Measurements (Errors) Practice Questions - DPP for NEET

 Page 1


1. (d) Density 
2
r==
p
MM
V
rL
Þ 2
Dr D DD
=++
r
M rL
M rL
= 
0.003 0.005 0.06
2
0.3 0.5 6
+´+
= 0.01 + 0.02 + 0.01 = 0.04
\ Percentage error = 
100 0.04 100 4%
Dr
´ = ´=
r
2. (a) In division (or multiplication), the number of significant
digits in the answer is equal to the number of
significant digits which is the minimum in the given
numbers.
3. (d) Percentage error in A
= 
1
2 1 3 3 1 2 2 % 14%
2
æö
´+´+´+´=
ç÷
èø
4. (a) Percentage error in X = aa+ bb + cg
5. (c) Errors in A and B will be added.
6. (c) Given, L = 2.331 cm
= 2.33 (correct upto two decimal places)
and B = 2.1 cm = 2.10 cm
\ L + B = 2.33 + 2.10 = 4.43 cm. = 4.4 cm
Since minimum significant figure is 2.
7. (d) The number of significant figures in all of the given
number is 4.
8. (c)
2
4
=
p
MgL
Y
Dl
 so maximum permissible error in Y
= 
2
100 100
æö D D D D DD
´ = + + + +´
ç÷
èø
Y M g L Dl
Y M g L Dl
= 
1 1 1 11
2 100
300 981 2820 41 87
æö
++ +´ +´
ç÷
èø
= 0.065 × 100 = 6.5%
9. (d) a = 
ab
gd
bc
de
So, maximum error in a is given by
max
100 . 100 . 100
D DD æö
´ =a ´ +b´
ç÷
èø
a bc
a bc
. 100 . 100
DD
+g ´ +d ´
de
de
= (ab
1
+ bc
1
+ gd
1
+ de
1
)%
10. (c) 2 =p
l
T
g
Þ 
22
4 =p
l
T
g
 Þ
2
2
4p
=
l
g
T
Here % error in 
1 0.1
100 100 0.1%
100 100
= ´= ´=
mm
l
cm
and % error in 
0.1
100 0.05%
2 100
= ´=
´
T
\ % error in g
  = % error in l + 2(% error in T)
  = 0.1 + 2 × 0.05 = 0.2%
11. (c) Mean time period T = 2.00 sec
& Mean absolute error DT = 0.05 sec.
To express maximum estimate of error, the time period
should be written as (2.00 ± 0.05) sec
12. (b)
3
4
3
=p Vr
\ % error in volume = 3 × % error in radius
= 3 × 1 = 3%
13. (a) W eight in air = (5.00 ± 0.05)N
W eight in water = (4.00 ± 0.05)N
Loss of weight in water = (1.00 ± 0.1)N
Now relative density = 
weight in air
weight loss in water
i.e. R.D = 
5.00 0.05
1.00 0.1
±
±
Now relative density with max permissible error
= 
5.00 0.05 0.1
100 5.0 (1 10)%
1.00 5.00 1.00
æö
± ± ´ = ±+
ç÷
èø
= 5.0 ± 11%
14. (b)
max
100 100 100
D DD æö
\ ´ = ´ +´
ç÷
èø
R Vl
R Vl
= 
5 0.2
100 100 (5 2)% 7%
100 10
´ + ´ = +=
15. (c) V olume of cylinder V = pr
2
l
Percentage error in volume =
2
100 100 100
D DD
´ = ´ +´
V rl
V rl
=
0.01 0.1
2 100 100 (1 2)% 3%
2.0 5.0
æö
´ ´ + ´ =+=
ç÷
èø
Page 2


