1 Marks Question

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:

Atoms are very small objects
A jet plane moves with great speed
The mass of Jupiter is very large
The air inside this room contains a large number of molecules
A proton is much more massive than an electron
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.

An atom is a very small object in comparison to a soccer ball.
A jet plane moves with a speed greater than that of a bicycle.
Mass of Jupiter is very large as compared to the mass of a cricket ball.
The air inside this room contains a large number of molecules as compared to that present in a geometry box.
A proton is more massive than an electron.
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

It does not undergo any change with different places.
The wavelength of the light is not affected by time and environment.

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Question 221 Mark

What does RADAR stand for?

Answer

RADAR stands for 'Radio detection and ranging'.

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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 241 Mark

How many dynes make a newton?

Answer

1 newton $=10^5$ dynes.

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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?

$2.0\ cm$
$2.00\ cm$
$2.000\ cm$

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:

Momentum per unit area.
Momentum per unit area per unit time.
Momentum per unit volume.
Energy per unit volume.

A
$b$ and $d$
B
$a$ and $b$
C
$b$ and $c$
D
$c$ and $d$

Answer

Momentum per unit area per unit time

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

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

Pressure energy.
Potential energy.
Kinetic energy.
Work.
Torque.
Moment of force.
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 341 Mark

Do mass and weight have the same dimensions?

Answer

No.

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Question 351 Mark

Write the dimensional formula of torque.

Answer

$\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]$

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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 411 Mark

How many newton make one kg wt.?

Answer

1 kgwt = 9.8N.

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Question 421 Mark

Write dimensional formula for compressibility.

Answer

$\left[M^{-1} L^1 T^2\right]$

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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 491 Mark

Are all constants dimensionless?

Answer

No, it is not possible.

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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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Question 511 Mark

Do specific heat and latent heat have the same dimensions?

Answer

No.

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Question 521 Mark

Precisions describe the limitation of the measuring instrument. Is the statement false?

Answer

No, the statement is true.

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Question 531 Mark

State the number of significant figures in the following: $2.64 \times 10^{24} \mathrm{~kg}$

Answer

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Question 541 Mark

Force $(F)$ and density $(d)$ are related as: $\text{F}=\frac{\alpha}{\beta+\sqrt{\text{d}}}$

Then the dimensions of $\alpha$ are.
Then the dimensions of $\beta$ are.

Answer

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

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Question 551 Mark

What does LASER stand for?

Answer

LASER stands for 'Light Amplification by Stimulated Emission of Radiation'.

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Question 561 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 571 Mark

How many dynes make a newton?

Answer

1 newton $=10^5$ dynes.

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MCQ 581 Mark

Which of the following set have different dimensions?

A
Pressure, Young's modulus, Stress.
B
Emf, potential difference, Electric potential.
C
Heat, Work done, Energy.
✓
Dipole moment, Electric flux, Electric field.

Answer

Correct option: D.

Dipole moment, Electric flux, Electric field.

$\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 591 Mark

If $f=x^2$, relative error in $f$, then how many times the relative error would be in $x$ ?

Answer

It would be two times.

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Question 601 Mark

Write the dimensional formula corresponding to :

Photon.
Calorie.

Answer

i. Photon $- \left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]$
ii. Calorie $- \left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]$

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Question 611 Mark

What are derived units?

Answer

Units of those physical quantities which are derived from the fundamental units are called derived units.

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Question 621 Mark

Name at least six physical quantities whose dimensions are$\mathrm{ML}^2 \mathrm{~T}^{-2}$.

Answer

Work.
Torque.
Moment of force.
Couple.
Potential energy.
Kinetic energy.

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Question 631 Mark

Obtain the dimensional formula for coefficient of viscosity.

Answer

As $\text{F}=\eta\text{A}\frac{\text{dv}}{\text{dx}}$ Hence, $\eta=\frac{\text{F dx}}{\text{A dv}}$ $\therefore[\eta]=\frac{[\text{F}][\text{dx}]}{[\text{A}][\text{dv}]}$ $=\frac{[\text{MLT}^{-2}.\text{L}]}{[\text{L}^2.\text{LT}^{-1}]}=[\text{M}^1\text{L}^{-1}\text{T}^{-1}]$

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Question 641 Mark

Assuming that the mass (m) of the largest stone that can be moved by a flowing river depends only upon the velocity v, the density $\rho$ of water and the acceleration due to gravity g. Show that m varies, with the sixth power of the velocity of the flow.

