Scientific Notation and rounding off

(18) Value of the gravitational constant is correctly written as

(a) 0.0000000000667 Nm²kg^-2      (b) 667*10^-13 Nm²kg^-2      (c) both a and b                 (d) none of these

Solution: The standard method is, first, there must be only one non-zero digit before the decimal. So the decimal has to be placed at 6.67.  Second, if we move the decimal to the right, then 10 will be raised to the power equal to the number of digits moved through with a negative sign. Since we move the decimal to the right by 11 digits, therefore, power of 10 will be -11. The correct answer will be 6.67*10^-11 Nm²kg^-2.

(19) The scientific notation of a number 0.0023 is expressed as

(a) 2.3*10^-3                (b) 0.023*10^-2                 (c) 2.3*10^-4                    (d) 0.2*10^-3           

 (20) The scientific notation of the number 0.2300 is expressed as

(a) 2.3*10^-1                        (b) 0.23*10^-2                      (c) 2.3*10^-3                         (d) 23.00*10^-2

(21) A beaker contains 656 ml of water. What is the volume in liters?

(a) 6.56*10² liters                            (b) 0.656*10³ liters         

  (c) 6.56*10^-1 liters            (d)65.6*10^-3 liters

(22) 11.1 and 11.11 added and round off gives

(a) 22.21                             (b) 22.2                               (c) 22                                   (d) 22.22

(Hint) In addition and subtraction the result is limited by the measurement with most uncertainty.  Here 11.1 have more uncertainty than 11.11, so the result is rounded off to one decimal place.

(23) 22.1 – 20.09 when solved and rounded off gives

(a) 2.01                               (b) 2                                     (c) 2                                     (d) 2.0

Solution: 22.1 – 20.09 = 2.01

The result has to be rounded off to one decimal place because 22.1 have more uncertainty than 20.09.

 (24) The sum (5.3 +5.33+5.306) cm up to correct decimal place is

(a) 15.936                           (b) 15.9                               (c) 15                    (d) 15.93

(25) 50.5 * 12 when solved and rounded off gives

(a)606                                  (b) 610                                (c) 600                                 (d) 606.0

Hint: In multiplication and division the answer is limited by the measurement with the least number of significant figures. Here 12 has least number of sig figs, 2, therefore, the answer is limited to 2 sig figs.

(26) 100.50 was divided by 25 and the answer obtained was round off as

(a) 4                                     (b) 4.02                               (c) 4.0                                 (d) 4.1

Solution:   4.02 have 3 sig figs. When rounded to 2 figures, give 4.0

(27)If π=3.14. The length of the circumference of a circle of radius 0.1 m is

(a) 628 m                            (b) 6.3 m                             (c) 62.8 m            (d) 0.6 m

Hint: C =2πr

(28) The area of a square of side length 1.5 cm up to correct sig fig is

(a) 225 cm²                         (b) 22.5 cm²        (c) 2.2 cm²                                        (d) 2.3 cm²

(29) 12274 rounded off to three sig figs is

(a)122                                  (b) 122.74                           (c) 123                                 (d) 12300

(30) Rounded off 422.49325 to three digits is

(a) 422                                (b) 423                                 (c) 422.493                         (d) none of these

Sphere

(31) Volume of a sphere is given by

(a) 4/3 πr³           (2) 2πr                                 (c) 4πr²                                (d)4/3 πr²

(32) Surface area of a sphere is given by

(a) 4/3 πr³            (2) 2πr                                 (c) 4πr²                               (d) 4/3 πr²