[5 Mark Question Answer] - PHYSICS STD 9 Questions - Vidyadip

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Question 15 Marks

How does the time period (T) of a simple pendulum depend on its length (l)? Draw a graph showing the variation of $T^2$ with l. How will you use this graph to determine the value of g (acceleration due to gravity)?

Answer

The time period of a simple pendulum is directly proportional to the square root of its effective length.
$T \propto \sqrt{l}$

From this graph, the value of acceleration due to gravity ( $g$ ) can be calculated as follows.
The slope of the straight line can be found by taking two points $P$ and $Q$ on the straight line and drawing normals from these points on the $X$ - and $Y$-axis, respectively. Then, the value of T2 is to be noted at $a$ and $b$, the value of $I$ at $c$ and $d$. Then,
$\text { Slope }=\frac{ PR }{ QR }=\frac{ ab }{ cd }=\frac{T_1^2-T_2^2}{l_1-l_2}$
This slope is found to be constant at a place and is equal to $\frac{4 \pi^2}{ g }$, where $g$ is the acceleration due to gravity at that place. Thus, $g$ can be determined at a place from these measurements using the following relation:
$g =\frac{4 \pi^2}{\text { Slope of } T^2 Vs l \text { graph }}$

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Question 25 Marks

How is the time period of a simple pendulum affected, if at all, in the following situations:
(a) The length is made four times.
(b) The acceleration due to gravity is reduced to one-fourth.

Answer

We know that, $T =2 \pi \sqrt{\frac{l}{ g }}$
(a) If length quadruples then,
$T \prime=2 \pi \sqrt{\frac{4 l}{ g }}$
or , $\frac{T}{T \prime}=\frac{2 \pi \sqrt{\frac{l}{ g }}}{2 \pi \sqrt{\frac{4 l}{ g }}}=\frac{1}{2}$
or,$T=2 T^{\prime}$
Therefore, the time period is doubled.
(b) If the acceleration due to gravity is reduced to one-fourth,
$ T \prime=2 \pi \sqrt{\frac{l}{\frac{1}{4} g }}$
$\text { or }, \frac{ T }{ T \prime}=\frac{2 \pi \sqrt{\frac{l}{ g }}}{2 \pi \sqrt{\frac{l}{\frac{1}{4} g }}}=\frac{1}{2} $
or, $T=2 T^{\prime}$
Therefore, the time period is doubled.

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Question 35 Marks

A seconds pendulum is taken to a place where acceleration due to the gravity falls to one-fourth. How is the time period of the pendulum affected, if at all? Give reasons. What will be its new time period?

Answer

Time period of 'a' is inversely proportional to the square root of acceleration due to gravity. i.e $T \propto \frac{1}{\sqrt{ g }}$
Now, if the acceleration due to gravity falls to one-fourth, the time period will be doubled.
Let the new time period be $T ^{\prime}$ and let $g$ ' be the acceleration due to gravity.
Then,
$
\frac{T}{T^{\prime}}=\sqrt{\frac{ g ^{\prime}}{ g }}
$
or , $\frac{T}{T \prime}=\sqrt{\frac{\frac{1}{4} g }{ g }}$
or , $\frac{T}{T \prime}=\frac{1}{2}$
or, $T^{\prime}=2 T$
or, $T =2 \times 2=4 s$

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Question 45 Marks

Draw a neat and labelled diagram of a vernier callipers. Name its main parts and state their functions.

Answer

Diagram of vernier callipers:

Main parts and their functions:
Main scale: It is used to measure length correct up to 1 mm.
Vernier scale: It helps to measure length correct up to 0.1 mm.
Outside jaws: It helps to measure length of a rod, diameter of a sphere, external diameter of a hollow cylinder.
Inside jaws: It helps to measure the internal diameter of a hollow cylinder or pipe.
Strip: It helps to measure the depth of a beaker or a bottle.

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Question 55 Marks

In the vernier callipers, there are $10$ divisions on the vernier scale and $1 \ cm$ on the main scale is divided into $10$ parts. While measuring the length, the zero of the vernier lies just ahead of the $1.8 \ cm$ mark and the $4th$ division of vernier coincides with a main scale division.
(a) Find the length.
(b) If zero error of vernier callipers is -$0.02 cm,$
What is the correct length ?

