A block and a pan of equal masses are connected by a string going over a

smooth light pulley as shown in the figure. Initially the system is at rest. A particle of mass m falls on the pan and strikes the pan with a speed v. Find the speed with which the system moved just after the collision.

(a) v

3 (b) v

5 (c) v

9 (d) v

7

30. The half-life of

²¹⁵

At is 100 µs. If a sample initially contains 6 mg of the element, what is its activity initially and after 200 µs ?

(a) 1.4 × 10¹⁷ Bq and 2 8. ×10¹⁵ Bq (b) 1.7 × 10¹⁸ Bq and 2 5. ×10¹⁷ Bq (c) 1.16 × 10²³ Bq and 2 9. ×10²² Bq (d) 1.9 × 10²⁴ Bq and 3 4. ×10²⁰ Bq

Detailed solutions of these questions are available on http://www.arihantbooks.com/Physics%20Spectrum.pdf

Answers

1. (b) 2. (d) 3. (b) 4. (c) 5. (d) 6. (d) 7. (b) 8. (b) 9. (d) 10. (b)

11. (c) 12. (d) 13. (c) 14. (c) 15. (c) 16. (a) 17. (a) 18. (c) 19. (c) 20. (b)

21. (c) 22. (b) 23. (b) 24. (b) 25. (c) 26. (a) 27. (b) 28. (b) 29. (a) 30. (c)

S R Q P log I

t

θ

v₂ 60°

400 m/s

m m m

1. (b) As, potential energy U of a particle varies with distance x from a fixed origin is given by =

+ A x x² B

here, B must have the dimension of x² i.e. [ ]L² Dimensions of A x

= x =

4 ² [ML T ][L ]−

L

2 2 2

1/ 2 =[ML^{7/ 2}T ]⁻²

∴ AB =[ML^{7/ 2}T ] [L ] [M L^{−2 2} = ^{1 11/ 2}T ]⁻²

2. (d) Let t is the time in which the car is acquiring maximum speed 35 7

t = ⇒ t = 5 s

In this time the car will cover a distance of s=1at = × × =

2 1

2 7 5 87 5

2 2 . m. If the car starts deaccelerating at t = 5 s, then it will travel a total distance of 2s = 175m, before coming to rest.

Since, this distance is not the 500 m distance required, the car will travel for some distance with constant maximum speed. Let this time be t₁.

Then the velocity-time graph can be drawn as shown in figure.

The distance travelled in time t₀ =500 175 325 m− = time taken, t₁ 325

35 9 28

= = . s

Hence, total time of travel = +t t₁+t = +5 9.28+5 = 19.28 s 3. (b) Let the initial amplitude a decreases to a₁, to the other side i.e.

after the first sweep,

decreases in elastic potential energy = work done against friction 1

2 1 2

2 12

ka − ka =µmg a( +a1)

or 1

2k a( +a¹)(a−a¹)=µmg a( +a¹)

or a a mg

− ₁=2µk …(i)

Similarly a a mg

1 2 k

− =2µ …(ii)

M

a a mg

n−₁− n=2µk …(iii)

On adding Eqs. (i), (ii) and (iii) we get,

a a mg

n k

− =2µ

the block stops when µmg=ka_n or a mg

n=µk Substituting in the above equation,

(2n 1) mg

k a

+ 

 

 = µ

or ( ) .

2 1 .20 0 3

0 04 1 10 15

n ka

+ = mg= ×

× × =

µ

2n + =1 15

∴ n = 7

4. (c) Let I₁ and I₂ be the current drawn from cells of emf 6V and 4 V in the circuit.

Then, I₁ 6

2 3 1 1

= + + = A

I₂ 4 6 4 0 4

= + = . A V_A−V_B= ×1 3=3V, V_B−V_C= 5 V

V_C −V_D =0 4. ×6 2 4= . V

∴ V_A−V_D=(V_A−V_B) (+ V_B−V_C) (+ V_C−V_D) = + +3 5 2 4 10 4. = . V

5. (d) Electric field within the plates E=E_Q1 +E_Q2

or E=E₁−E₂ = Q −

A Q

A

1 0

2

2 ε 2 ε0

E Q Q

= ₁A− ₂ 2 ε0

∴ Potential difference between the plates.

