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Physics 1 Marks Question

Units and Measurements · MP Board (MPBSE) STD 11 Science (Madhya Pradesh - English Medium). 149 questions.

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Question 11 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 21 Mark
Fill in the blanks: The volume of a cube of side 1cm is equal to .....$m^3$.
Answer
The volume of a cube of side 1cm is equal to $10^{-6} m^3$.
Length of edge $ = 1\text{cm} = \frac{1}{100\text{m}} $ Volume of the cube = $side^3$ 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 31 Mark
Fill in the blanks by suitable conversion of units: $3.0 \mathrm{~m} \mathrm{~s}^{-2}=\ldots . . \mathrm{km} \mathrm{h}-{ }^2$
Answer
$3.0 \mathrm{~m} \mathrm{~s}^{-2}=\mathbf{3 . 8 8} \times 10^4 \mathrm{~km} \mathrm{~h}^{-2} 1$ hour $=3600 \mathrm{sec}$ so that $1 \mathrm{sec}=1 / 3600$ hour $1 \mathrm{~km}=1000 \mathrm{~m}$ so that $1 \mathrm{~m}=1 / 1000 \mathrm{~km}$
$3.0 \mathrm{~m} \mathrm{~s}^{-2}=3.0(1 / 1000 \mathrm{~km})\left(1 / 3600 \mathrm{hour}^{-2}=3.0 \times 10^{-3} \mathrm{~km} \times\left((1 / 3600)^{-2} \mathrm{~h}^{-2}\right)=3.0 \times 10^{-3} \mathrm{~km} \times(3600)^2 \mathrm{~h}^{-2}=3.88 \times\right.$ $10^4 \mathrm{~km} \mathrm{~h}^{-2}$
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Question 41 Mark
Fill in the blanks by suitable conversion of units: 1m = ..... ly
Answer
$1 \mathrm{~m}=1 / 9.46 \times 10^{15} \mathrm{ly}=1.06 \times 10^{-16} \mathrm{ly} \text { Distance }=\text { Speed } \times \text { Time Speed of light }=3 \times 10^8 \mathrm{~m} / \mathrm{s} \text { Time }=1 \text { year }=365$
$\text { days }=365 \times 24 \text { hours }=365 \times 24 \times 60 \times 60 \mathrm{sec} \text { Putting these values in above formula we get } 1 \text { light year distance }$
$=\left(3 \times 10^8 \mathrm{~m} / \mathrm{s}\right) \times(365 \times 24 \times 60 \times 60 \mathrm{~s})=9.46 \times 10^{15} \mathrm{~m} 9.46 \times 10^{15} \mathrm{~m}=1 \mathrm{ly} \text { So that } 1 \mathrm{~m}=1 / 9.46 \times 10^{15} \mathrm{ly}=1.06$
$\times 10^{-16} \mathrm{ly}$
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Question 51 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 $11.3 g cm^{-3}or 1.13 \times 10^3kg m^{–3}.$
Density of lead = Relative density of lead × Density of water Density of water $= 1g/ cm^3$
Putting the values, we get Density of lead $= 11.3 \times 1g/ cm^3 = 11.3g cm^{-3} 1cm$
$= (1/100m) =10^{–2}m^3 1g$
$= 1/1000kg = 10^{-3}kg$
Density of lead $= 11.3g cm^{-3} = 11.3$
Putting the value of 1cm and $1$ gram $11.3g/ cm^3$
$= 11.3 \times 10^{-3}kg (10^{-2}m)^{-3}$
$= 11.3 \times 10^{–3} \times 10^6kg m^{-3}$
$​​​​​​​=1.13 \times 10^3kg m^{–3}​​​​​​​$​​​​​​​
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Question 61 Mark
State the number of significant figures in the following: $0.2370 \mathrm{~g} \mathrm{~cm}^{-3}$
Answer
Explanation:
The given quantity is $0.2370 \mathrm{~g} \mathrm{~cm}^{-3}$. 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 71 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 81 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 2.0 cm and height 10.0 cm is equal to $1.5 \times 10^4 \mathrm{~mm}^2$ Given, Radius, $\mathrm{r}=$ $2.0 \mathrm{~cm}=20 \mathrm{~mm}$ (convert cm to mm ) Height, $\mathrm{h}=10.0 \mathrm{~cm}=100 \mathrm{~mm}$ 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 91 Mark