Answer

(a) L.C. of vernier callipers $=0.01 cm$
Main scale reading $=1.8 cm$
Since $4^{\text {th }}$ division of the main scale coincides with the main scale, i.e. $p=4$.
Therefore, the vernier scale reading $=4 \times 0.01 cm=0.04 cm$
Total reading $=$ Main scale reading + vernier scale reading
$=(1.8+0.04) cm$
$=1.84 cm$
(b) Observed reading $=1.84 cm$
Zero error $= -0.02 cm$
Correct reading = Observed reading - Zero error (with sign)
$= [1.84 - (-0.02)] cm$
$= 1.86 cm$

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Question 65 Marks

What is meant by zero error of vernier callipers ? How is it determined? Draw neat diagrams to explain it. How is it taken in account to get the correct measurement?

Answer

Due to mechanical errors, sometimes the zero mark of the vernier scale does not coincide with the zero mark of the main scale, the vernier callipers is said to have zero error.
It is determined by measuring the distance between the zero mark of the main scale and the zero mark of the vernier scale.
The zero error is of two kinds
(i) Positive zero error
(ii) Negative zero error
(i) Positive zero error: On bringing the two jaws together, if the zero mark of the vernier scale is on the right of the zero mark of the main scale, the error is said to be positive.

To find this error, we note the division of the vernier scale, which coincides with any division of the main scale. The number of this vernier division when multiplied by the least count of the vernier callipers, gives the zero error.
For example, for the scales shown, the least count is 0.01 cm and the $6^{th} $division of the vernier scale coincides with a main scale division.
Zero error $= +6 \times L.C. = +6 \times 0.01 cm$
$= +0.06 cm$
(ii) Negative zero error: On bringing the two jaws together, if the zero mark of the vernier scale is on the left of the zero mark of the main scale, then the error is said to be negative.

To find this error, we note the division of the vernier scale coinciding with any division of the main scale. The number of this vernier division is subtracted from the total number of divisions on the vernier scale and then the difference is multiplied by the least count.
For example, for the scales shown, the least count is 0.01 cm and the sixth division of the vernier scale coincides with a certain division of the main scale. The total number of divisions on vernier callipers is 10.
Zero error $= - (10 - 6) \times L.C.$
$= - 4 \times 0.01 cm = - 0.04 cm$
Correction:
To get correct measurement with vernier callipers having a zero error, the zero error with its proper sign is always subtracted from the observed reading.
Correct reading = Observed reading - zero error (with sign)

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Question 75 Marks

The main scale of a vernier callipers is calibrated in mm and $19$ divisions of main scale are equal in length to $20$ divisions of the vernier scale. In measuring the diameter of a cylinder by this instrument, the main scale reads $35$ divisions and $4$th division of vernier scale coincides with a main scale division. Find
(i) Least count and (ii) Radius of a cylinder.

Answer

$ \text { (i) Value of } 1 m . s . d =1 mm =0.1 cm$
$20 \text { vernier divisions }=19 m . s . d \text {. }$
$\text { L.C. }=\text { Value of } 1 m . s . \text { d. } / \text { number of divisions on the vernier scale }$
$=1 mm / 20$
$=(0.1 / 20) cm$
$=0.005 cm $
(ii) Main scale reading $=35 mm =3.5 cm$
Since $4^{\text {th }}$ division of the main scale coincides with the main scale, i.e. $p=4$.
Therefore, the vernier scale reading $=4 \times 0.005 cm =0.02 cm$
Total reading $=$ Main scale reading + vernier scale reading
$ =(3.5+0.02) cm$
$=3.52 cm $
Radius of the cylinder $=\frac{\text { Diameter }(\text { Total reading) }}{2}$
$ =\left(\frac{3.52}{2}\right) cm$
$=1.76 cm $

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Question 85 Marks

A vernier callipers has its main scale graduated in mm and 10 divisions on its vernier scale are equal in length to $9 \ mm$. When the two jaws are in contact, the zero of the vernier scale is ahead of the zero of the main scale and the 3rd division of the vernier
scale coincides with a main scale division.
Find : (i) The least count and
(ii) The zero error of the vernier callipers.

