paper2 physics

8/4/2019 Paper2 Physics

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Head Ofce : Andheri : 26245223/09 : Dadar / Thane /Chembur/ Nerul / Powai/Borivali / Churchgate

Physics Paper - 2 ( Solution)

1. (b)

Q values of reaction is,

Q = (2 × 4 × 7.06 – 7 × 5.6) MeV = 17.28 MeV

Applying conservation of energy for collision,

...... (1)

(Here K P

and are the kinetic energies of proton and -particle respectively)

For conservation of linear momentum,

....... (2)

......... (3)

Solving equation (1) and (3) with Q = 17.28 MeV, we get

2. (D) The following forces act on the particle.

• Force T acting radially inwards.

• Centrifugal force

2mv

r acting radially outwards.

• Magnetic force qvB acting radially inwards.

⇒

2mv

T qvBr

+ =

⇒

2

0mv

qvB T r

− − =

⇒ ⇒ v =

2 2 2

2

1 4

2

qBr q B r Tr

m m m

+ +




3. (C)

dx

cx

0I

Consideraninnitesimalelementoflengthb.thicknessdxatadistancexfromthewire.

Since dφ = BdA

⇒ ( )0

2

I d Bdx

x

µ φ

π =

⇒

0

2

c a

c

IB dxd

x

µ φ φ

π

+

= =∫ ∫

⇒

0 log2

e

bI c a

c

µ φ

π

+ =

⇒

00 0log 1 1

2e

b c a t t I I I

c

µ φ

π τ τ

+ = − = −

Q

Since

d

dt

φ ε = −

⇒

0 0 1log

2e

bI c a

c

µ ε

π τ

+ = − −

⇒

0 0 log2

e

bI c a

c

µ ε

πτ

+ =

0 0 log2

e

bI c a I

R R c

µ ε

πτ

+ = =

⇒

0 0 log2

e

bI dq c a

dt R c

µ

πτ

+ =

⇒

0 0 log2

e

bI c adq dt

R c

µ

πτ

+ =

⇒

0 0

0

log2

e

bI c aq dt

R c

τ µ

πτ

+ =

∫ ⇒

4. (A) Energy density,

1u stress strain

2= ×




5. (D)

Drawing F.B.D. of rod PQ

P

Q

MgN

N µN

1m 1m

ΣFx=0

ΣFy = 0 ⇒ µ N = mg ….. (1)

Στ about centre of mass = 0

[ ] N sin 2 N cos tan

2 2 2

l l µ × θ = µ × θ ⇒ θ =

Hence( )2sec 4 / 2θ = µ +

Thus,

224/ 2

41 2

µ + = ⇒ = µ +

l l

⇒

6. (D)

Common mistake:Theπ/8phaseleadat x2

might lead the student to believe that the

wave equation is of the form

sinx

y A t v

ω

= + (a wave travelling in the negative

x-direction). But the given equation is equivalent to y2 = 0.2 sin (3 πt – 15π/8) and this

would lead to a wave equation of the form

sinx

y A t v

ω = − (a wave travelling in the

negative x-direction).

It is possible to determine the velocity of the wave from particle velocity only

if

dy

dx is known. Knowledge of

dy

dt is irrelevant. Since the relation between

displacement and particle position is not given, it is impossible to determine the wave

velocity or direction.

7. (2)

Afterreceiving10.2eVenergyinrstexcitedstateatomsareioniseditmeansatoms

hasenergyintherstexcitedstate10.2eVorless.

Onlyhydrogenhas3.4eV(<10.2eV)inrstexcitedstate.

So one element is hydrogen.

Forsecondelementsixradiationarepossible.




So

or n = 4

Itmeansafterreceiving10.2eVenergyatomsexcitedton=4.

So

or

or

or Z = 2

So it is Helium. so Z1

Z2

= 1 × 2 = 2.

8. (1) A satellite is in force free space so momentum will remain constant

Using formula , we get

or

or or

So acceleration

or or or

or




9.

Currentsintheloopswillbeasshowninthegure.Slider can be replaced by battery of voltage induced in the slider i.e.,

Now the circuit is reduced to as given below

From second loop

.......... (1)

From Ist loop

or or

or or

or or

or

Current in the slider

I = I1 + I2

or

or

On substituting value of V i.e. we get,

or




or,

or, x=2.

10.

or or

or

or .......... (A)

or

or or

or ............ (1)

For second case

or or .......... (B)

or or

or or ......... (2)

From equation (1) substituting value of in equation (2), we get

or




11. (8)

Potential difference across 40 cm of this wire = 100

5

x40=2volt.

∴ Potential difference across 20 cm of wire CD = 2 volt.

∴ Potential difference across wire CD = 20

2

× 80 = 8 volt.

Potential difference across 2 Ω resistor = 2 × 2 = 4 volt

∴ Emf of the cell = 12 volt.

12. (A) 13. (D) 14. (B)

Substituting this value into equation (3)

V’ = 6 − 12 × 1/3 ⇒ V’ = 2V (Ans. of Question 14: B)

From Eq, (4)

V = 4V (Ans. of Question 13: D)

15. (B)

16. (B)

17. (B)

18. A - q, r , B- p, r C - q, t D - q, r

When the source is at C and the detector is at E, the relative velocity between detector and

source will be zero. In all other cases, there will be a net relative velocity between S and D.

Doppler effect is observable when there is a net relative velocity between source and detector

along the line of sight.

When source is at A and detector is at E, the net relative velocity of detector with respect to

source is perpendicular to the line of sight. Hence there is no observed Doppler Effect.

When source is at C and detector is at G, the net relative velocity of detector with respect to

source is perpendicular to the line of sight. Hence there is no observed Doppler Effect.

When source is at C and detector is at E, the net relative velocity of detector with respect to

source is zero. Hence there is no observed Doppler Effect.

When source is at B and detector is at F, the net relative velocity of detector with respect to

source is along the line of sight. Hence there is an observed Doppler Effect. Since the source

and detector are effectively approaching each other, the detected frequency is higher than the

source frequency.




19. A : PQRS B : PRS C : S D :RS

Since in all processes temperature changes from T to 2T.

or [n = 1 given]

or

or

In process A and B volume increases so dW > 0 in process D, volume is constant.

So dW = 0.

and in process C, dW < 0.

dQ = dU + dW

Since dU > 0 and dW is either zero or positive for A, B and D

So dQ > 0 i.e., energy is absorbed.

For C,

since dU > 0 while dW < 0

So dQ may be greater or less than zero

ForprocessA,workdoneismaximumsodQwillbemaximumforprocess

Aandhencespecicheatwillbemaximum.

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