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Question 12 Marks
Compare the time periods of the two pendulums of lengths $1\ m$ and $9\ m.$
Answer
Let $T_1$ and $T_2$ be the time periods of the two pendulums of lengths $1\ m$ and $9\ m$, respectively.
$ \frac{T_1}{T_2}=\frac{2 \pi \sqrt{\frac{1}{g}}}{2 \pi \sqrt{\frac{9}{g}}}$
$\text { or , } \frac{T_1}{T_2}=\sqrt{\frac{1}{9}}$
$\text { or , } \frac{T_1}{T_2}=\frac{1}{3}$
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Question 22 Marks
Name two factors on which the time period of a simple pendulum depends. Write the relation for the time period in terms of the above named factors.
Answer
Two factors on which the time period of a simple pendulum depends are
(i) Length of pendulum (l)
(ii) Acceleration due to gravity (g)
$T=2 \pi \sqrt{\frac{l}{g}}$
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Question 32 Marks
The time period of a simple pendulum is $2 s$. What is its frequency? What name is given to such a pedulum ?
Answer
Time period $= 2 s$
Frequency $= 1$ /time period
$= \frac{1}{2}s^{-1}$
$= 0.5 s^{-1}$
Such a pendulum is called the seconds' pendulum.
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Question 42 Marks
A simple pendulum completes $40$ oscillations in one minute. Find its (a) Frequency and (b) Time period.
Answer
(a) Frequency = Oscillations per second
$=\left(\frac{40}{60}\right) s ^{-1} $
$=0.67 s ^{-1}$
(b) Time period $=1 /$ frequency
$=\left(\frac{1}{0.67}\right) s $
$=1.5 s$
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Question 52 Marks
What is a simple pendulum? Is the pendulum used in a pendulum clock simple pendulum? Give reason to your answer.
Answer
A simple pendulum is a heavy point mass (known as bob) suspended from a rigid support by a massless and inextensible string.
No, the pendulum used in pendulum clock is not a simple pendulum because the simple pendulum is an ideal case. We cannot have a heavy mass having the size of a point and string having no mass.
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Question 62 Marks
What is a second's pendulum?
Answer
A pendulum with the time period of oscillation equal to two seconds is known as a seconds pendulum.
OR
“A pendulum which has a time period of two seconds is called seconds’ pendulum”
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Question 72 Marks
Two simple pendulums A and B have lengths 1.0 m and 4.0 m, respectively, at a certain place. Which pendulum will make more oscillations in 1 min? Explain your answer.
Answer
Pendulum A will take more time (twice) in a given time because the time period of oscillation is directly proportional to the square root of the length of the pendulum. Therefore, the pendulum B will have a greater time period and thus making lesser oscillations.
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Question 82 Marks
Two simple pendulums A and B have equal lengths, but their bobs weigh 50 gf and 100 gf, respectively. What would be the ratio of their time periods? Give reasons for your answer.
Answer
The ratio of their time periods would be 1:1 because the time period does not depend on the weight of the bob.
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Question 92 Marks
The pitch of a screw gauge is 0.5 mm and the head scale is divided in 100 parts. What is the least count of a screw gauge?
Answer
Pitch of a screw gauge = 0.5 mm
No. of divisions on the circular scale = 100
L.C. = $\left(\frac{0.5}{100}\right)$ mm
= 0.005 mm or 0.0005 cm
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Question 102 Marks
When is a vernier Callipers said to be free from zero error?
Answer
A vernier Callipers are said to be free from zero error if the zero marks of the vernier scale coincides with the zero marks of the main scale.
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Question 112 Marks
Define the term 'Vernier constant'.
Answer
Vernier constant is equal to the difference between the values of one main scale division and one vernier scale division. It is the least count of vernier callipers.
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Question 122 Marks
A boy uses a vernier callipers to measure the thickness of his pencil. He measures it to be 1.4 mm. If the zero error of vernier callipers is +0.02 cm, what is the correct thickness of the pencil?