Answer

$\text{Let}\text{ m}\propto\text{v}^{\text{a}}\rho^{\text{b}}\text{g}^\text{c}$ $\therefore\text{m}=\text{k}\text{v}^{\text{a}}\rho^{\text{b}}\text{g}^{\text{c}},$ where k is constant. where k is constant Taking the dimensions of various physical quantities on both the sides, We have, $[\text{M}]=-[\text{LT}^{-1}]^{\text{a}}[\text{ML}^{-3}]^{\text{b}}[\text{LT}^{-2}]^{\text{c}}$ $=[\text{M}^{\text{b}}\text{L}^{\text{a}-3\text{b}+\text{c}}\text{T}^{-\text{a}-2\text{c}}]$ Comparing the powers of M, L and T on both the sides, We have: $\text{b} = 1 \dots(\text{i})$ $\text{a} - 3\text{b} + \text{c} = 0 ....(\text{ii})$ $-\text{a} - 2\text{c} = 0 \dots(\text{iii})$ Solving these equation We get: $\text{b} = 1, \text{a} = 6 \text{ and c} = -3$ $\therefore\text{m}=\text{k}\text{v}^{6}\rho^{1}\text{g}^{-3}$ $\text{m}\propto\text{v}^6$

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Question 651 Mark

Name four units used in the measurement of extremely short distances.

Answer

$1\text{ micron}(1\mu)=10^{-6}\text{m}$ $1\text{ nanometre}(1\text{nm})=10^{-9}\text{m}$ $1\text{ angstrom}(1\mathring{\text{A}})=10^{-10}\text{m}$ $1\text{ fermi}(1\text{f})=10^{-15}\text{m}$

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Question 661 Mark

What is the difference between nm, mN and Nm?

Answer

nm stands for nanometre, $1 \mathrm{~nm}=10^{-9} \mathrm{~m}, \mathrm{mN}$ stands for milli-newton, $1 \mathrm{mN}=10^{-3} \mathrm{~N}, \mathrm{Nm}$ stands for newton metre.

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Question 671 Mark

How many amu make 1kg?

Answer

$1 \mathrm{amu}=1.66 \times 10^{-27} \mathrm{~kg}$
$\therefore1\text{kg}=\frac{1}{1.66\times10^{-27}}\text{ amu}$ $=0.6\times10^{27}\text{ amu}$

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Question 681 Mark

Obtain the dimensional formula for coefficient of viscosity.

Answer

Coefficient of viscosity $(\eta)=\frac{\text{fdx}}{\text{A.dv}}=\frac{[\text{MLT}]^{-2}[\text{L}]}{[\text{L}^2][\text{LT}^{-1}]}$ $=[\text{M}^{-1}\text{L}^{-1}\text{T}^{-1}]$

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Question 691 Mark

Write the order of following length in metres:

Radius of earth.
The height of average man.
Thickness of sheet of paper.
The radius of hydrogen atom.

Answer

i. $6.4 \times 10^6 \mathrm{~m}$
ii. $1.8 \times 10^0 \mathrm{~m}$
iii. $1 \times 10^{-4} \mathrm{~m}$
iv. $5 \times 10^{-11} \mathrm{~m}$

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Question 701 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 711 Mark

Define parsec.

Answer

The distance at which a star would have annual parallax of 1 second of arc. 1 parsec $=3.08 \times 10^{16} \mathrm{~m}$.

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Question 721 Mark

Define Atomic mass unit (a.m.u.).

Answer

1a.m.u. $=\frac{1}{12}$ th mass of carbon -12 atom, i.e., $1.66 \times 10^{-27} \mathrm{~kg}$.

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Question 731 Mark

Write three pairs of physical quantities, which have same dimensional formula.

Answer

Work and energy.
Energy and torque.
Pressure and stress.

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Question 741 Mark

What is the dimensional formula for torque?

Answer

$\left[\mathrm{ML}^{-2} \mathrm{~T}^{-2}\right]$

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Question 751 Mark

Which of the following measurements is more accurate and why?

$0.0002g$
$20.0g$

Answer

$0.0002g$ is more accurate. Since it measures upto $4$ places of decimal.

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Question 761 Mark

What does SONAR stand for?

Answer

SONAR stands for 'sound navigation and ranging’.

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Question 771 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 781 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 791 Mark

Give an example of: A constant which has a unit.

Answer

Gravitational constant $(\text{G})=6.67\times10^{-11}\text{N-m}^2/\text{kg}^2$

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Question 801 Mark

Which of the length measurements is the most accurate and why?

$500.0\ cm.$
$0.0005\ cm.$
$6.00\ cm.$

Answer

The length measurement $500.0\ cm$ is most accurate as it has $4$ significant figures.