V V E d Q Q

A d

A− B= ⋅ = −

 



1 2

2 ε0 = −



 



= −

Q Q

A d

Q Q

C

1 2

0

1 2

2 ε 2

6. (d) Velocity vector is perpendicular to magnetic field. Therefore path of the particle is a circle of radius r mv

= Bq⁰

v=v₀ as the speed of the particle does not change in magnetic field.

Centre of the circle is C

CP=CQ

∴ ∠CPQ= ∠CQP

or 90−α=90−β ∴ (α=β)

For the PQ=2PR=2rcos(90−α = 2 r sin α)

∴ PQ mv

=2 Bq₀sin α

Answer with Explanations

t t1 t t(s)

7 7

35 v(m/s)

+Q2

+Q1 E2 E₁

R C S 90°

90°

Q v β

v₀ α P

∠PCR= ∠RCQ=α

∴ ∠PCQ=2 α

arc PSQ r mv

=(2 −2 ) =2 ₀Bq( − )

π α π α

∴ t PSQ

V m

PSQ= = Bq−

0

2 (π α)

7. (b) As in L-C-R circuit, resistance R is given by

R L

= C

∴ R L

C

2= or CR L

=R

Hence, time constants of both the circuits are equal τ_C=τ_L = τ

Now, i V

R e

L= (1− ⁻t^/τ) …(i)

i V

R e

C= (1− ⁻t^/τ) …(ii)

Equating Eqs. (i) and (ii), we get

(1 −e^−t/τ)=e^−t/τ or 2e^{−t/ τ}=1 e^{−t/ τ}=1

2, t τ= ln (2) or t = τ ln (2) = CR ln (2) 8. (b) Magnetic field inside the solenoid is

B=µ₀nI=(4π×10⁻⁷)×2000 2 5× . =2π×10⁻³ T Bev mv

max= rmax 2

or v Ber

max= m

= × × × ×

×

− −

−

( ) ( . ) .

.

2 10 1 6 10 0 01

9 1 10

3 19

31

π

=1.112×10⁷m/s hc

mv W

λ =1 +

2

2 0

max

=1× × ⁻ × × +

2 ( .9 1 10³¹) (1.112 10^{7 2}) 0 5.

×1.6 10× ⁻¹⁹

=5.64 10× ⁻¹⁷ λ =

× ⁻ hc

5.64 10¹⁷= × × ×

×

−

−

( .6 63 10 ) (3 10 ) 10

34 8

5.64 17 =3.52×10⁻⁹ m 9. (d) At t = 0, suppose number of atoms of X is N0

At t = 4 hours (i.e. two half lives)

Number of atoms of X left undecayed N N

x= ⁰ 4 Number of atoms of Y formed, N N

y=3 4

0

∴ N

N

x y

=1 3 At, t = 6 hours (i.e. three half lives),

N N

x= ⁰

8 and N N

y=7 8

0

N N

x y

=1 7

As the given ratio (1/4) lies between (1/3) and (1/7), therefore, time t lies between 4 and 6 hours.

10. (b) here, h_FE= forward current ratio i.e. β =90 , V° _CE= 4 V If I_C is the collector current, then

9−4= I R_{C C} or I_C= 2 5. mA

I I

B= C

β=2 5 90

. mA=2 78 10. × ⁻⁵ A

Since, the transistor operates in active region therefore, V_BE= 0 7. V

∴ R

B I

b

= −

=

× ⁻

3 0 7 2 3

2 78 10 ⁵

. .

. =82×10³ Ω R_B= 82 kΩ 11. (c) From the theorem

L₀=L_com+M(r×v) …(i)

Angular momentum about O = Angular momentum about center of mass + Angular momentum of center of mass about origin

=L₀=I_ω+MRv =1 + 2

MR2ω MR R( ω) =3 2

MR ω2

In this case both the terms in Eq. (i) i.e. L_com and M(r×v) have the same direction X. That’s why we have to use L₀=I_ω+mrv. 12. (d) Let r be the radius of the satellite from centre of earth.

Then, orbital velocity is given

v GM

0 r

2= …(i)

If, R = radius of earth

Applying conservation of mechanical energy between points A and B,

kinetic energy = change in potential energy

∴ 1

2

mv2 GMm r

GMm

= − − − R

 



v Gm

R Gm

r

2 2 2

= − =v_e²−2v₀²= v_e²−2v₀² 13. (c) Let the displacement equation of particle is

v=R_ω

x y

O ω

v

x y

O ω

R

r

v B A

u=0

O P

B t=0

a → x a−x

x=asinωt a A

Time period of particle would be

14. (c) Applying Bernoullis theorem between point 1and2

v₂²=v₁²+2gh …(i)

Applying continuity equation between 1 and point we get A v_{1 1}=A v_{2 2}

Substituting the given values, A₂

As, cross-sectional area of stream of water 0.015 m below the tap is A₂=5.0×10⁻⁵ m².

