JEE 2011 :: General Questions :: Physics

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A composite block is made of slabs A, B, C, D and E of different thermal conductivities (given in terms of a constant, K) and sizes (given in terms of length, L) as shown in the figure. All slabs are of the same width. Heat Q flows only from left to right through the blocks. Then, in steady-state

Answer:

Explanation:

 Thermal  resistance

 $ R= \frac{l}{KA}$

 $\therefore$   $R_{A}= \frac{ L}{(2K)(4L \omega)}=\frac{1}{ 8K \omega}$ 

    (here , $\omega$= width)

 $R_{B}= \frac{ 4L}{3K(L \omega)}= \frac{4}{ 3K \omega}$

 $R_{C}= \frac{4L}{(4K)(2L \omega)}=\frac{1}{2K \omega}$

 $R_{D}=\frac{4L}{(5K)(L \omega)}=\frac{4}{5 K \omega}$

 $R_{E}= \frac{L}{(6K)(L \omega)}=\frac{1}{6 K \omega}$

 $R_{A}:R_{B}:R_{C}:R_{D}:R_{E}$

 =15:160:60:96:12

So, Let us write , $R_{A}=15 R$,$R_{B}=160 R$ etc and draw a simple electrical circuit as shown in figure.

 H= Heat current = Rate of heat flow

 $H_{A}=H_{E}=H$      (let)

$\therefore$ option (a) is correct

 in parallel  current distributes in inverse ratio of resistance

 $\therefore$     $H_{B}:H_{C}:H_{D}= \frac{1}{R_{B}}:\frac{1}{R_{C}}:\frac{1}{R_{D}}$

   = $\frac{1}{160}:\frac{1}{60}:\frac{1}{96}$

=9:24-:15

  $\therefore$     $H_{B}=\left(\frac{9}{9+24+15}\right)H= \frac{3}{16} H$

     $H_{C}=\left(\frac{24}{9+24+15}\right)H= \frac{1}{2} H$

 and    $H_{D}=\left(\frac{15}{9+24+15}\right)H= \frac{5}{16} H$

 $H_{C}=H_{B}+H_{D}$

 $\therefore$ option (d) is correct 

 Temperature  difference (let us call it T)= (Heat current ) x (Thermal resistance) 

 $T_{A}= H_{A} R_{A}= (H) (15 R)  =15 HR$

 $T_{B}=H_{B}R_{B}==\left(\frac{3}{16}H\right) (160 R)=30HR$

 $T_{C}=H_{C}R_{C}==\left(\frac{1}{2}H\right) (60 R)=30HR$

 $T_{D}=H_{D}R_{D}==\left(\frac{5}{16}H\right) (96 R)=30HR$

 $T_{E}=H_{E}R_{E}=(H) (12 R)=12 HR$

 Here, $T_{E}$ is minimum . Therefore option (c) is also correct

 $\therefore$ Correct options are (a) (c) and (d) 

Analysis of Question
(i) From calculation point of view, question is difficult otherwise the question is simple.
(ii) In heat transfer, questions are mainly asked from conduction and radiation topic.

An electron and a proton are moving on shaight parallel paths with same velocity. They enter a semi-infinite region of uniform magnetic field perpendicular to the velocity. Which of the following statement(s) is/are true?

Answer:

Explanation:

 $ r= \frac{mv}{Bq}$

 or   $ r \propto m$

 $\therefore$    $r_{e} <r_{p}$  as $m_{e}< m_{p}$

 Further  , $T = \frac{2 \pi m}{Bq}$

 or  $ T \propto m$

 $\therefore$   $T_{e} <T_{p}$

 $t_{e} = \frac{ T_{e}}{2}$

 and $t_{p}=\frac{ T_{P}}{2}$

 or   $ t_{e}  < t_{p}$

 $\therefore$   Correct options are (b) and (d) 

Note In JEE 2011, official answer key' correct oPtions were given bc, bd, bcd, here student should realise that both the electron and Proton enter the magnetic field at the same time or at different times