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Question 811 Mark

Which unit is used to measure size of a nucleus?

Answer

The size of nucleus is measured in fermi. 1 fermi $=10^{-15} \mathrm{~m}$

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Question 821 Mark

Write the dimensional formula of wavelength and frequency of a wave.

Answer

Wavelength ${\lambda}=[\text{L}]$ Frequency $[\text{v}]=[\text{T}^{-1}]$

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Question 831 Mark

The rotational kinetic energy of a body is given by $\text{F}=\frac{1}{2}\text{l}\omega^2$ where w is the angular velocity of the body. Use the equation to obtain dimensional formula for moment of inertia I. Also write its SI unit

Answer

The given relation is $\text{F}=\frac{1}{2}\text{l}\omega^2$ $\text{I}=\frac{[\text{E}]}{[\omega]^2}=\frac{[\text{ML}^2\text{T}^{-2}]}{[\text{T}^{-1}]^2}\Big[\frac{\text{ML}^2\text{T}^{-2}}{\text{T}^{-2}}\Big]=[\text{ML}^2]$ Its si unit is joule.

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Question 841 Mark

Express a joule in terms of fundamental units.

Answer

$[\text{Energy}]=[\text{ML}^2\text{T}^{-2}],$ hence 1 joule $=1 \mathrm{~g} \times 1 \mathrm{~m}^2 \times 1 \mathrm{~s}^{-2}=1 \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-2}$

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Question 851 Mark

If velocity of light $c$, Planck’s constant $h$ and gravitational contant $G$ are taken as fundamental quantities then express mass, length and time in terms of dimensions of these quantities.

Answer

We have to apply principle of homogeneity to solve this problem. Principle of homogeneity states that in a correct equation, the dimensions of each term added or subtracted must be same, i.e., dimensions of $\text{LHS}$ and $\text{RHS}$ should be equal, We know that, dimensions of, $[\text{h}]=[\text{ML}^2\text{T}^{-1}],[\text{c}]=[\text{LT}^{-1}],\text[{G}]=[\text{M}^{-1}\text{L}^3\text{T}^{-2}]$

Let $\text{m}\propto\text{c}^\text{x}\text{h}^\text{v}\text{G}^\text{z}$

$\Rightarrow\text{m}=\text{kc}^\text{a}\text{h}^\text{b}\text{G}^\text{c}\ \ \ \ ...(\text{i})$
Where, $k$ is a dimensionless constant of proportionality.
Substituting dimensions of each term in Eq. $(i)$, we get,
$[\text{ML}^0\text{T}^0]=[\text{LT}^{-1}]^\text{a}\times[\text{ML}^2\text{T}^{-1}]^\text{b}[\text{M}^{-1}\text{L}^3\text{T}^{-2}]^\text{c}$
Comparing powers of same terms on both sides, we get,
$\text{b}-\text{c}=1\ \ \ ...(\text{ii})$
$\text{a}+2\text{b}+3\text{c}=0\ \ \ \ ...(\text{iii})$
$-\text{a}-\text{b}-2\text{c}=0\ \ \ \ ...(\text{iv})$
Adding Eqs. $(ii), (iii)$ and $(iv)$, we get,
$2\text{b}=1\Rightarrow\text{b}=\frac{1}{2}$
Substituting value of $b$ in Eq. $(ii),$ we get,
$\text{c}=-\frac{1}{2}$
From Eq. $(iv)$
$\text{a}=-\text{b}-2\text{c}$
Substituting values of $b$ and $c,$ we get,
$\text{a}=-\frac{1}{2}-2\Big(-\frac{1}{2}\Big)=\frac{1}{2}$
Putting values of $a, b$ and $c$ in Eq. $(i)$, we get,
$\text{m}=\text{kc}^\frac{1}{2}\text{h}^\frac{1}{2}\text{G}^{-\frac{1}{2}}=\text{k}\sqrt{\frac{\text{ch}}{\text{G}}}$

