Screw Gauge

Q. If the number of divisions on the circular scale is 100 and the number of full rotations given to screw is 8 and dis†an ce moved by the screw is 4mm then what will be the least count of the screw gauge

Q.

Which of the following is the most precise device for measuring length : (a) a vernier callipers with 20 divisions on the sliding scale (b) a screw gauge of pitch 1 mm and 100 divisions on the circular scale (c) an optical instrument that can measure length to within a wavelength of light?

Q. A screw gauge having $100$ equal divisions and a pitch length of $1\text{mm}$ is used to measure the diameter of a wire of length $5.6\text{cm}$. The main scale reading is $1\text{mm}$ and ${47}^{\text{th}}$ circular division coincides with the main scale. The curved surface area of wire (in ${\text{cm}}^2$) to appropriate significant figures is

1. $1.60$
2. $3.20$
3. $1.25$
4. $2.6$

Q. Answer the following : (a)You are given a thread and a metre scale. How will you estimate the diameter of the thread ? (b)A screw gauge has a pitch of 1.0 mm and 200 divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale ? (c) The mean diameter of a thin brass rod is to be measured by vernier callipers. Why is a set of 100 measurements of the diameter expected to yield a more reliable estimate than a set of 5 measurements only ?

Q. Formula for calculating zero error in micrometer screw gauge

Q.

When an object is placed between the jaws of a vernier caliper, the zero of the vernier scale is found to be on the right of $7^{th}$the division of the main scale. The $8^{th}$division of the vernier scale is found to coincide with one of the divisions of the main scale. Arrange the following steps in sequential order to determine the length of the object. [$1MSD=1mm,numberofVSD=10$]

(A) Determine the length of the object
(B) Use the formula $MSR+(VSR\times LC)$
(C) Note the M.S.R., V.C.D., 1 M.S.D, and number of V.S.D.
(D) By using the information given determine the least count.

1. ACBD

2. CDBA

3. DBCA

4. ACDB

Q. A student measured the length of a rod and wrote it as 3.50 cm. Which instrument did he use to measure it? Options :a) A meter scale with least count 0.1 cm Should have chosen b)A vernier calliper where the 10 divisions in vernier scale matches with 9 divisions in main scale and main scale has 10 divisions in 1 cm c) A screw gauge having 100 divisions in the circular scale and pitch as 1 mm d) A screw gauge having 50 divisions in the circular scale and pitch as 1 mm

Q.

Which is the most accurate atomic clock?

1. Stop clock

2. Carbon-12 clock

3. Carbon atom clock

4. Caesium atom clock

Q. A screw gauge gives the following reading for measuring the diameter of a wire.
Main scale reading : $0\text{mm}$
Circular scale reading : $60$ divisions
Given, that $1\text{mm}$ on main scale corresponds to $100$ divisions on circular scale. The diameter of wire is,

1. $0.050\text{cm}$
2. $0.054\text{cm}$
3. $0.060\text{cm}$
4. $0.062\text{cm}$

Q.

Students I, II and III perform an experiment for measuring the acceleration due to gravity (g) using a simple pendulum.They use different lenghts of the pendulum and record time for different number of oscillations.The observations are shown in the table.Least count for length = 0.1cm, Least count for time = 0.1s

$\begin{array}{ccccc}\text{Student}& \text{Length of the pendulum(cm)}& \text{Number of oscillations(n)}& \text{Total time for (n) oscillation(s)}& \text{Time period(s)}\\ \text{I}& 64.0& 8& 128.0& 16.0\\ \text{II}& 64.0& 4& 64.0& 16.0\\ \text{III}& 20.0& 4& 36.0& 9.0\end{array}$

If $E_I$, $E_{II}$ and $E_{III}$ are the percentage errors in g, i.e. $(\frac{△g}{g}\times 100)$ for student I, II and III, respectively, t = $2\pi \sqrt{\frac{l}{g}}$ , then

1. $E_l$ = 0

2. $E_l$ is minimum

3. $E_l$ and $E_{ll}$ = 0

4. $E_{ll}$ is maximum

Q. 15.How to take a reading on vernier calliper and screw gauge

Q. The thickness of a glass sheet measured using a screw gauge is given by 1.52 mm, 1.56 men. 1 mm respectively.Find the thickness of the glass sheet, mean absolute error and percentage

Q. A screw gauge has a negative error of 8 divisions. While measuring the diameter of a wire, the reading on the main scale is 3 divisions and the 24th circular scale division coincides with the base line. If the number of divisions on the main scale are 20 to a centimeter and the circular scale has 50 divisions, then the correct diameter of the wire is

1. 0.182 cm
2. 0.166 cm
3. 0.174 cm
4. 3.32 mm

Q. A student measured the diameter of a small steelball using a screw gauge of least count 0.001 cm.The main scale reading is 5 mm and zero ofcircular scale division coincides with 25 divisionsabove the reference level. If screw gauge has azero error of -0.004 cm, the correct diameter of theball is(1) 0.521 cm(2) 0.525 cm(3) 0.529 cm(4) 0.053 cm

Q. The pitch of a screw gauge is $1\text{mm}$ and there are $100$ divisions on circular scale. When the faces $A$ and $B$ are just touching each without putting anything between the studs, the ${32}^{\text{nd}}$ division of the circular scale coincides with the reference line. The error in measurement due to screw gauge will be (zero of linear scale is not hidden from circular scale when $A$ and $B$ touches)

1. $-0.04\text{mm}$
2. $+0.32\text{mm}$
3. $+0.16\text{mm}$
4. $+0.04\text{mm}$

Q. Least count of a screw gauge is 0.01mm. 6th division of circular scale coincise with reference line when the gap is closed. What is the 0 error?

