Q: वर्नियर कैलीपर्स के मुख्य स्केल का अल्पतमांक 1 mm है। इसका वर्नियर स्केल 10 खाने में विभक्त है। वर्नियर कैलिपर्स में मुख्य स्केल का 1 खाना वर्नियर स्केल के 7 वें खाना से सम्पाती है। वर्नियर स्केल का शून्य, मुख्य स्केल के दायीं ओर है। जब यह वर्नियर एक बेलन की लम्बाई मापन के लिये उपयोग किया जाता है तो वर्नियर स्केल का शून्य 3.1 cm तथा 3.2 cm के बीच एवं चौथा खाना वर्नियर स्केल का मुख्य स्केल का सम्पाती है। बेलन की लम्बाई है

वर्नियर कैलीपर्स का अल्पतमांकL.C. =1MSD−1VSD=(1−109)×1=0.1 mm=0.01 cmयहाँ वार्नियर कैलिपर्स में, मुख्य स्केल का 1 खाना वर्नियर स्केल के 7 वें खानों से सम्पाती है। वर्नियर स्केल का शून्य, मुख्य स्केल के दांयी ओर है।शून्य त्रुटि =7×0.1=0.7 mm=0.07 cmबेलन की लम्बाई = मापा गया मान - शून्य त्रुटि =(3.1+4×0.01)−0.07=3.07 cm

Q: A spectrometer gives the following reading when used to measure the angle of a prism Main scale reading : 58.5 degree Vernier scale reading : 09 divisions Given that 1 division on main scale corresponds to 0.5 degree. Total divisions on the vernier scale is 30 and match with 29 divisions of the main scale. The angle of the prism from the above data

30VSD=29MSD1VSD=3029MSDLeast count of yernier =1M.S.D.−1V.S.D. =0.5∘−3029×0.5∘=300.5∘Reading of vernier = M.S. reading + V.S. reading ×L. C.=58.5∘+9×300.5∘=58.65

Question 1

Which of the following is the most precise device for measuring length?

Accepted Answer

Most precise device is the one with the imum least count1. Vernier Calliper20 divisions on the sliding scaleL.C=201​mm=0.05mm2. Screw Gaugepitch 1 mm and 100 divisions on the circular scaleL.C= 1001mm​=0.01mm3.optical instrument within a wavelength of lightWavelenght of light ≈ 589nmHence Optical instrument is the most precise

Question 2

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 1.0mm 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?

Accepted Answer

(a) Wrap the thread on a uniform smooth rod in such a way that the coils thus formed are very close to each other. Measure the length of the thread using a metre scale. The diameter of the thread is given by the relation,Diameter = Length of rod/Number of turns(b) It is not possible to increase the accuracy of a screw gauge by increasing the number of divisions of the circular scale. Increasing the number of divisions of the circular scale will increase its accuracy to a certain extent only.(c) A set of 100 measurements is more reliable than a set of 5 measurements because random errors involved in the former are very less as compared to the latter.

Question 3

The Vernier scale of a travelling microscope has 50 divisions which coincide with 49 main scale divisions. If each main scale division is 0.5 mm , calculate the minimum inaccuracy in the measurement of distance(in mm).

Accepted Answer

According to the problem, 50 divisions of Vernier scale which coincide with 49 main scale divisions.50VSD=49MSD⇒1MSD=4950​VSD or 1VSD=5049​MSD where MSD= Main scale division and VSD=Vernier scale division.We know thatMinimum inaccuracy = Vernier constant=1MSD−1VSD=1MSD−5049​MSD=501​MSD It is given in the problem that 1MSD=0.5 mm Hence, minimum inaccuracy =501​×0.5 mm=1001​=0.01 mm

Question 4

Least count of Vernier callipers is 1×10−4 m .The main scale reading before zero is 9 and the zeroth division of Vernier scale division coincides with main scale division. Each main scale division is 1×10−3 m. then the measured value is ____ mm.

Accepted Answer

In Vernier callipersLength measured =  Reading before zero +  number of coinciding Vernier division ×  Vernier constant=9×10−3+0×1×10−4 =0.009 m  or 9 mm

Question 5

वर्नियर कैलीपर्स के मुख्य स्केल का अल्पतमांक 1 mm है। इसका वर्नियर स्केल 10 खाने में विभक्त है। वर्नियर कैलिपर्स में मुख्य स्केल का 1 खाना वर्नियर स्केल के 7 वें खाना से सम्पाती है। वर्नियर स्केल का शून्य, मुख्य स्केल के दायीं ओर है। जब यह वर्नियर एक बेलन की लम्बाई मापन के लिये उपयोग किया जाता है तो वर्नियर स्केल का शून्य 3.1 cm तथा 3.2 cm के बीच एवं चौथा खाना वर्नियर स्केल का मुख्य स्केल का सम्पाती है। बेलन की लम्बाई है

Accepted Answer

वर्नियर कैलीपर्स का अल्पतमांकL.C. =1MSD−1VSD=(1−109​)×1=0.1 mm=0.01 cmयहाँ वार्नियर कैलिपर्स में, मुख्य स्केल का 1 खाना वर्नियर स्केल के 7 वें खानों से सम्पाती है। वर्नियर स्केल का शून्य, मुख्य स्केल के दांयी ओर है।शून्य त्रुटि =7×0.1=0.7 mm=0.07 cmबेलन की लम्बाई = मापा गया मान - शून्य त्रुटि =(3.1+4×0.01)−0.07=3.07 cm

Question 6

एक विद्यार्थी ने एक छड़ की लम्बाई मापकर 3.50 cm लिखी। इसको मापने में उसने किस उपकरण का प्रयोग किया?

