test -1 physics held on 20-april-14
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Test -1 Physics Held on 20-April-14TRANSCRIPT
UG – 1 & 2, Concorde Complex, Above OBC Bank. R.C. Dutt Road., Alkapuri Baroda. Ph. (0265) 6509041, 6534152, 6625979 Page # 2
Solution Physics FRESH ER BAT CH T est - 1 Date: 20-4-2014
1. c
2. c
3. c
4. d
5. c
6. c
7. a
8. b
9. a
10.b
11.a
12.a
13.a
14.b
15.a
16.a
17. (b) 2
2
1mvE
% Error in K.E. = % error in mass + 2 × % error in velocity
= 2 + 2 × 3 = 8 % 18. (b) 19. (b) Number of significant figures are 3, because 103 is decimal multiplier.
20.(b) 3
3
4rV
% error is volume %3 error in radius 13 = 3% 21.(c) Mean time period T = 2.00 sec & Mean absolute error T = 0.05 sec.
To express maximum estimate of error, the time period should be written as )05.000.2( sec
22.(b) Here, )2.08.13( S m
and )3.00.4( t sec
1
1
13.83.45
4.0
0.3
SV ms
t
V S t
V S t
S VV V
S V
ms
23.(c)
24.(a) 05.020
1
Decimal equivalent upto 3 significant figures is 0.0500 25.(b)
26.(b) 3
3
4rV
% error in volume %3 error in radius.
UG – 1 & 2, Concorde Complex, Above OBC Bank. R.C. Dutt Road., Alkapuri Baroda. Ph. (0265) 6509041, 6534152, 6625979 Page # 3
1003.5
1.03
27.(c) Since for 50.14 cm, significant number = 4 and for 0.00025, significant numbers = 2 28.(d) edcba /
So maximum error in a is given by
100.100.100max
c
c
b
b
a
a
100.100. e
e
d
d
%1111 edcb
29. (b) 100100100max
I
I
V
V
R
R
10010
2.0100
100
5 )%25( = 7%
30.(b) Average value 5
80.271.242.256.263.2
sec62.2
Now 01.062.263.2|| 1 T 06.056.262.2|| 2 T 20.042.262.2|| 3 T 09.062.271.2|| 4 T 18.062.280.2|| 5 T
Mean absolute error
5
|||||||||| 54321 TTTTTT
sec11.0108.05
54.0