Answer with Explanations

4. (a,b,c,d) As conservation of linear momentum and energy, maximum height and velocity attained by block moving on smooth horizontal surface

i.e. P_i m v

where, P is the momentum of N-particle system.

Taking block and wedge as a system, we can analyse that there is no horizontal force acting on system. So, horizontal momentum is conserved.

So, con ser va tion of lin ear mo men tum,

mu=mv₁−mv₂ …(i)

and conservation of kinetic energy, then

C B

Answer with Explanations

u

mu mv mv

Solving Eqs. (i) and (ii), we get v₁= , vu ₂=0

∴ From beginning to maximum height mu=m v(2 ⇒ v) u

=2 So, conservation of mechanical energy upto maximum height, we get

5. (a,c) We have to use the steady state concept, i.e. in steady state capacitor will work as open circuit. We will calculate the potential difference across R₂ and accordingly charge will be calculated The charge will be = C (potential difference across R₂) Q₀ is the steady state charge stored in the capacitor Q₀= (potential difference across capacitor in steady state)C

= C(steady state current through R₂) (R₂)

6. (a,b,c) The mass of the portion of the ring, determine field and potential due to each part and then find the net potential and net field at point O due to elementary masses dm_a dm_a

1, 2 while comparing, we get

Elementary masses, dm_a1 = λ(r d₁ θ) and dm_a2 = λ(r d₂ θ)

Gravitational field at A

dE Gdm

∴ Gravitational potential

dV Gdm i.e. element a, produces equal potential to element a₂

7. (a,b,c,d) First conservation of mechanical energy is used i.e. loss in potential energy mgh is converted in to KE

mgh=1mv 2

2

⇒ v= 2gh

When the cylinder starts moving on the plank rolling as well as rotational motion will be retarted. Initially block is having translation as well as rotation and after sometime it will start pure rolling on the plank.

Form conservation of energy we can say that the velocity of cylinder just before getting onto plank is

mgh=1mv

2 ⁰

2

where, v₀= 2gh

FBD for motion on rough plank.

So, cylinder is having retardation of linear motion.

v u= ₀−µgt and it is having angular acceleration.

α µ Plank will be having accelerated motion

v′ =0 µ+ gt

For minimum length of plank let us consider cylinder (w.r.t.) plank, for linear motion distance moved by COM in time of pure rolling S=v t_o −1 g t

8. (a,b,c) As first law of thermodynamics and its processes i.e. isochoric and isobaric

Q=U+ ∆W where, Q is heat energy, U is potential energy and ∆W is the work done by the system.

In AB process, W > 0 and ∆U > 0 because due to volume increment temperature also increases so, ∆Q for AB will also be positive. So heat is given in process AB, in BC work done is zero and temperature decreases due

to ∆U and ∆Q will be negative that means heat is rejected by gas.

Q₁=W₁+ ∆U₁

9. (a,c) According to radioactive law, we know R= λN₀ and t_{1 2} 2

16 (4 half-lives) ( )

10. (a,b) Ap ply Bernoulli’s equation of con ti nu ity at two lev els con sid ered. The height of the con sid ered level will be taken as zero, hence the term con tain ing height i.e. ρgh = 0

Apply Bernoulli’s equation between P and Q points

p₀ gh 1 v₁² p₀ v₂² From equation of continuity, we get

Av av A

Acceleration of top layer of liquid, a dv

11. (2) It is based on Bernoulli’s equation for flowing liquid for steady or streamline flow.

i.e. p+1 v + gh

2

ρ 2 ρ = constant

where, p is pressure energy, 1 2

ρv is kinetic energy per unit2

volume and ρgh is the potential energy per unit volume Apply Bernoulli’s equation at pipe 1, 2, and 3

p₃ 1 v₃² p₂ v₂² p₁ v₁²

Equation of continuity at point 3 and 2 are v A_{3 3}=v A_{2 2}

12. (0) According to impulse and conservation of energy in elastic collision.

