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CPT-13 JEE Mains Physics Held on 19-Oct-14
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UG 1 & 2 Concorde Complex, Above, OBC Bank. R.C. Dutt Road. Alkapuri Baroda, Ph. 6625979, 6534152 Cell 9426503864 Page # 1
JEE ADVANCED | | JEE MAINS | | AIPMT
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