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COMMON PRACTICE TEST 13 XIth-CBSE MAINS PHYSICS SOLUTION Date : 19-10-2014

1. (B) 2. (A) 3. (D) 4. (A) 5. (A) 6. (C) 7. (B) 8. (D) 9. (D) 10. (A) 11. (A) 12. (D) 13. (D) 14. (A) 15. (A)

16. (b) The displacement of the particle is )k6j2i3()k9j13i14(rrr 12 )k15j11i11( Work done = Fr )k3ji4()k15j11i11( J10015311114 Hence the correct choice is (b).

17. (d) Work done is 50

5

0

2 dx)x3x27(FdxW

=5

032 xxx7

= 7 5 (5)2 + (5)3 = 135 J 18. (c) Power .dt

m

Pdvvorvdtdv

mmavFvP

UG 1 & 2 Concorde Complex, Above, OBC Bank. R.C. Dutt Road. Alkapuri Baroda, Ph. 6625979, 6534152 Cell 9426503864 Page # 2

Integrating, we have

)constantP(dtmPvdv

2/12

tm

P2vor

m

Pt2

vor

dttm

P2dxortm

P2dtdx

or 2/12/1 Integrating again, we have

2/32/1 tm

P232

xordttm

P2dx i.e. x t3/2. Hence the correct choice is (c) 19. (c) In going from (0, 0) to (a, 0), the x-coordinate varies from 0 to a while the y-coordinate remains zero. Work done by force F along this path is )0ij( a

0

a

01 )0ij(0idx)jKx(dxFW

In going from (a, 0) to (a, a), the x-coordinate remains constant at x = a while the y-coordinate changes from 0 to a.

Work done by force F along this path is )ax(

a0

a

02 jdyjaiyKdyFW

)1jj,0ji(KadyKa 2a

0

Since work is a scalar quantity, the total work done is W = W1 + W2 = 0 Ka2 = - Ka2 Hence the correct choice is (c). 20. (b) Power delivered in time t1 is P1 = Fv1 = m a v1 Now, acceleration vector is

1

1tv

a )vvv( 2111 Power delivered at time 2

1

21

1

1

t

tmvt

tP

t Hence the correct choice is (b). notice that choices (a), (c) and (d) do not have the dimensions. 21. (a) 22. (c) Let u be the speed of the ball before the collision after the collision, its sped will be

22

2eu

2u

v 21eu858u2u 22

UG 1 & 2 Concorde Complex, Above, OBC Bank. R.C. Dutt Road. Alkapuri Baroda, Ph. 6625979, 6534152 Cell 9426503864 Page # 3

Fraction of K.E. lost 2

22

mu21

mv21

mu21

83

851

u

v1 22

23. (d) Refer to figure. Here p = mv and P = MV. The resultant of p and P is

2222r )MV()mv(Ppp

Which is choice (a). The angle which the resultant momentum pr subtends with the x-axis is given by

mv

MVpP

tan , which is choice (b). Loss of KE

)mM( VMvm21MV21mv212222

22

)vV()mM(Mm

21 22 , which is choice (c).

24. (D) 25. (C) 26. (D)

27. (d) Condition for stable equilibrium 0dxdUF

0612 xbxadxd 0612 713 bxax

713612x

bx

a 62 xba 6 2

ba

x 28. (c) mEP 2 mP const.)if ( E

2

1

2

1

m

m

PP

29. (d) m

E2P2 2PE

i.e. if P is increased n times then E will increase n2 30. (c) P.E. of bob at point A = mgl This amount of energy will be converted into kinetic energy K.E. of bob at point B = mgl

and as the collision between bob and block (of same mass) is elastic so after collision bob will come to rest and total Kinetic energy will be transferred to block. So kinetic energy of block = mgl

A

B m m

m