Ball with respect to surface upward impulse on ball

Ndt=P_f−P_i

The coefficient of restitution e along normal direction

e v_y

So,

∫

Ndt = + =^{4 8 12} Horizontal impulse

−

∫

^µNdt=^Pf−^Pi −^{µ Ndt}

∫

=^2vx−²×²

−1× = −

3 12 2v_x 4 i.e. v_x= 0 13. (2) Con sider the rays as

shown in fig ure, for ray 1 with an gle of in ci dence i₁, (i₁ < θ_C) where, (θ_C is crit i cal an gle). The ray will get re fracted, but H₁ is not the min i mum height.

For ray 3 with an gle of

in ci dence i i_{3 3}( < θ the ray will get re _C) flected (due to phe nom e non of TIR), now for light ray 2 with an gle of in ci dence i₂ (i₂= θ the ray will just re fracted (lim it ing ray) and cor re spond ing _C) H₂ will be min i mum height so, that light will just refracted. For TIR at that point of in ci dence on curved sur face. We can early say that, as h increases i de creases and for TIP at the point of in ci dence min i mum value of i should be θC and cor re spond ing to that h will be minimum for refraction.

According to Snell’s law, we get 3sini =1sin90°

sin i =1 3 So, slope of tangent will be 90° − i

dy

dx=tan(90° −i) =cot i=2 2

1

At, y=h

dy dx= 2 x 2x =2 2 x= 2m y=h=( 2)²=2 m

The minimum distance of incident ray from surface AD is 2 m.

14. (3) It is based on conservation of mechanical energy for the two springs and the non-conservative force i.e. friction will dissipate same part of mechanical energy.

Velocity of block at A is according to conservation of energy mv² Kx²

2 = 2

1 2

1350 0 1 2

2 2

×v_A = ×( . )

v_A² 1350

= 100 ⇒ v_A= 1350 100m/s During motion of a block from A to B,

v_B²=v_A²+2as a= −µg= −0 3 10. × = −3 m/s =1350− × ×

100 2 3 2 =1350− 100 12

= −

1350 1200= 100

150 100m/s²

Now, with this KE, it will compress right spring, we get mv_B² kx_f²

2 = 2

1 2

150 100

1350 2

× = × x_f2

x_f² 150 1300

1 100

1

= × =900

x_f= 1 = = − 30

100

30 3 33. cm~3

15. (3) It is based on the thermal conductivity of the material and bulk modulus of elasticity. We can directly relate change in length of steel bar with stress applied in bar and easily.

Consider an element of the bar of the length dx shown in figure.

Change in length of element dx due to temperature difference is dU=dx×α×∆ T

dU dx T x

= ×α× ⁰L

2

2

Integrating both sides, we get

dU T x

L dx

∫

⁼

∫

L^{α 0} 2

0 2

U T L

=αL₀ ³ 3 2

Bar is rigidly field stress in given by σ = E × strain =U×

L E

σ=α α

⋅ × =

T L

L₀L³ E E T

2

0

3 3

Stress in bar, σ =EαT₀ 3 As per given in question, we get

E T x

E T

α ₀ α ₀

= 3

So, x = 3

16. (5) Con sider small seg ment that will be disc at dis tance x as shown

dm= ⋅ρ dV

For the moment of inertia of the disc about then given axis dI=1dm y⋅

2

2

where, y is the radius of a solid spherical body

The solid sphere is generated by revolving circle about diameter.

Mass of spherical segment

M = density × volume of segment 1

2

3 H₁

H₂ H3

i3

i₂ i₁

h

y=x2

i (90° – i) (90° – i)

A D

x dx

O x

x y

dx A

R

C

2 2 2

x +y =R

ρ

Consider a small disc of radius y and thickness dx as shown is figure.

Mass of small disc dm= ⋅ρ πy dx² Moment of inertia of disc about O, x-axis

dI=1dm y dl⋅ = y dx×y = y dx 2

1

2 2

2 ρπ 2 2 ρπ 4

Total moment of inertia

I dl y dx

17. (7) Net work done for the process, Efficiency = Total work done

Total heat supplied

W_AB=p(∆V)=nR T∆ (isochoric process)

ln ln (isothermal process)

= × × ×

Heat supplied to the monoatomic gas from A to B,

Q nC T R

1500 1200 2

R

18. (2) It is based on photoelectric effect and ionisation energy of an atom. So, using de-Broglie relation

i.e. λ = h

mE 2

where, λ is wavelength and E is energy of an electron.

