Units and Measurements Online Objective Test | Units and Measurements Online Objective Test

# Units and Measurements # CBSE 11TH PHYSICS Prepare / Learn

Q1. Approximately How many light years is a meter?

A. 10−12

B. 10−11

C. 10−10

D. 10−16

Answer : Option D
Explaination / Solution:

A light-year is a unit of distance. It is the distance that light can travel in one year. Light moves at a velocity of about 300,000 kilometers (km) each second. So in one year, it can travel about 10 trillion km. More p recisely, one light-year is equal to

$9.5 \times 10^{15} m$

=> 1 Light Year = $9.5 \times 10^{15} m$

=> 1 m = $\frac{1}{9.5 \times 10^{15}} L i g h t y e a r$

$=> 1 m \approx 10^{- 16} L i g h t y e a r$

# Units and Measurements # CBSE 11TH PHYSICS Prepare / Learn

Q2. Approximately How many km

h^{- 2}

is 3 m

s^{- 2}

A. 3.9

\times

1 0^{3}

B. 3.9

\times

1 0^{3}

C. 3.5

\times

1 0^{4}

D. 3.6

\times

1 0^{4}

Answer : Option B
Explaination / Solution:

1 km h-2 = $\frac{1000 m}{3600 \times 3600 s^{2}} = \frac{1}{360 \times 36} m s^{- 2}$

=> $1 m s^{- 2} = 360 \times 36 k m h^{- 2}$

=> $3 m s^{- 2} = 3 \times 360 \times 36 k m h^{- 2}$

=> $3 m s^{- 2} = 38880 k m h^{- 2}$

$=> 3 m s^{- 2} \approx 3.9 \times 10^{4} k m h^{- 2}$

# Units and Measurements # CBSE 11TH PHYSICS Prepare / Learn

Q3. A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the Sun and the Earth in terms of the new unit if light takes 8 min and 20 s to cover this distance?

A. 550

B. 450

C. 600

D. 500

Answer : Option D
Explaination / Solution:

Distance between the Sun and the Earth = Speed of light ×Time taken by light to cover the distance Given that in the new unit, speed of light = 1 unit Time taken, t = 8 min 20 s = 500 s Distance between the Sun and the Earth = 1 x 500 = 500 units

# Units and Measurements # CBSE 11TH PHYSICS Prepare / Learn

Q4. Which of the following is the most precise device for measuring length?

A. an optical instrument that can measure length to within a wavelength of light

B. a screw gauge of pitch 1 mm and 100 divisions on the circular scale

C. a vernier calipers with 20 divisions on the sliding scale

D. a screw gauge of pitch 1 mm and 50 divisions on the circular scale

Answer : Option A
Explaination / Solution:

A device with minimum count is the most suitable to measure length.

Least count of vernier callipers = 1 standard division (SD) – 1 vernier division (VD) = $1 - \frac{19}{20} = \frac{1}{20} = 0.05$ cm

Least count of screw gauge = $\frac{P i t c h}{N u m b e r o f d i v i s i o n s}$ = $\frac{1}{1000}$ = 0.001cm

Least count of an optical device = Wavelength of light $\approx$ 10–5 cm = 0.00001 cm

Hence, it can be inferred that an optical instrument is the most suitable device to measure length.

# Units and Measurements # CBSE 11TH PHYSICS Prepare / Learn

Q5. A student measures the thickness of a human hair by looking at it through a microscope of magnification 100. He makes 20 observations and finds that the average width of the hair in the field of view of the microscope is 3.5 mm. What is the estimate on the thickness of hair ?

A. 0.025 mm

B. 0.035 mm

C. 0.040 mm

D. 0.030 mm

Answer : Option B
Explaination / Solution:

Magnification of microscope = 100 Observed width of the hair = 3.5 mm Estimate on the thickness of hair is given by

Magnification = $\frac{O b s e r v e d w i d t h}{R e a l w i d t h}$

=> Real width = $\frac{O b s e r v e d w i d t h}{M a g n i f i c a t i o n}$ = $\frac{3.5}{100}$ = 0.035 mm

# Units and Measurements # CBSE 11TH PHYSICS Prepare / Learn

Q6.

The photograph of a house occupies an area of 1.75 $c m^{2}$ on a 35 mm slide. The slide is projected on to a screen, and the area of the house on the screen is 1.55 $m^{2}$ . What is the linear magnification of the projector-screen arrangement?

A. 94.1

B. 88.4

C. 84.1

D. 90.2

Answer : Option A
Explaination / Solution:

Area of the house on the slide = 1.75 cm2

Area of the image of the house formed on the screen = 1.55 m2 = 1.55 × 104 cm2

Arial magnification am = $\frac{A r e a o f i m a g e}{a r e a o f o b j e c t}$

= $\frac{1.55 \times 10^{4}}{1.75}$

Linear magnification = $\sqrt{a_{m}} = \sqrt{8857}$

= 94.1

# Units and Measurements # CBSE 11TH PHYSICS Prepare / Learn

Q7. The number of significant digits in 0.007 is

A. 4

B. 1

C. 2

D. 3

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 1 significant digit.

# Units and Measurements # CBSE 11TH PHYSICS Prepare / Learn

Q8. The number of significant digits in 2.64 ×

1 0^{24}

A. 4

B. 1

C. 2

D. 3

Answer : Option D
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 3 significant digit.

# Units and Measurements # CBSE 11TH PHYSICS Prepare / Learn

Q9. The number of significant digits in 0.2370 is

A. 6

B. 3

C. 4

D. 5

Answer : Option C
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.

# Units and Measurements # CBSE 11TH PHYSICS Prepare / Learn

Q10. The number of significant digits in 6.320 J is

A. 5

B. 3

C. 4

D. 6

Answer : Option C
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.

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