Chap No.1 Measurements MCQs with solution
Chap No.1 Measurements
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1)
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Which expression using SI base units is
equivalent to the volt; a. kg m2
s-1 A-1 b.kg
m s-2 A c. kg m-2
s-1 A d.
kg m2 s-3 A-1 |
D |
V=W/q=fd/q=mad/It=mvd/tIt=msd/ttIt= kgmm/ssAs=
kg m2 /s3 A1= kg m2 s-3
A-1 |
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2)
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What is the
circumfrence of the circle whos area is 100p : a. 10p b. 20p c. 10 d.290 |
B |
Circumfrnce
of circle(c)= 2pr and area of circle(a) =pr2 NOW a=pr2=100p→r2=100→r=10 to find C=2pr = 2p10=20p |
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3)
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The force of one Newton per meter square is
equal to one. (a) Bar (b)
Atm (c) Pascal (d)
Erg. |
C |
P=F/A |
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4)
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Which
of following is unit of Pressure? (a) Kg m s-1 (b)
Kg m-1 s-2 (c) Kg m2
s-2 (d) Kg m-2 s-1
|
B |
Kg m-1
s-2Hints: |
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5)
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If p is a
pressure and d is a density then p/dha units of: (a) m2/s2 b)
N/m2 (c) Kg/m2 (d)
m3/Kg |
A |
P =
F/A=ma/A= kg. ms-2/m2 = kg/m s and dha=(m/V)ha= kg. m.m2/m3
= kg now p/dha= kg/ms /kg =m s |
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6)
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Which of
the following is closest to a yard: (a) 0.01 m (b)
0.1 m (c) 1 m (d) 100 m |
C |
1 m is
closed to year |
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7)
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A student
measure current as 0.5A. which of the following correctly expresses the
result: a. 50Ma b. 50MA c.500mA d.
500MA |
C |
0.5 A= 0.5
mA/m = 0.5/10-3 mA= 0.5 x 103 mA= 5x 1000 mA=500 mA |
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8)
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The prefix
“tetra” stands for; a. 106
b. 109 c. 10-9 d. 106 |
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exa |
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9)
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What is the
ratio of 1 Gm/1µm? (a) 10-3 (b) 10-7 (c) 10-18 (d) 1015 |
D |
1Gm =Gaga
meter = 109,1µm = micro meter = 10-6 Thus; 1Gm/1µm = |
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10) |
The prefix ‘Pico’ stands for: (a) 106 (b) 109 (c) 10–12 (d) 1012
|
C |
Pice = 10-12 |
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11) |
9.5 × 1015m
when rounded off 40 is 1016 m which is equal to (a) Tera meter (b) Atto meter (c) Exa
meter (d)
Light year |
D |
as Peta =
1015 so Light year = 9.5 × 1015 m = peta meter |
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12) |
The
measurement of physical quantity may be subject to random errors and to
systematic errors. Which statement is
correct? (a) Random
errors are always caused by the person taking the measurement. (b) A
systematic error cannot be reduced (c) Random
errors can be reduced by taking the average of several measurements (d) A
systematic error results in a different reading each time the measurement is
taken. |
C |
Random
errors can be reduced by taking average of several measurements |
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13) |
The number
of significant fiures in the measurement of 5.005x 10-5 s is; a. 2 b. 3 c. 4 d 5 |
c |
The zero
between two significant figure is also significant and power is not counted
in significant figures. |
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14) |
The
scientific notation of a number 0.0023 is expressed as: A) 2.3 × 10-3
B) 0.023 × 10-2 C)2.3 × 10-4 D) 0.2 × 103 |
A |
2.3 × 10-3.
