IIT JEE past Question vernier calliper

The side of a cube is measured by vernier callipers. (10 divisions of a vernier scale coincide with 9 divisions on main scale where 1 division on main scale is 1 mm.)

The main scale reads 10 mm. first division of the vernier scale coincides with the main scale.

mass of cube is 2.736 g.

Find the density of cube in appropriate significant figures.

9 Answers

106
Asish Mahapatra ·

LC = 0.1mm

Side of cube = 10+1*0.1 = 10.1mm = 1.01 cm
volume of cube = 1.03 cm3

So, density = 2.736/1.03 = 2.66 gm/cm3

1
ANKIT MAHATO ·

reading is 10.1 mm = 1.01 cm
Volume = (1.01)3 = 1.03 cm3 (the answer should have same no. of sig. fig as each individual units ).
Density = 2.736 g/1.03 cm3 = 2.6563 g / cm3

Density = 2.66 g / cm3 (roundoff to 2 sig fig .. because the no. of sig fig is same as in the no. having the smallest no . of sig fig)

24
eureka123 ·

how did u get """LC = 0.1mm"""????

21
tapanmast Vora ·

same doubt........ i thot it was sumthin lyk 0.9 [11]

1
vector ·

LC=1main scale reading-1v.scale reading i v s d=.9 msd

21
tapanmast Vora ·

sum1 tell yaar !!! cmon!!!!!

how is LC = 0.1mm [11]

106
Asish Mahapatra ·

1 MSD = 1mm
9 MSD = 9mm
10 VSD = 9mm
1 VSD = 0.9mm

LC = 1 MSD - 1 VSD = 0.1 mm

21
tapanmast Vora ·

LC = 1 MSD - 1 VSD ??

is this by definition ??

21
tapanmast Vora ·

Ya... thats true...[1]... chkd out in Google.........

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