Physical Quantities And Measurement

Unit.1 Physical Quantities and Measurements

1. Express the following quantities using prefixes.

(a) 5000g  (b) 2000 000 W (c)52×10^-10kg

(d) 225×10^-8s  

Solution

(a) As it is given that 5000g now we write it in a suitable prefix!

 = 5000g = 5×1000g = 5kg

(b) As it is given that 2000 000 W now we write it in a suitable prefix!

2000, 000W = 2×1000000W = 5MW

(c) As it is given that 52×10^-10kg now we write it in a suitable prefix!

= 52×10^-10kg = 5.2×10^-9×10³g = 5.2×10^-6g

                                               = 5.2µg

(d) As it is given that  225×10^-8s now we write it in a suitable prefix!

= 225×10^-8s =2.25×10²×10^-8s = 2.25×10^-6s

                                               = 2.25µs

2. How do the prefixes micro, nano and pico related to each other?

As we know that

                      1 micro = 10^-6 ,   1 nano = 10^-9, 1 pico = 10^-12

                      1 pico = 10^-6 micro

                      1 pico = 10^-3 nano

                 Hence

                      1 pico = 10^-6 micro = 10^-3 nano

3. Your hair grow at the rate of 1mm per day. Find their growth rate in nms^-1.

Hence the Growth rate will be 11.57nms^-1.

4. Rewrite the following in the standard form.

(a)1168×10^-27 (b) 32×10⁵  (c) 725×10^-5kg (d) 0.02×10^-8

Solution

Part -a

Part- b

Part- c

Part -d

5.Write the following quantities in standard form.

(a) 6400 km  (b) 380 000 km (c) 300 000 000 ms^-1

(d) Seconds in  a day?

Solution

5.Write the following quantities in standard form.

(a) 6400 km  (b) 380 000 km (c) 300 000 000 ms^-1

(d) Seconds in  a day?

Solution

Part-a                         6400 km 

6400km = 6400 × 1000/1000 = 6.4 × 10³

Part-b                         380 000km

380000km = 380000 × 100000/100000

                    = 3.8 × 10⁵km

Part-c                         300 000 000 ms^-1

300000000km = 6400 × 10000000/10000000

 = 6.4 × 10⁸ms^-1

Part-d                         Seconds in a day

= 24 × 60 × 60 = 86400s

= 86400s = 86400 × 10000/10000

= 8.64 × 10⁴s

6. On Closing the Jaws of Vernier Callipers, zero of the Vernier scale is on

the right to its main scale such that 4^th division of its Vernier scale coincides 

with one of the main scale division. Find its zero Error and Zero Correction? 

Solution

Zero Error

Conceding Division = 4^th

Zero error = 4 × 0.01cm = 0.04 cm

Zero Error = 0.04 cm

Zero Correction

As it is positive zero error so it will be subtracted from the total measurement.

Total measurement – 0.04 = Final measurement. 

7.A screw gauge has 50 division on its circular scale. The pitch of the screw gauge is 0.5mm. what is its least count?

Solution

Given Data

                 

                  No of divisions =50

                  Pitch of screw gauge = 0.5 mm

To Find

                   Least Count = L.C =?

Formula

                  L.C = Pitch/No.of divisions         

     

Calculation

                   As we know that

                             L.C = Pitch/No. of divisions

                   By putting the values

                             L.C = 0.5/50  = 0.01mm or 0.001cm                     

8. Which of the following quantities have three significant figures?

(a) 3.0066 m (b) 0.00309 kg (c) 5.05 × 10^-27kg

(d) 301.0s

                                     Solution

	Quantity	No of significant Digits
A	3.0066 m	5
B	0.00309 kg	3
C	5.05 × 10-27kg	3
D	301.0s	4

9. What are the significant figures In the following measurements.

(a) 1.009m (b)0.00450kg (c) 1.66×10^-27kg (d)2001s

                                     Solution

	Quantity	No of significant Digits
A	1.009m	4
B	0.00450kg	3
C	1.66×10-27kg	3
D	2001s	4

10. A chocolate wrapper is 6.7 cm long and 5.4 cm wide. Calculate its area up to reasonable number of significant figures.

Solution

Given Data

                  Length = 6.7 cm

                  Width = 5.4 cm

To find

                  Area =?

Formula

                 Area = Length ×Width

Calculation

                 Area = Length ×Width

                 Area = 6.7cm ×5.4cm

                 Area = 36.18

                 Area = 36.2 cm²