Decimal moved to right after first non zero digit and sign of power will be
negative while moving to right |
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15) |
If 7.635
& 4.81 are two significant numbers, their multiplication in significant
digit is: (a)
36.72435 (b) 36.724 (c) 36.72 (d)
36.7 |
B |
Answer
should be carried to least significant figure operation i.e. 4.81 which are
to be multiplied. |
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16) |
The number
of significant figures in 4.0030 is; (a)
Four (b) Five (c) Two (d)
Three |
B |
Zero after
decimal to the right are also significant |
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17) |
The number
of significant figures in the measurement x = 10.00300; (a) 7 (b) 8 (c) 5 (d) 3 |
A |
Zero after
decimal to the right are also significant because it sow least count of
measuring instrument |
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18) |
The number
of significant figures in the measurement of 5.05 ´ 10-3 m/s is; (a) 2 (b) 3 (c) 4 (d) 8 |
b |
Number in
powers are in included in significant figures |
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19) |
During the
experiment one measured the mass of Mosquito and fount it 1.20×10–5
Kg. The numbers of significant figures in this case are: (a) Five (b) One (c) Two (d) Three |
D |
Number in
powers are in included in significant figures |
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20) |
In a
cricket match 500 spectators are counted one by one. How many significant
figures will be there in the final result? (a) 0 (b) 1 (c) 2 (d) 3 |
D |
If L.C is 100 than 1 significant figure. If L.C is 10 than 2
significant figure. As there are 500 spectators and are counted one by one means L.C is 1
So there will be 3 significant figure |
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21) |
A value for
the acceleration of free fall on earth is given (10+2) ms2. Which
statement is the most correct a. the
value is accurate but not precise b. the
value is both accurate and precise c. the
value is neither precise nor accurate d. the
value is precise but not accurate |
A |
Accurate
because value is near to 9.8 but not precise because least count is little
more. |
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22) |
The maximum
error in measurement of mass and length of the sides of the cube are 3% and
2% respectively. The maximum error in the measurement of its density is; a. 3% b. 5% c. 6% d. 9% |
d |
D=m/V=m/LLL=3/2222=3+2+2+2=9% |
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23) |
In simple
electrical circuit the current in a resistor is measured as 2.50+0.05mA. the
resistor is marked as having a value of 4.7 a. 2% b. 4% c. 6% d.
8% |
c |
P=VI=IRI=I2%I=2+2+2=6% Uncertainty
in I=0.05/2.50 x100=2% |
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24) |
The power
loss in resistor is culates by formula
P=V2/R. the uncertaniuty in V s 3% and in R is 2%.
Uncertainty in P is: a. 4% b.
7% c. 8% d.
11% |
|
P=V2/R=VV/R=3%3%/2%=3+3+2=8% |
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25) |
A quantity
x is to be determined by the equation x=P-Q. P is measured as 1.27 a. 0.04% b.2% c. 3% d.
7% |
D |
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26) |
The quantity
x is to be determined form the equation x = p-Q.nP is measured as (1.27+
0.02)m and Q is measured as (0.03 + 0.01)m. What is the percentage
uncertainty in x to one significant figure.
(a) 4% (b) 2% (c) 3% (d)7% |
B |
Hints; x=P-Q=1.27-0.03=1.24 , Error =
0.02+0.01=0.03 Percentage
uncertainty |
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27) |
The density of the steel ball was determined
by measuring the mass and diameter. The mass was measured with 1% and
diameter 3% of the error. In the calculated density of the steel ball is at
most. (a) 2% (b) 4% (c) 8% (d) 10% |
D |
By formula
of density D=m/V ,the error is 1%+3% = 4% |
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28) |
The power loss, P in resistor is calculated
using the formula P = V2/R. The uncertainty in the potential
difference V is 3% and the uncertainty in the resistance R is 2%, what is the
uncertainty in P? (a) 4% (b) 7% (c) 8% (d) 11% |
C |
P = Note; power
is multliped to the error |
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29) |
The uncertainty recorded in the radius of a
sphere is 1.6%. The uncertainty in the area of that sphere is; (a) 4.8% (b) 3.2% (c) 1.6% d) 0.8% |
B |
Area of
sphere = 4pr2 => Thus uncertainty in
area = (1.6%)2 = 1.6% × 2 = 3.2% (In uncertainty power is multiplied) |
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30) |
The percentage error in the measurement of
mass and speed are 5% or 6% respectively the maximum error in the measurement
of K.E is:
A) 17% B)
30% C) 15% D)
90% |
A |
Maximum
Error in K.E = |
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