Find the thickness of the wire. The least count is \(0.01~\text{mm}\). The main scale reads (in mm):
1. \(7.62\)
2. \(7.63\)
3. \(7.64\)
4. \(7.65\)
The main scale of a vernier callipers reads 10 mm in 10 divisions. 10 divisions of Vernier scale coincide with 9 divisions of the main scale. When the two jaws of the callipers touch each other, the fifth division of the vernier coincides with 9 main scale divisions and the zero of the vernier is to the right of zero of main scale. When a cylinder is tightly placed between the two jaws, the zero of vernier scale lies slightly behind 3.2 cm and the fourth vernier division coincides with a main scale division. The diameter of the cylinder is.
(1) 3.10 cm
(2) 3.8 cm
(3) 3.09 cm
(4) -3.09 cm
