A screw gauge has $50$ divisions on its circular scale. The circular scale is $4$ units ahead of the pitch scale marking, prior to use. Upon one complete rotation of the circular scale, a displacement of $0.5\, mm$ is noticed on the pitch scale. The nature of zero error involved, and the least count of the screw gauge, are respectively
A screw gauge has $50$ divisions on its circular scale. The circular scale is $4$ units ahead of the pitch scale marking, prior to use. Upon one complete rotation of the circular scale, a displacement of $0.5\, mm$ is noticed on the pitch scale. The nature of zero error involved, and the least count of the screw gauge, are respectively
- [JEE MAIN 2020]
- A
Negative, $2\, \mu m$
- B
Positive, $10\, \mu m$
- C
Positive, $0.1$ $\mu m$
- D
Positive, $0.1$ $mm$
Similar Questions
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 ? .......... $mm$
$1\,M.S.D. = 100\, C.S.D. = 1\, 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 ? .......... $mm$
$1\,M.S.D. = 100\, C.S.D. = 1\, mm $
Least count of a vernier caliper is $\frac{1}{20 \mathrm{~N}} \mathrm{~cm}$. The value of one division on the main scale is $1 \mathrm{~mm}$. Then the number of divisions of main scale that coincide with $\mathrm{N}$ divisions of vernier scale is :
Least count of a vernier caliper is $\frac{1}{20 \mathrm{~N}} \mathrm{~cm}$. The value of one division on the main scale is $1 \mathrm{~mm}$. Then the number of divisions of main scale that coincide with $\mathrm{N}$ divisions of vernier scale is :
- [JEE MAIN 2024]
In a screw gauge, $5$ complete rotations of circular scale give $1.5\, mm$ reading on linear scale. Circular scale has $50$ divisions. Least count of the screw gauge is
In a screw gauge, $5$ complete rotations of circular scale give $1.5\, mm$ reading on linear scale. Circular scale has $50$ divisions. Least count of the screw gauge is
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 ?
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 ?
The diameter of a spherical bob is measured using a vernier callipers. $9$ divisions of the main scale, in the vernier callipers, are equal to $10$ divisions of vernier scale. One main scale division is $1\, {mm}$. The main scale reading is $10\, {mm}$ and $8^{\text {th }}$ division of vernier scale was found to coincide exactly with one of the main scale division. If the given vernier callipers has positive zero error of $0.04\, {cm}$, then the radius of the bob is $...... \,\times 10^{-2} \,{cm}$
The diameter of a spherical bob is measured using a vernier callipers. $9$ divisions of the main scale, in the vernier callipers, are equal to $10$ divisions of vernier scale. One main scale division is $1\, {mm}$. The main scale reading is $10\, {mm}$ and $8^{\text {th }}$ division of vernier scale was found to coincide exactly with one of the main scale division. If the given vernier callipers has positive zero error of $0.04\, {cm}$, then the radius of the bob is $...... \,\times 10^{-2} \,{cm}$
- [JEE MAIN 2021]