Units and Measurements Test

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Units and Measurements Test - 61

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Weekly Quiz Competition

Question 1
1 / -0

The mass of a box measured by a grocer's balance is 2.3 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. The total mass of the box is:

A
2.3 kg

B
2.34 kg

C
2.340 kg

D
2.3403 kg

Solution

Here, mass of the box, $$m = 2.3 kg$$
Mass of one gold piece, $$m_1 = 20.15 g = 0.02015 kg$$
Mass of other gold piece, $$m_2 = 20.17 g = 0.02017 kg$$
$$\therefore$$ Total mass = $$m + m_1 +m_2 = 2.3 kg + 0.02015 kg +0.02017 kg = 2.34032 kg$$
As the result is correct only upto one place of decimal, therefore, on rounding off, we get Total mass $$= 2.3 kg$$
Question 2
1 / -0

The numbers $$2.745$$ and $$2.735$$ on rounding off to $$3$$ significant figures will give:

A
$$2.75$$ and $$2.74$$

B
$$2.74$$ and $$2.73$$

C
$$2.75$$ and $$2.73$$

D
$$2.74$$ and $$2.74$$
Solution
Hint: While rounding off measurements, we use the following rules by convention:
- If the digit to be dropped is 5 or 5 followed by zeros, then the preceding digit is left unchanged if it is even.
- If the digit to be dropped is 5 or 5 followed by zeros, then the preceding digit is raised by one if it is odd.
Step 1: finding the rounded off digits
Let us round off $$2.745$$ to $$3$$ significant figures.
Here the digit to be dropped is $$5$$, then the preceding digit is left unchanged because it is even.
Hence on rounding off $$2.745$$, it would be $$2.74$$.

Now consider $$2.737$$, here also the digit to be dropped is $$5$$, then the preceding digit is raised by one if it is odd.
Hence on rounding off $$2.735$$ to 3 significant figures, it would be $$2.74$$.
Question 3
1 / -0

A cube has a side of length 1.2 x $$10^{-2}$$. Its volume upto correct significant figures is

A
1.7 x $$10^{-6}m^3$$

B
1.73 x $$10^{-6}m^3$$

C
1.78 x $$10^{-6}m^3$$

D
1.732 x $$10^{-6}m^3$$

Solution

Here
Length of the cube, $$L =12 x10^{-2} m$$
Volume of the cube, $$V= (1.2 \times10^{-2} m)^3= 1.728 \times 10^{-6}m^3$$
As the result can have only two significant figures, therefore, on rounding off, we get, $$V. 1.7 \times 10^{-6} m^3$$

Question 4

1 / -0

Match the Column I with Column II

	Column I (Units)		Column II (Dimensional formulae)
(A)	Pa s	(p)	$$[M^0L^2T^{-2}K^{-1}]$$
(B)	N m K$$^{-1}$$	(q)	$$[MLT^{-3}K^{-1}]$$
(C)	J kg$$^{-1}$$ K$$^{-1}$$	(r)	$$[ML^{-1}T^{-1}]$$
(D)	W m$$^{-1}$$ K$$^{-1}$$	(s)	$$[ML^2T^{-2}K^{-1}]$$

A - q, B - p, C - r, D - s

A - p, B - q, C - s, D - r

A - r, B - s, C - p, D - q

A - s, B - r, C - q, D - p

Solution

(A) $$\displaystyle Pa \,s = [ML^{-1}T^{-2}][T] = [ML^{-1}T^{-1}]$$
A - r
(B) $$\displaystyle N\,m\,k^{-1} = \frac{[MLT^{-2}][T]}{[K]} = [ML^2T^{-2} K^{-1}]$$
B - s
(C) $$\displaystyle J\,kg^{-1}\,k^{-1} = \frac{[ML^2T^{-2}]}{[M][K]} = [M^0L^2T^{-2} K^{-1}]$$
C - p
(D) $$\displaystyle W\,m^{-1}\,k^{-1} = \frac{[ML^2T^{-3}]}{[L][K]} = [MLT^{-3} K^{-1}]$$
D - q

Question 5
1 / -0

The radius of a sphere is 1.41 cm. Its volume to an appropriate number of significant figures is then

A
11.73 $$cm^3$$

B
11.736 $$cm^3$$

C
11.7 $$cm^3$$

D
117 $$cm^3$$

Solution

Radius of the sphere, $$r = 1.41 cm$$ ($$3$$ significant figures)
Volume of the sphere,
$$\displaystyle V = \dfrac {4}{3} \pi r^3 = \dfrac{4}{3} \times 3.14 \times (1.41)^3\,cm^3 = 11.736\, cm^3$$
Rounded off upto $$3$$ significant figures $$= 11.7 cm^3$$.
Question 6
1 / -0

If E, m, l and G denote energy, mass, angular momentum and gravitational constant respectively, then the quantity $$\displaystyle (\frac {El^2}{m^5G^2})$$ has the dimensions of then

A
mass

B
length

C
time

D
angle.

