Experimental Skills Notes and MCQs

Notes 75 Pages

Joint Entrance Examination (Main)

Physics

Contributed by

Yash Kuruvilla

360
EXPERIMENT -1
Vernier Callipers
SUMMARY
The least count of venier callipers = M-V ......(1)
Where,
M = The distance between two consecutiue divisions on the main scale.
V = The distance between two consecutive divisions on the vernier scale.
The length of x divisions on vernier scale = the length of y divisions on the main scale.
V M
x y
 
yM
V
x

From equation (1) The LC of vernier callipers =
yM -y
= M - = M -
x
x x
   
   
   
M
x

* Zero Error :
- When the jaws are made to touch eachother, the zero mark of the main scale and vernier scale
may not be with the straight line.
- This gives rise to an error called the zero error.
- Positive zero erro = (No. of the vernier division coinciding with the main scale) X (L.C.)
- Corrected reading = Obeserved reading - Positive zero error.
- Negative Zero error = (No. of vernier division coinciding with the main scale) X (L.C.) 0
(Smallest main scale units)
- Corrected reading = Observed reading + Negavit zero error)
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361
MCQ
1. What is the least count of the vernier callipers ?
(A) Smallest division on the vernier scale.
(B) differenceof the smallest division on the mai scale and the smallest division on the vernier
scale.
(C) sum of the smallest division on the main scale and the smallest division on the vernier scale.
(D) smallest division on the main scale.
2. What is the least count of commonly available vernier ?
(A) 0.01 cm (B) 0.001 cm
(C) 0.0001 cm (D) 0.1 cm
3. When the zero mark on the vernier scale lies towards the left side of the zero mark of the main scale,
when the jaws are connect, then what will be the zero error ?
(A) zero error is positive (B) zero error is negative
(C) zero correction is positive (D) zero error does not exist
4. When the zero mark on the vernier scale lies towards the right side of the zero mark of the main
scale, when the jaws are in contact, then what will be the zero error ?
(A) zero correction in positive (B) zero correction is negative
(C) zero error in positive (D) zero error does not exist
5. If observed reading is OR, corrected reading is CR, zero error in ZE and zero correction in ZC,
then what will be the possibility ?
(A) CR = OR + ZC and ZE = CR-OR (B) CR = OR + ZE and ZC = CR-OR
(C) CR = OR - ZC and ZE = OR-CR (D) CR = OR - ZE and ZC = CR-OR
6. When the jaws of a standard vernier are together, the 6
th
vernier scale division coincides with the 6
th
main scale division, then what in the zero error ?
(A) -0.4 mm (B) + 0.6 mm
(C) -0.6 mm (D) + 0.4 mm
7. When the jaws of a standard vernier are together, the 6
th
main scale division coincides with the 7
th
vernier scale division, then what is the zero error ?
(A) -0.7 mm (B) +0.3 mm
(C) -0.3 mm (D) +0.7 mm
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8. In an anusual vernier, 9 vernier scale divisions coincide with 8 main scale division, then what is the
least count of the vernier ?
(A)
9
8
mm (B)
9
1
mm
(C)
17
1
mm (D)
8
1
mm
9. In an unsual vernier, 10 vernier scale divisions, coinside with 8 main scale divisions, then what is the
least count of the vernier ?
(A) 0.1 mm (B) 0.2 mm
(C) 0.8 mm (D)
8
1
mm
10. Match the two columns :
Column - I Column - II
(a) Jaws CD (p) Slide and fix position of vernier scale
(b) Strip N (q) depth of a calerimeter
(c) Screw S (r) external diameter of a cylindrical vessel
(d) Jaws AB (s) Internal diameter of a cylindrical vessel.
(A) a

r, b

q, c

p, d

s
(B) a

s, b

p, c

q, d

r
(C) a

r, b

q, c

s, d

p
(D) a

p, b

s, c

q, d

r
11. N divisions on the main scale of a vernier callipers coincides with (N+1) divisions on the vernier
scale. If each division on the main scale is of a units, the least of count of instrument is.................
(A)
1N
a

