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Question 11 Mark
The unit of length convenient on the atomic scale is known as an angstrom and is denoted by $\mathring{\text{A}}:1\mathring{\text{A}}=10^{-10}\text{m}.$ The size of a hydrogen atom is about $0.5\mathring{\text{A}}.$ What is the total atomic volume in m3 of a mole of hydrogen atoms?
Answer
Radius of hydrogen atom, $\text{r}=0.5\mathring{\text{A}}=0.5\times10-10\text{m}$

Volume of hydrogen atom $=4/3\pi\text{r}^3$

= 4/3 × 22/7 × (0.5 × 10-10)3

= 0.524 × 10-30m3

1 mole of hydrogen contains 6.023 × 1023 hydrogen atoms.

$\therefore$ Volume of 1 mole of hydrogen atoms = 6.023 × 1023 × 0.524 × 10-30

= 3.16 × 10-7m3

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Question 21 Mark
Which of the following is the most precise device for measuring length:
an optical instrument that can measure length to within a wavelength of light?
Answer
Wavelength of light, $\lambda\approx10^{-5}\text{cm}=0.00001\text{cm}$
Hence, it can be inferred that an optical instrument is the most suitable device to measure length.
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Question 31 Mark
Fill in the blanks:
The surface area of a solid cylinder of radius 2.0cm and height 10.0cm is equal to ...(mm)2
Answer
The surface area of a solid cylinder of radius 2.0cm and height 10.0cm is equal to 1.5 × 104mm2

Given,

Radius, r = 2.0cm = 20mm (convert cm to mm)

Height, h = 10.0cm =100mm

The formula of total surface area of a cylinder $\text{S}=2\pi\text{r}(\text{r}+\text{h})$

Putting the values in this formula, we get

Surface area of a cylinder $\text{S} = 2\pi\text{r} (\text{r} + \text{h}) = 2\times3.14 \times 20 (20+100)$

$= 15072 = 1.5\times104\text{mm}^2$

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Question 41 Mark
Fill in the blanks:
The volume of a cube of side 1cm is equal to .....m3.
Answer
The volume of a cube of side 1cm is equal to 10-6 m3.

Length of edge $ = 1\text{cm} = \frac{1}{100\text{m}} $

Volume of the cube = side3

Putting the value of side, we get

Volume of the cube $=\Big(\frac{1}{100\text{m}}\Big)^3$

$\Big(\frac{1}{100}\Big)\Big(\frac{1}{100}\Big)\Big(\frac{1}{100}\Big)\text{m}^3=\frac{1}{6^6}\text{m}^3=10^{-6}\text{m}^3$