Solution

$$\displaystyle [E] = [ML^2T^{-2}], [m] $$

$$= [M][l] = [ML^2T^{-1}], [G] = [M^{-1}L^3T^{-2}]$$

$$\displaystyle \therefore \left [ \frac {El^2}{m^5G^2} \right ] = \frac {[ML^2T^{-2}][M^2L^4T^{-2}]}{[M^5][M^{-2}L^6T^{-4}]}$$

$$= [M^0L^0T^0]$$
As angle has no dimensions, therefore $$\displaystyle \frac {El^5}{m^5G^2}$$ has the same dimensions as that of angle.
Question 7
1 / -0

Which of the following pairs of physical quantities have same dimensions?

A
Force and power

B
Torque and energy

C
Torque and power

D
Force and torque

Solution

Torque is the moment of force given by:$$ \vec\tau=\vec r\times \vec F$$

It’s unit is Newton-meter.

The unit of work/energy is also Newton-meter or Joule.

Thus, both of these physical quantities have the same units/dimensions.

$$[Torque]=[M][L]^{2}[T]^{−2}=[Energy]$$
Question 8
1 / -0

The number of significant figure In the numbers $$4.8000 \times 10^4$$ and $$48000.50 $$ are respectively then

A
$$5$$ and $$6.$$

B
$$5$$ and $$ 7$$.

C
$$2$$ and $$7.$$

D
$$2$$ and $$6$$.

Solution

If any number has more than one digit after decimal place are significant. Power of 10 does not affect number of significant figures.
$$4.8000 \times 10^{4} = 5$$ significant figure $$(4,8,0,0,0)$$
$$48000.50 = 7$$ significant figure $$(4,8,0,0,0,5, 0)$$
Question 9
1 / -0

The numbers 3.845 and 3.835 on rounding off to 3 significant figures will give then

A
3.85 and 3.84

B
3.84 and 3.83

C
3.85 and 3.83

D
3.84 and 3.84

Solution

The number 3.845 rounded off to three significant figures becomes 3.84 since the preceding digit is even. On the other hand, the number 3.835 rounded off to three significant figures becomes 3.84 since the preceding digit is odd.
Question 10
1 / -0

The dimensions of Planck's constant are the same as that of the

A
linear impulse

B
work

C
linear momentum

D
angular

Solution

Energy of a photon, $$E = h\upsilon$$

where h is the Planck's constant and $$\upsilon$$ is the frequency.

$$\displaystyle \therefore [h] = \frac {[E]}{[v]} = \frac {[ML^2T^{-2}]}{[T^{-1}]}= [ML^2T^{-1}]$$

Angular momentum = Moment of inertia x Angular velocity (Angular momentum)
= $$\displaystyle [ML^2][T^{-1}] = [ML^2T^{-1}]$$

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Re-Attempt Weekly Quiz Competition

Units and Measurements Test - 61

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Units and Measurements Test - 61

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SHARING IS CARING

Question 1

2.3 kg

2.34 kg

2.340 kg

2.3403 kg

Solution

Question 2

$$2.75$$ and $$2.74$$

$$2.74$$ and $$2.73$$

$$2.75$$ and $$2.73$$

$$2.74$$ and $$2.74$$

Solution

Question 3

1.7 x $$10^{-6}m^3$$

1.73 x $$10^{-6}m^3$$

1.78 x $$10^{-6}m^3$$

1.732 x $$10^{-6}m^3$$

Solution

Question 4

A - q, B - p, C - r, D - s

A - p, B - q, C - s, D - r

A - r, B - s, C - p, D - q

A - s, B - r, C - q, D - p

Solution

Question 5

11.73 $$cm^3$$

11.736 $$cm^3$$

11.7 $$cm^3$$

117 $$cm^3$$

Solution

Question 6

mass

length

time

angle.

Solution

Question 7

Force and power

Torque and energy

Torque and power

Force and torque

Solution

Question 8

$$5$$ and $$6.$$

$$5$$ and $$ 7$$.

$$2$$ and $$7.$$

$$2$$ and $$6$$.

Solution

Question 9

3.85 and 3.84

3.84 and 3.83

3.85 and 3.83

3.84 and 3.84

Solution

Question 10

linear impulse

work

linear momentum

angular

Solution

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