E₁=W₀+ KE₁ …(i) Energy of first line in Lyman series

=  − and energy of series limit of Balmer series

= × −

19. (1) As the charge density vary with sine function, then half part of ring will be positive and half part will be negative. We can consider positive element of find the dipole moment of this small part.

Then,

p=

∫

0 ²dp φ −

2 sin [ ]

/ direction j

π

20. (0) The concept of complex number in which we calculate impedance of circuit by treating as pure real number and capacitance, inductance as pure imaginary number

W₀

For C X According to Newton’s law of cooling, we get

⇒ T⁴−T₀⁴=[(T₀+∆ )T ⁴−T_o⁴)]

T (using binomial theorem)

=  + −

By Stefan’s law dT Thus, curve will be given as showin in figure

2. (b) Let the thickness of ice at 0°C is x. If the thickness is increased by dx in time dt. Then amount of heat flowing through the slab in time dt is given by.

Q KA T dt

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

x = ρA L dx ⇒ dt L KT xdx

=ρ By integration, we get,

dt L

Time taken to increase the thickness of the layer

t L

3. (a) As, percentage change in radius of a metal disc, dR

Rotational kinetic energy K I

I I where, R is radius of disc.

⇒ K L

On differentiating, we get dK Thus, % change in rotational kinetic energy,

K dK

Now, the work done, W pdV

v Work done in complete cycle

W=W_ab+W_bc +W_cd +W_da

=1000Rln2+0−600Rln2+0 = 400Rln2 This is the value of heat involved in the given cyclic process.

5. (a) Equipartition of en ergy tells that energy associated with each de gree of free dom is 1

2KT. As the di atomic gas mol e cule has two rot ational de gree of free dom so, fi nal tem per a ture of the system.

T f n T f n T For monoatomic gas,

f₂=3,n₂=5 and T₂=470 K

∴ T_f= × × + × ×

× + × =

5 3 250 3 5 470

5 3 3 5 360 K

Suppose, the final angular velocity of diatomic gas molecule is ω_{rms, f} then according to law of equipartition of energy

1 2

Iω2_rms,f=KT

Angular velocity of diatomic molecule, ωrms. f

KT

6. (b) The circular spring is given below.

The two springs can be assumed to be parallel between two masses m₁ and m₂.

The equivalent spring constant K_eq=K₁+K₂

If the spring cut, the force constant of spring K

∝1l.

⇒ K l_{2 2}=K l_{1 1}=Kl Substituting l l

1 The reduced mass,

µ₁ ^{1 2}

Now, frequency of oscillation.

ω= kµ = k

7. (b) Let the density of liquid be ρ. The total mechanical energy of liquid column is

E A h x A h x dx

On differentiating, we get d x dt

Thus, the frequency of oscillation of the liquid,

or ω

8. (d) If the potential of sphere is raised to V, the electron should have a minimum energy φ + eV to be able to come out. Thus, emission of photoelectron will stop when

hc eV

The change on the sphere needed to take its potential to V is φ= (4πε₀a V)

The number of electron emitted is n

The emf in induced between the ends of wire.

E=vB l As rails and wire are of zero resistance then, charge on capacitor.

q=C_E=CvBl The current through the circuit i dq

dt CBldv

= = dt = CBla The current is the cause to force on the wire.

Thus, the magnetic force F′ =i lB CB l a= ^{2 2}

10. (c) Let the magnetic field produced by short bar magnet be B. As short bar magnet is oscillating, hence the resultant horizontal field is B_H+ .B

Now, suppose, M and M′ denote the magnetic moments of oscillating magnet and other magnet, respectively.

Then, ν As the magnetic field can be written as

B m

the ca ble. The mag netic field at each point of the cir cle will have the same mag ni tude and will be tan gen tial to it.

