JEE 2011 :: General Questions :: Physics

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Column I shows four systems, each of the same length L, for producing standing waves. The lowest possible natural frequency of a system is called its fundamental frequency, whose wavelength is denoted as $\lambda_{f}$ . Match each system with statements given in Column II describing the nature and wavelength of the standing waves.

Answer:

Explanation:

In organ pipes, longitudinal waves are formed. In string, transverse waves are formed. Open-end of pipe is displacement antinode and closed end is the displacement node. In case of string fixed end of a string is node.

Further, least distance between a node and an antinode is  $\frac{\lambda}{4}$ and between two  nodes is $\frac{\lambda}{2}$  Keeping these Points in mind answer to this question is as under

 (A) $\rightarrow$  (p,t)

(B) $\rightarrow$  (p,s)

(C) $\rightarrow$  (q,s)

(D) $\rightarrow$  (q,r)

Analysis of Question
(i) Unlike the Previous years few match the column Problems, this question seems to be more logical' If you know the subject, You can select
correct matching without confusion.
(ii) In option (d) of column I, the string is fixed from centre, therefore at lowest frequency, two loops will be formed as shown below

One mole of a monatomic ideal gas is taken through a cycle ABCDA as shown in the p-V diagram. column II gives the characteristics involved in the cycle. Match them with each of the processes given in Column I

Answer:

Explanation:

 Internal energy $\propto T \propto pV$

 This is because 

 $U= \frac{nf}{2}RT=\frac{f}{2}pV$

 Here, n= number of moles

 f= degree of freedom

If the product pV increases, then internal energy will increase and if the product decreases, the internal energy will decrease.

Further, work is done on the gas, if the volume of gas decreases. For heat exchange

 $Q= W+\triangle U$

Work done is area under p-V graph. If volume increases work done by gas is positive and if volume decreases work done by gas is negative. Further  $\triangle U$ is 

positive if product of pV is increasing and $\triangle U$ is negative if the product of pV is decreasing. If heat is taken by the gas Q is positive and if heat is lost by the gas Q is negative.
Keeping the above points in mind the answer to this question is as under.

 (A) $\rightarrow$ (p,r,t)

 (B) $\rightarrow$ (p,r)

 (C) $\rightarrow$ (q,s)

 (D) $\rightarrow$ (r,t)

A silver sphere of radius 1 cm and work function 4.7 eV is suspended from an insulating thread in free-space. It is under continuous illumination of 200 nm wavelentth light. As phottelectrons are emitted, the sphere gets charged and acquires a potential. The maximum number of photoelectrons emitted from the sphere  is $A \times 10^{z}$

(where 1< A < 10). The value of Z is 

Answer:

Explanation:

Photoemission will stop when the potential on silver sphere becomes equal to the stopping potential.

$\therefore$   $\frac{hc}{\lambda}-W=eV_{0}$

 Here, $V_{0}=\frac{1}{4 \pi \epsilon_{0}} \frac{ne}{r}$

$\therefore$  $\left(\frac{1240}{1200}eV\right)-(4.7 eV)=\frac{9\times10^{9}\times n \times1.6 \times10^{-19}}{10^{-2}}$

$(6.2-4.7)=\frac{9\times10^{9}\times n \times1.6 \times10^{-19}}{10^{-2}}$

or   $n=\frac{1.5\times10^{-2}}{9\times 1.6\times10^{-10}}=1.04\times10^{7}$

 $\therefore$  Answer is 7

A block of mass 0.18 kg is attached to a spring of force constant 2 N/m. The coefficient of friction between the block and the floor is 0.1. Initially, the block is at rest and the spring is unstretched. An impulse is given to the block as shown in the figure The block slides a distance of 0.06 m and comes to rest for the first time. The initial velocity of the block in m/s is, V=$\frac{N}{10}$. Then, N is

Answer:

Explanation:

Decrease in mechanical energy = Work done against friction

$\therefore$    $\frac{1}{2}mv^{2}-\frac{1}{2}kx^{2}=\mu ,mgx$

 or  $v=\sqrt{\frac{2\mu mgx+kx^{2}}{m}}$

 substituting the values , we get

 $v=0.4 ms^{-1}=(\frac{4}{10})ms^{-1}$

$\therefore$ answer is 4

Water (with refractive index =$\frac{4}{3}$) in a tank is 18 cm deep. Oil of refractive index  $\frac{7}{4}$  lies on the water making a convex surface of radius of curvature R = 6 cm as shown. Consider oil to act as a thin lens. An object S is placed 24 cm above the water surface. The location of its image is at x cm above the bottom of the tank. Then, x is

Answer:

Explanation:

 Two refractions will take place, first from the spherical surface and the other from the plane surface. So, applying 

   $\frac{\mu_{2}}{v}-\frac{\mu_{1}}{u}= \frac{ \mu_{2}-\mu_{1}}{R}$

two times with proper sign convention. Ray of light is travelling downwards.
Therefore, downward direction is taken as positive direction.

$\frac{7/4}{v}-\frac{1.0}{-24}= \frac{ 7/4-1.0}{+6}$.....(i)

$\frac{4/3}{(18-x)}-\frac{7/4}{v}= \frac{ 4/3-7/4}{\propto}$

Solving these equations, we get 

                  x=2 cm

$\therefore$   Answer is 2

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