Answer

(i) 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 the vernier scale
$= 1 mm/10$
$= 0.1 mm or 0.01 cm$
(ii) On bringing the jaws together, the zero of the vernier scale is ahead of zero of the main scale, the error is positive.
$3^{rd}$ vernier division coincides with a main scale division.
Total no. of vernier divisions $= 10$
Zero error $= +3 \times L.C.$
$= +3 \times 0.01 cm$
$= +0.03 cm$

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Question 95 Marks

Define the least count of vernier callipers. How do you determine it?

Answer

The least count of vernier callipers is equal to the difference between the values of one main scale division and one vernier scale division.
Let $n$ divisions on vernier callipers be of length equal to that of ( $n-1$ ) divisions on the main scale and the value of 1 main scale division be $x$. Then,
Value of $n$ divisions on vernier $=(n-1) x$
Alternatively, value of 1 division on vernier $=\frac{(n-1) x }{n}$
Least count $=x-\frac{(n-1) x }{n}=\frac{x}{n}$
L.C. $=\frac{\text { (Value of one main scale division) }}{\text { (Total no. of divisions on vernier callipers) }}$
Value of one main scale division $=1 mm$
Total no. of divisions on vernier $=10$
Therefore, L.C. $=\frac{1 mm }{10}=0.1 mm =0.01 cm$

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Question 105 Marks

Describe the procedure to measure the diameter of a wire with the help of a screw gauge.

Answer

Measurement of diameter of wire with a screw gauge:
The wire whose thickness is to be determined is placed between the anvil and spindle end, the thimble is rotated until the wire is firmly held between the anvil and spindle. The rachet is provided to avoid excessive pressure on the wire. It prevents the spindle from further movement. The thickness of the wire could be determined from the reading as shown in the figure below.

The pitch of the screw $=1 mm$
L.C. of screw gauge $=0.01 mm$
Main scale reading $=2.5 mm$
$46^{\text {th }}$ division of circular scale coincides with the base line.
Therefore, circular scale reading $=46 \times 0.01=0.46 mm$
Total reading $=$ Main scale reading + circular scale reading $=(2.5+0.46) mm$
$=2.96 mm$

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Question 115 Marks

Draw a neat and labelled diagram of a screw gauge.
Name its main parts and state their functions.

Answer

Diagram of screw gauge:

Main parts and their functions:

Ratchet:It advances the screw by turning it until the object is gently held between the stud and spindle of screw.
Sleeve:It marks the main scale and base line.
Thimble:It marks the circular scale.
Main scale:It helps to read the length correct up to 1 mm.
Circular scale:It helps to read length correct up to 0.01 mm.

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Question 125 Marks

Fig., below shows the reading obtained while measuring the diameter of a wire with a screw gauge. The screw advances by 1 division on the main scale when circular head is rotated once.

Find:
(i) Pitch of the screw gauge,
(ii) Least count of the screw gauge and
(iii) The diameter of the wire.

Answer

No. of divisions on the circular scale = 50
(i) Pitch = Distance moved ahead in one revolution
= 1 mm/1 = 1 mm.
(ii) L.C. = Pitch/No. of divisions on the circular head
= (1/50) mm
= 0.02 mm
(iii) Main scale reading = 4 mm
No. of circular division coinciding with m.s.d. (p) = 47
Circular scale reading = p × L.C.
= (47 × 0.02) mm
= 0.94 mm
Diameter (Total reading) = M.s.r. + circular scale reading
= (4 + 0.94) mm
= 4.94 mm

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Question 135 Marks

A screw gauge has 50 divisions on its circular scale and its screw moves by 1 mm on turning it by two revolutions. When the flat end of the screw is in contact with the stud, the zero of the circular scale lies below the base line and 4th division of the circular scale is in line with the base line. Find
(i) The pitch,
(ii) The least count and
(iii) The zero error 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
(iii) Because the zero of the circular scale lies below the base line, when the flat end of the screw is in contact with the stud, the error is positive.
No. of circular division coinciding with m.s.d. = 4
Zero error = + (4 × L.C.)
= + (4 × 0.01) mm
= + 0.04 mm

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Question 145 Marks

When a screw gauge with a least count of 0.01 mm is used to measure the diameter of a wire, the reading on the sleeve is found to be 1 mm and the reading on the thimble is found to be 27 divisions.
(i) What is the diameter of the wire in cm?
(ii) If the zero error is +0.005 cm, what is the correct diameter?