(B)
1N
a

(C)
N+1
a
(D)
a
1N 
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363
12. The edge of a cube is measured using a vernier caliper (9 divisions of the main scale is equal to 10
divisions of vernier scale and 1 main scale division is 1mm). The main scale division reading is 10
and 1 division of vernier scale was found to be coinciding with the main scale. The mass of the cube
is 2.736 g. What will be the density in
3
cm
g
upto correct significant figures ?
(A)
3
3
cm
g
1066.2


(B)
3
3
cm
g
1066.2 
(C)
3
cm
g
66.2
(D)
3
6
cm
g
1066.2


KEY NOTE
1 (C) 2 (A) 3 (A) 4 (A)
5 (D) 6 (B) 7 (C) 8 (B)
9 (B) 10 (A) 11(A) 12 (C)
HINT
1. If main scale division = M and
vernier scale division = V
9M = 10 V
(10-1) M = 10V
10M - M = 10V
   
10
M
VM,MVM10 
= least count.
2. Least count
cm01.0mm1.0
10
mm1
n
M

3. Here zero error is positive, so zero. correction is negative, as zero correction = - (zero error)
4. Here zero error is negative, so zero correction is positive.
5. Corrected reading = Observed reading - zero error
ZEORCR 
CR = OR + ZC
ORCRZC 
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364
6. As vernier divisions are smaller then the main scale dividions (9M = 10V), the zero of the vernier
must be on the right side of the zero of the main scale. Here zero error is positive.
.

Zero error
   
mm6.0mm1.06VM6V6M6 
7. zero error
 
MV7M7V7M6 

 
mm3.0mm1mm1.07 
8.
V9MM9V9M8 
 
MVM9 
or
 
mm
9
1
9
M
VM 
9.
 
M2VM10,V10M2M10V10M8 
 
mm2.0
10
M2
VM 
11. (N+1) divisions on the vernier scale = N divisions on main scale

1 division on vernier scale
1N
N


divisions on main scale
Each division on the main scale in of a units

1 division on vernier scale








 1N
N
a units = a' (say)
Least count = 1 main scale division -1 vernier scale division
1N
a
a
1N
N
a'aa












12. 1 MSD = 1 mm
9 MSD = 10 VSD

Least count,
mm
10
9
mm1VSO1MSO1LC 
mm
10
1

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365
Measure reading of edge
 
LCVSRMSR 
1
= 10 + 1 = 10.1 mm
10
Volume of cube V = (101)
3
cm
3
= 1.03 cm
3
[After rounding off upto 3 significant digits, as edge length is measured upto 3 significant digits]
Density of cube
3
cm
g
6563.2
03.1
2736


3
cm
g
66.2
(After rounding off to 3 significant digits)
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366
EXPERIMENT -2
Micrometer screw gauge :
SUMMARY
 Least count of screw gauge
pitch
=
Total number of divisions on circular scale
 Negative zero error :
If the zero error is negative and the n
th
division of the circular scale coincides with the line of graduation
then
 
LC100nz 
corrected reading = observed reading + zero correction
 If the zero error is positive and the n
th
division of the circular scale division concides with the line of
graduation then
LCnz 
corrected reading = observed reading - zero error.
 If the zero mark of the venier upwards of the line of the graduation then megative zero error arise
and zero mark of the verniers downwards then positive zero error aris.
MCQ
1. When the edge of the circular scale lies to the left of O mark on the main scale, when the stud and
spindle touch each other, Then what will be the zero error ?
(A) zero error is negative (B) zero error is positive
(C) zero error does not exist (D) zero correction is negative
2. Whe the edge of the circular scale lies to the right of the O mark on the main scale, when the stud
and the spindle touch each other,then what will be the zero error ?
(A) zero correction is negative (B) zero error is negative
(C) zero error does not exist (D) zero correction is positive
3. When the screw and stud touch each other, the edges of a certain screw gauge is on left of the O
mark on the main scale and the 96
th
division of the circular scale coincides with the circular line of
graduation then what is the value of zero error ?
(A) zero error = + 0.96 mm (B) zero error = – 0.96 mm
(C) zero error = + 0.04 mm (D) zero error = – 0.04 mm
4. When the screw and stud touch each other, the edge of a certain screw gauge is to the right of the O
mark on the main scale and 5
th
division of the circular scale coincides with the line of graduation,
then what is the value of zero error ?
(A) zero error = + 0.95 mm (B) zero error = – 0.95 mm
(C) zero error = + 0.05 mm (D) zero error = – 0.05 mm
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5. Screw guage A has a pitch of 1 mm and 50 division on its circular scale screw guage B has a pitch
of 0.5 mm and 100 divisions on its circular scale. If (LxC) is least count, then which posibility is
true ? What is the parilrility ?
(A)