Let $\text{L}\propto\text{c}^\text{a}\text{h}^\text{b}\text{G}^\text{c}$

$\Rightarrow\text{L}=\text{kc}^\text{a}\text{h}^\text{b}\text{G}^\text{c}\ \ \ \ ...(\text{v})$
Where $k$ is a dimensionless constant.
Substituting dimensions of each term in Eq. $(v)$, we get
$[\text{M}^0\text{LT}^0]=[\text{LT}^{-1}]^\text{a}\times[\text{ML}^2\text{T}^{-1}]^\text{b}\times[\text{M}^{-1}\text{L}^3\text{T}^{-2}]^\text{c}$
$=[\text{M}^{\text{b}-\text{c}\ }\text{L}^{\text{a}+2\text{b}+3\text{c}}\ \text{T}^{-\text{a}-\text{b}-2\text{c}}]$
On comparing powers of same terms, we get,
$\text{b}-\text{c}=0\ \ \ ...(\text{vi})$
$\text{a}+2\text{b}+3\text{c}=1\ \ \ ...(\text{vii})$
$-\text{a}-\text{b}-2\text{c}=0\ \ \ \ ...(\text{viii})$
Adding Eqs. $(vi), (vii)$ and $(viii)$, we get,
$2\text{b}=1\Rightarrow\text{b}=\frac{1}{2}$
Substituting value of $b$ in Eq. $(vi)$, we get,
$\text{c}=\frac{1}{2}$
From Eq. $(viii), \text{a}=-\text{b}-2\text{c}$
Substituting values of $b$ and $c$, we get,
$\text{a}=-\frac{1}{2}-2\Big(\frac{1}{2}\Big)=-\frac{3}{2}$
Putting values of $a, b$ and $c$ in Eq. $(v)$, we get,
$\text{L}=\text{kc}^{-\frac{3}{2}}\text{h}^\frac{1}{2}\text{G}^\frac{1}{2}=\text{k}\sqrt{\frac{\text{hG}}{\text{c}^3}}$

Let $\text{T}\propto\text{c}^\text{a}\text{h}^\text{b}\text{G}^\text{c}$

$\Rightarrow\text{T}=\text{c}^\text{a}\text{h}^\text{b}\text{G}^\text{c}\ \ \ \ ...(\text{ix})$
Where, $k$ is a dimensionless constant.
Substituting dimensions of each term in Eq. $(ix)$, we get
$[\text{M}^0\text{L}^0\text{T}^1]=[\text{LT}^{-1}]^\text{a}\times[\text{ML}^2\text{T}^{-1}]^\text{b}\times[\text{M}^{-1}\text{L}^3\text{T}^{-2}]^\text{c}$
$=[\text{M}^{\text{b}-\text{c}}\text{ L}^{\text{a}+2\text{b}+3\text{c}}\ \text{T}^{-\text{a}-\text{b}-2\text{c}}]$
On comparing powers of same terms, we get,
$\text{b}-\text{c}=0 \ \ \ ...(\text{x})$
$\text{a}+2\text{b}+3\text{c}=1\ \ \ ...(\text{xi})$
$-\text{a}-\text{b}-2\text{c}=1\ \ \ \ ...(\text{xii})$
Adding Eqs. $(x), (xi)$ and $(xii)$, we get,
$2\text{b}=1\Rightarrow\text{b}=\frac{1}{2}$
Substituting the value of $b$ in Eq. $(x)$, we get,
$\text{c}=\text{b}=\frac{1}{2}$
From Eq. $(xii),$
$\text{a}=-\text{b}-2\text{c}-1$
Substituting values of $b$ and $c$, we get,
$\text{a}=-\frac{1}{2}-2\Big(\frac{1}{2}\Big)-1=-\frac{5}{2}$
Putting values of $a, b$ and $c$ in Eq. $(ix)$, we get,
$\text{T}=\text{kc}^{\frac{-5}{2}}\text{h}^\frac{1}{2}\text{G}^\frac{1}{2}=\text{k}\sqrt{\frac{\text{hG}}{\text{c}^5}}$

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Question 861 Mark

Using the relation E = hv, obtain the dimensions of Planck's constant.

Answer

We know that dimensional formula of energy E of photon is [$M^1L^2T^{-2}$) and dimensional formula of frequency v is [$T^{-1}$] $[\text{h}]=\frac{[\text{E}]}{[\text{V}]}=\frac{\text{M}^{1}\text{L}^2\text{T}^{-2}}{[\text{T}^{-2}]}=[\text{M}^1\text{L}^2\text{T}^{-1}]$

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Question 871 Mark

If $\mathrm{x}=\mathrm{a}+\mathrm{bt}+\mathrm{ct}^2$, where x is in metres andt is second, what is the dimensional formula of c ?