Analysis of Question
(i) Question is simPle'
(ii) In the situation given above particle
will never complete the circle unless
the magnetic field is all around

A meter bridge is set up as shown in the figure, to determine an unknown resistance X using a standard $10 \Omega$  resistor. The galvanometer shows a null point when the tapping key is at 52 cm mark. The end-corrections are 1 cm and 2 cm respectively for ends A and B. The determined value of X is 

Answer:

Explanation:

 Using the concept  of balanced 

 Wheatstone bridge , we have 

 $\frac{p}{Q}=\frac{R}{S}$

 $\therefore$   $\frac{ X}{(52+1)}=  \frac{10}{48+2}$

 $\therefore$    X= $\frac{10 \times 53}{50}$ =10.6 $\Omega$ 

 $\therefore$  Correct option is (b) 

Analysis of Question
Question is moderately tough because normally end corrections are not taught
in the coaching schools in this type of problem.

The wavelength of the first spectral line in the Balmer series of a hydrogen atoms is 6561 Å The wavelength of the second  spectral line in the Balmer series of singly ionized helium atoms is 

Answer:

Explanation:

  For hydrogen or hydrogen type atoms

 $\frac{1}{\lambda}=RZ^{2}\left(\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}}\right)$

 In the transition from $n_{i} \rightarrow n_{f}$

 $\therefore$    $\lambda \propto \frac{1}{Z^{2}\left(\frac{1}{n^{2}_{f}}-\frac{1}{n_{i}^{2}}\right)}$

  $\therefore$   $\frac{\lambda_{2}}{\lambda_{1}}=\frac{Z_{1}^{2}\left(\frac{1}{n^{2}_{f}}-\frac{1}{n_{i}^{2}}\right)_{1}}{Z_{2}^{2}\left(\frac{1}{n^{2}_{f}}-\frac{1}{n_{i}^{2}}\right)_{2}}$

$\lambda_{2}=\frac{\lambda_{1}Z_{1}^{2}\left(\frac{1}{n^{2}_{f}}-\frac{1}{n_{i}^{2}}\right)_{1}}{Z_{2}^{2}\left(\frac{1}{n^{2}_{f}}-\frac{1}{n_{i}^{2}}\right)_{2}}$

 Substituting the values , we have 

$Å \frac{(6561 Å)(1)^{2}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)}{(2)^{2}\left(\frac{1}{2^{2}}-\frac{1}{4^{2}}\right)}=1215$Å

$\therefore$   correct option is (a)

Analysis of Question

(i) Question is simple.
(ii) In modern physics, mostly questions are asked on the emission of a photon by the transition of electrons from some higher energy state to some lower energy state

(iii) For hydrogen and hydrogen-like atoms which have onlY single

electron, we can use the formula

   $\frac{1}{\lambda}=RZ^{2}\left(\frac{1}{n_{f}^{2}}-\frac{1}{n^{2}_{i}}\right)$

 (iv)  Further, the student should also remember different series, as shown

in figure below

A police car with a siren of frequency 8 kHz is moving with uniform velocity 36 km/h towards a tall building which reflects the sound waves. The speed of sound in air is 320 m/s. The frequency of the siren heard by the car driver is

Answer:

Explanation:

$36 km/h^{-1}=36 \times \frac{5}{18} =10 ms ^{-1}$

Apparent frequency of sound heard by car driver (observer) reflected from the
building will be

 $f'= f\left(\frac{v+v_{0}}{v-v_{s}}\right)$

 $= 8\left(\frac{320+10}{320-10}\right)$

 =8.5 kHz

 $\therefore$  Correct option is (a) 

Analysis of question

 (i) Question is simple 

(ii) Driver will listen to two sounds, direct and reflected. Direct sound will be of 8 kHz as the driver has no relative motion with the car. But the reflected sound is of increased frequency because driver and image of car both are approaching each other

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