The circulation of field B along the circle is B ⋅ =

∫

^{d l B2 πx}

The current enclosed within the circle is i

Ampere’s law tells that B⋅ l=

∫

^{d µ0i}

Thus, B x i x a

2 ^{0 0}

2

π µ 2

=

⇒ B i x

=µ a π

0 0

2 2

The direction is along the tangent to the circle.

12. (a) The area of x-sec tion of the outer shell is πc²−πb². The area of x-sec tion of outer shell within part c is πx²−πb². Thus, cur rent through this part is i x b

c b

0 2 2

2 2

( − )

− . The net cur rent en closed by the cir cle is

i i x b

c b

i c x

c b

0 0 2 2

2 2

0 2 2

2 2

− −

− = −

−

( ) ( )

From Ampere’s law, B x i c x

c b

2 ⁰

2 2

2 2

π µ

= −

−

( )

⇒ B i c x

x c b

= −

− µ

π

0 0 2 2

2 2

2

( )

( )

13. (a) Con sider the di a gram be low.

Let the bigger mass of 2 m accelerates towards right with acceleration be a. Then, free body diagram can be given as

For mass along M₂ 2mω²R T− =2ma ...(i) For mass along M₁, T−mω²R=ma ...(ii) Adding Eqs (i) and (ii), we get,

mω²R=3ma ⇒ a R

=ω² 3 14. (b) Now, sub sti tut ing the value of a in Eq (i), we get

2 2

3

2 2

m R T m R

ω ω

− = . ⇒ Tension developed in the string T=4m R

3 ω2

15. (c) Con sider the di a gram.

The hollow sphere is released from the top of incline having angle of inclination θ.

To prevent the sliding, the sphere will make only pure rolling in this condition.

mgsin θ − =f ma ...(i)

where, f is friction force

Now, torque produced by the friction about the centre of sphere (let radius of sphere is R)

f R mR a

× = ×R

 

 2

3

2 ⇒ f=2ma

3 ...(ii)

where, a = acceleration Now, from Eqs. (i) and (ii), we get

mgsin θ −2ma=ma

3 ⇒ a=3g 5 sin θ

Therefore, Eq (i) becomes mgsinθ− =f 3mgsinθ

5

⇒ f=2mg

5 sin θ

⇒ µN=µmgcosθ=2mgsinθ 5 Minimum coefficient of friction, µ=2 θ

5tan

16. (a) The fric tion be comes its half when the co ef fi cient of fric tion be comes half of its orig i nal value, then

µ=  θ θ θ

 

 = =

1 2

2 5

2 10

1

tan tan 5tan

Then, net torque = Iα (given) µmgcosθR=2mR α

3

2

⇒ 1 10

2 3

×tanθ×(mgcos )θ ×R= mR2α Angular acceleration of a sphere

⇒ α θ

= 

 

 3 10

g R sin

17. (a) (A) Emission of radiation → Surface property.

(B) Automatic open and closing of doors → Photoelectric effect (C) Radio activity → Nucleon property

(D) Breader reactor → Generatiuon of fuel itself 18. (a) (i) Potential energy of a simple pendulum

(ii) Range of a projectile when projected at a fixed angle

(iii) As, square of time period of a simple pendulum as a function of its length

19. (d) (A) Aluminium → 0.16 (B) Iron → 0.27

(C) Steel → 0.19 (D) Tungsten → 0.2 4

20. (d) (A) Coefficient of viscosity → [ML T^{−1 −1}] (B) Gravitational consatant → [M L T^{-1 3 −2}] (C) Thermal conductivity → [MLT K^{−3 −1}]

(D) A unit vector in a particular direction → [M L T K° ° ° °] m

M₂ 2m M₁

m T

a

mω R2

2m T

a

2mω R2

θ mg

mg sin θ a

mg cos θ θ

y

x

y

x

y

x

1. The potential energy U of a particle varies with distance x from a fixed origin as U

A x

x B

=

²

+ , where A

and B are dimensional constants.

In document Physics Spectrum (Page 28-36)