PHYSICS CLASS 9TH CHAPTER 1 ENGLISH MEDIUM
MCQS AND SHORT QUESTIONS WITH NUMERICALS
IQ SCHOOL 18 HAZARI CLASS 9TH NOTES PHYSICS
PRINCIPAL: NEELUM MALIK
VP: ASIA MALIK
M.D: FAREED FIDA
CONTACT NO:03066734325
Chapter no 1
PHYSICAL QUANTITIES AND MEASUREMENTS
MCQ’S
1. Number of base units in SI a) 3 b) 6 c) 7 d) 9
2. Which is not derived unit (a) watt (b)Newton (c) kg (d) Pascal
3. Unit to find the amount of substance in terms of numbers (a)mole (b)Newton (c) kg (d)gram
4. 200u second is equal to (a)0.2 s (b)0.02s (c)2×10-4s (d)2×10-6s
5. Which is smallest one quantity(a)0.01g (b)2mg (c)100mg (d) 5000ng
6. The most accurate instrument to determine the internal diameter of a test tube is
(a)v. calipers (b)screw gauge (c)meter rod (d)measuring tape
7. A student claimed the diameter of wire is 1.032cm using v. calipers. Up to what extent do you agree with him? (a)1cm (b)1.0cm (c)1.03cm (d) 1.032cm
8. A measuring cylinder is used to measure (a) mass (b) area (c) volume (d) level of liquid
9. A student noted the thickness of glass sheet using a screw gauge. On the main scale it reads 3 div. while 8th division on circular scale coincides with index line. Its thickness is? (a) 3.8cm (b) 3.08mm (c) 3.8mm (d) 3.08m
10. Significant figures in an expression are (a) all the digits (b)all the accurately known digits (c) all accurately known digits & the first doubtful digit (d) all accurately known and all doubtful digits
SHORT QUESTIONS WITH ANSWERS
Q No; 1.5 Estimate your age in seconds?
Ans: My age of 14 years in seconds is 14x365x24x60x60s= 441504000s
Q:1.6 What role SI units have played in the development of science?
Ans: system International (SI) is adopted since 1960. It has played an important role in the field of science & technology especially to exchange the scientific and technical information across the nations.
Q#1.7What is meant by vernier constant?
ANS. least count of verniercalipers is called vernier constantof verniercalipers which can be found as given below
L.C =
L.C = 
mm
mm = o.1mm
= o.o1cm.
Q#1.8: What do you understand by the zero error of a measuring instrument?
Zero error in vernier callipers
ANS: if an instrument has an error in start in such a way that its initial reading is not starting with zero, its reading is either above or below the zero reading then its error is called zero error.
Q: #1.9 why is the use of zero error necessary in a measuring instrument?
Ans. To calculate accurate measurement we need to measure zero error if exists, zero error in any measurement is calculated and corrected it is called zero correction.
Q: 1.10What is a stopwatch?
ANS: A stopwatch is an instrument used to measure Intervals of time. Least count of a mechanical stopwatch that we use in laboratories is 0.1sec.
Q#1.11; Why do we need to measure extremely small intervals of time?
ANS: In Laboratory experiments we often need to measure very much small intervals of time like the time interval of the falling ball, the time interval of vibration of simple pendulum etc.
Q:1.12What is meant by significant figures of a measurement?
ANS: All the accurately known digits and the first doubtful digit are called significant figures.
Q:1.13. How is precision related to the significant figures in a measured quantity?
Ans: The measurement is more precise if it has more number of significant figures. There will be more number of significant figures if the least count of measuring instrument is smaller.
Numerical
Q1.1: Express the following quantities using prefixes
A: 5000 g
B: 2000 000 w
C: 52x10-10 kg
D: 22510 s
(A); 5000 g
= 5 × 103 g = 5kg
(B) = 2000 000 w
=2 × 10 6 w
= 2 Mw
(C) = 52 x 10-10kg
= 52 x 10-10kg
= 52 x 10-10× 103 g
= 5.2 x 10 1 x10-7g
= 5.2 x10 - 6g
= 5.2 µg.
Q:1.2How all prefixes identify with one another?
( smaller scale and nano )
1nano = 10-9
1micro = 10-6
1 nano = 10-6 × 10-3
1 nano = 1µ x 1/1000
1000 nano = 1u
(pico and nano)
1pico =10-12
1pico =10-9 x 10-3
1 pico=1nano x
1000pico = 1nano
(Smaller scale and Pico)
1pico = 10-12
1pico= 10-6 x10-6
1pico= 1micro x
1000000pico= 1 smaller scale
Q#1.3Your hair develops at the pace of 1 mm for each day .discover your development rate in nanometer every second.
Arrangement;
Your hair development rate = as
1nm = 10-9m
1nm = 10-3× 10-6m
1nm= 1mm×
1mm=1000000 nm
: 1 day= 24x60x60s= 86400 s
Your hair development rate
=1 000000 nm/86400 s
Your hair development rate = 11.57 NMS-1
Q#1.4: Rewrite the accompanying in standard structure.
A: 1168 x10-27
= 1.168 x103 x 10-27
= 1.168 x10-24
= 1.168 x10-24
(B): 32 x105
= 3.2 x101 x 105
= 3.2 x106
(C): 725 x10-5 kg
=7.25 x 102 x10-5
= 7.25 x 10-3
(D); 0.02 x 10-8
= 2 x10-2
=2 x10-10
Q:#1.5 Write the accompanying amounts in standard structure.
(A): 6400 km
= 6.4 × 103 km
(B) : 380000km
= 3.8 × 105 km
(C): 3 00 000, 000 ms-1
= 3 ×108 ms-1
(D):
One day = 24 ×60×60 s
= 8.64 × 104 s
Q#1.6; On shutting the jaws of the vernier callipers .0 of the vernier scale is on the privilege to its primary scale to such an extent that fourth division of its vernier scale agrees to one of the MASTER scale LINE figure its zero adjustments and zero mistakes?
Ans.: zero mistakes = 4 × least tally
Zero mistake = 4 × (0.01 cm)
Zero mistake = 0.04 cm
Zero remedy = - 0.04cm
Q#1.7: A screw check has 50 divisions on its round scale. with 5mm screw measure pitch. compute the least check?
Least count=?
Absolute no of roundabout scale div.= 50
Pitch of screw measure = 0.5 mm
Pitch = cmPitch = 0.05 cm
L.C =
L.C =
L.C = 0.001cm
Q:#1.8Which of the accompanying amounts have three noteworthy figures?
Answers;
(b) 0.00309 kg has 3 quantities of noteworthy figures.
(c) 5.05× 10-27kg has 3 quantities of noteworthy figures.
Q:#1.9; What are the noteworthy figures in the accompanying estimations?
(A); 1.009 m
No. of noteworthy figures = 4
(B); 0.00450 kg
No. Of noteworthy figures =3
(C); 1.66 × 10-27 Kg
No. Of noteworthy figures=3
(D) ; 2001 S
No. Of noteworthy figures = 4
Q #1.10: A chocolate wrapper is 6.7 cm long and 5.4 cm width figure its region up to sensible no. of critical figures.
ANS:
Information;
Length =6.7 cm
Width = 5.4 cm
Territory =6.7 cm
Territory = 6.7 cm × 5.4 cm
Territory = 36.18 cm2
Territory = 36 cm2
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