State the number of significant figures in the following: $2.64 \times 10^{24} \mathrm{~kg}$
Answer
3 Explanation: The given quantity is $2.64 \times 10^{24} \mathrm{~kg}$. 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 $18 \mathrm{~km} \mathrm{~h}^{-1}$ covers ....m in 1 s
Answer
A vehicle moving with a speed of $18 \mathrm{~km} \mathrm{~h}^{-1}$ covers $\mathbf{5 m}$ in 1 s . Using the conversion, Given, Time, $\mathrm{t}=1 \mathrm{sec}$ speed $=$ $18 \mathrm{~km} \mathrm{~h}^{-1}=18 \mathrm{~km} /$ hour $1 \mathrm{~km}=1000 \mathrm{~m}$ and 1 hour $=3600 \mathrm{sec}$ Speed $=18 \times 1000 / 3600 \mathrm{sec}=5 \mathrm{~m} / \mathrm{sec}$ Use formula Speed $=$ distance/time Cross multiply it, we get Distance $=$ Speed $\times$ Time $=5 \times 1=5 \mathrm{~m}$
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Question 111 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 121 Mark
State the number of significant figures in the following:6.032 $\mathrm{N} \mathrm{m}^{-2}$
Answer
4 Explanation: The given quantity is $6.032 \mathrm{~N} \mathrm{~m}^{-2}$. All zeroes between two non-zero digits are always significant.
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Question 131 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 141 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, $\mathbf{r}=0.5 \mathring{\text{A}}=0.5 \times 10-10 \mathrm{~m}$ Volume of hydrogen atom $=4 / 3 \pi \mathrm{r}^3=4 / 3 \times 22 / 7 \times(0.5$ $\left.\times 10^{-10}\right)^3=0.524 \times 10^{-30} \mathrm{~m}^3 1 \mathrm{~mole}$ of hydrogen contains $6.023 \times 10^{23}$ hydrogen atoms.
$\therefore$ Volume of 1 mole of hydrogen atoms $=6.023 \times 10^{23} \times 0.524 \times 10^{-30}$
$=3.16 \times 10^{-7} \mathrm{~m}^3$
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Question 151 Mark
Fill in the blanks by suitable conversion of units: $G = 6.67 \times 10^{-11} N m^2 (kg)^{-2} = .... (cm)^3 s^{-2} g^{-1}$​​​​​​​
Answer
$G = 6.67 \times 10^{–11} N m^2 (kg)^{–2} = 6.67 \times 10^{–8} (cm)^3s^{–2} g^{–1}.$
Given, $G = 6.67 \times 10^{–11} N m^2 (kg)^{–2} $ We know that $1N = 1kg m s^{-2}$
$1kg = 10^3g 1m = 100cm = 10^2cm$ Putting above values,
we get $6.67 \times 10^{–11} N m^2kg^{–2}$
$= 6.67 \times 10^{–11} \times (1kg m s^{–2}) (1m^2) (1Kg^{–2})$ Solve and cancel out the units we get
$\Rightarrow 6.67 \times 10^{–11} \times (1kg^{–1} \times 1m^3 \times 1s^{–2})$ Putting above values to convert Kg to g and m to cm
$\Rightarrow 6.67 \times 10^{–11} \times (10^3g)^{-1} \times (10^2cm)^3 \times (1s^{–2})$
$\Rightarrow 6.67 \times 10^{–11} \times 10^{-3}g^{-1} \times 10^6cm^3 \times (1s^{–2})$
$\Rightarrow 6.67 \times 10^{–8}cm^3 s^{–2} g^{–1}$
$G = 6.67 \times 10^{–11} N m^2 (kg)^{–2}$
$= 6.67 \times 10^{–8} (cm)^3s^{–2} g^{–1}.$
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Question 161 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 171 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 $=2.56 \mathrm{~s}$ Speed of light $=3 \times 10^8 \mathrm{~m} / \mathrm{s}$ Time taken by the laser beam to reach Moon $=1 / 2 \times 2.56=1.28$ s Radius of the lunar orbit = Distance between the Earth and the Moon $=1.28 \times 3 \times 10^8=3.84 \times 10^8 \mathrm{~m}=3.84 \times 10^5 \mathrm{~km}$
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Question 181 Mark
State the number of significant figures in the following: $0.0006032 \mathrm{~m}^2$
