Question:
In a Vernier calipers, when both jaws touch each other, zero of the Vernier scale is shifted to the right of zero of the main scale and $7^{th}$ Vernier division coincides with a main scale reading. If the value of 1 main scale division is 1 mm and there are 10 Vernier scale divisions, then the Vernier caliper has:
In a Vernier calipers, when both jaws touch each other, zero of the Vernier scale is shifted to the right of zero of the main scale and $7^{th}$ Vernier division coincides with a main scale reading. If the value of 1 main scale division is 1 mm and there are 10 Vernier scale divisions, then the Vernier caliper has:
Updated On: Apr 12, 2026
- 0.07 cm negative zero error
- 0.7 cm negative zero error
- 0.07 cm positive zero error
- 0.7 cm positive zero error
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Understanding the Question:
The question describes a zero error in a measuring instrument. When the Vernier zero is to the right of the Main Scale zero, the error is positive.
Step 2: Key Formula or Approach:
Least Count (LC) = 1 Main Scale Division (MSD) - 1 Vernier Scale Division (VSD).
Zero Error = $\pm (n \times LC)$, where $n$ is the coinciding division.
Step 3: Detailed Explanation:
Given: 1 MSD = 1 mm.
Standard Vernier Caliper configuration (unless stated otherwise): 10 VSD = 9 MSD.
1 VSD = 0.9 MSD = 0.9 mm.
LC = 1 MSD - 1 VSD = 1.0 mm - 0.9 mm = 0.1 mm = 0.01 cm.
Since zero of Vernier scale is shifted to the right of the zero of main scale, the error is positive.
Coinciding division $n = 7$.
Positive Zero Error $= + (n \times LC) = + (7 \times 0.01 \text{ cm}) = + 0.07$ cm.
Step 4: Final Answer:
The caliper has 0.07 cm positive zero error.
The question describes a zero error in a measuring instrument. When the Vernier zero is to the right of the Main Scale zero, the error is positive.
Step 2: Key Formula or Approach:
Least Count (LC) = 1 Main Scale Division (MSD) - 1 Vernier Scale Division (VSD).
Zero Error = $\pm (n \times LC)$, where $n$ is the coinciding division.
Step 3: Detailed Explanation:
Given: 1 MSD = 1 mm.
Standard Vernier Caliper configuration (unless stated otherwise): 10 VSD = 9 MSD.
1 VSD = 0.9 MSD = 0.9 mm.
LC = 1 MSD - 1 VSD = 1.0 mm - 0.9 mm = 0.1 mm = 0.01 cm.
Since zero of Vernier scale is shifted to the right of the zero of main scale, the error is positive.
Coinciding division $n = 7$.
Positive Zero Error $= + (n \times LC) = + (7 \times 0.01 \text{ cm}) = + 0.07$ cm.
Step 4: Final Answer:
The caliper has 0.07 cm positive zero error.
Was this answer helpful?
0
0
Top JEE Main Physics Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
- $I _{ CM }$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc $I_{A B}$ is it's moment of inertia about an axis $AB$ perpendicular to plane and parallel to axis $CM$ at a distance $\frac{2}{3} R$ from center Where $R$ is the radius of the dise The ratio of $I _{ AB }$ and $I _{ CM }$ is $x: 9$ The value of $x$ is______
- A nucleus disintegrates into two smaller parts, which have their velocities in the ratio $3: 2$ The ratio of their nuclear sizes will be $\left(\frac{x}{3}\right)^{\frac{1}{3}}$ The value of ' $x$ ' is:-
View More Questions
Top JEE Main Units, Dimensions and Measurements Questions
- Match List-I with List-II.
- Find the dimension of \( \frac{E}{B} \) where, E represents electric field and B represents magnetic field.
- Which of the following quantity has the same dimensions as \( \sqrt{\frac{\mu_0}{\epsilon_0}} \)?
- A fractional errors in $x$, $y$, and $z$ are $0.1$, $0.2$, and $0.5$ respectively. Find the maximum fractional error in $x^{-2} y^3 z^{-2}$.
- Match the columns:
View More Questions
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
View More Questions