15. (c) When there is equal level of liquid in two arms of U-tube, then height of liquid in each arms of U-tube =h₁+h₂ liquid has been transferred from left arm to right arm of U-tube.

The mass of the liquid transferred from left arm to right arm of

U-tube = −

 



h h

1 2 A

2 where, A = cross-sectional area of tube, ρ = density of liquid. The decrease in height of this liquid

= −

 

 h₁ h₂

2

loss in potential energy of this liquid = −

  If this liquid moves with velocity v, then

KE =1 + +

2 ^{1 2}

(h h h Av)ρ 2

using law of conservation of energy.

1 if ρ is coefficient of volume expansion of liquid, then

Density at temperature t₁is, d d

1 0t

According to Archimede’s principle f Vd_{1 1}=m=f Vd_{2 2} or d The coefficient of volumetric expansion of liquid is

ρ = −

17. (a) In the processes AB and CD, volume is constant therefore, no work is done

W_DA=RT₀log_e2

Total work done by the gas W=W_DA+W_BC

=RT₀log_e2−2RT₀log_e2 = − RT0loge2

| |W =RT₀log_e2

18. (c) Let d ′ be the diameter of refracted beam

Then, d=PQcos 60°

sin sin

r i

= = =

µ 3 2 3 2

1 3

cosr = 1−sin²µ = 2 3

∴ d ′ = ( )( )2 2 2 3 The diameter of refracted beam is = 4 2

3 cm ≈ 3.26 cm

19. (c) For understanding, let us assume that the two loops are lying in the plane of paper as shown. The current in loop 1 will produced, • magnetic field in loop 2. Therefore,

increase in current loop 1 will produced an induced current in loop 2 which produces ⊗ magnetic field passing through it i.e.

induced current in loop 2 will also be clockwise as shown above.

The loop will now repel each other as the current at the nearest and farthest points of the two loops.

20. (b) If the current flows out of the paper, the magnetic field at points to the right of the wire will be upwards and to the left will be downwards as shown in figure.

Now, let us come to the problem, magnetic field at C = 0 Magnetic field in region BX′ will be upwards (+ve) because all points lying in this region are to the right of both the wires.

Similarly, magnetic field in region AX will be downwards (–ve) magnetic field in region AC will be upwards (+ve), because points are closer to A compared to B. Similarly magnetic field in region, BC will be downwards (–ve) graph B satisfies all the conditions.

21. (c) For first resoance,

30 7. =4λ+n …(i)

for second resonance, 63 2 3

. = 4λ+n …(ii)

On solving Eqs. (i) and (ii), we get λ = 65 cm

The error in measuring the length using metric scale would be 0.1 cm which is least count of metric scale.

Therefore, λ =(65.0±0 1 cm. ) As, velocity of sound is

v = ×ν λ =512(65.2±0 1. ) =(33980±51.2 cm/s) Therefore, error in velocity = 51.2 cm/s

22. (b) While charging in a RC circuit, I E Re ^{t RC}

= ^{− /} Taking log of both sides

logI logE R

t

= −RC This is the equation of a straight line, with slope = − 1

RC

When R is doubled, slope becomes less negative i.e. more also at t = 0, current will be less, graph Q represents these facts.

23. (b) The voltage amplification A R

V RL

C

=β

=60×6 10× 600

3

= 600

Now, power amplification, A R

P R L

C

=β²

⇒ A_P=(60) ×6000

600

2

= 36000

The required ratio, A A

P V

=36000= 600 60 24. (b) The minimum deviation formula

Snell’s law applied to the first surface

sin sin

θ θ

2 1 1

2

=n

n = °

1.00 = 1.65

sin65 . 0 549

or θ₂=33 3. °

from geometry

φ₂=A−θ₂=60−33 3. ° =26 7. ° Snells law applied to the second surfaces thus gives

sin sin

φ φ

1 2 2

1

=n

n = × °

1.65 26.7 = 1.00

sin 0 741.