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Question 51 Mark
Which of the following is the most precise device for measuring length:
a screw gauge of pitch 1mm and 100 divisions on the circular scale.
Answer
Least count of screw gauge = Pitch/Number of divisions = 1/1000 = 0.001 cm.
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Question 61 Mark
Fill in the blanks by suitable conversion of units:
3.0m s-2 = .... km h-2
Answer
3.0m s–2 = 3.88 × 104 km h–2
1 hour = 3600 sec so that 1 sec = 1/3600 hour
1km = 1000m so that 1m = 1/1000km
3.0m s–2 = 3.0 (1/1000km)( 1/3600 hour)-2 = 3.0 × 10–3km × ((1/3600)-2 h–2)
= 3.0 × 10–3km × (3600)2 h–2 = 3.88 × 104 km h–2
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Question 71 Mark
Fill in the blanks by suitable conversion of units:
1m = ..... ly
Answer
1m = 1/9.46 × 1015 ly = 1.06 × 10-16 ly
Distance = Speed × Time
Speed of light = 3 × 108 m/s
Time = 1 year = 365 days = 365 × 24 hours = 365 × 24 × 60 × 60 sec
Putting these values in above formula we get
1 light year distance = (3 × 108 m/s) × (365 × 24 × 60 × 60 s) = 9.46 × 1015 m
9.46 × 1015 m = 1 ly
So that 1m = 1/9.46 × 1015 ly = 1.06 × 10-16 ly
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Question 81 Mark
Fill in the blanks:
The relative density of lead is 11.3. Its density is ....g cm-3 or ....kg m-3.
Answer
The relative density of lead is 11.3. Its density is 11.3 g cm-3 or 1.13 × 103kg m–3.
Density of lead = Relative density of lead × Density of water
Density of water = 1g/ cm3
Putting the values, we get
Density of lead = 11.3 × 1g/ cm3
= 11.3g cm-3
1cm = (1/100m) =10–2m3
1g = 1/1000kg = 10-3kg
Density of lead = 11.3g cm-3 = 11.3
Putting the value of 1cm and 1 gram
11.3g/ cm3 = 11.3 × 10-3kg (10-2m)-3 = 11.3 × 10–3 × 106kg m-3 =1.13 × 103kg m–3
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Question 91 Mark
State the number of significant figures in the following:
2.64 × 1024kg
Answer
 3
Explanation:
The given quantity is 2.64 × 1024kg.
Here, the power of 10 is irrelevant for the determination of significant figures. Hence, all digits i.e., 2, 6 and 4 are significant figures
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Question 101 Mark
Fill in the blanks:
A vehicle moving with a speed of 18km h-1 covers ....m in 1 s
Answer
A vehicle moving with a speed of 18km h–1covers 5m in 1 s.
Using the conversion,
Given,
Time, t = 1 sec
speed = 18km h-1 = 18km/hour
1km = 1000m and 1hour = 3600 sec
Speed = 18 × 1000/3600 sec = 5m/sec
Use formula
Speed = distance/time
Cross multiply it, we get
Distance = Speed × Time = 5 × 1 = 5m
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Question 111 Mark
State the number of significant figures in the following:

6.032 N m-2

Answer
4
Explanation:
The given quantity is 6.032 N m-2.
All zeroes between two non-zero digits are always significant.
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Question 121 Mark
Explain this statement clearly:
“To call a dimensional quantity ‘large’ or ‘small’ is meaningless without specifying a standard for comparison”. In view of this, reframe the following statements wherever necessary:
  1. Atoms are very small objects
  2. A jet plane moves with great speed
  3. The mass of Jupiter is very large
  4. The air inside this room contains a large number of molecules
  5. A proton is much more massive than an electron
  6. The speed of sound is much smaller than the speed of light.
Answer
The given statement is true because a dimensionless quantity may be large or small in comparision to some standard reference. For example, the coefficient of friction is dimensionless. The coefficient of sliding friction is greater than the coefficient of rolling friction, but less than static friction.
  1. An atom is a very small object in comparison to a soccer ball.
  2. A jet plane moves with a speed greater than that of a bicycle.
  3. Mass of Jupiter is very large as compared to the mass of a cricket ball.
  4. The air inside this room contains a large number of molecules as compared to that present in a geometry box.
  5. A proton is more massive than an electron.
  6. Speed of sound is less than the speed of light.
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Question 131 Mark
A LASER is a source of very intense, monochromatic, and unidirectional beam of light. These properties of a laser light can be exploited to measure long distances. The distance of the Moon from the Earth has been already determined very precisely using a laser as a source of light. A laser light beamed at the Moon takes 2.56s to return after reflection at the Moon’s surface. How much is the radius of the lunar orbit around the Earth?
Answer
Time taken by the laser beam to return to Earth after reflection from the Moon = 2.56s
Speed of light = 3 × 108m/s
Time taken by the laser beam to reach Moon = 1/2 × 2.56 = 1.28s
Radius of the lunar orbit = Distance between the Earth and the Moon = 1.28 × 3 × 108 = 3.84 × 108m = 3.84 × 105km
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Question 141 Mark
Fill in the blanks by suitable conversion of units:
1kg m2 s-2 = ....g cm2 s-2
Answer
1kg m2s–2107g m2 s–2
1kg = 103g
1m2 = 104cm2
1kg m2 s–2 = 1kg × 1m2 × 1s–2
= 103g × 104cm× 1s–2 = 107 g cm2 s–2
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Question 151 Mark
State the number of significant figures in the following:
0.007m2
Answer
1
Explanation:
The given quantity is 0.007m2.
If the number is less than one, then all zeros on the right of the decimal point (but left to the first non-zero) are insignificant. This means that here, two zeros after the decimal are not significant. Hence, only 7 is a significant figure in this quantity.
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Question 161 Mark
State the number of significant figures in the following:
0.0006032m2
Answer
4

