TARGET : PRE-MEDICAL 2016

HINT – SHEET

ANSWER KEY

LTS/HS - 1/6

Leader Test Series/Joint Package Course

0999DM310315010

1. Moment of inertia (= S mr2) for a given body

depends on the axis of rotation. So if axis of

rotation changes, usually moment of inertia

will changes. Moment of inertia for a given

axis depends on mass, shapes with same axis,

moment of inertia will be different.

For a given shape, size, mass and axis it also

depends on the distribution of mass within

the body. Farther the constituent particles of

a body are from the axis of rotation, larger

will be its moment of inertia. So moment of

inertia does not depend only on angular

velocity.

[Note: Due to above reason it follows that

in case of hollow and solid bodies of same

mass, radius and shape for a given axis, the

moment of inertia of the hollow body is

greater than that of the solid body ]

TEST # 08 DATE : 18 - 10 - 2015

Test Type : Unit Test Test Pattern : AIPMT

2. According to theorem of parallel axis,

I = ICG + M(2R)2

where, ICG = MI about an axis through the

centre of gravity.

I = 25

MR2 + 4MR2 = 222MR

5

or MK2 = 222MR

5

\ 22K R.

5=

3. Let M be the mass of each disc. Let RA and

RB be the radii of discs A and B respectively.

Then

M = pR2AtdA = pR2

BtdB

As dA > dB

\ R2A < R2

B

Que 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20Ans. 2 2 3 3 4 3 3 3 4 2 3 3 4 2 1 1 3 4 2 2Que 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40Ans. 4 1 4 4 3 3 3 2 1 4 3 1 2 2 1 1 1 3 3 1Que 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60Ans. 3 4 2 2 3 2 4 3 2 1 3 1 3 3 3 3 3 2 1 3Que 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80Ans. 4 2 3 4 3 3 2 3 2 1 3 4 3 4 2 4 2 1 3 4Que 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100Ans. 1 1 3 4 3 3 3 1 1 1 4 1 4 4 3 2 3 1 1 4Que 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120Ans. 2 4 1 2 3 1 4 4 3 2 4 2 4 4 4 4 1 2 4 1Que 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140Ans. 4 2 4 4 2 3 1 1 1 4 3 2 4 3 2 2 2 2 2 1Que 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160Ans. 1 1 3 1 4 4 2 4 3 4 3 4 3 2 1 2 2 4 4 2Que 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180Ans. 1 4 4 3 2 2 3 3 1 3 3 1 2 4 1 2 4 3 2 1

LTS/HS - 2/6

Target : Pre-Medical 2016/AIPMT/18-10-2015

0999DM310315010

Now, IA = 2 2A B B

1 1MR , I MR

2 2=

\2

A A2

B B

I R1

I R= <

i.e., IA < IB.

4. ICOD = S m2

= 2 2A A Bm r mBr+

l2

l1

= ma2 + m(a cos 60°)2

= ma2 + m 2

2a 5ma

4 4=

5. Imedian line = IA + IB + IC + ID

= 22Ml l

2 2Ml2 2

æ ö´ + ç ÷è ø

= 2 2Ml Ml

6 2+

l

l

A

B

C

D

Median

= 22Ml

3.

6. The distance OC = L L L2 3 6

- =

Applying the theorem of parallel axes,IC = IO + M(OC)2

= 22ML L

M12 6

æ ö+ ç ÷è ø

= 2ML

9.

L/2 L/2

L/3

OA C B

7. M = Mass of the square plate before cuttingthe holesMass of one hole,

22

M Mm R

1616Rpæ ö= p =ç ÷

è ø

\ Moment of iner tia of the remainingportion,

I = Isquare – 4Ihole

= 2

2 2 2M mR(16R 16R ) 4 m(2k )

12 2

é ù+ - +ê ú

ë û

= 2 28MR 10mR

3-

= 28 10

MR3 16

pæ ö-ç ÷è ø

.

