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# uptu paper

Jul 06, 2018

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UPSEE 2016

Paper 1 Code AA Solutions

Physics

Ans.1: (D) 2 R

By energy conservation between points A and B

2 21 1(2 ) (0) (0) 2 2

Mg R m mgH m     2 H R 

Ans.2: (D) 40 sec

4 2 4(60)t t     40t     Ans.3: (A) Towards the left

Point of contact of wheel has velocity towards left.

Ans.4: (C) b  and m  alone

2 dU

F bx dx

      2b

m   

Ans.5: (D) light is absorbed in quanta of energy  E h    Ans.6: (B) 954 kg/m

3

    5

724 1000 6 6

V V  V g g g      

3954 /Kg m   

Ans.7: (C) 144 cm

(18)n l   where length of string is l

( 1)(16)n l 

Gives n=8 and l =144cm

Ans.8:(A) 44.8 10   C 

  4 420 10 2.4 4.8 10 10

Q C

R

    

 

   

Ans.9: (B)

0

3

2 2

q

a

0 0 0

( ) (3 )

sin 45 sin 45 cos 45

kq k q k q V

a a a

   

0   0

1 3 3

4   2 2

2

q q

a   a       

Ans.10:(D) 

Resistance = 1 1 1

0

dV

dI dI Slope

dV

          

Ans.11: (C) It moves back and forth (oscillating) towards the wolf

Sound wave is longitudinal wave .

Ans.12:(B)Silver

Ans.13: (D) 4V

  2 2

1 1 2 2   2

B  AV A V R V R V     

4  B

V V  

Ans.14: (B)6 minutes

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 0av

d  k

dt

      

(59 61) 61 59 30

4 2 k

          1 1

30 2 60

k k   

(49 51) 51 49 30

2 k

          2

(20) 6k t  t

  

Ans.15:(C)18000C

7.5

9 i 

7.5 (6)(60)(60) 18000

9 Q it C     

Ans.16: (B)

2

2

B l

Ans.17: (A) 3  E 

0

E

q

   

0 0

( 3 ) 5 3 3

E

q q q q

 

       

Ans.18:(B)27A

IV=P 1 +P

2 +P

3

I(120)=1800+1300+100 ∴ I=26.67A

Ans.19: (A)2A

 0 10 8 20 0 2 (0.1)

B I  

      

∴I=2

Ans.20:(C)80V

6

6

6 0 (40 10 ) 80

3 10

di  L V

dt      

      

    Ans.21: (D) 12.1eV

3 1   1.5 ( 13.6) 12.1 E E E eV        

Ans.22: (D) There is no change

0  

F qv B     So velocity is constant

Ans.23: (B) 510

Ans.24: (A) 1

,

GM  U m

r   

. 2 2

GM GM GM    E K E U m m m

r r r        

Alternative:

we know that  E K E K    

2

21 1 1. 2 2 2

GM GM   K E mv m m

r r

     

 

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Ans.25:(B)

216 / , 4 /m s m s

0 232 cos 60 16 / c

a m s 

2 2

16 1c

v v a

R   

4 /v m s 

Ans.26:(D) 2

10  m

s  upwards the incline

0 275 5 sin 3 0 (75 25) / 5 10 /

5

g a m s

    

Ans.27:(A) 60J

f i W KE KE   

1 1 (3)(64 16) (3)(36 4)

2 2    

1 (3)(80 40)

2     60 J 

Ans.28: (C) 335J

W=Q A−QR

25=360−Q R ∴

Q R  =335J

Ans.29: (A)

0

3

2





0 0 0 0

2 4 3

2 2 2 2  E

   

   

    

Ans .30: (C) Three in parallel

21

2 U CV 

For U maximum, C must be maximum

Ans.31:(D) 20

3 

By Wheatstone bridge   

   

4 6 8 12   20 R

4 6 8 12 3eq

    

     Ans.32: (C) ,a b b c 

Ans.33: (B)  2 f

  1 1 1

1  f R R

     

     

  1

1 1 1 1

f R 

      

1   2 f f  

Ans.34: (B) 26V

(2 4)4 2 26V volt     

Ans.35: (C) 2 2

2 2

h h h

p   mKE mqV       

2 21

2   1 1

4 2 2 2

p

p

m em q

m q m e



    

Ans.36:(A)  6

sin 5

g

L 

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2 2

254 2 2 4

L L  I m m mL

         

   

3 4 sin sin sin

2 2 2

L L L mg mg mg       

I      6

sin 5

g

I L

      

Ans.37: (C)  ˆ ˆ4 5i j   Horizontal component remains constant, whereas vertical component changes its sign.

Ans.38: (C) 5 %

2   l

T  g

 

2 2

T l g

T l g

     

3 7 % % 5%

2 2

         

    Ans.39: (D) 100W

Work per cycle      1

30 10 8 2 60 2

J     

60 100 100

60 P W

   

Ans.40: (A)Path –I

Ans.41: (A) 3Hz

2 , 30300 / 101 300 Hz     

1 2   3 Hz    

Ans.42: (C)  00.75 I

2 0

0 0 cos 30 0.75 I I I  

Ans.43: (B) laser light is highly coherent

Ans.44:(B) 19%

2   2 2

2

2

(0.9 ) 0.81

2 2 2

p   p p KE

m m m   

Ans.45: (A) Magnification of microscope is inversely proportional to the least distance of distinct vision.

Magnification   1   D

M   f

 

Ans.46: (C)

264   SR    2 2 28 (3 ) 8 ( ) 64 ( )W S S R S R S R       

Ans.47: (C)Less than 300 km/hr

1 2

200 200 800 267 /

200 200   3

400 200

d d  v km hr

t t

      

 

Ans.48: (C) remains constant

0 dQ

dS  T

 

∴S=constant

Ans.49:(C) 0, 1, 1 A B C      Output C A AB    Ans.50: (C) chromatic aberration

1   30300 /100 303 Hz    

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Chemistry

Ans.51: (C) He +

Ionization Potential = E   – E1

54.4 = 0 – E 1

or E 1  = – 54.4 eV

But E 1  = –13.6 ×

2

2 (1)

Z  eV

or –54.4eV = –13.6 × Z 2   or Z=2 ,So He

+  ion

Ans.52:(C) ,3n   ,2l   ,1m 2 1s

Energy    ( )n l

For Options: (A) ( ) 3 0 3n l   

(B) ( ) 3 1 4n l   

(C) ( ) 3 2 5n l   

(D) ( ) 4 0 4n l   

So 3,n    2,l    1,m    1

2 s   Set of quantum number has highest energy.

Ans.53: (C) sp 3

OF2 :-

6O  1s22s22p4

or 6O

or

sp 3 , Two lone pairs of electron V-shape

Ans.54:. (D) 23 3,SO ClO    and 33 BO



3 NO   sp

2  Trigonal planar

3 3 AsO     sp

3  Pyramidal (onelone pair)

2 2 3CO sp    Trigonal planar

2 3ClO sp    Pyramidal(one lone pair)

2 3 3SO sp    Pyramidal(one lone pair)

3 3 3 BO sp

   Pyramidal(one lone pair)

So 2

3 3,SO ClO  

&   3

3 BO   

all are non-planar Ans.55: (B) stronger 2p(B)–2p(F)      bonding

Size of Cl is more than the size of F so in case of BF 3  strong 2p(B)–2p(F)   -bonding occurs so le

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