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JEE ADVANCED – 2015 (PT – 1)PHYSICS SOLUTION Dt. 10-04-2015

Section – I (Single Correct Type)

1. (a) ( )vdv

a velocity slopeds

For accelerated motion both acceleration & velocity must have same sign

2. (b) 2 3 2ds

v t tdt for accelerated motion both acceleration and velocity must have same sign.

3. (b) 0 2( / 8) / 2u g H gH

20

1

2H u T gT

4. (b) Along vertical

2

0 2

1

21

10 5 2 sin 45 ( 10)2

2sec

yy u ayt

t t

t

5. (d) 21 1(5 3)4 (2 )

2r

dsA d v vdv

a vdv

16 2 8ds s m

6. (c) avg2

atV u

Section – II (Multiple Correct Type)

1. (a, d)

00

0

5sin 53 10tan( 45 ) 1

5cos53y

X

v t

v

3 4 10t

1 7sec sec

10 10t or

2. (a, c)

2

2 21 1 2 cos0 cos

2 2x x

ux u t a t u t at t

a

……..(1)

2 21 1 2 sin0 sin

2 2y y

uy u t a t u t gt t

g

……..(2)

3. (b, c)

0

2 2tan 37 y

x z

u gt

u u

4. (b)

Initially &a ve u ve

Finally &a ve u ve

5. (a, b, c, d)Conceptual

6. (a, c, d)

6 3dv

vdt

0 06 3

v tdvdt

v

Section – III (Paragraph Type Questions)

Passage (Question No. 1 – 2)

1. (a) Let t be time gap

Along horizontal 1 2x x

05 3 cos 60 5 3( )t t t ……………(1)

Along vertical 1 2y y

0 21 15 3 cos 60 ( )

2 2t gt g t t

22 3 4t t l t ……..(2)

From (1) and (2): 1sec,& 2sect t

2. (c) 5 3( ) 5 3x t t m

2110 10 ( ) 5

2y t t m

Passage – 2 (Questions No. 3 – 4)

3. (a)

along horizontal 213 (1.5) .............(1)

2xu t t

3

Along vertical 1.25= 21(10)

2yu t t …………(2)

0 10tan 45 y

x

u t

u

……………….(3)

From 1,2,3

T=1sec

4. (c) From pervious , 3.75& 6.25x yu u

Section – IV (Matrix-match Correct Type)

1. A-R, B-S, C-P, D-T

0tan 1 45 & 80R m

120 2u ms

A] 045 is angle of projection

B] 120 10(1) 1tan tan

20 2

C] 2

020 2 10 (20)tan 0 0

20

D] . 0r v has no real solution

2. A-R, B-S, C-P, D-Q

Section – V (Integer Type Question)

1. (8)

2. (1)

3. (5) 2 12n

aS u n

4. (5)cos 3u

sin 4u 5. (3)

2 6 0a t 3secAt t

2 6v t t