Least count of a standard vernier callipers is 0.01cm. When the two jaws are in contact, the 5^th division of vernier scale coincides with main scale division and the zero of the vernier scale lies to the left of the zero of the main scale. While measuring diameter of the sphere, the zero mark of vernier scale lies between 2.4cm and 2.5cm and 6^th vernier division coincides with a main scale division. The diameter of the sphere is $d=x \times 10^{- 2} cm$ then x is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

answer is 251.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$L e a s t c o u n t, L C = 0.01 c m H e r e t h e n e g a t i v e z e r o e r r o r o f c a l l i p e r s = 5 \times 0.01 = 0.05 c m a n d i t s h o u l d b e a d d e d t o f i n a l r e a d i n g . M a i n s c a l e r e a d i n g, M S R = 2.4 c m a n d V S D = 6 S o t o t a l r e a d i n g m e a n s t h e d i a m e t e r o f s p h e r e R = M S R + (V S D \times L C) + z e r o e r r o r = 2.4 + (6 \times 0.01) + 0.05 = 2.51 c m$