66I think there is some data missing in Q1). Even the closest star Proxima Centauri subtends as small as 0.77 sec of arc. The closer the star the more the angle. And the options given are quite large.
25 Answers
66
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1please tell the topics to prepare for kvpy exam 2011 and which books should I follow???????????
11Application Form 2011
Available from 15 July 2011
visit http://www.kvpy.org.in for details
106appearing for KVPY exam doesnt require forgoing of NTSE scholarship. You need to forgo NTSE only if you get selected for receiving the KVPY scholarship.
Take KVPY scholarship, it gives more (Rs 64K [1]if im not mistaken) whereas in NTSE (if it still gives Rs 500 per month) then you will require atleast 10 years more to get that much.
11rickde sir,
i am preparing for kvpy. what were the books you referred? help me
P.S: i am also a NTSE scholar. should i forgo my scholarship for appearing in KVPY?
66Q2) The energy density Ï = E/V (the symbols have meanings as in the problem)
So, we have
\mathrm d\rho = \mathrm d(E/V)=\dfrac{V\,\mathrm dE-E\,\mathrm dV}{V^2}=\dfrac{\mathrm dE}{V}-\dfrac{E}{V}\,\dfrac{\mathrm dV}{V}=-(p+\rho)\,\dfrac{\mathrm dV}{V}
But p=Ï/3 so we have
\dfrac{\mathrm d\rho}{\rho}=-\dfrac{4}{3}\dfrac{\mathrm dV}{V}\quad \Rightarrow \ \rho V^{4/3} = \text{ constant}
Now V \sim a^3(t)\ \Rightarrow V^{4/3}\sim a^4(t)
And hence
\rho \sim \dfrac{1}{a^4(t)}
So the required option is (A).
1sir could u give some suggestions for the xam
the questions in physics seem a bit hard
1i also think so tried q1 for weeks
i have copy pasted as such from the sample paper
1q1 A star overhead appears to rotate in a cone of a very small angle over a year. If the orbital speed of the earth is 30 km/s, the value of θ is approximately
A 50.0 sec of arc
B 20.5 sec of arc
C 35.0 sec of arc
D 70.5 sec of arc
66Q7) The Doppler shift for light is
\dfrac{1}{\lambda}=\left(1-\dfrac{v}{c}\right)\dfrac{1}{\lambda_0}
where v is the speed of the source relative to the receiver: it is negative when the source is moving towards the receiver, positive when moving away
Applying this, we get
v=\left(1-\dfrac{\lambda_0}{\lambda}\right)c
Applying this, we get the answer as (B)
66Q5) If the intensities of polarized and unpolarized light be Ip and Iu then during rotation the maximum intensity transmitted
Imax = Ip + Iu2
while the minimum intensity transmitted is
Imin = Iu2
Given
Imax=5 Imin
which gives Ip=2 Iu
So the required fraction
= IpIp + Iu = 23
So the answer is (D)
66Q4) The angular resolution \Delta \theta of an eye (or a device like telescope) is given by
\Delta \theta =1.22 \dfrac{\lambda}{D}
where \lambda is the light wavelength and D the aperture of te device.
For a yellow light (around 570 nm), this resolution is
\Delta \theta =1.22 \dfrac{570\ \mathrm{ nm}}{1 \ \mathrm{mm}}\approx .70\ \mathrm{mrad}
(mrad is milli-radian). So the required distance
d_\mathrm{max}=\dfrac{2\ \mathrm{m}}{.70 \times 10^{-3}}\approx 2.8\ \mathrm{km}
66Q3) If the final temperature is T0, we have
ms(T0-T) = ms(4T-T0)
giving T0 = 52 T
The entropy change for water which gets heated from T to T0 is
\Delta S_1=\int_T^{T_0}\dfrac{\delta Q}{T}=\int_T^{T_0}ms\dfrac{\mathrm dT}{T}=ms\ln \dfrac{T_0}{T}=ms\ln(5/2)
Similarly the entropy change for water which cools from 4T to T0 is
\Delta S_2=\int_{4T}^{T_0}\dfrac{\delta Q}{T}=\int_{4T}^{T_0}ms\dfrac{\mathrm dT}{T}=ms\ln \dfrac{T_0}{4T}=ms\ln(5/8)
Hence, the entropy change of the universe
\Delta S=\Delta S_1 +\Delta S_2 =ms\ln(25/16) =2ms\ln(5/4)
Hence, option (b)
1q7
The K line of singly ionized calcium has a wavelength of 393.3 nm on earth. In the spectrum of an observed galaxy, this line has a wavelength 401.8 nm. The speed with which the galaxy is moving away from us is
A 3 x 10^8 m/s
B 6480 km/s
C 30 km/s
D 3240 km/s
1q6
Phosphorous is doped in silicon with concentration 1018 cm-3. Its donor level is 0.045 eV below the conduction band edge. At room temperature (300 K), the number of donor electrons per cm3 in the silicon conduction band is
A 10^14
B 10^13
C 10^12
D 10^17
1q5
A partially polarized light i.e. a mixture of polarized and unpolarized light is passed through a polarizer. When the polarizer is rotated through 360 while keeping it perpendicular to the beam, the transmitted intensity varies by a factor of 5 during rotation. The fraction of the intensity of the original beam associated with polarized light is o
A 0.20
B 0.33
C 0.50
D 0.66
1q4
The headlights of a truck are separated by 2m and emit yellow light. The maximum distance from where our eye (aperture of the eye is approximately 1 mm) can resolve the two headlights is close to
A 1.4 km
B 2.8 km
C 3.3 km
D 3.8 km
1q3
A mass m of water at temperature T is mixed adiabatically with an equal mass of water at temperature 4T. The entropy change of the universe is (s: specific heat)
A zero
B 2ms ln(5/4)
C ms 3T/2
D -ms 3T/2
1q2
Observation indicates that the Universe is continuously expanding. The distance between two objects is therefore a function of time d(t), which may be written as
d(t)=a(t)d° ,
where d° is the distance at some fixed time t° and the scale factor a(t) gives the expansion. Using the first law of thermodynamics dE=-pdV and the equation of state for a gas of photons (radiation) p=Ï/3 , the dependence of the (energy) density Ï as a function of
a(t) is of the type:
A Ï~ 1/a4(t)
B Ï~ 1/a3(t)
C Ï~ a3(t)
D Ï~ a4(t)