JEE Advanced 2015Paper 1
- Exam
- JEE Advanced
- Year
- 2015
- Paper
- Paper 1
- Subject
- Physics
- Chapter
- Experimental Physics
JEE Advanced 2015 · Paper 1 · Physics · Experimental Physics
Consider a Vernier callipers in which each 1 cm on the main scale...
Consider a Vernier callipers in which each 1 cm on the main scale is divided into 8 equal divisions and a screw gauge with 100 divisions on its circular scale. In the Vernier callipers, 5 divisions of the Vernier scale coincide with 4 divisions on the main scale and in the screw gauge, one complete rotation of the circular scale moves it by two divisions on the linear scale. Then:
Options
- AIf the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.01 mm
- BIf the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.005 mm
- CIf the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.01 mm
- DIf the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.005 mm
Correct answer
B. If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.005 mm
Step-by-step solution
1 main scale division (M.S.D)
5 Vernier scale division (V.S.D)
Least count of Vernier scale \left(L . C .\right) = 1M . S . D . - 1V . S . D .
\dfrac{4}{5}
\left(L . C\right) = \dfrac{1M . S . D}{5} = \dfrac{1}{40}cm
For option A of Vernier scale
=\dfrac{1}{20}cm= \dfrac{Pitch}{100} = \dfrac{0.50}{100}
For option C and D
Least count of linear scale of screw gauge
\dfrac{1}{40} = \dfrac{1}{20}cm= 2 \times \dfrac{1}{20} = \dfrac{1}{10}cm
Least count of screw gauge =\dfrac{1mm}{100} =0.01mm
Hence answer is (B, C)
5 Vernier scale division (V.S.D)
Least count of Vernier scale \left(L . C .\right) = 1M . S . D . - 1V . S . D .
\dfrac{4}{5}
\left(L . C\right) = \dfrac{1M . S . D}{5} = \dfrac{1}{40}cm
For option A of Vernier scale
=\dfrac{1}{20}cm= \dfrac{Pitch}{100} = \dfrac{0.50}{100}
For option C and D
Least count of linear scale of screw gauge
\dfrac{1}{40} = \dfrac{1}{20}cm= 2 \times \dfrac{1}{20} = \dfrac{1}{10}cm
Least count of screw gauge =\dfrac{1mm}{100} =0.01mm
Hence answer is (B, C)