1. (d) Density 
2
r==
p
MM
V
rL
Þ 2
Dr D DD
=++
r
M rL
M rL
= 
0.003 0.005 0.06
2
0.3 0.5 6
+´+
= 0.01 + 0.02 + 0.01 = 0.04
\ Percentage error = 
100 0.04 100 4%
Dr
´ = ´=
r
2. (a) In division (or multiplication), the number of significant
digits in the answer is equal to the number of
significant digits which is the minimum in the given
numbers.
3. (d) Percentage error in A
= 
1
2 1 3 3 1 2 2 % 14%
2
æö
´+´+´+´=
ç÷
èø
4. (a) Percentage error in X = aa+ bb + cg
5. (c) Errors in A and B will be added.
6. (c) Given, L = 2.331 cm
= 2.33 (correct upto two decimal places)
and B = 2.1 cm = 2.10 cm
\ L + B = 2.33 + 2.10 = 4.43 cm. = 4.4 cm
Since minimum significant figure is 2.
7. (d) The number of significant figures in all of the given
number is 4.
8. (c)
2
4
=
p
MgL
Y
Dl
 so maximum permissible error in Y
= 
2
100 100
æö D D D D DD
´ = + + + +´
ç÷
èø
Y M g L Dl
Y M g L Dl
= 
1 1 1 11
2 100
300 981 2820 41 87
æö
++ +´ +´
ç÷
èø
= 0.065 × 100 = 6.5%
9. (d) a = 
ab
gd
bc
de
So, maximum error in a is given by
max
100 . 100 . 100
D DD æö
´ =a ´ +b´
ç÷
èø
a bc
a bc
. 100 . 100
DD
+g ´ +d ´
de
de
= (ab
1
+ bc
1
+ gd
1
+ de
1
)%
10. (c) 2 =p
l
T
g
Þ 
22
4 =p
l
T
g
 Þ
2
2
4p
=
l
g
T
Here % error in 
1 0.1
100 100 0.1%
100 100
= ´= ´=
mm
l
cm
and % error in 
0.1
100 0.05%
2 100
= ´=
´
T
\ % error in g
  = % error in l + 2(% error in T)
  = 0.1 + 2 × 0.05 = 0.2%
11. (c) Mean time period T = 2.00 sec
& Mean absolute error DT = 0.05 sec.
To express maximum estimate of error, the time period
should be written as (2.00 ± 0.05) sec
12. (b)
3
4
3
=p Vr
\ % error in volume = 3 × % error in radius
= 3 × 1 = 3%
13. (a) W eight in air = (5.00 ± 0.05)N
W eight in water = (4.00 ± 0.05)N
Loss of weight in water = (1.00 ± 0.1)N
Now relative density = 
weight in air
weight loss in water
i.e. R.D = 
5.00 0.05
1.00 0.1
±
±
Now relative density with max permissible error
= 
5.00 0.05 0.1
100 5.0 (1 10)%
1.00 5.00 1.00
æö
± ± ´ = ±+
ç÷
èø
= 5.0 ± 11%
14. (b)
max
100 100 100
D DD æö
\ ´ = ´ +´
ç÷
èø
R Vl
R Vl
= 
5 0.2
100 100 (5 2)% 7%
100 10
´ + ´ = +=
15. (c) V olume of cylinder V = pr
2
l
Percentage error in volume =
2
100 100 100
D DD
´ = ´ +´
V rl
V rl
=
0.01 0.1
2 100 100 (1 2)% 3%
2.0 5.0
æö
´ ´ + ´ =+=
ç÷
èø
4
DPP/ P 02
2
100 100 100
D DD
´ = ´ +´
V rl
V rl
= 
0.01 0.1
2 100 100 (1 2)% 3%
2.0 5.0
æö
´ ´ + ´ =+=
ç÷
èø
16. (b) H = I
2
Rt
2
100 100
D D DD æö
´ = + +´
ç÷
èø
H I Rt
H I Rt
= (2 × 3 + 4 + 6)% = 16%
17. (c) Quantity C has maximum power. So it brings maximum
error in P.
18. (d) Kinetic energy 
2
1
2
= E mv
22
2
'
100 100
D-
\ ´=´
E vv
E
v
 = [(1.5)
2
 – 1] × 100
100 125%
D
\ ´=
E
E
19. (b) Required random error = 
4
x
20. (b)
2
1
2
\= E mv
\ % Error in K.E.
= % error in mass + 2 × % error in velocity
= 2 + 2 × 3 = 8%
21. (c)
22. (d) Since for 50.14 cm, significant number = 4 and for
0.00025, significant numbers = 2
23. (a) Since percentage increase in length = 2%
Hence, percentage increase in area of square sheet
= 2 × 2% = 4%
24. (d) Here, s = (13.8 ± 0.2) m
t = (4.0 ± 0.3) s
velocity, 
s 13.8
v
t 4.0
== = 3.45 ms
–1 
 = 3.5 ms
–1
(rounding off to two significant figures)
v st
v st
D DD æö
=±+
ç÷
èø
( ) 0.8 4.14 0.2 0.3
13.8 4.0 13.8 4.0
+
æö
=± + =±
ç÷
´
èø
Þ 
v 4.94
0.0895
v 13.8 4.0
D
= ± =±
´
D v = ± 0.0895 × v  = ± 0.0895 × 3.45 = ± 0.3087 = ± 0.31
(rounding off to two significant fig.)
Hence,  v = (3.5 ± 0.31) ms
–1
% age error in velocity = 
v
100
v
D
´ = ± 0.0895 × 100 = ± 8.95
% = ± 9%
25. (a) Maximum percentage error in measurement of e, as given
by Reyleigh’s formula.
(Given error is measurement of radius is 0.1 cm)
De = 0.6 DR = 0.6 × 0.1 = 0.06 cm.
Percentage error is 
e 0.06
100 100 3.33%
e 0.63
D
´= ´=
´
26. (b) Speed of sound at the room temperature.
l
1 
= 4.6 cm,  l
2 
= 14.0 cm.,
l = 2 (l
2
 – l
1
) = 2 (14.0 – 4.6) = 18.8 cm.
v = f l = 
18.8
2000 376 m / s
100
´=
27. (c) End correction obtained in the experiment.
21
3 14.0 3 4.6
e 0.1 cm.
22
- -´
= ==
ll
28. (d) Since zeros placed to the left of the number are never
significant, but zeros placed to right of the number are
significant.
29. (b) The last number is most accurate because it has
greatest significant figure (3).
30. (a) As the distance of star increases, the parallex angle
decreases, and great degree of accuracy is required for
its measurement. Keeping in view the practical
limitation in measuring the parallex angle, the maximum
distance of a star we can measure is limited to 100 light
year.