Explanation:

The given quantity is 0.0006032m2.

If the number is less than one, then the zeroes on the right of the decimal point (but left to the first non-zero) are insignificant. Hence, all three zeroes appearing before 6 are not significant figures. All zeros between two non-zero digits are always significant. Hence, the remaining four digits are significant figures.

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Question 171 Mark
Fill in the blanks by suitable conversion of units:
G = 6.67 × 10-11 N m2 (kg)-2 = .... (cm)3 s-2 g-1
Answer
G = 6.67 × 10–11 N m2 (kg)–2 = 6.67 × 10–8 (cm)3s–2 g–1.

Given,

G = 6.67 × 10–11 N m2 (kg)–2

We know that

1N = 1kg m s-2

1kg = 103g

1m = 100cm = 102cm

Putting above values, we get

 6.67 × 10–11 N m2kg–2 = 6.67 × 10–11 × (1kg m s–2) (1m2) (1Kg–2)

Solve and cancel out the units we get

⇒ 6.67 × 10–11 × (1kg–1 × 1m3 × 1s–2)

Putting above values to convert Kg to g and m to cm

⇒ 6.67 × 10–11 × (103g)-1 × (102cm)3 × (1s–2)

⇒ 6.67 × 10–11 × 10-3g-1 × 106cm3 × (1s–2)

⇒ 6.67 × 10–8cm3 s–2 g–1

G = 6.67 × 10–11 N m2 (kg)–2 = 6.67 × 10–8 (cm)3s–2 g–1.

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Question 181 Mark
State the number of significant figures in the following:

0.2370g cm-3

Answer
4

Explanation:

The given quantity is 0.2370g cm-3.

For a number with decimals, the trailing zeroes are significant. Hence, besides digits 2, 3 and 7, 0 that appears after the decimal point is also a significant figure.