8. Torque exerted on the disc,t = TR

Now, t - Ia

\2

2

TR 2TR 2T1I MRMRMR2

ta = = = =

RM

T

9.2

T

2 2

1mvK 2

1 1K mv I2 2

=+ w

= 2

2

2 2 2 2 2

1mv 22 ( I mr v r )

1 1 2 5mv mr v v / r2 2 5

= = w+ ´ ´ ´

Q

\ TK 1 52K 715

= =+

or 5 : 7

10. In vertical direction,0 – v2 sin2 45° = –2gh

or h = 2v

4g

At the highest point momentum = mv cos 45°

= mv

2

LTS/HS - 3/6

Leader Test Series/Joint Package Course/AIPMT/18-10-2015

0999DM310315010

(because vertical component of velocitybecomes zero at the highest point)

\ Angular momentum = mv

h2

´

= 2 3mv v mv

4g2 4 2g´ =

11. Total kinetic energy - KR + KT

= 2 2 2 2 2 21 1 1 1I Mv Mk MR

2 2 2 2w + = w + w

= 2 2 21M (k R )

2w +

KR = 2 21M k

2w and 2 2

T

1K M R

2= w

2 22

R2 2

2 2 2

1M kK k2

1Total KE (k R )M (k R )2

w\ = =

+w +

12. The rolling sphere has rotation as well astranslation kinetic energy,

2 21 1KE mu I

2 2= + w

2 2 21 1 2mu mr

2 2 5æ ö= + wç ÷è ø

= 2

2 21 mu 7mu mu

2 5 10+ =

Potential energy = Kinetic energy

\ mgh = 27mu

10

h = 27u

10g

13. From the figure in question 191, we findh = s sin q

So, v2 = (2gs sin q)/bDifferentiating it w.r.t. time, we get;

dv 2gsin ds2v.

dt dtq

= ´b

But,ds

vdt

= and dv

adt

=

\ a = gsin q

b.

15. When the gymnast lower his hands, hismoment of inertia about the axis of rotationdecreases (because mass is now locatednearer the axis of rotation).Now, angular momentum of the system willremain constant (i.e., Iw = constt.), henceangular velocity gets increased.

16. n = 90060

rps = 15 rps

w0 = 2pn = 2p × 15 = 30p rad/sec t = 60 sec, w = 0Now, w = w0 + at 0 = 30p + a × 60

\ a = – 3060 2

p p= -

\ Retardation = 2p

rad/sec2.

17. For sliding motion, V = V0 + at = V0 –mgt...(i)

as a = –f Mg

gM M

-m= = -m

It is worth noting here that when a body is giveninitial velocity V0 on a rough surface, the velocityis reduced because sliding is opposed by friction.On the other hand, rolling is caused by friction.When the relation V = Rw is satisfied, pure rollingoccurs.Now, angular velocity of the body after t sec is,

w = w0 + 9 at = at (Q w0 = 0) ...(ii)

Also, t = Ia = = fR = mMgR

\ a = MgR

Im

....(iii)

Form eqn. (ii) and (iii),

w = MgR

tI

m

LTS/HS - 4/6

Target : Pre-Medical 2016/AIPMT/18-10-2015

0999DM310315010

\2

MgR gtt

2 2MR R

5 5

m mw = =

Hence, Rw = 5

gt2

m

or mgt = 2 2

R V5 5

w =

(For pure rolling, V = Rw)Putting in eqn. (i), We have,

V = V0 – 2

V5

or 0

7V V

5=

V = 0

5V

718. When the separation between the lines of

action of the forces is zero, they cannotconstribute a couple and the body cannotrotate.

20. d = 1m, r = 0.5 m

w = 120 rpm = 120 2

60´ p

= 4p rad/sec

\ L =Iw = 2Mr

2w

= 220 (0.5)

42

´´ p

= 10 × 0.25 × 4× 3.14 = 31.4 kg-m2/sec.