Answer
4 Explanation:The given quantity is $0.0006032 \mathrm{~m}^2$.
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 191 Mark
Fill in the blanks by suitable conversion of units: $1 \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-2}=\ldots . . \mathrm{g} \mathrm{cm}^2 \mathrm{~s}^{-2}$
Answer
$1 \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-2}=10^7 \mathrm{~g} \mathrm{~m}^2 \mathrm{~s}^{-2} 1 \mathrm{~kg}=10^3 \mathrm{~g} 1 \mathrm{~m}^2=10^4 \mathrm{~cm}^2 1 \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-2}=1 \mathrm{~kg} \times 1 \mathrm{~m}^2 \times 1 \mathrm{~s}^{-2}=10^3 \mathrm{~g} \times 10^4 \mathrm{~cm}^2 \times 1 \mathrm{~s}^{-2}=10^7 \mathrm{~g}$ $\mathrm{cm}^2 \mathrm{~s}^{-2}$
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Question 201 Mark
State the number of significant figures in the following: $0.007 \mathrm{~m}^2$
Answer
1 Explanation: The given quantity is $0.007 \mathrm{~m}^2$. 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 211 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 231 Mark
Do $\mathring{\text{A}}$ and AU stand for same length?
Answer
No, $1\mathring{\text{A}}$ (Angstrom) = $10^{-10} \mathrm{~m}$ and $1 \mathrm{~A} . \mathrm{U} .=1.496 \times 10^{11} \mathrm{~m}$.
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Question 251 Mark
Solve with due regard to significant figures:
Answer
$\sqrt{6.5-6.32}=\sqrt{0.18}=\sqrt{0.4242}$ upto one decimal place = 0.43 (having 2 significant figures).
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Question 261 Mark
Which of the following length measurement is most accurate and why?
  1. $2.0\ cm$
  2. $2.00\ cm$
  3. $2.000\ cm$
Answer
$2.000\ сm$ is most accurate because it is correct upto third place of a decimal.
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MCQ 271 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.
  • A
    $b$ and $d$
  • B
    $a$ and $b$
  • C
    $b$ and $c$
  • D
    $c$ and $d$
Answer
  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 281 Mark
Name at least seven physical quantities whose dimensions are $\mathrm{ML}^2 \mathrm{~T}^{-2}$.
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 291 Mark
In a number without decimal, what is the significance of zeros on the right of non-zero digits?
Answer
All such zeros are not significant. e.g. x = 678000 has only three significant figures.
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Question 301 Mark
Is it possible to have length and velocity both as fundamental quantities? Why?
Answer
No, since length is fundamental quantity and velocity is the derived quantity.
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Question 311 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 321 Mark
Given that the value of G in the CGS system as $6.67 \times 10^{-8}$ dyne $\mathrm{cm}^2 \mathrm{~g}^{-2}$, find the value in MKS system.
Answer
$6.67 \times 10^{-8}$ dyne $\mathrm{cm}^2 \mathrm{~g}^{-2}=6.67 \times 10^{-11} \mathrm{Nm}^2 / \mathrm{kg}^2$
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Question 331 Mark
Do all physical quantities have dimensions? If no, name four physical quantities which are dimensionless.
Answer
No, all physical quantities do not possess dimensions. Angle, specific gravity, Poisson's ratio and Strain are four examples of dimensionless quantities.
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Question 361 Mark
Does AU and $\mathring{\text{A}}$ represent the same unit of length?
Answer
No, AU and $\mathring{\text{A}}$ represent two different units of length. 1 AU = 1 astronomical unit $=1.496 \times 10^{11} \mathrm{~m}$ $1\mathring{\text{A}} =10^{-10} \mathrm{~m}$
 
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Question 371 Mark
Find the dimensional formulae of (i) Kinetic energy and (ii) Pressure.