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Question 191 Mark
State the number of significant figures in the following:
6.320 J
Answer
4
Explanation:
The given quantity is 6.320 J.
For a number with decimals, the trailing zeroes are significant. Hence, all four digits appearing in the given quantity are significant figures.
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Question 201 Mark
Which of the following is the most precise device for measuring length:
a vernier callipers with 20 divisions on the sliding scale.
Answer
Least count of this vernier callipers = 1SD - 1 VD = 1 SD - 19/20 SD = 1/20 SD
= 1.20mm = 1/200cm = 0.005cm
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Question 211 Mark
Let us consider an equation
$
\frac{1}{2} m v^2=m g h
$
where $m$ is the mass of the body, $v$ its velocity. $g$ is the acceleration due to gravity and $h$ is the helght. Check whether this equation is dimensionally correct.
Answer
The dimensions of LHS are
$
\begin{array}{c}
{[ M ]\left[ L T ^{-1}\right]^2=[ M ]\left[ L ^2 T ^{-2}\right]} \\
=\left[ M L ^2 T ^{-2}\right]
\end{array}
$
The dimensions of RHS are
$
\begin{aligned}
{[ M ]\left[ L T ^{-2}\right][ L ] } & =[ M ]\left[ L ^2 T ^{-2}\right] \\
& =\left[ M L ^2 T ^{-2}\right]
\end{aligned}
$
The dimensions of LHS and RHS are the same and hence the equation is dimensionally correct.
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Question 221 Mark
5.74 g of a substance occupies $1.2 cm ^3$. Express its density by keeping the significant figures in view.
Answer
There are 3 significant figures in the measured mass whereas there are only 2 significant figures in the measured volume. Hence the density should be expressed to only 2 significant figures.
$
\begin{aligned}
\text { Density } & =\frac{5.74}{1.2} g cm ^{-3} \\
& =4.8 g cm ^{-3} .
\end{aligned}
$
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Question 231 Mark
Each side of a cube is measured to be 7.203 m. What are the total surface area and the volume of the cube to appropriate significant figures?
Answer
The number of significant figures in the measured length is 4 . The calculated area and the volume should therefore be rounded off to 4 significant figures.
$
\begin{aligned}
\text { Surface area of the cube } & =6(7.203)^2 m ^2 \\
& =311.299254 m ^2 \\
& =311.3 m ^2 \\
& =(7.203)^3 m ^3 \\
& =373.714754 m ^3 \\
& =373.7 m ^3
\end{aligned}
$
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Question 241 Mark
Write the dimensional formula corresponding to :
  1. Photon.
  2. Calorie.
Answer
  1. Photon - [ML2T-2]
  2. Calorie - [ML2T-2]
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Question 271 Mark
Are inertial and gravitational mass of a body different from one another?
Answer
No, the inertial and gravitational mass of a body are equivalent.
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Question 281 Mark
What is the difference between nm, mN and Nm?
Answer
nm stands for nanometre, 1nm = 10-9m, mN stands for milli-newton,
1 mN = 10-3N, Nm stands for newton metre.
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Question 301 Mark
Calculate the length of the arc of a circle of radius 31.0cm which subtends an angle of at the centre.
Answer
Hence, length of the arc = ?

Radius $31.0\text{cm},\theta=\frac{\pi}{6}$

From length of the arc of a circle $\text{l}=\text{r}\theta$

$=31.0\times\frac{\pi}{6}$

$=16.2\text{cm}$

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Question 311 Mark
Write three pairs of physical quantities, which have same dimensional formula.
Answer
  1. Work and energy.
  2. Energy and torque.
  3. Pressure and stress.
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Question 321 Mark
What is the difference between 4.0 and 4.000?
Answer
4.0 has two significant figures whereas 4.000 has four significant figures.
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Question 331 Mark
Write the two physical quantities whose dimensions are same.
Answer
Work and torque have the same dimension [ML2T-2].
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Question 361 Mark
Write two advantages in choosing the wavelength of a light radiation as a standard of length.
Answer
  1. It does not undergo any change with different places.
  2. The wavelength of the light is not affected by time and environment.
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Question 371 Mark
Name at least seven physical quantities whose dimensions are ML2T-2.
Answer
  1. Pressure energy.
  2. Potential energy.
  3. Kinetic energy.
  4. Work.
  5. Torque.
  6. Moment of force.
  7. Couple.
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Question 381 Mark
Force (F) and density (d) are related as:
$\text{F}=\frac{\alpha}{\beta+\sqrt{\text{d}}}$
  1. Then the dimensions of $\alpha$ are.
  2. Then the dimensions of $\beta$ are.
Answer
  1. $[\text{M}^{\frac{3}{2}}\text{L}^{-1}\text{T}^{-2}]$

  2. $[\text{M}^{\frac{1}{2}}\text{L}^{\frac{-3}{2}}\text{T}^{0}]$

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Question 391 Mark
Which of the following set have different dimensions?
  1. Pressure, Young's modulus, Stress.
  2. Emf, potential difference, Electric potential.
  3. Heat, Work done, Energy.
  4. Dipole moment, Electric flux, Electric field.
Answer
  1. Dipole moment, Electric flux, Electric field.