Answer

Here, $\text{x}=[\text{L}]$ $\text{t}=[\text{T}],\text{x}=\text{ct}^2$ $[\text{L}]=\text{c}\times[\text{T}^2]$ $\Rightarrow\frac{\text{L}}{\text{T}^2}=\text{c}\Rightarrow\text{c}=[\text{LT}^{-2}]$

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Question 881 Mark

Find the relative error in Z if $\text{Z}=\frac{\text{A}^4\text{B}^{\frac{1}{3}}}{\text{CD}^{\frac{3}{2}}}$

Answer

Here, $\text{Z}=\frac{\text{A}^4\text{B}^{\frac{1}{3}}}{\text{CD}^{\frac{3}{2}}}$ Relative error, $=\frac{\Delta\text{Z}}{\text{Z}}$ $=\pm\Big[4\Big(\frac{\Delta\text{A}}{\text{A}}\Big)+\frac{1}{3}\Big(\frac{\Delta\text{B}}{\text{B}}\Big)+\Big(\frac{\Delta\text{C}}{\text{C}}\Big)+\frac{3}{2}\Big(\frac{\Delta\text{D}}{\text{D}}\Big)\Big]$

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Question 891 Mark

Obtain the dimensions of relative density.

Answer

As relative density is defined as the ratio of the density of given substance and the density of standard distance (water), it is a dimensionless quantity.

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Question 901 Mark

Find the area of the circle of radius 3.458cm upto correct significant figures.

Answer

$=\pi^2 = 3.141\times(3.458)^2$ $= 3.141\times11.96$ $=37.5664$ $=37.57 \text{cm}^2$

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Question 911 Mark

What do you mean by order of magnitude of a length? Write the order of magnitude of following lenghts.

Size of a hydrogen atom.
Diameter of earth.
Light year.
Size of milky way.

Answer

a. $10^{-10} \mathrm{~m}$
b. $10^7 \mathrm{~m}$
c. $10^{16} \mathrm{~m}$
d. $10^{21} \mathrm{~m}$

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Question 921 Mark

Are all constants dimensionless?

Answer

No, it is not true.

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Question 931 Mark

Atoms are very small objects
A jet plane moves with great speed
The mass of Jupiter is very large
The air inside this room contains a large number of molecules
A proton is much more massive than an electron
The speed of sound is much smaller than the speed of light.

Answer

An atom is a very small object in comparison to a soccer ball.
A jet plane moves with a speed greater than that of a bicycle.
Mass of Jupiter is very large as compared to the mass of a cricket ball.
The air inside this room contains a large number of molecules as compared to that present in a geometry box.
A proton is more massive than an electron.
Speed of sound is less than the speed of light.

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Question 941 Mark

Human heart is an inbuilt clock. Comment.

Answer

True. Explanation: because human heart beats at a regular rate.

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Question 951 Mark

What is meant by angular diameter of moon?

Answer

Angular diameter of moon is the angle subtended at a point on the earth, by two diameterically opposite ends of the moon. Its value is about $0.5^\circ$.

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MCQ 961 Mark

The number of significant figures in the numbers $4.8000 \times 10^4$ and 48000.50 are respectively:

A
5 and 6
✓
5 and 7
C
2 and 7
D
2 and 6

Answer

Correct option: B.

5 and 7

$4.8000 \times 10^4$ has 4, 8, 0, 0, 0 = 5 significant digits. 48000.50 has 4, 8, 0, 0, 0, 5, 0 = 7 significant digits.

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Question 971 Mark

Name two quantities with:

Same dimensions.
Constant value having dimension.

Answer

Work and torque.
Gravitational constant $G$ and Planck's constant $h.$

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Question 981 Mark

Express a joule in terms of fundamental unit.

Answer

Energy $=\left[\mathrm{ML}^{-2} \mathrm{~T}^{-2}\right]$ Hence, 1 Joule $=1 \mathrm{~kg} \times 1 \mathrm{~m}^2 \times 1 \mathrm{~s}^{-2}=1 \mathrm{kgm}^2 \mathrm{~s}^{-2}$

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Question 991 Mark

What is percentage error in volume of a sphere, when error in measuring its radius is 2%?

Answer

As Volume of a sphere (V), i.e. $\frac{\Delta\text{V}}{\text{V}}\times100=3\frac{\Delta\text{r}}{\text{r}}\times100$ $=3\times(\pm2\%)=\pm6\%$

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Question 1001 Mark

Magnitude of force F experienced by a certain object moving with speed v is given by $\text{F}=\text{k}\text{v}^2$ where k is constant. Find the dimensions of K.

Answer

Since, $\text{F}=\text{k}\text{v}^2$ Here, $[\text{k}]=\frac{[\text{F}]}{[\text{v}^2]}$ $=\frac{[\text{MLT}^{-2}]}{[\text{LT}^{-1}]^2}$ $=[\text{M}^1\text{L}^{-1}]$

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Question 1011 Mark

State dimensional formulae for stress, strain and Young's modulus.

