$A$ tiny metallic rectangular sheet has a length and breadth of $5 \ mm$ and $2.5 \ mm$,respectively. Using a specially designed screw gauge which has a pitch of $0.75 \ mm$ and $15$ divisions on the circular scale,you are asked to find the area of the sheet. In this measurement,the maximum fractional error will be $\frac{x}{100}$ where $x$ is . . . . . .

  • A
    $3$
  • B
    $13$
  • C
    $5$
  • D
    $14$

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Similar Questions

In a vernier callipers,each $cm$ on the main scale is divided into $20$ equal parts. If the $10^{th}$ vernier scale division coincides with the $9^{th}$ main scale division,then the value of the vernier constant will be $\dots \; \times 10^{-2} \; mm$.

$A$ 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 gauge,the state of the instrument is shown by diagram $(I)$. When both the rods are inserted together in series,the state is shown by diagram $(II)$. What is the zero error of the instrument in $mm$? Given: $1 \, M.S.D. = 100 \, C.S.D. = 1 \, mm$.

$A$ steel wire of diameter $0.5 \text{ mm}$ and Young's modulus $2 \times 10^{11} \text{ N m}^{-2}$ carries a load of mass $M$. The length of the wire with the load is $1.0 \text{ m}$. $A$ vernier scale with $10$ divisions is attached to the end of this wire. Next to the steel wire is a reference wire to which a main scale,of least count $1.0 \text{ mm}$,is attached. The $10$ divisions of the vernier scale correspond to $9$ divisions of the main scale. Initially,the zero of vernier scale coincides with the zero of main scale. If the load on the steel wire is increased by $1.2 \text{ kg}$,the vernier scale division which coincides with a main scale division is. . . . Take $g = 10 \text{ m s}^{-2}$ and $\pi = 3.2$.

$10$ divisions on the main scale of a Vernier calliper coincide with $11$ divisions on the Vernier scale. If each division on the main scale is of $5$ units,the least count of the instrument is :

The vernier calliper used for measurement has a positive zero error of $0.3 \ mm$. While taking measurement for the internal diameter of a vessel,it was observed that the zero of the vernier scale lies between $9.5 \ cm$ and $9.6 \ cm$ of the main scale and the $6^{th}$ division of the vernier scale coincides with a main scale division. If the least count of the vernier calliper is $0.01 \ cm$,the correct value of the diameter will be: (in $cm$)

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