Units and Measurements - Online Test

Answer : Option B
Explaination / Solution:

There are three rules on determining how many significant figures are in a number:

• Non-zero digits are always significant.
• Any zeros between two significant digits are significant.
• A final zero or trailing zeros in the decimal portion ONLY are significant.

So keeping these rules in mind, there are 4 significant digit

Answer : Option A
Explaination / Solution:

There are three rules on determining how many significant figures are in a number:

• Non-zero digits are always significant.
• Any zeros between two significant digits are significant.
• A final zero or trailing zeros in the decimal portion ONLY are significant.

So keeping these rules in mind, there are 4 significant digit

Answer : Option D
Explaination / Solution:

Difference in masses = 20.17 – 20.15 = 0.02 g In subtraction, the final result should retain as many decimal places as there are in the number with the least decimal places.

Answer : Option A
Explaination / Solution:

length, l = 4.234 m

breadth,b = 1.005 m

thickness, t = 2.01 cm = 2.01 × 10-2 m Area of the sheet = 2 (l × b + b × t + t × l) = 2 (4.234 × 1.005 + 1.005 × 0.0201 + 0.0201 × 4.234) = 2 (4.3604739) = 8.7209478 m​​​​​​2 

As area can contain a maximum of three significant digits, therefore, rounding off, we get Area = 8.72 m​​​​​​2

Answer : Option C
Explaination / Solution:

length, l = 4.234 m

breadth,b = 1.005 m

thickness, t = 2.01 cm = 2.01 × 10-2 m

volume = l × b × t

=> V = 4.234 × 1.005 × 0.0201 = 0.0855289 = 0.0855 m​​​​​​3 (Significant Figures = 3)

Answer : Option A
Explaination / Solution:

Value of P is given as 3.763. By rounding off the given value to the first decimal place, we get P = 3.8.

Answer : Option D
Explaination / Solution:

Dimension of y = M​​​​​​0 L​​​​​​1 T0

Dimension of a = M0L​​1T​​​​​​0 

Dimension of sin2πtT$sin\frac{2\pi t}{T}$= M0L0T0$M^0{L}^0{T}^0$

Dimension of L.H.S = Dimension of R.H.S Hence, the given formula is dimensionally correct.

Answer : Option A
Explaination / Solution:

Radius of hydrogen atom, r = 0.5 Å = 0.5 × 10-10 m
Volume of hydrogen atom = (4/3) π r3
= (4/3) × (22/7) × (0.5 × 10-10)3
= 0.524 × 10-30 m3
1 mole of hydrogen contains 6.023 × 1023 hydrogen atoms.
∴ Volume of 1 mole of hydrogen atoms = 6.023 × 1023 × 0.524 × 10–30
= 3.16 × 10–7 m3

≈3×10−7m3

Answer : Option D
Explaination / Solution:

Radius of hydrogen atom, r = 0.5 = 0.5 × 10–10 m
Volume of hydrogen atom = 43×227×(0.5×10−10)3

= 0.524×10−30m3

Now, 1 mole of hydrogen contains 6.023 × 1023 hydrogen atoms.
Volume of 1 mole of hydrogen atoms, Va = 6.023 × 1023 × 0.524 × 10–30
= 3.16 × 10–7 m3
Molar volume of 1 mole of hydrogen atoms at STP,
Vm = 22.4 L = 22.4 × 10–3 m3

=> VmVa=22.4×10−33.16×10−7= 7.08 ×10−4

= ≅10−4

Answer : Option B
Explaination / Solution:

The horsepower (hp) is a unit in the foot-pound-second ( fps ) or English system, sometimes used to express the rate at which mechanical energy is expended. It was originally defined as 550 foot-pounds per second (ft-lb/s). A power level of 1 hp is approximately equivalent to 746 watt.