The diameter of a sphere is measured using a Vernier calliper whose 9 divisions of main scale are equal to 10 divisions of Vernier scale. The shortest division on the main scale is equal to 1 mm. The main scale reading is 2 cm and second division of Verni

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
The diameter of a sphere is measured using a Vernier calliper whose 9 divisions of main scale are equal to 10 divisions of Vernier scale. The shortest division on the main scale is equal to 1 mm. The main scale reading is 2 cm and second division of Vernier scale coincides with a division on main scale. If mass of the sphere is 8.635 g, the density of the sphere is [2024]

1 $2.5$ $g / c m^{3}$

2 $1.7$ $g / c m^{3}$

3 $2.0$ $g / c m^{3}$

4 $2.2$ $g / c m^{3}$

Ans.
(3)

$Given : 9$ $M S D = 10$ $V S D, m = 8.635$ $g$

$L C = 1$ $M S D - V S D$

$L C = 1$ $M S D - \frac{9}{10} M S D = \frac{1}{10} M S D \Rightarrow LC = 0.01$ $cm$

$Reading of diameter = M S R + L C \times V S R$

$= 2 cm + (0.01) \times (2) = 2.02$ $c m$

$Volume of sphere = \frac{4}{3} π {(\frac{d}{2})}^{3} = \frac{4}{3} π {(\frac{2.02}{2})}^{3} = 4.32 {cm}^{3}$

$Density = \frac{mass}{volume} = \frac{8.635}{4.32} = 1.998 ~ 2.00$ ${g / cm}^{3}$

Want Us to Email you About Special Offers & Updates?