BA
CLCL2  (B)




BA
CL4CL 
(C)




BA
CLCL  (D)




BA
CL2CL 
6. The screw gauge shown above has a zero error of -0.02 mm and 100 divisions on the circular scale.
What is the diameter of wire ?
(A) 0.28 mm (B) 0.22 mm
(C) 0.24 mm (D) 0.26 mm
7. The pitch of screw gauge is 1 mm and there are 100 divisions on the circular scale. What measuring
the diameter of a wire, the linear scale reads, 1 mm and 47
th
division on the circular scale coincides
with the reference line. The length of the wire is 5.6 cm.What will be the curved surface area (in cm
2
)
of the wire in appropriate number of significant figures.
(A) 2.6 cm
2
(B) 2.5848 cm
2
(C) 2.585 cm
2
(D) 2.5 cm
2
8. Match column A and B
Column - I Column - II
(a) A (p) Main scale
(b) B (q) circular scale
(c) C (r) stud
(d) D (s) spindle
(e) E (t) Ratchet
(f) E (u) Thimble
(g) G (v) Screen
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(a) a

r, b

s, c

v, d, d

p, e

q, f

u, g

t
(b) a

u, b

t, c

v, d, d

p, e

q, f

r, g, g

s
(c) a

v, b

p, c

r, d, d

s, e

t, f

q, g

u
(d) a

p, b

q, e

u, d

t, e

r, f, f

s, g

v
KEY NOTE
1 (A) 2(A) 3 (D) 4(C)
5 (B) 6 (D) 7(A) 8(A)
HINT
2. Here zero error is positive, so zero correction is negative.
3. Here the zero error is negative.
 
LC10096Z 
mm01.04  mm04.0
4. Here the zero error is positive
mm01.05C.L5z  mm05.0
5. Least count
scalecircularondivisionsof.No
Pithc

6. The observed reading is 0.24 mm. The corrected reading = observed reading - zero error.
 
mm02.0mm24.0 

mm26.0
7. Curved surface area =
2 rl
Page 9
369
Experiment - 3
SIMPLE PENDULUM
SUMMARY
 Simple pendulum oscillation consider as a undamped oscillation.
periodic time
l 2π
T = 2π =
g ω
Mechanical energy
 
2
0
1
E = k A
2
 Simple pendulum oscillation consider as a damped oscillatien,
Periodic time
 
2 2
2
mr + ml
I
5
T = 2π = 2π I = M.I
mgl mgl
Here
2
l
T
g
 
 Mechanical energy
2
–bt
2
2m
0
1
E = kA e
2
 
 
 
displacement
 
–bt
2m
0
A e cos 't + φx  
where
2
2
k b
' = =
m 4m

MCQ
1. Complete the following sentance.
Time period of oscillatin of a simple pendulum is dependent on............
(A) Length of thread (B) initial phase
(C) amplitude (D) mass of bob
2. Complete the following seutence
In a damped oscillation of a pendulam .........
(A) the sum of potential energy and kinetic energy is conserved.
(B) mechanical energy is not conserved
(C) the kinetic energy is conserved
(D) the potential energy is conserved
Page 10

Experimental Skills Notes and MCQs

Experimental Skills Notes and MCQs

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