Accepted Answer

छड़ की मापी गई लम्बाई =3.50 cmवर्नियर स्केल के लिए 1 भाग मुख्य स्केल =1 mm9 भाग मुख्य स्केल =10 भाग वर्नियर स्केलअल्पतमांक =1MSD−1VSD=0.1 mm

Question 7

In a vernier callipers, one main scale division is x cm and n division of the vernier scale coincide with (n−1) divisions of the main scale. The least count (in cm ) of the callipers is

Accepted Answer

One main scale division, 1 M.S.D. =x cmOne vernier scale division, 1 V.S. D. =n(n−1)x​Least count =1 M.S.D. −1 V.S. D.=nnx−nx+x​=nx​ cm

Question 8

The internal and external diameters of a hollow cylinder are measured with the help of a vernier calipers. Their values are 4.23±0.01 cm and 3.87±0.01 cm respectively. The thickness of the wall of the cylinder is

Accepted Answer

Subtract 3.87 from 4.23 and then divide by 2 .

Question 9

A vernier callipers has 1 mm marks on the main scale. It has 20 equal divisions on the Vernier scale which match with 16 main scale divisions. For this Vernier callipers, the least count is

Accepted Answer

20VSD=16MSD1VSD=0.8MSDLeast count =MSD−VSD=1 mm−0.8 mm=0.2 mm

Question 10

A spectrometer gives the following reading when used to measure the angle of a prism
Main scale reading : 58.5 degree
Vernier scale reading : 09 divisions
Given that 1 division on main scale corresponds to 0.5 degree. Total divisions on the vernier scale is 30 and match with 29 divisions of the main scale. The angle of the prism from the above data

Accepted Answer

30VSD=29MSD1VSD=3029​MSDLeast count of yernier =1M.S.D.−1V.S.D. =0.5∘−3029​×0.5∘=300.5∘​Reading of vernier = M.S. reading + V.S. reading ×L. C.=58.5∘+9×300.5∘​=58.65

Question 11

In a vernier callipers, ten smallest divisions of the vernier scale are equal to nine smallest division on the main scale. If the smallest division on the main scale is half millimeter, then the vernier constant is

Accepted Answer

We have given,10VD=9MD∴ 1 VD=109​MDVernier constant =1MD−1VD =(1−109​)MD=101​MD=101​×21​=0.05 mm

Question 12

N divisions on the main scale of a vernier calliper coincide with (N+1) divisions of the vernier scale. If each division of main scale is ' a ' units, then the least count of the instrument is

Accepted Answer

No of divisions on main scale =NNo of divisions on vernier scale =N+1 size of main scale division =aLet size of vernier scale division be b then we haveaN=b(N+1)⇒b=N+1aN​Least count is a−b=a−N+1aN​ =a[N+1N+1−N​]=N+1a​

Question 13

The diameter of a cylinder is measured using a Vernier callipers with no zero error. It is found that, the zero of the Vernier scale lies between 6.10 cm and 5.95 cm of the main scale and there is no other main scale division in between there two. The Vernier scale has 50 division equivalent to 1.75 cm. The 25th division of the Vernier scale exactly coincides with one of the main scale division, the diameter of the cylinder is

Accepted Answer

Given, 50VSD=1.75 cm⇒∴ Least count of Vernier ​1VSD=501.75​=0.035 cm1MSD=6.10−5.95=0.15 cm=1MSD−1VSD=0.15−0.035=0.115 cm​The diameter of cylinder​=MSR+(VSR×LC)=5.95+(25×0.115)=8.825 cm​

Question 14

The main scale of a Vernier calipers has n divisions /cm.n divisions of the Vernier scale coincide with (n−1) divisions of the main scale. The least count of the Vernier calipers is

Accepted Answer

n(VSD)=(n−1)MSD⇒1VSD=n(n−1)​MSDLeast count =1MSD−1VSD=[1−n(n−1)​]MSD=n1​MSD=n1​(n1​)cm=n21​cm

Question 15

The figure below shows the final reading on a Vernier caliper with a zero error of −0.3 mm. What is the actual measurement?

Accepted Answer

Negative zего еггог =−0.3 mmMain scale reading =10 cmVernier scale reading =2×0.01=0.02 cmFinal reading =10+0.02−0.03=9.99 cm

Question 16

A student measured the length of a rod and wrote it as 3.50 cm. Which instrument did he use to measure it?(a) A meter scale(b) A vernier caliper 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

Accepted Answer

If student measure 3.50 cm, it means that there in an uncertainly of order 0.01 cm.For vernier scale with 1MSD=1 mm and 9MSD =10VSD∴ LC of VC=1MSD−1VSD=101​(1−109​)=1001​ cm

Question 17

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 8th  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 0.004 cm , then the radius of the bob is ______ ×10−2 cm.

Accepted Answer

​9MSD=10VSD1VSD=0.9 mm1MSD=1 mm∴ L.C. =MSDMSD−VSD​=11−0.9​=0.1​ Reading = Main scale reading + L.C. × vernier scale reading= 10 + 0.1 × 8= 10.8Reading of diameter = 10.8 mm – 0.04 cm= 10.8 mm – 0.4 mm= 10.4 mmradius of the bob = 5.2 mm= 0.52 cm =52×10−2 cm

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Solution for Class 11, Physics, Units and Measurements

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