or φ₁=47 8. °

Figure shows a ray traced through the prism. The angle of deviation is

θ_D=θ₁+φ₁−A

=65 47 8+ . ° −60°=52 8. ° The difference in θ_D for blue and red light is

∆θ_D=557. ° −52 8. =° 2 9. °

25. (c) Each fission yield 200 MeV =(200 10× ⁶) ( .×1 6 10× ⁻¹⁹) J of energy

Only 20% of this is utilised efficiently and so,

Energy generated per fission =(200 10× ⁶)×1.6×10⁻¹⁹×0.20 =6 4 10. × ⁻¹² J

Because the reactors output is 700 106× J/s, the number of fission required per second is

1

F

F F

2 Perpendicular

to paper outwards

Perpendicular to paper inwards

B i°

B

B

X A C B X′

i

A

θ₁ θ2

θD

φ1

φ2

180 –A

fission/s = ×

× ⁻ 7 10 6 4 10

8

12

J s/

. =1.1 10× ²⁰ s⁻¹

and fission/day = 86400 S/d ×11 10. × ²⁰ s⁻¹ =9 5 10. × ²⁴ d⁻¹ There are 6 02. ×10²⁶ atoms in 235 kg of Uranium-235. Therefore the mass of uranium 235 consumed in one day is

Mass = ×

×



 

 × 9 5 10

6 02 10 235

24

26

.

. = 3 7. kg

26. (a) Let S and S′ refer to object and image distances. The nearest distance S occer when S′ is largest, So

1 1 1

S=f−S

′ = 1 − 1 = 12

1 60

5.0cm cm cm

or S = 8 57. cm

The farthest distance S occurs when S′ is smallest, so

1 1 1

S=f−S

′ = − 1

1 = 7

2 35

5.0cm cm cm

or S = 17 5. cm

magnification of the images of an object at the nearest

m S

=S′= 12 = cm 1 4

cm

8.57 .

At the farthest distance

m S

=S′

= 7 =

17 5cm 0 4 cm

. .

27. (b) Over a very small time interval surrounding the moment of explosion, the effect of gravity, (an external force) can be neglected. Then, all forces are internal and momentum is conserved.

The vector diagram for momentum conservation is shown in figure.

3

4 ² 4 ¹

Mv =Mv−Mv

or v₂ 4v v₁

3 1

= −3

Then, v₂²=v₂⋅v₂=16 ⋅ + ⋅ + ⋅ 9

8 9

1

1 9 1 1

(v v) (v v) (v v)

=16 − ° +

9 200 8

9200 400 60 1

9 400

2 2

( ) ( )( )cos ( )

=48 10× 9

4

and v₂ 400

3 231

= = m/s

28. Since, the wavelength ( )λ is increasing, we can say that the galaxy is receding. Doppler effect can be given by

λ′ =λ +

− C v

C v …(i)

or 706=656 +

− C v C v

or C v

C v +

− =

 

 = 706 656

2

1.16

∴ C+ =v 1.16C−1.16v

v C

=0.16

2 16. =0 16×3 10× 2 16 . 8

.

v =0.22×10⁸ m/s ≈ 2×10⁷ m/s 29. (a) Let N = magnitude of the contact force between the particle

and the pan.

T = tension in the string

Consider the impulse imparted to the particle, the force is N in upward direction and the impulse is N dt

∫

. This should be equal to the change in its momentum.

Thus,

∫

N dt⁼mv⁻mV ^…(i)

Similarly considering the impusle imparted to the pan, (N−T dt) =mV

∫

^…(ii)

and that to the block,

Tdt mV

∫

^{= …(iii)}

On adding Eqs. (ii) and (iii), we get N dt mV

∫

^{= 2}

Comparing with Eq. (i)

mv−mV= 2mV or V v

=3 30. (c) The number of radioactive atoms initially present is

N₀

3 23

6 10

215 6 03 10

= ×

×

( − )

( . )

g

g / mol atoms mol/ =1.68 10× ¹⁹ atoms

and the decay constant of ²¹⁵ At is

λ = =

× =

−

ln .

/

2 0 693

100 10 6930

1 2 6

T s⁻¹

Hence, the initial activity is

A₀=λN₀=6930×1.68×10¹⁹ =1.6 10× ²³ Bq At t=T_{1 2/}

A=A e₀ ^−λt = × ×

 



1.16 10 1

2

23 2

=2.9 10× ²²Bq Mv

M4v1

3M 4 v₂ θ

60°

1. The heat produced in a wire carrying an electric

In document Physics Spectrum (Page 38-44)