21.1 1

Tf 60

= =

60º3p

f = =

So, time difference T

T3 2 6

p= ´ =

´ p

160 6 360

p= =

´22. Circuit is resistive

so,V 120

I 1 ampR 120

= = =

23. at half power frequency Z 2R=25. VL = VC

so, circuit is in resonanceso, voltage drop at combination of L & C willbe zeroso, VR = 200 V

26. I = 10 sin 314 t< I2 > = < 102 sin2 314 t >

= 100 < sin2 314 t >

10050 amp

2= =

27. 2 2R L CV V (V V )= + -

2 2V 80 (60 120)= -

V = 100 V28. Frequency of D.C. source f = 0

so, reactance by capacitor

C

1X

2 fc= = ¥

p

so, 22

CZ R X= +

Z = ¥29. average power Pavg = Vrms Irms cos f

100 R100

10 Z= ´ ´

100 6100

10 10= ´ ´

= 600 W30. Z = 100 f = 30º

R 3cos

2 2f = =

R 3R 50 3

100 2= Þ = W

31. XL = 10000 W given f1 = 10000 H2

1L 1X 2 f L 10000= p = W

at f2 = 20000 Hz

2L 2X 2 f L 2 20000L= p = p ´

1

2

L 1

L 2

X 2 f LX 2 f L

p=

p

1

2

L 1

L 2

X f 10000 1X f 20000 2

= = =

2 1L LX 2X 20000= = W

LTS/HS - 5/6

Leader Test Series/Joint Package Course/AIPMT/18-10-2015

0999DM310315010

32. V = –10 sin wt, I = 5 cos (wt + p/6)Þ V = 10 cos (p/2 + wt)

so, phase difference 2 6 3p p p

= - =

33. XL µ fso, when freq. is doubled then XL will alsobe doubled.so, when switch B is closed then current willbe half.

34. frequency of LC oscillator.

1f

2 LC=

p

6 3

1 1w

LC 30 10 27 10- -= =

´ ´ ´

8

1w

27 3 10-=

´ ´

410w rad / s

9=

35.1 1

f f2 LC C

= Þ µp

for min. value of capacitance frequencyshould be max. so fmax. = 1200 k Hz

3 11200 10

2 LC´ =

p

2

3 6

1 1C

1200 10 2 200 10-

æ ö= ´ç ÷´ ´ p ´è ø

C = 87.9 pf36. in LC oscillator energy will be completely

magnetic when current is max so

T 3Tt ,

4 4= .. .. etc.

3T 6.28 10t 1.59ms

4 4

-´= = =

37. in L–C–R circuit at resonance frequencyXL = XC

C

1X

fµ & XL µ f

so, when frequency of applied emf is lessthan the resonant frequency then XC > XL

so, circuit is capacitive.

38. quality factor LQ

Rw

=

40. r.m.s. value of electric field 0E

2=

0E= 720 N/C

2 (given)

0E = 720 2 N/C

Average total energy density 20 0

1E

2= Î

( )2121

8.854 10 720 22

-= ´ ´ ´

= 4.58 × 10–6 J/m3

42. ˆB i[ (dir)]® ´ur

direction of prppagation is along

y-dir. so ˆC j®ur

for E-M wave

C E B® ´ur ur ur

( )ˆ ˆB i,C j® ®ur ur

so, ˆE k®ur

so that 0

2 ˆE E cos wt y zpæ ö= -ç ÷lè ø

ur

43. NCERT XII part-I Pg.# 27444. NCERT XII part-I Pg.# 27745. NCERT XII part-I Pg.# 27246. NCERT P.g. # 159-16047. 2Na[Al(OH)4] aq. + CO2(g) ® Al2O3 . xH2O(s)

Sod. Aluminate + 2 NaHCO3(aq.)48. NCERT P.g. # 14949. NCERT P.g. # 16150. NCERT P.g. # 149

