1LC =( No. of VSDs - corresponding no. of MSDs )*1 MSD/no. of VSDs= ((n-2) - n)/n-2 mm = 1 mm
-2/n-2 mm = 1 mm
... n = 0
ghatak ..concept sikhaye hai .. bhaiya .. . :)
n main scale divisions corespond to n-2 vernier scale divisions
1 main scale division is 1 mm.
if the least count of the vernier is .001 m find n
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17 Answers
62haha :D
so why this answer??
someone explain?
what if the question was
least count is 0.00001
1
62But then, the reason for you all not to solve this is that you dont think in terms of concepts...
Think conceptually this has an answer too !!
62ok ok.. enuf bullying :P :D
I was trying to make questions myself.. you know what happens :P
13how can both main scale division and least count be same?[7]
1[5] ...yup...mujhe v kuch aisa hi lg raha....n -2 main scale divisions should corespond to n vernier scale divisions....
den ...
n= 2. [4]
1akand i have observed another thing .... this question is correct ..
1yup ankit........nice observation....................so QUESTION WRONG.............grace marks for everybody..............
or rather grace pinks for everybody hehehe
1not possible...
it should be the otherway round .. n-2 main scale div' coincides with n vernier scale divisions ... eg .. vernier calliper ko hi dekh lo ... [320]
I think i am correct because the least count bhaiya mentioned is equal to the 1 main scale division .. so remember that no. of div. coinciding with with MSD is more if both the main scale and vernier scale begin from 0 ..
