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where on the vernier caliper would you read to obtain the vernier scale reading
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Where on the vernier caliper would you read to obtain the vernier scale reading?
You would read the main scale reading on the vernier caliper from the scale that is closest to the object being measured. The vernier scale reading is obtained by noting which vernier scale division aligns with a main scale division. The combined main scale reading and vernier scale reading provide the precise measurement of the object.
Where on the vernier caliper would you read to obtain the main scale reading?
The main scale is before the zero line of the vernier scale.
What is the unit of vernier scale reading of vernier caliper?
No unit for vernier scale coincidence. But as we multiply it by least count which has unit then we get vernier scale reading with the same unit. Of course main scale reading and vernier scale reading would have the same unit
Differentiate between main scale reading and vernier scale reading?
main scale is the least accurate reading on the device. the vernier scale uses an offset of divions that only allows one "line" or division to line up with the scale below. This scale is more accurate than the main scale. for example, the main scale may produce an accuracy of 0.125 of and inch plus or minus 0.125 of an inch, whereas the vernier would produce an accuracy of 0.025 of and inch. therefore if two divivions of the main scale and one division of the vernier scale were combined it would prodice and reading of (0.125+0.125+0.025 = 0.275) inches.
What is the reading uncertainty for vernier caliper?
The reading uncertainty for a vernier caliper is typically equal to half of the smallest division on the main scale. For example, if the smallest division on the main scale is 0.1 mm, the reading uncertainty would be 0.05 mm. This uncertainty represents the smallest increment that can be reliably measured with the vernier caliper.
How do you find radius of cylinder when the main scale reading is 35 division and 4 division of vernier scale coincides with a main scale division?
The vernier scale provides a more precise measurement by showing how much the object has moved past a whole number on the main scale. Since 4 divisions of the vernier scale coincide with one less division on the main scale, each division on the vernier scale represents 1/4 of a division on the main scale. In this case, the total reading would be 35 main scale divisions plus 4 divisions on the vernier scale, giving a total of 35.25 divisions on the main scale. To find the radius of the cylinder, you would measure the circumference with a tape measure or string and then divide that measurement by 2π to find the radius.
How do you take reading in traveling microscope?
let us think with all our common sense.The indicator which reads the reading on the main scale is the zero mark on the vernier.On the main scale each mm is divided into half mm. Hence the reading would be read for divisions on the main scale. ie say I start from 3 cm.3 cm, 3.05 cm, 3.1 cm, 3.15 cm, 3.2 cm, 3.25 cm and so on.But for our convenience let us read as 3.000 cm, 3.050 cm, 3.100 cm, 3.150 cm, 3.200 cm, 3.250 cm and so on,Now say your indicator ie zero marking on the vernier is lying in between say 3.200 and 3.250 cm. Then main scale reading would be recorded by taking the lower value namely 3.200 cm.Now the vernier would give the last two decimal values for accuracy. If the 43rd division on the vernier get coincided with some line on the main scale, then the vernier reading will be got by multiplying this vernier coincidence 43 by the least count. The least count will usually be 0.001 cm.So the vernier scale reading will be 0.043 cmNow by adding this along with the main scale reading ie 3.200 cm we get 3.243 cm.Another example: If suppose the indicator lies in between 6.450 and 6.500. The vernier division 17 get coincided.Main scale reading is 6.450 cmVernier scale reading is 0.17 cmSo the total reading ie accurate reading is 6.467 cmUNNATI CHANDANESTD:XII
Why is the vernier caliper so called?
A Vernier allows a precise reading of some value. In the figure to the right, the Vernier moves up and down to measure a position on the Scale. This could be part of a barometer which reads atmospheric pressure. The "pointer" is the line on the vernier labelled "0". Thus the measured position is almost exactly 756 in whatever units the scale is calibrated in. If you look closely you will see that the distance between the divisions on the vernier are not the same as the divisions on the scale. The 0 line on the vernier lines up at 756 on the scale, but the 10 line on the vernier lines up at 765 on the scale. Thus the distance between the divisions on the vernier are 90% of the distance between the divisions on the scale. If we do another reading with the vernier at a different position, the pointer, the line marked 0, may not line up exactly with one of the lines on the scale. Here the "pointer" lines up at approximately 756.5 on the scale. If you look you will see that only one line on the vernier lines up exactly with one of the lines on the scale, the 5 line. This means that our first guess was correct: the reading is 756.5. Here is a final example, with the vernier at yet another position. The pointer points to a value that is obviously greater than 756.5 and also less than 757.0. Looking for divisions on the vernier that match a division on the scale, the 7 line matches fairly closely. So the reading is about 756.7. In fact, the 7 line on the vernier appears to be a little bit above the corresponding line on the scale. The 8 line on the vernier is clearly somewhat below the corresponding line of the scale. So with sharp eyes one might report this reading as 756.73 ± 0.02. This "reading error" of ± 0.02 is probably the correct error of precision to specify for all measurements done with this apparatus. Now we shall use a simulation of a Vernier Caliper. A caliper measures a length, and in the following figure we show a caliper being used to measure the length of an Object. The Object will be placed between the "jaws" of the caliper. The Object is almost exactly 75 mm (2.95 in) long. In the above photograph, you can see that on the top of the caliper are two "prongs" which can be used to measure an interior dimension. There is also a piece of metal sticking out from the right side of the caliper, which is a depth gauge. Calipers commonly use a vernier scale. In the simulation below, you may "grab" the jaw of the caliper with the left button of the mouse and move it to some position. When you click on the Show button the distance between the jaws will be shown. Note that there is a small difference between the simulation and a real caliper: in the simulation the distance between the jaws is always an even multiple of a tenth of a millimeter. It does not allow readings between these values, for which we would have to estimate the value. The Java applet to simulate the vernier caliper was written by Fu-Kwan Hwang, Department of Physics, National Taiwan Normal Univ., and is used by permission. See related link for the applet.
Can you use the tail of the vernier caliper to measure the depth of the test tube?
No, the tail of a vernier caliper is used to measure inside dimensions like holes or pipes, not depth. To measure the depth of the test tube, you would need to use the main scale and the vernier scale on the caliper's jaws.
What is least count of vernier caliper?
Least count of vernier caliper= 1 MSD - 1 VSD. Main scale usually would have 1 mm as the minimum division. Such 9 divisions, ie 9 mm would be divided into ten equal parts in the vernier. So VSD = 0.9 mm Hence the least count = 1 - 0.9 = 0.1 mm If the same is to be given in centimetre, then it will be 0.01 cm.
If a vernier caliper has zero reading of 0.025 cm and the caliper reads 5.565 cm what should be the correct reading?
If a vernier caliper has zero reading of 0.025 cm and the caliper reads 5.565 cm what should be the correct reading? 5.565 - 0.025 = 5.54 so caliper reads 5.54 cm If the zero reading on the vernier calipers is -0.025 cm with the reading still 5.565cm, it is 5.565 - (-0.025) = 5.565 + 0.025 = 5.59 cm
At what temperature does the reading of Fahrenheit scale would Celsius scale coincide?
At -40.