Answer

Strain is dimensionless quantity. Dimensional formula for stress is $\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]$ Young's modulus has same dimensional formula as stress i.e. $\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]$

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Question 1021 Mark

What do you mean by order of magnitude? Explain.

Answer

The order of magnitude of a numerical quantity $(\mathrm{N})$ is the nearest power of 10 to which its value can be written. For example: Order of magnitude of nuclear radius $1.5 \times 10^{-14} \mathrm{~m}$ is -14 .

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Question 1031 Mark

Give the relationship between light year and metre.

Answer

1 light year $=9.467 \times 10^{15} \mathrm{~m}$.

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Question 1041 Mark

How many light years make 1 par sec?

Answer

3.26 light years make 1 par sec.

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Question 1051 Mark

Write the two physical quantities whose dimensions are same.

Answer

Work and torque have the same dimension $\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]$.

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Question 1061 Mark

A physical quantity $\text{X}=\frac{\text{a}^2\text{b}^{\frac{-3}{2}}}{\text{c}^4}$ A student says that the relative error in $2\frac{\Delta\text{a}}{\text{a}}-\frac{3}{2}\frac{\Delta\text{b}}{\text{b}}-4\frac{\Delta\text{c}}{\text{c}}.$ Do you agree with the student? If not, what is the relative error in X?

Answer

Errors are always additive. Therefore, the relative error in: $\text{X}=2\frac{\Delta\text{a}}{\text{a}}-\frac{3}{2}\frac{\Delta\text{b}}{\text{b}}-4\frac{\Delta\text{c}}{\text{c}}$

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Question 1071 Mark

Name some physical quantities which are dimensionless.

Answer

Solid angle, relative density, strain, Reynold's number and Poisson's ratio.

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Question 1081 Mark

The pairs of physical quantities that have the same dimensions are:

Reynolds number and coefficient of friction.
Latent heat and gravitational potential.
Curie and frequency of light wave.
Planck's constant and torque.

Answer

Reynolds number and coefficient of friction.
Latent heat and gravitational potential.
Curie and frequency of light wave.

Explanation:

Reynolds number and coefficient of friction, both are dimensionless.
$\text{L}=\frac{\text{Q}}{\text{m}}=\frac{\text{ML}^2\text{T}^{-2}}{\text{M}}=[\text{L}^{2}\text{T}^{-2}]$

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

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

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

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Question 1091 Mark

Do specific heat and latent heat have the same dimensions?

Answer

No.

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Question 1101 Mark

Answer

Radius of hydrogen atom, $\mathbf{r}=0.5 \mathring{\text{A}}=0.5 \times 10-10 \mathrm{~m}$ Volume of hydrogen atom $=4 / 3 \pi \mathrm{r}^3=4 / 3 \times 22 / 7 \times(0.5$ $\left.\times 10^{-10}\right)^3=0.524 \times 10^{-30} \mathrm{~m}^3 1$ 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 1111 Mark

Which of these is largest : astronomical unit, light year and par sec?

Answer

Par sec is larger than light year which in turn is larger than an astronomical unit.

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Question 1121 Mark

What is the order of precision of an atomic clock?

Answer

About 1 in $10^{12}$ to $10^{13}$ s.

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Question 1131 Mark

Round off to four significant figures:

$36.879$
$1.0084$

Answer

$36.88$
$1.008$

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Question 1141 Mark

Express an acceleration of $10m/ s^2$ in $km/ h^2$.

Answer

Acceleration$=\frac{10\text{m}}{(1\text{s)}^2}=\frac{10\times10^{-3}}{\Big[\frac{1}{60\times60}\text{h}\Big]^2}$ $=(3600)^2\times10^{-2}\text{km}/\text{ h}^2$ $=1.29\times10^5\text{km}/\text{ h}^2$

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Question 1151 Mark

If $f = x^2$, then what is the relative error in f?

Answer

$\frac{2\Delta\text{x}}{\text{x}}$

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Question 1161 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 1171 Mark

If $x = a + bt + ct^2$ where x is in metre and t in second, then what is the unit of e?

Answer

According to the principle of dimensions. $[\text{ct}^20]=[\text{L}]\text{ or }[\text{c}]=[\text{LT}^{-2}]$ So, the units of c is $ms^{-2}$

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Question 1181 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 1191 Mark

How many unit systems are there?

Answer

Unit systems are:

$\text{FPS}$ system.
$\text{MKS}$ system.
$\text{CGS}$ system.

