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