NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI02:UNITS AND MEASUREMENTS
367550
One centimeter on the main scale of vernier callipers is divided into ten equal parts. If 20 divisions of vernier scale coincide with 8 small divisions of the main scale. What will be the least count of callipers ?
367551
One centimetre on the main scale of vernier callipers is divided into ten equal parts. If 10 divisions of the vernier scale coincide with 8 small divisions of the main scale, the least count of the callipers is
1 \(0.01\,cm\)
2 \(0.02\,cm\)
3 \(0.05\,cm\)
4 \(0.005\,cm\)
Explanation:
Least count \( = \,1\,M.S.D - 1\,V.S.D\) From the given information \(8\,M.S.D = 10\,V.S.D\) \( \Rightarrow 1\,V.S.D = \frac{8}{{10}}M.S.D\) \( \Rightarrow LC = M.S.D\left( {1 - \frac{8}{{10}}} \right)\) \( = \left( {1 - \frac{8}{{10}}} \right) = \frac{2}{{10}}mm = 0.02\,cm\)
PHXI02:UNITS AND MEASUREMENTS
367552
The main scale of vernier calliper is divided into \(0.5\,mm\) and its least count is \(0.005\,cm\). The number of vernier divisions are
1 10
2 20
3 30
4 40
Explanation:
Given, L.C. \({=\dfrac{S}{n}}\) \({\therefore}\) No. of divisions on main scale \({(n)}\) \({=\dfrac{S}{\text { L.C. }}=\dfrac{0.5 {~mm}}{0.05 {~mm}}=10}\)
367550
One centimeter on the main scale of vernier callipers is divided into ten equal parts. If 20 divisions of vernier scale coincide with 8 small divisions of the main scale. What will be the least count of callipers ?
367551
One centimetre on the main scale of vernier callipers is divided into ten equal parts. If 10 divisions of the vernier scale coincide with 8 small divisions of the main scale, the least count of the callipers is
1 \(0.01\,cm\)
2 \(0.02\,cm\)
3 \(0.05\,cm\)
4 \(0.005\,cm\)
Explanation:
Least count \( = \,1\,M.S.D - 1\,V.S.D\) From the given information \(8\,M.S.D = 10\,V.S.D\) \( \Rightarrow 1\,V.S.D = \frac{8}{{10}}M.S.D\) \( \Rightarrow LC = M.S.D\left( {1 - \frac{8}{{10}}} \right)\) \( = \left( {1 - \frac{8}{{10}}} \right) = \frac{2}{{10}}mm = 0.02\,cm\)
PHXI02:UNITS AND MEASUREMENTS
367552
The main scale of vernier calliper is divided into \(0.5\,mm\) and its least count is \(0.005\,cm\). The number of vernier divisions are
1 10
2 20
3 30
4 40
Explanation:
Given, L.C. \({=\dfrac{S}{n}}\) \({\therefore}\) No. of divisions on main scale \({(n)}\) \({=\dfrac{S}{\text { L.C. }}=\dfrac{0.5 {~mm}}{0.05 {~mm}}=10}\)
367550
One centimeter on the main scale of vernier callipers is divided into ten equal parts. If 20 divisions of vernier scale coincide with 8 small divisions of the main scale. What will be the least count of callipers ?
367551
One centimetre on the main scale of vernier callipers is divided into ten equal parts. If 10 divisions of the vernier scale coincide with 8 small divisions of the main scale, the least count of the callipers is
1 \(0.01\,cm\)
2 \(0.02\,cm\)
3 \(0.05\,cm\)
4 \(0.005\,cm\)
Explanation:
Least count \( = \,1\,M.S.D - 1\,V.S.D\) From the given information \(8\,M.S.D = 10\,V.S.D\) \( \Rightarrow 1\,V.S.D = \frac{8}{{10}}M.S.D\) \( \Rightarrow LC = M.S.D\left( {1 - \frac{8}{{10}}} \right)\) \( = \left( {1 - \frac{8}{{10}}} \right) = \frac{2}{{10}}mm = 0.02\,cm\)
PHXI02:UNITS AND MEASUREMENTS
367552
The main scale of vernier calliper is divided into \(0.5\,mm\) and its least count is \(0.005\,cm\). The number of vernier divisions are
1 10
2 20
3 30
4 40
Explanation:
Given, L.C. \({=\dfrac{S}{n}}\) \({\therefore}\) No. of divisions on main scale \({(n)}\) \({=\dfrac{S}{\text { L.C. }}=\dfrac{0.5 {~mm}}{0.05 {~mm}}=10}\)