JIPMER Previous Paper 2012 - NEET MCQ

 De-Broglie wavelength 

 i.e. graph will be a rectangular hyperbola.

The depth of the image of an object submerged in a transperent medium; it is reduced from the real depth of the object by a factor equal to the relative index of refraction of the medium with respect to air.

Refractive index μ = 

For 1st liquid  

Similarly, for 2nd liquid, 

Total apparent depth = x₁  + x₂

Total mass of the shell = 20 kg

Ratio of the masses of the fragments = 2: 3 

∴ Masses of the fragments are 8 kg and 12 kg Now, according to the conservation of momentum , m₁v₁ = m₂v₂

∴ 8 × 6 = 12 × v

 v (velocity of the larger fragment) = 4 m/s

Kinetic energy 

In Young's double slit experiment half angular width is given by 

Capacitor with air as the dielectric has capacitance

C1 = ɛo/d (3A/4) = 3 ɛo_A/4d

Similarly the capacitor with k as dielectric constant has capacitance

We can clearly see that the first two forces 1N,2N if they are in the same direction, then they would equal to the third force 3N. But it is given that the three forces are in different directions. So there is no possibility of equilibrium with these forces in different directions.

Word done W = Area ABCEFDA = Area ABCD + Area CEFD

According to principle of continuity, for streamline flow of fluid 
through a tube of non-uniform cross-section the rate of flow of fluid 
(Q) is same at every point in the tube. 
i.e, Av = constant
Therefore, the rate of flow of fluid is same at M and N.  

Number of turns per unit width =  

Consider an elemental ring of radius x and with thickness dx Number of turns in the ring  

Magnetic field at the centre due to the ring element 

∴ Field at the centre

Let the police party catch the thief after t 

Distance travelled by police party in t  = v t

and distance travelled by thief    

Since by using the Quadratic equation formula

The excess pressure inside the soap bubble is inversely proportional to the radius of soap bubble i.e. P ∝1/r, r being the radius of the bubble. It follows that pressure inside a smaller bubble is greater than that inside a bigger bubble. Thus, if these two bubbles are connected by a tube, air will flow from smaller bubble to bigger bubble and the bigger bubble grows at the expense of the smaller one.

The thermo emf is given by E = aθ + bθ²

For neutral temperature 

⇒ a + 2bθ_n = 0

But neutral temperature can never be negative (less than zero) i.e.,  θ_n< |0^oC|

 Hence, no neutral temperature is possible for this thermocouple.

The two lenses of an achromatic doublet should have, sum of the product of their powers and dispersive power equal to zero.

Free body diagram of the two blocks are

Let Young's modulus of steel is Y₁  and of brass is Y₂

Dividing Eq. (i) by Eq. (ii), we get

Force on steel wire from free body diagram 

    T = F₁ = 2 g N

Force on brass wire from free body diagram

     F₂ = T' = T + 2g

          = 4 g N

Now, putting the values of F₁ , F₂   Eq. (iii) we get

Initially area of soap bubble  

Under isothermal condition radius becomes 2 r, Then, area A₂ = 

= 

Increase in surface area    

= 

Energy spent W = T × ΔA

 = T . 24

W = 24 

Given:- Current = 1.6 A

 t = 1 × 60s = 60 s

Charge flowing in one minute, q = I × t 

= 1.6 × 60

Charge required to liberate one copper ion 2C

Number of Cu²⁺ liberated 

A moving electric charge has a magnetic field associated with it. So, it's magnetic field interacts with the external magnetic field and a force is exerted on the moving charge.
But, a fixed charge does not have any magnetic field associated with it, so id does not experience any force in a magnetic field.

Image is not sharp because of deviation of light from straight line path account of diffraction of light.

 The reflected sound appears to propagate in a direction opposite to that of moving engine. Thus, the source and the observer can be presumed to approach each other with same velocity.

Wein's displacement law:-

Relationship between the temperature of a black body ( an ideal substance that emits and absorbs all frequencies of light) and the wavelength at which it emits the most light.

By using Wein's Displacement law,

⇒     500 × 6000 = 400 × T₂

As the block A moves  with velocity 0.15ms−1, it compresses the  spring which pushes B towards right. A goes on compressing the spring till the velocity acquired by B becomes equal to the velocity of A, ie, 0.15ms−1,

Let this velocity be v. Now, spring is in a state of maximum compression. Let x be the maximum compression at this stage. 

According to the law of conservation of linear momentum, we get  m_Au = (m_A+m_B)v or v = 

According to the law of conservation of energy, 

Let T°C be final temperature of the mixture.
According to principle of calorimetry,
Heat lost = Heat gained
∴ 0.1× 103 × swater × (80−T)
= 0.3 × 103 × swater × (T−60)
⇒ T = 65^∘C