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Question 1201 Mark

The mass of a body is measured by two persons is 10.2kg and 10.23kg. Which one is more accurate and why?

Answer

The value m = 10.23kg is more accurate, being correct upto $2^{nd}$ place of decimal.

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Question 1211 Mark

Answer

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Question 1221 Mark

Is Avogadro's number a dimensionless quantity?

Answer

No, it has dimensions. In fact its dimensional formula is $[mol^{‑1}]$.

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Question 1231 Mark

What is the dimensional formula for torque?

Answer

$[\text{ML}^2\text{T}^{-2}]$

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Question 1241 Mark

Do A and A.U. stand for same length?

Answer

No, $1\mathring{\text{A}}=10^{-10}\text{m}$ $1\text{ A.U}=1.496\times10^{11}\text{m}$

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Question 1251 Mark

Determine $\pi$ with due regard to significant figures. $[\text{Given }\pi=3.14]$

Answer

$\pi= 3.14 \times 3.14 = 9.8596 = 9.86$

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Question 1261 Mark

Calculate the length of the arc of a circle of radius 31.0cm which subtends an angle of $\frac{\pi}{6}$ at the centre.

Answer

$\text{Angle}=\frac{\text{length of arc}}{\text{radius of arc}}$ $\frac{\pi}{6}=\frac{\text{x}}{31}$ $\text{x}=\frac{31\times\pi}{6}=\frac{31\times3.14}{6}=16.22\text{cm}.$

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Question 1271 Mark

Write the number of significant figures in each of the following measurements:

$1.67 \times 10^{-27}kg.$
$0.270\ cm.$

Answer

Three
Three

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Question 1281 Mark

Is Avogadro's number a dimensionless quantity.

Answer

No, it has dimensions. In fact, its dimensional formula I $[mol^{-1}]$.

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Question 1291 Mark

Define one Barn. How it is related with metre?

Answer

One Barn is a small unit of area used to measure area of nuclear cross-section. $\therefore1\text{ barn}=10^{-28}\text{m}^2$

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Question 1301 Mark

Can a physical quantity have dimensions but still have no units?

Answer

No, it is not possible.

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Question 1311 Mark

Write the dimensional formula of the following physical quantities:

Stress.
Coefficient of viscosity.

Answer

Dimensional formula for stress $= ML^{-1}T^{-2}$
Dimensional formula of coefficient of viscosity $= ML^{-1}T^{-1}$

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Question 1321 Mark

If all measurements in an experiment are taken upto same number of significant figures, then which measurement is responsible for maximum error?

Answer

The maximum error will be due to:

Measurement which is least accurate.
Measurement of the quantity which has maximum power in the formula.

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Question 1331 Mark

A jeweller put a diamond weighing 5.42g in a box weighing 1.2kg. Find the total weight of the box and the diamond to correct number of significant figures.

Answer

Weight of diamond = 5.42g = 0.00542kg Total weight = 1.2 + 0.00542 = 1.20542kg = 1.2kg

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MCQ 1341 Mark

If $3.8 \times 10^{-6}$ is added to $4.2 \times 10^{-5}$ giving due regard to significant figures, then the result will be:

A
$4.58 \times 10^{-5}$
✓
$4.6 \times 10^{-5}$
C
$45 \times 10^{-5}$
D
None of these.

Answer

Correct option: B.

$4.6 \times 10^{-5}$

By adding $3.8 \times 10^{-6}$ and $42 \times 10^{-6}$, We get: $=45.8 \times 10^{-6}=4.58 \times 10^{-5}$ As least number of significant figures in given values are 2 , so we round off the result to $4.6 \times 10^{-5}$.

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Question 1351 Mark

If $x = a + bt + ct^2$, where x is in metres and t in second, what is the dimensional formula of c.

Answer

$[M^0L^1T^{-2}]$

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Question 1361 Mark

Percentage error in the measurement of height and radius of a cylinder are X and Y respectively. Find percentage error in the measurement of its volume. Which of the measurement height or radius need more attention?

Answer

Height of cylinder = X, radius of cylinder = Y, Volume of cylinder $\text{V}=\pi\text{Y}^2\text{X}$ Percentage error in measurement of volume: $\frac{\Delta\text{V}}{\text{V}}\times100=\pm\Big(2\frac{\Delta\text{Y}}{\text{Y}}+\frac{\Delta\text{X}}{\text{ X}}\Big)\times100$ Hence, radius needs more attention because any error in its measurement is multiplied two times.

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Question 1371 Mark

Name two physical quantities whose dimensions are same.

