Question

(V
)
IU
Type III.
4. Diameter of a wire on a computer chip is 0.000003m and thickness of a piece of paper is 0.0016 cm.
Find the sum of diameters of 1000 computer chips and thickness of 20 pieces of paper.
10.
Thdim​

Answer 1

Answer:

Total required sum of difference = 3.32 × 10^(-3) m

Diameter of a wire on a computer chip = 0.000003

= 3 × 10^(-6) m

No. of computer chips = 1000

So, Diameter of 1000 chips

= 3 × 10^(-6) × 1000 m

= 3 × 10^(-3) m

Thickness of a piece of paper

= 0.0016 cm

1 meter = 100 cm

So, thickness of a piece of paper is = 0.0016 × 10^(-2) m

= 1.6 × 10^(-5) m

No. of pieces of paper = 20

⇒ Total thickness

= 1.6 × 10^(-5) × 20

= 3.2 × 10^(-4) m

= 0.32 × 10^(-3) m

Total required sum of difference

= 3 × 10^(-3) + 0.32 × 10^(-3) m

= 3.32 × 10^(-3) m

hope it helps......