Q. Diameter of a wire measured by a screw guage in four different readings is found to be 2.20 mm, 2.10 mm, 2.40 mm and 1.90 mm.

1. Number of significant figures in all readings is two.

2. Error in $1^{st}$ reading is negative.
3. Mean error of readings is zero.
4. Error in $4^{th}$ reading is 0.25 mm.

Q. Formula for calculating zero error in micrometer screw gauge

Q.

In a Searle's experiment, the diameter of the wire as measured by a screw gauge with least count 0.001 cm is 0.050 cm. The length, measured by a scale of least count 0.1 cm, is 110.0 cm. When a weight of 50 N is suspended from the wire, the extension is measured to be 0.125 cm by a micrometer of least count 0.001 cm. Find the maximum error in the measurement of Young's modulus of the material of the wire from these data.

$y=\frac{W}{A}\times \frac{L}{X}$
Where W is the weight suspended from the wire, A is the cross section area, L is the length and X is the extension in the wire.

[IIT JEE 2004]

1. 4.89 %

2. 1.26 %

3. 9.67 %

4. 15.56 %

Q. Diameter of a wire measured by a screw guage in four different readings is found to be 2.20 mm, 2.10 mm, 2.40 mm and 1.90 mm.

1. Number of significant figures in all readings is two.

2. Error in $1^{st}$ reading is negative.
3. Mean error of readings is zero.
4. Error in $4^{th}$ reading is 0.25 mm.

Q. A screw gauge with a pitch of $0.5\text{mm}$ and a circular scale with 50 divisions is used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement, it is found that when the two jaws of the screw gauge are brought in contact, the 45th division coincides with the main scale line and that the zero of the main scale is barely visible. What is the thickness of the sheet if the main scale reading is $0.5\text{mm}$ and the 25th division coincides with the main scale line?

1. $0.50\text{mm}$
2. $0.75\text{mm}$
3. $0.80\text{mm}$
4. $0.70\text{mm}$

Q.

Students I, II and III perform an experiment for measuring the acceleration due to gravity (g) using a simple pendulum.They use different lenghts of the pendulum and record time for different number of oscillations.The observations are shown in the table.Least count for length = 0.1cm, Least count for time = 0.1s

$\begin{array}{ccccc}\text{Student}& \text{Length of the pendulum(cm)}& \text{Number of oscillations(n)}& \text{Total time for (n) oscillation(s)}& \text{Time period(s)}\\ \text{I}& 64.0& 8& 128.0& 16.0\\ \text{II}& 64.0& 4& 64.0& 16.0\\ \text{III}& 20.0& 4& 36.0& 9.0\end{array}$

If $E_I$, $E_{II}$ and $E_{III}$ are the percentage errors in g, i.e. $(\frac{△g}{g}\times 100)$ for student I, II and III, respectively, t = $2\pi \sqrt{\frac{l}{g}}$ , then

1. and = 0

2. = 0

3. is minimum

4. is maximum

Q. In an experiment the angles are required to be measured using an instrument. 29 divisions of the main scale exactly coincide with the 30 divisions of the vernier scale. If the smallest division of the main scale is half-a-degree $\left(0.5°\right)$ then the least count of the instrument is

1. one minute
2. half minute
3. one degree
4. half degree

Q. Two full turns of the circular scale of a screw gauge cover a distance of 1 mm on its main scale. The total number of divisions on the circular scale is 50. Further, it is found that the 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 divisions 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

Q. In a vernier caliper, 10 vernier scale divisions are equal to 9 main scale divisions. One main scale division is 1 mm. The device has some zero error as shown in the figure (a). When a cylinder of diameter D is inserted between them, the reading of scale is shown in figure (b). Then, the diameter of the cylinder in cm is

Q. a circular scale of screw gauge has 50 divisions and pitch is 0.5mm. it coincides at 25th division and is main scale reading is 2. find the diameter of the sphere.

Q. The pitch of a screw gauge is $2\text{mm}$ and there are $100$ divisions on the circular scale. In measuring the diameter of a sphere, there are $6$ divisions on the linear scale and $40$ divisions on circular scale coinciding with the reference line. The diameter of the sphere is:

1. $6.4\text{mm}$
2. $12.8\text{mm}$
3. $3.2\text{mm}$
4. $20\text{mm}$

Q. A screw gauge gives the following reading for measuring the diameter of a wire.
Main scale reading : $0\text{mm}$
Circular scale reading : $60$ divisions
Given, that $1\text{mm}$ on main scale corresponds to $100$ divisions on circular scale. The diameter of wire is,

1. $0.050\text{cm}$
2. $0.054\text{cm}$
3. $0.060\text{cm}$
4. $0.062\text{cm}$

Q. The main scale reading of a screw gauge is 8 mm and the circular scale coincides with 25th division of the main scale. If the pitch of the screw gauge is 0.5 mm and the number of circular scale division is 100, then the observed reading will be

(A) 8.005 mm
(B) 8.025 mm
(C) 8.125 mm
(D) 8.000 mm

Q.

In a Searle's experiment, the diameter of the wire as measured by a screw gauge with least count 0.001 cm is 0.050 cm. The length, measured by a scale of least count 0.1 cm, is 110.0 cm. When a weight of 50 N is suspended from the wire, the extension is measured to be 0.125 cm by a micrometer of least count 0.001 cm. Find the maximum error in the measurement of Young's modulus of the material of the wire from these data.

$y=\frac{W}{A}\times \frac{L}{X}$
Where W is the weight suspended from the wire, A is the cross section area, L is the length and X is the extension in the wire.

[IIT JEE 2004]

1. 4.89 %

2. 1.26 %

3. 9.67 %

4. 15.56 %