Answer
Thickness of the pencil (observed reading) = 1.4 mm
Zero error = + 0.02 cm = + 0.2 mm
Correct reading = observed reading - zero error (with sign)
= 1.4 mm - 0.2 mm
= 1.2 mm
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Question 132 Marks
A boy measures the length of a pencil and expresses it to be 2.6 cm. what is the accuracy of his measurement ? Can he write it as 2.60 cm ?
Answer
The least count of a metre rule is 1 cm.
The length cannot be expressed as 2.60 cm because a metre scale measures length correctly only up to one decimal place of a centimeter.
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Question 142 Marks
A vernier has 10 divisions and they are equal to 9 divisions of the main scale in length. If the main scale is calibrated in mm, what is its least count?
Answer
Value of 1 m.s.d. = 1 mm
10 vernier divisions = 9 m.s.d.
L.C. = Value of 1 m.s.d./number of divisions on vernier scale
= 1 mm/10
= 0.1 mm or 0.01 cm
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Question 152 Marks
A boy makes a ruler with graduations in cm on it, (i.e. 100 divisions in 1 m). To what accuracy this ruler can measure? How can this accuracy be increased?
Answer
Total length of the ruler = 1 m = 100 cm
No. of divisions = 100
Length of each division = $\frac{\text { Total length }}{\text { total no. of divisions }}$
= 100 cm/100
= 1 cm
Thus, this ruler can measure with an accuracy of 1 cm.
To increase the accuracy, the total number of divisions on the ruler must be increased.
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Question 162 Marks
A stopwatch has 10 divisions graduated between the 0 and 5 s marks. What is its least count?
Answer
Range of the stop watch $=5 s$
Total number of divisions $=10$
$
\text { L.C. }=\frac{5}{10}=0.5 s
$
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Question 172 Marks
Explain the meaning of the term 'least count of an instrument' by taking a suitable example.
Answer
The least count of an instrument is the smallest measurement that can be taken accurately with it. For example, if an ammeter has 5 divisions between the marks 0 and 1A, then its least count is 1/5 = 0.2 A or it can measure current up to the value 0.2 accurately.
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Question 182 Marks
A screw has a pitch equal to 0.5 mm. What should be the number of divisions on its head so as to read correctly up to 0.001 mm with its help?
Answer
Pitch of the screw gauge = 0.5 mm
L.C. of the screw gauge = 0.001 mm
No. of divisions on circular scale = Pitch / L.C.
= 0.5 / 0.001
= 500
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Question 192 Marks
How can the least count of a screw gauge be decreased?
Answer
The least count of a screw gauge can be decreased by
  1. Decreasing the pitch and
  2. Increasing the total number of divisions on the circular scale.
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Question 202 Marks
Explain the terms : Least count of a screw gauge.
How are they determined?
Answer
Least count (L.C.) of a screw gauge : The L.C. of a screw gauge is the distance moved by it in rotating the circular scale by one division.
Thus, L.C. = Pitch of the screw gauge/total no. of divisions on its circular scale.
If a screw moves by 1 mm in one rotation and it has 100 divisions on its circular scale, then pitch of screw = 1 mm.
Thus, L.C. = 1 mm / 100 = 0.01 mm = 0.001 cm
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Question 212 Marks
The thimble of a screw gauge has 50 divisions. The spindle advances 1 mm when the screw is turned through two revolutions.
(i) What is the pitch of the screw gauge?
(ii) What is the least count of the screw gauge?
Answer
No. of divisions on the circular scale = 50
(i) Pitch = Distance moved ahead in one revolution
= 1 mm/2 = 0.5 mm
(ii) L.C. = Pitch/No. of divisions on the circular head
= (0.5/50) mm
= 0.01 mm
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Question 222 Marks
State three uses of the vernier callipers.
Answer
Three uses of vernier callipers are
(a) Measuring the internal diameter of a tube or a cylinder.
(b) Measuring the length of an object.
(c) Measuring the depth of a beaker or a bottle.
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Question 232 Marks
Define Standard metre.
Answer
The standard metre is defined in terms of speed of light, according to which, one metre is the distance travelled by light in $\frac{1}{299,792,458}$ of a second in air (or vacuum).
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Question 242 Marks
Define a fundamental unit.