Answer

Stress and Young's modulus or Work and Energy have same dimension.

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Question 1381 Mark

When white light travels through glass the refractive: index $\mu=\Big(\frac{\text{Velocity of light in air}}{\text{Velocity of light in glass}}\Big)$ is found to vary with wavelength as $\mu=\text{A}+\frac{\text{B}}{\lambda^2}$ where A and B are constants. Using the principle of homogeneity of dimensions, determine the SI unit in which A and B must be expressed.

Answer

A is a constant and have no unit and SI unit of B is $m^2$.

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Question 1391 Mark

Which is a bigger unit-light year or parsec?

Answer

Parsec is bigger unit than light year (1 parsec = 3.26 light year).

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Question 1401 Mark

Give an example of: A physical quantity which has a unit but no dimensions.

Answer

Solid angle $\Omega=\frac{\text{A}}{\text{r}^2}$ steradian and a plane angle $\theta=\frac{\text{L}}{\text{r}}$ radian. Both are dimensionless but have units.

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Question 1411 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 1421 Mark

The period of oscillation of a simple pendulum is $\text{T}=2\pi\sqrt{\frac{\text{L}}{\text{g}}}.$ Measured value of L is 20cm known to 1mm accuracy and time for 100 oscillations of the pendulum is found to be 90s using a wrist watch of 1s resolution. What is the accuracy in the determination of g?OR
The length, breadth and thickness of a rectangular sheet of metal are 4.234m, 1.005m and 2.01cm respectively. Find the area and volume of the sheet to correct significant figures.

Answer

$\text{T}=2\pi\sqrt{\frac{\text{L}}{\text{g}}},$ Squaring both sides and rearranging for G, We have $\text{g}=\frac{4\pi^2\text{L}}{\text{T}^2}$ $\therefore\frac{\Delta\text{g}}{\text{g}}=\frac{\Delta\text{L}}{\text{L}}+\frac{2\Delta\text{T}}{\text{T}}$ $\Delta\text{L}=1\text{mm}=0.1\text{cm},$ To calculate $\frac{\Delta\text{T}}{\text{T}}$ $\text{t}=\text{nt}$ $\therefore\frac{\Delta\text{T}}{\text{T}}=\frac{\Delta\text{t}}{\text{t}}$ Putting, this value $\therefore\frac{\Delta\text{g}}{\text{g}}=\frac{\Delta\text{L}}{\text{L}}+\frac{2\Delta\text{t}}{\text{t}}$ $=\frac{0.1}{20}+2\frac{1}{90}=0.027=0.03$ Hence accurcy is 3%.OR
$\mathrm{L}=4.234 \mathrm{~m}, \mathrm{~B}=1.005 \mathrm{~m}, \mathrm{~d}=2.01 \mathrm{~cm}=2.01 \times 10^{-2} \mathrm{~m}$ Area of metal sheet $=\mathrm{L} \times \mathrm{B}=4.234 \times 1.005=4.25517 \mathrm{~m}^2$. Since both length and breadth have four significant figure area of metal sheet is given by $4.255 \mathrm{~m}^2$. Volume $=$ area $<$ thickness $=0.0855 \mathrm{~m}^3$

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Question 1431 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 1441 Mark

State the number of significant figures in the following: $0.007m^2$

Answer

1 Explanation: The given quantity is $0.007m^2$. 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 1451 Mark

The time of oscillation t of a small drop of liquid under surface tension depends upon the density p, radius r and surface tension o. Show dimensionally that: $\text{t}=\sqrt{\frac{\rho\text{r}^3}{\sigma}}$

Answer

Given: $\text{t}=\sqrt{\frac{\rho\text{r}^3}{\sigma}}$ Where, $\rho=\text{density},\text{r}=\text{radius},$ $\sigma=\text{surface tesion}$ $\because[\rho]=[\text{ML}^{-3}],[\sigma]=[\text{MT}^{-2}],[\text{r}]=[\text{L}]$ $\therefore\text{R.H.S}=\Big[\frac{\text{M}^{1}\text{L}^{-3}\text{L}^3}{\text{M}^{1}\text{L}^0\text{T}^{-2}{}}\Big]^{\frac{1}{2}}$ $\Big[\frac{1}{\text{T}^{-2}}\Big]^{\frac{1}{2}}=\text{T}^{2\times\frac{1}{2}}=\text{T}$ L.H.S. = time of oscillation t of a small drop of liquid = T L.H.S. = R.H.S.

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Question 1461 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 1471 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 1481 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 1491 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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