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+1 vote
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in Physics by (44.7k points)

The edge of a cube is measured using a vernier caliper (9 divisions of the main scale is equal to 10 divisions of vernier scale and 1 main scale division is 1mm). The main scale division reading is 10 and 1 division of vernier scale was found to be coinciding with the main scale. The mass of the cube is 2.736 g. What will be the density in {g / (cm3)} upto correct significant figures ? 

(A) 2.66 × 10–3 {g / (cm3)} 

(B) 2.66 × 103 {g / (cm3)} 

(C) 2.66 {g / (cm3)} 

(D) 2.66 × 10–6 {g / (cm3)}

1 Answer

+1 vote
by (69.7k points)
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Best answer

The correct option is (C) 2.66 {g / (cm3)}.

Explanation:

9 division of main scale = 10 division of vernier scale 

9 M = 10 V 

Also 1 MSD = 1 mm 

We know Least count = M – V 

= M – (9/10) M 

= (1 / 10) M 

= (1 / 10) × 1 mm 

= (1 / 10) mm 

Measured reading of edge = MSR + VSR (LC) 

= 10 + {1 × (1 / 10)} 

= 10 + 0.1 

= 10.1 mm 

Volume of cube = (side)3 

∴ V = (10.1)3 mm3 

= (1.01)3 cm3 

= 1.030 cm3 

∴ As edge length is measured up to 3 significant digits, after rounding off up to 3 significant digits. 

V = 1.03 cm3 

Density = ρ = {(mass) / (volume)} 

Hence ρ = {(2.736) / (1.03)} {g / (cm3)} 

= 2.6563 = 2.66 g/cm3 

∴ ρ = 2.66 g/cm3 (after rounding off to 3 significant digits)

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