(E.g. s in froth floatation) Aniline ® |

NH2

LTS/HS - 6/6

Target : Pre-Medical 2016/AIPMT/18-10-2015

0999DM310315010

51. NCERT P.g. # 1572 Cu2S + 3O2 ® 2 Cu2O + 2SO2

53. NCERT P.g. # 16055. NCERT P.g. # 154 (Ex. 6.3)57. NCERT P.g. # 201

Flourine is most reactive.58. NCERT P.g. # 17959. NCERT P.g. # 17460. N2 does not support combustion. It is filled in

electric bulb.62. C12H22O11 + 18[O] ¾¾® 6 (COOH)2 + 5 H2O

From HNO3 Oxalic acid

63. BaO2 + O3 ¾¾® BaO + 2 O2

65. Oxidising agent are electron acceptore, there fore,I.E. which involves removal of e– has no relevancein oxidising property.

81. N = wE

× 1000

V(mL) = 15.8

158/ 5 ×

100050

= 10 N

82. Eq. of O2 = eq. of Fe2O3

w 16008 160 / 6

= 2 3

f

Fe O Fe

V 2(3 0) 6

®æ öç ÷ç ÷= - =è ø

w = 480 gm84. Cr2O4

2– + 8H+ + 3HCOOH ® 2Cr+3 + 3CO2 + 7H2O3 moles of formic acid reduces = 1 mol K2Cr2O7

1 mole of formic acid reduce = 1/3 mole of K2Cr2O7

85. NO3– + 2H+ + e– ® NO2 + H2O ...(1)

NO3– + 6H+ + 5e– ® 0.5N2 + 3H2O ...(2)

NO3– + 5H+ + 4e– ® 0.5N2O + 2.5H2O ...(4)

86. CrO

O

O

OO

Two peroxide linkage = –4 one double bond = –2

Total = –6

87. O = C = C = C = O+2 +20

Avg. ON Þ 4/3100. NCERT-XI Pg. # 212, 213101. NCERT-XI Pg. # 158102. NCERT-XI Pg. # 159103. NCERT-XI Pg. # 155104. NCERT-XI Pg. # 201 Fig.-12.3105. NCERT-XI Pg. # 197106. NCERT-XI Pg. # 231,232107. NCERT-XI Pg. # 236

108. NCERT-XI Pg. # 176

109. NCERT-XI Pg. # 189

110. NCERT-XI Pg. # 179

120. NCERT-XI Pg. # 158

127. NCERT-XI Pg. # 217

129. NCERT-XI Pg. # 217

130. NCERT Pg. # 208

131. NCERT-XI Pg. # 232

133. NCERT-XI Pg. # 222

134. NCERT Pg. # 182

136. NCERT Pg.# 218 (E), 219 (H)

138. NCERT Pg.# 213 (E), (H)140. NCERT Pg.# 223 (E), (H)142. NCERT Pg.# 212 (E), (H)

143. NCERT Pg.# 191 (E), (H)

144. NCERT Pg.# 178 (E), (H)

145. NCERT Pg.# 189, 190 (E)

146. NCERT Pg.# 196, (E), (H)

147. NCERT Pg.# 198, (E), (H)

148. NCERT Pg.# 250, (E), 250-251 (H)

149. NCERT Pg.# 221 (E), (H)

150. NCERT Pg.# 251 (E), (H), Fig. 15.12

156. NCERT Pg.# 240, 245, 246, 247 (E), (H)

157. NCERT Pg.# 250 (E),(H)

158. NCERT Pg.# 252

159. NCERT Pg.# 250

160. NCERT Pg.# 249, (E), (H)

163. NCERT Pg.# 189, (E), 190 (H)

164. NCERT Pg.# 185,186,187,177

165. NCERT Pg.# 201, 198, 197

166. NCERT Pg.# 199

167. 6CO2 produced in aerobic respiration [264gms]

Each Co2 produces ® 2646

=44. In alcoholic

fermentation 2CO2 produced so, 2 × 44 = 88gms. CO2

produced

171. NCERT Pg.# 184,179,185

173. NCERT Pg.# 232

174. NCERT Pg.# 208, (E)(H)

175. NCERT Pg.# 235, 236 (H)

179. NCERT Pg.# 187(E), 188(H)