5 Mock Test 5 Mains Hs

PHYSICS

mk MOCK TEST (MAINS) PAGE # 1

http://manishkumarphysics.in/

MOCK TEST-5PART-A

Q.1

[Sol: =g

u

u

2

2

=u

g

2

Q.2

[Sol: SBu ,

= juiu ˆ60sinˆ60cos

gSu ,

= i10

gBu ,

= i

u ˆ2

10

+ j

u ˆ2

3

u = 20

Hmax =

g2

3102

= 15 m ]

Q.3[Sol: V = x2/3

a =dx

dVV = x2/3

3/13

2

x=

3/1

3

2x

at x = 8, a =3

4m/s2 ]

Q.4[Sol: Path 1 = S1P + ( – 1)t

Path 2 = S2Px = Path 2 – Path 1

=D

xd– ( – 1)t

But x =d

D

l = ( – 1)t =

1

t

= 1.5 (B) ]

Q.5

[Sol: E =

hCEV > EI > EB > EG

Energy of photon corresponding to blue is greater than green (A) ]

PHYSICS


http://manishkumarphysics.in/Q.6

[Sol: t =a

l2

(trough)m = tsmooth

mKgg

l

cossin

2=

sin

2

g

l

sin = (sin – cos) m2

1 = ( 1 – k) m2

k = 1 – 2

1

m]

Q.8[Sol: Considering them as one mass

1600 – 900 = 340 a

a =34

70m/s2

Block A

800 – 300 – T = (170)

34

7

T = 150 N

Q.9

[Sol:ndT

dw

ndT

dV

ndT

dQ

C = Cv +ndT

dw

T

a– Cv =

ndT

dw ndTC

T

adw v

w = an lnT –0

0

2T

TvnC (n = 1)

= aln – Cv (h – 1) T0

= aln – 1

1

r

R TT0 (B) ]

Q.11[Sol. F = 6rv

F

6 rv

=

1 1 2

1 1 1

M L T

[L ] [L T ]

= [M1L–1T–1]. ]

PHYSICS



[Sol. Initially s = ut +1

2at2 ; h = 0 +

1

2gT2

at time = T/3

221 T 1 1

h ' 0 g gT2 3 9 2

;h

h '9

So height above ground =h 8h

h9 9

. ]

Q.13

[Sol. 1 22u sin 2u sin(90 )

T ; Tg g

2 2

1 2 2 2

4u sin cos 2u sin 2T T

g g

; TT1T2 =

2R

g. ]

Q.14[Sol. Force applied on person

If person fires n bullets/sec

nmuF

1

144 = n(40 × 10–3) (1200) ; n = 3

Q.15[Sol. In equilibrium F = mg sin

10 = m × 10 × sin 30ºm = 2kg. ]

Q.16[Sol. In uniform circular motion its kinetic energy will constant. ]

Q.17

[Sol. Solid sphere IA =2

5mR2

Hollow sphere IB =2

3mR2 ; IB > IA. ]

Q.18[Sol. Time period does not depend on mass of satellite. ]

Q.19

[Sol. F n n

1 kF

r r

2

n

k mvF

rr ; n 1

k 1v

m r

Time period n 12 r mT T 2 r (r )

v k

; T n 1

2r

. ]

PHYSICS



[Sol. 1 21 2

m mt 2 ; t 2

k k

In series combination1 2

1 1T 2 m

k k

2 21 2T t t . ]

Q.21

[Sol. Time displacement 2 20

1

m( ) . ]

Q.22[Sol. Work done = (1/2) × force × extension

W =1

2× F × . ]

Q.23[Sol. F = 6 rv. ]

Q.24[Sol. (T2 > T1)

eq2 1

K AdQ(T T )

dt 5x 2KK

x 4xT1T2

Keq in series combination, eq5

K K3

2 1dQ KA

(T T )dt 3x

Compare with given expression, f = 1/3. ]

Q.25

[Sol. Power,2

1 1 12

1 1 1

v v mv tP F.v mav m t

t t t

. ]

PHYSICS



[Sol.

–Q

–Q –Q

q

–Q

( )2

KQq

r 2

( )

2

2

KQ

2r

2

2

KQ

r

2

2

KQ

r

For system equilibrium

Fnet = 02 2

2 2 2

KQ KQ 2KQq2 0

r 2r r

Q

q 1 2 24

. ]

Q.28[Sol. Rcircuit = 1.5 I = 4A. ]

Q.29

[Sol. BCoil = 0 NI

2R

For same length of wire1

NR

BCoil N2

B' = n2B. ]

Q.30

[Sol. 0 1 21

I IF I

2 r

r

I2I1

FF

0 1 2(2I ) I 2F ' F

2 3r 3

3r

I22I1

F'F'

. ]

Q.31[Sol. For electromagnet, soft ferromagnetic material is required which has low retentivity and low

coercivity. ]

PHYSICS


http://manishkumarphysics.in/Q.32[Sol. Mean power per period

=2rmsV

R

22

2 2

r B2

2

(B r )

R 8R

. ]

Q.33[Sol. D.C. Ammeter measures average value and average value for one cycle of A.C. is zero. ]

Q.34

[Sol.1

Frequency2 LC

C 2CL L/2. ]

Q.35[Sol. (A). The effective focal length is given by

m

1 2 1

f f f

But1 1 1 1

(1.5 1)f 30 60

or2 1

f 30

. Again, R = 30 cm, fm =R

2= 15 cm

Now,1 1 1

f 30 15 or

1 1 2 3 1

f 30 30 10

or f = 10 cm

To have a real image of the size of the object, the object must be placed at the centre of curvatureof the equivalent mirror. So, the required distance is 2 × 10 cm = 20 cm. ]

Q.36[Sol. Distance of closest approach

2 9 19 2

0 6 19

2KZe 2 9 10 92 (1.6 10 )r

E 5 10 1.6 10

= 10–14 meter = 10–12 cm. ]

Q.37[Sol. 1H

2 + 1H2 2He4

(4 × 7) – [(1.1) × 2 + (1.1) × 2] = 23.6 MeV. ]

PHYSICS


http://manishkumarphysics.in/Q.38[Sol. h = + EK

EK = h – This is equation of straight line, whose slope is 'h'. ]

Q.39[Sol. When npn transistor is used as an amplifier electron move emitter to base and then base to collector.]

Q.40[Sol. (B). Pauli's exclusion principle

Q.41

[Sol. v 5 m / sk

. ]

Q.42[Sol. Brewster’s law: According to this law the ordinary light is completely polarised in the plane of

incidence when it gets reflected from transparent medium at a particular angle known as the angleof polarisation.n = tan ip. ]

Q.43

[Sol.2

0

1 (q / 2) (3q / 4) 3FF

4 8d

. ]

Q.44[Sol. m = Zit,

m = 3.3 × 10–7 × 3 × 2 = 19.8 × 10–7 kg. ]

Q.45

[Sol. In CE configuration, fe1

0e L

hA

1 h R

6 3

5048.78

1 25 10 1 10

. ]

5 Mock Test 5 Mains Hs

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