Answer
$\text{KE}=\frac{1}{2}\text{mv}^2\text{ i.e}.,$ Dimensional formula of KE is $[\text{ML}^2\text{T}^{-2}]$ Pressure $=\frac{\text{Force}}{\text{Area}}=\frac{[\text{MLT}^{-2}]}{[\text{L}^2]}=[\text{ML}^{-1}\text{T}^{-2}]$
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Question 381 Mark
What are the dimensions of a and b in the relation: F = a + bx , where F is force and (x) is distance?
Answer
$[\text{a}]=[\text{F}]=[\text{MLT}^{-2}]$ $[\text{b}]=\Big[\frac{\text{F}}{\text{X}}\Big]=\Big[\frac{\text{MLT}^{-2}}{\text{L}}\Big]$
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Question 391 Mark
Why do spring balances show incorrect reading after long use?
Answer
A spring balance shows incorrect reading after long use due to repeated stress and strain as the spring loses its elastic behaviour.
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Question 401 Mark
State the number of significant figures in the following:$0.2370g ~cm^{-3}$
Answer
4Explanation:
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 431 Mark
Are all dimensionally correct equations numerically correct? Give one example.
Answer
No. e.g., $v^2=u^2+2 a s$.
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Question 441 Mark
What is $\mathrm{Nm}^{-1} \mathrm{~s}^2$ equal to?
Answer
$\mathrm{Nm}^{-1} \mathrm{~s}^2$ is nothing but SI unit of mass i.e., the kilogram.
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Question 451 Mark
The orbital velocity v of a satellite may depend on its mass m, distance r from the centre of earth and acceleration due to gravity g. Obtain an expression for orbital velocity.
Answer
Suppose orbital velocity of satellite be given by the relation: $\text{v}=\text{km}^{\text{a}}\text{r}^{\text{b}}\text{g}^\text{c}$ E v =kmorogo where, k is a dimensionless constant and a, b, c are unknown powers. Writing dimensions on two sides of equation, We have: $[\text{M}^{0}\text{L}^{1}\text{T}^{-1}]=[\text{M}]^{\text{a}}[\text{L}]^{\text{b}}[\text{LT}^{-2}]^{\text{C}}=[\text{M}^{\text{a}}\text{L}^{\text{b}+\text{c}}\text{T}^{-2\text{c}}]$ Applying principle of homogeneity of dimensional equation, We find that: $\text{a}=0\Rightarrow\text{b}+\text{c}=1\Rightarrow-2\text{c}=-1$ On solving these equations We find that: $\text{a}=0,\text{b}=+\frac{1}{2}\text{ and }\text{c}=+\frac{1}{2}$ $\text{v}=\text{kr}^{\frac{1}{2}}\text{g}^{\frac{1}{2}}\Rightarrow\text{v}=\text{k}\sqrt{\text{rg}}$
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Question 461 Mark
What do you understand by fundamental physical quantities?
Answer
Fundamental physical quantities are those quantities which are independent of each other. For example, mass, length, time, temperature, electric current, luminous intensity and amount of substance are seven fundamental physical quantities.
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Question 471 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 481 Mark
Which is the most accurate clock?
Answer
A cesium clock is most accurate. Two cesium clocks may differ only by 1s after running for 5000 years.
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Question 501 Mark
In CGS system, the value of Stefan's constant $(\sigma)$ is $5.67 \times 10^{-5} \mathrm{erg} \mathrm{~s}^{-1} \mathrm{~cm}^{-2} \mathrm{~K}^{-4}$. Write down its value in SI units.
Answer
$\sigma=5.67\times10^{-5}\text{ erg}\text{ s}^{-1}\text{ cm}^{-2}\text{ K}^{-4}$ $=5.67\times10^{-5}(10^{-7}\text{J})(\text{s}^{-1})(10^{-2}\text{m})\text{K}^{-4}$ $=5.67\times10^{-8}\text{Js}^{-1}\text{m}^{-2}\text{K}^{-4}$
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1 Marks Question - Physics STD 11 Science Questions - Vidyadip