MOCK TEST FOR CPET

 

Q/1. An object is dropped on a cushion from a height 10m above it. On being hit, the cushion is depressed by 0.1 m . Assuming that the cushion provides a constant resistive force, the deceleration of the object after hitting the cushion, in terms of the acceleration due to gravity g is ?

 (a) 10 g               (b) 50 g             (c) 100 g               (d) g

Ans :  From conservation at energy

                         mgh =  mv²/2

             ⇒  v = √(20g)

 The equation at motion when partition the cushion

                              mv(dv/dx) = mg – k

             _√(20g)∫⁰ vdv = _0.1∫⁰ (g-k/m) dx

                  V²/2 |_√(20g)⁰  = (g-k/m)-0.1

                   By sloving we get ; a= 100g

 

Q/2. Assume that the earth revolves in a circular orbit around the sun. Suppose the gravitational constant G varies slowly as a function of time. In particular, it decreases to half its initial value in the course of one million years. Then during this time the?

 (a) Radius of the earth’s orbit will increase by a factor of two

(b) Total energy of the earth remains constant

(c) Orbital angular momentum of the earth will increase

(d) Radius of the earth’s orbit remains the same.

 Ans : G = G(t)      ;    M =  Mass of sun , m = mass of Earth

The given Problem is central force problem so angular momentum of system is conserved. 

 Total Energy of system is

            E = mr ̇²/2 +J²/2mr² – GMm/r

Hence G = G(t)    so  dE/dt ¹ 0

So total Energy is not conserve

 Condition for circular motion  ;     J²/mr³ = GMm/r²

                   ⇒  r ∝ 1/G

  ⇒  r₂/r₁ = G₁/G₂

      ⇒   r₂ = (G₁/G₂)r₁

  By putting ; G₂ =G₁/2

              ⇒   r₂ = 2r₁

 

 

Q/3. A particle of mass m moves in One dimension in the potential V(x)= kx⁴ ; (k > 0) . at time t = 0 the particle starts from rest at x = A. For bounded motion, the time period of its motion is  ?

(a) Proportional to  A^-1/2

(b) Proportional to A^-1

(c) Independent of A

 (d) Not well-defined

Ans: J = ∫pdx           

        J = 4 ₀∫ √{2m(E -kx⁴)} dx          ; upper limit (E/k)^1/4

             Again  P²/2m +kx⁴ = E ;    J ∝ 4√(2mE)* (E/k)^1/4

           P=0, x=A  ; E= kA⁴    ,J ∝ E^3/4

           T= dJ/dE ∝ E^-1/4

            T=kA^-1     T∝ 1/A

 

Q/4. The wavefunction of a free particle of mass m , constrained to move in the interval -L ≤ x ≤  L , is Ψ(x) =A(L+x)(L- x)  , where A is the normalization constant. The probability that the particle will be found to have the energy p²ђ²/2mL²  is ?

(a) 0             (b) 1/√2            (c) 1/2√3           (d) 1/π

 Ans :       E_n = n²π²ђ²/2mL²  

         E₁ =π²ђ²/2mL²  ;     |Φ₁> =√(2/2L) cos(πx/2L)   ; -L ≤ x  ≤  L

         E₂ = 4π²ђ²/2mL²  ;     |Φ₂ > =√(2/2L) sin(2πx/2L)   ; -L ≤  x ≤  L

        P(E₂) = |< Φ₂|Ψ >|²/<Ψ|Ψ >

        = |_-L∫^L √(2/2L) sin(2πx/2L)A(L+x)(L- x)dx|²/_-L∫^L A²(L+x)²(L- x)² dx

_-L∫^L sin(2πx/2L)A(L+x)(L- x)dx =0

        Where  (2πx/2L)A(L+x)(L- x) is odd

 

Q/5. A particle moving in a central potential is described by a wavefunction Ψ(x) = zf(r) where r = (x,y,z ) is the position vector of the particle and f (r) is a function of r = |r|. If L is the total angular momentum of the particle, the value of  L² must be ?

(a) 2ђ²                (b) ђ²          (c) 4ђ²            (d) (¾)ђ²

 Ans : Ψ(r) = zf(r) =  r cosθ f(r)  

cosθ = P₁(cosθ)      ⇒ l=1

   Ψ(r,θ,Φ) = P₁(cosθ)r f(r)         

 the measure at  L² have eigen value       l(l+1)ђ²   

put ; l=1      eigen value =2ђ²                

 

Q/6. The ratio of orbital magnetic dipole moment m_l to the orbital angular momentum L of an electron in an orbit is given by

(a)μ_L/L =μ_b/ђ     (b) μ_L/L = -μ_b/ђ     (c) μ_L/L = -μ_b/2ђ     (d) μ_L/L =μ_b/2ђ     

Ans :  i = -e/(2πr/v)      

           μ = -eVπr²/2pr  = -eVr/2

         μ =-eL/2m = -(eђ/2m)√{l(l+1)}                  (L= mVr)

         μ =-μ_b√{l(l+1)} ;     

        μ =-μ_bL/ђ ;         μ_L/L = -μ_b/ђ            

Q/7. The half-width of gain profile of a He-Ne laser is  2*10^-3nm . If the length of the cavity is30 cm , how many longitudinal modes can be excited? The emission wavelength is 6328 A⁰

(a) 1              (b) 2           (c) 3              (d) 4   

Ans : For longitudinal modules         nλ/2  = L

                   n= 2L/λ  = (2*0.3)/(6328*10^-10) = 10⁶

          again  λ= 2L/n ;   dλ = |-(2L/n²)dn|

                dλ  = (2*0.3/n²)*1 =0.6/10¹²= 0.6*10^-12

     Number of modes = Dλ/dλ = 2*10^-18/0.6*10^-12 = 3.33≈ 3

 

Q/8. The mean free path of molecules of a gas at pressure p and T temperature is 2*10^-5 cm  . The mean free path at pressure  p/2  and temperature 2T will be

 (a) 10^-5 cm          (b) 8´10^-5 cm       (c)10^-5 m           (d) 8´10^-5 m  

Ans :  λ₁ = k_BT/sP

          λ₂ = k_B*2T/s(P/2)

            λ₂ = 4l₁ = 4*2*10^-5 = 8*10^-5 cm

Q/9. For the adiabatic expansion of a blackbody radiation enclosure, which of the following is correct?

 (a) V^1/3T = constant

(b) V T.= constant

 (c) V^4/3T = constant

 (d) V.T =constant               where V is the volume and T is the temperature of the enclosure

Ans : For adiabatic expansion PV^ꝩ =k (constant)

g = 4/3   (Assuming photon as monoatomic gas)

    PV = RT ;  P=RT/V

   ⇒ (RT/V)V^ꝩ =k

    TV^ꝩ-1 = TV^1/3 = constant k

Q/10. At what temperature; pressure remaining unchanged, will the molecular velocity (root mean square velocity) of hydrogen will be double of its value at NTP?

  (a)1092⁰ C      (b)819⁰ C     (c)1092⁰ F       (d)819⁰ K

Ans : V_{rms ∝}√T   

         ⇒2V/V  = √(T₂/T₁ NTP)

                  T₂/T₁ = 4

T₂= 4*273 K =  1092⁰K

   T₂(⁰ C) = 1092-273 = 819⁰ C