Answer

(i) L.C. of screw gauge = 0.01 mm or 0.001 cm
Main scale reading = 1 mm or 0.1 cm
No. of division of circular head in line with the base line (p) = 27
Circular scale reading = (p) × L.C.
= 27 × 0.001 cm
= 0.027 cm
Diameter (Total reading) = M.s.r. + circular scale reading
= (0.1 + 0.027) cm
= 0.127 cm
(ii) Zero error = 0.005 cm
Correct reading = Observed reading - zero error (with sign)
= [0.127 - (+0.005)] cm
= 0.122 cm

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Question 155 Marks

Describe in steps, how would you use a vernier callipers to measure the length of a small rod ?

Answer

Measuring the length of a small rod using vernier calipers:
The rod whose length is to be measured is placed in between the fixed end and the vernier scale as shown in the figure.

In this position, the zero mark of the vernier scale is ahead of $1.2 \ cm$ mark on main scale. Thus the actual length of the rod is $1.2 \ cm$ plus the length ab (i.e., the length between the $1.2 \ cm$ mark on the main scale and 0 mark on vernier scale).
To measure the length $a b$, we note the $p ^{\text {th }}$ division of the vernier scale, which coincides with any division of main scale.
Now, $a b+$ length of $p$ divisions on vernier scale $=$ length of $p$ divisions on main scale
Alternatively, $a b=$ length of $p$ divisions on the main scale - length of $p$ divisions on the vernier scale.
$= p$ (length of 1 division on main scale - length of 1 division on vernier scale)
$= p \times L.C.$
Therefore, total reading = main scale reading + vernier scale reading
$= 1.2 cm + (p \times L.C.)$

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Question 165 Marks

The pitch of a screw gauge is 1 mm and the circular scale has 100 divisions. In measurement of the diameter of a wire, the main scale reads 2 mm and 45th mark on the circular scale coincides with the base line. Find
(i) The least count and
(ii) The diameter of the wire.

Answer

Pitch of the screw gauge = 1mm
No. of divisions on the circular scale = 100
(i) L.C. = Pitch/No. of divisions on the circular head
= (1/100) mm
= 0.01 mm or 0.001 cm
(ii) Main scale reading = 2mm = 0.2 cm
No. of division of circular head in line with the base line (p) = 45
Circular scale reading = (p) × L.C.
= 45 x 0.001 cm
= 0.045 cm
Total reading = M.s.r. + circular scale reading
= (0.2 + 0.045) cm
= 0.245 cm

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Question 175 Marks

What are the fundamental unit in S.I system? Name them along with their symbols.

Answer

Fundamental quantities, units and symbols in S.I. system are:

Quantity	Unit	Symbol
Length	Metre	m
Mass	kilogramme	kg
Time	second	s
Temperature	kelvin	K
Luminous intensity	candela	cd
Electric current	ampere	A
Amount of substance	mole	mol
Angle	radian	rd
Solid angle	steradian	st-rd

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Question 185 Marks

'The distance of a star from the earth is $8.33$ light minutes.' what do you mean by this statement? Express the distance in metre.

Answer

The distance of a star from the earth is 8.33 light minutes' implies, it takes $8.33$ minutes for light to reach the earth from the star.
As we know,
Distance = speed × time
Given,
Speed $= 3 \times 10^8ms^{-1}$
$Time = 8.33 min$
$= 8.33 \times 60s$
$= 499.8s$
$≈ 500s$
Substituting the values in the formula above we get,
Distance $= 3 \times 10^8 \times 500$
$= 1500 \times 10^8$
$= 1.5 \times 10^{11}$
Therefore, the distance from the star to the earth is $1.5 \times 10^{11} m.$

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[5 Mark Question Answer] - PHYSICS STD 9 Questions - Vidyadip