Answer
A fundamental (or basic) unit is that which is independent of any other unit or which can neither be changed nor can be related to any other fundamental unit.
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Question 252 Marks
The mass of an oxygen atom is $16.00\ u.$ Find its mass in $kg.$
Answer
$\text { MASS OF OXYGEN = 16 u. }$
$1 u =1.66 \times 10^{-27} kg$
$kg =16 \times 1.66 \times 10^{-27}$
$=26.56 \times 10^{-27}$
$=2.656 \times 10^{-26} kg $
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Question 262 Marks
The wavelength of light is 589 nm. what is its wavelength in Å?
Answer
Wavelength of light $=589 nm$
1 angstrom $=0.1 nm$
We have $589 nm$
$
\begin{aligned}
& \frac{589}{0.1}=5890 \text { Angstroms } \\
& = 5 8 9 0 Å
\end{aligned}
$
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Question 272 Marks
What are the three requirements for selecting a unit of a physical quantity?
Answer
The three requirements for selecting a unit of a physical quantity are
(1) It should be possible to define the unit without ambiguity.
(2) The unit should be reproducible.
(3) The value of units should not change with space and time.
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Question 282 Marks
The distance of a galaxy is $5.6 \times 10^{25} m$. Assuming the speed of light to be $3 \times 10^8 ms ^{-1}$. Find the time taken by light to travel this distance . $\left[\right.$ Hint: Time taken $\left.=\frac{\text { Distance travelled }}{\text { speed }}\right]$
Answer
Distance of galaxy $=5.6 \times 10^{25} m$
Speed of light $=3 \times 10^8 m / s$
$ \text { Time taken by light }=\frac{\text { Distance travelled }}{\text { speed of light }}$
$=\left(\frac{5.6 \times 10^{25}}{3 \times 10^8}\right) s$
$=1.87 \times 10^{17} s $
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Question 292 Marks
The size of bacteria is $1 µ$. Find the number of bacteria present in $1 \ m$ length.
Answer
Size of a bacteria = $1 µ$
Since $1 µ = 10^{-6} m$
$\therefore$ Number of the particle = Total length/size of one bacteria
$= 1 m/10^{-6} m$
$= 10^6$
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Question 302 Marks
What is a lunar month?
Answer
One lunar month is the time in which the moon completes one revolution around the earth. A lunar month is made of nearly 4 weeks.
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Question 312 Marks
Name two units of time bigger than a second. How are they related to the Second?
Answer
The units minute (min) and year (yr) are two units of time bigger than second(s).
$1 min = 60 s$
$1 yr = 3.1536 \times 10^7 s$
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Question 322 Marks
Name the S.I unit of time and define it.
Answer
The S.I. unit of time is second (s).
A second is defined as 1/86400th part of a mean solar day, i.e.
$1 s =\frac{1}{86400} \times$ one mean solar day
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Question 332 Marks
State two units of mass bigger than a kilogram. Give their relationship with the kilogram?
Answer
The units quintal and metric tonne are two units of mass bigger than the kilogramme.
1 quintal = 100 kg
1 metric tonne = 1000 kg
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Question 342 Marks
State two units of mass bigger than smaller than a kilogram. How are they related to the kilogram.
Answer
The two units of mass smaller than a kilogram (kg) are:
gram (g) - Relation between gram and kilogram is
1 g = 10-3 kg
milligram (mg) - Relation between milligram and kilogram is
1 mg = 10-6 kg
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Question 352 Marks
Name the S.I unit of mass and define it.
Answer
S.I. unit of mass is ‘kilogramme’.
In 1889, one kilogramme was defined as the mass of a cylindrical piece of a platinum-iridium alloy kept at the International Bureau of Weights and Measures at serves near Paris.
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Question 362 Marks
Name two units of length which are bigger than a metre. How are they related of the metre?
Answer
Astronomical unit (A.U.) and kilometre (km) are units of length which are bigger than a metre.
$1 km = 1000 m$
$1 A.U. = 1.496 \times 10^{11} m$
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[2 Mark Question Answer] - PHYSICS STD 9 Questions - Vidyadip