Explanation:

$\text{p} = \text{q} \times 2\text{a} = (\text{AT}) \times \text{L} = [\text{M}^{0}\text{L}^1\text{T}^1\text{A}^1] $

Dimensions of elecric field

$\text{e}=\frac{\text{F}}{\text{q}}=\frac{\text{MLT}^{-2}}{\text{AT}}=[\text{MLT}^{-3}\text{A}^{-1}]$

$\phi=\vec{\text{E}}.\vec{\text{A}}=\text{EA}\cos\theta$

$\frac{​​\text{F}}{\text{q}}\text{A}\cos\theta=\Big[\frac{\text{MLT}^{-2}}{\text{AT}}\times\text{L}^2\times1\Big]$

$=[\text{M}^{1}\text{L}^3\text{T}^{-3}\text{A}^{-1}]$

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Question 401 Mark
Name two quantities with:
  1. Same dimensions.
  2. Constant value having dimension.
Answer
  1. Work and torque.
  2. Gravitational constant G and Planck's constant h.
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Question 411 Mark
The pairs of physical quantities that have the same dimensions are:
  1. Reynolds number and coefficient of friction.
  2. Latent heat and gravitational potential.
  3. Curie and frequency of light wave.
  4. Planck's constant and torque.
Answer
  1. Reynolds number and coefficient of friction.
  2. Latent heat and gravitational potential.
  3. Curie and frequency of light wave.

Explanation:

  1. Reynolds number and coefficient of friction, both are dimensionless.

  2. $\text{L}=\frac{\text{Q}}{\text{m}}=\frac{\text{ML}^2\text{T}^{-2}}{\text{M}}=[\text{L}^{2}\text{T}^{-2}]$

Gravitational Potential $=\frac{​​\text{W}}{\text{m}}=\frac{\text{ML}^{2}\text{T}^{-2}}{\text{M}}=[\text{L}^2\text{T}^{-2}]$

  1. 1 curie $=3.7\times10^{10}$ disintegrations/ sec =

$\text{T}^{-1}$= Frequency $=\text{T}^{-1}$

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Question 431 Mark
Name two pairs of physical quantities whose dimensions are same.
Answer
  • Stress and Young's modulus.
  • Work and Energy.
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Question 441 Mark
Define parsec.
Answer
The distance at which a star would have annual parallax of 1 second of arc.
1 parsec = 3.08 × 1016m.
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Question 451 Mark
Pressure is defined as:
  1. Momentum per unit area.
  2. Momentum per unit area per unit time.
  3. Momentum per unit volume.
  4. Energy per unit volume.
Answer
  1. Momentum per unit area per unit time.
  1. Energy per unit volume.

Explanation:

  1. Momentum per unit area per unit time

$=\frac{\text{Momentum}}{\text{Area}\times\text{Time}}\frac{\text{MLT}^{-1}}{\text{L}^{2}\text{T}}=[\text{M}^{1}\text{L}^{-2}\text{T}^{-2}]$

  1. Energy per unit volume. $=\frac{\text{ML}^2\text{T}^{-2}}{\text{T}^3=[\text{M}^1\text{L}^{-1}\text{T}^{-2}]}$
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Question 461 Mark
Name at least six physical quantities whose dimensions are ML2T-2.
Answer
  1. Work.
  2. Torque.
  3. Moment of force.
  4. Couple.
  5. Potential energy.
  6. Kinetic energy.
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Question 471 Mark
Write the order of following length in metres:
  1. Radius of earth.
  2. The height of average man.
  3. Thickness of sheet of paper.
  4. The radius of hydrogen atom.
Answer
  1. 6.4 × 106m
  2. 1.8 × 100m
  3. 1 × 10-4m
  4. 5 × 10-11m
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Question 491 Mark
A screw gauge has a pitch of 1.0mm and 200 division on the circular scale. Is is possible to increase the accuracy of the screw gauge by increasing the number of divisions on the circular scale?
Answer
No.
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Question 501 Mark
What do you mean by order of magnitude of a length? Write the order of magnitude of following lenghts.
  1. Size of a hydrogen atom.
  2. Diameter of earth.
  3. Light year.
  4. Size of milky way.
Answer
  1. 10-10m
  2. 107m
  3. 1016m
  4. 1021m
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1 Marks Question - Physics STD 11 Science Questions - Vidyadip