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JEE physics MCQ

Universe · CBSE STD 12 Science (CBSE - English Medium). 67 questions.

Questions · Page 2 of 2

MCQ

MCQ 511 Mark
Fraunhofer lines of the solar system is an example of
  • A
    Emission spectrum
  • B
    Emission band spectrum
  • C
    Continuous emission spectrum
  • ✓
    Line absorption spectrum
Answer
Correct option: D.
Line absorption spectrum
Fraunhofer lines are produced by the absorption of rays of the Sun in the atmosphere. When white light from photosphere passes through chromosphere, the vapours and gases present in it absorbs certain wavelengths and produces dark lines (Fraunhofer lines).
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MCQ 521 Mark
Smaller pieces of heavy stones and metals which on entering earth's atmosphere burns out are
  • A
    Comets
  • ✓
    Meteorites
  • C
    Asteroids
  • D
    All of these
Answer
Correct option: B.
Meteorites
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MCQ 531 Mark
Which one of the following is known as Saptarishi
  • A
    Orion
  • ✓
    Ursa major
  • C
    Ursa minor
  • D
    Scorpion
Answer
Correct option: B.
Ursa major
Ursa major is known as saptarishi.
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MCQ 541 Mark
Which one of the following planet has the longest day
  • ✓
    Venus
  • B
    Mars
  • C
    Mercury
  • D
    Earth
Answer
Correct option: A.
Venus
Venus has the longest day.
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MCQ 551 Mark
A planet which is born sister of earth is
  • A
    Mercury
  • ✓
    Venus
  • C
    Mars
  • D
    Jupiter
Answer
Correct option: B.
Venus
Planet Venus is called Earth's sister.
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MCQ 561 Mark
$C O$ gas is found in which of the following pairs of the planet
  • A
    Earth and Mercury
  • B
    Mercury and Saturn
  • C
    Venus and Saturn
  • ✓
    Venus and Mars
Answer
Correct option: D.
Venus and Mars
Venus and Mars have both $C O$ present.
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MCQ 571 Mark
A group of bright and faint stars is called
  • A
    Galaxy
  • B
    Comet
  • C
    Black hole
  • ✓
    Constellation
Answer
Correct option: D.
Constellation
A group of bright and faint stars is called a constellation
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MCQ 581 Mark
The galaxies are moving away from each other. It is explained by
  • A
    White dwarf star
  • ✓
    Red shift
  • C
    Neutron star
  • D
    None of these
Answer
Correct option: B.
Red shift
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MCQ 591 Mark
The sun emits a light with maximum wavelength $510\ mm$ while another star $X$ emits a light with maximum wavelength of $350\  nm$. What is the ratio of surface temperature of sun and the star $X$
  • A
    $2.1$
  • ✓
    $0.68$
  • C
    $0.46$
  • D
    $1.45$
Answer
Correct option: B.
$0.68$
As $\lambda \propto \frac{1}{T}$; so $\frac{T_1}{T_2}=\frac{\lambda_2}{\lambda_1}=\frac{350}{510}=0.68$
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MCQ 601 Mark
Suppose a planet goes around Sun with a linear speed twice as fast that of earth. What will be it's orbit size as compared to that of earth ? (Radius of earth $=R$ )
  • ✓
    $R / 4$
  • B
    $R / 2$
  • C
    $R$
  • D
    $2 R$
Answer
Correct option: A.
$R / 4$
From Kepler's law $T \propto R^{3 / 2}$ and also $T=\frac{2 \pi R}{v}$
$ \Rightarrow v \propto \frac{1}{R^{1 / 2}} \Rightarrow \frac{v_1}{v_2}=\left(\frac{R_2}{R_1}\right)^{1 / 2} $
$\Rightarrow \frac{v_1}{2 v_1}=\left(\frac{R_2}{R_1}\right)^{1 / 2} $
$ \Rightarrow R_2=\frac{R_1}{4}=\frac{R}{4}$
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MCQ 611 Mark
The percentage of Sun's total energy which reaches the earth's surface is
  • ✓
    $10^{-7} \%$
  • B
    $10^{-9} \%$
  • C
    $10^{-8} \%$
  • D
    $10^{-6} \%$
Answer
Correct option: A.
$10^{-7} \%$
If $S$ is the total energy emitted by Sun per second and $r$ is the distance of earth from Sun; then energy reaching earth of radius $R$ per second $=\frac{S}{4 \pi r^2} \times 2 \pi R^2=\frac{S R^2}{2 r^2}$.
$\therefore$ Percentage of energy reaching earth$=\frac{S R^2}{2 r^2 S} \times 100=\frac{\left(6.4 \times 10^6\right)^2 \times 100}{2\times\left(1.5 \times 10^{11}\right)^2} \simeq 10^{-7} \%$
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MCQ 621 Mark
A planet of mass $m$ moves in an ellipse around the sun of mass $M_S$ so that its maximum and minimum distances are $r_1$ and $r_2$ respectively. The angular momentum of the planet relative to the centre of the sun is
  • A
    $\sqrt{\frac{2 G M_S r_1}{\left(r_1+r_2\right)}}$
  • ✓
    $\sqrt{\frac{2 G M_S m^2 r_1 r_2}{\left(r_1+r_2\right)}}$
  • C
    $\sqrt{\frac{G M_S r_1 r_2}{\left(r_1+r_2\right)}}$
  • D
    $\sqrt{\frac{2 G M_S}{r_1 r_2\left(r_1+r_2\right)}}$
Answer
Correct option: B.
$\sqrt{\frac{2 G M_S m^2 r_1 r_2}{\left(r_1+r_2\right)}}$
From conservation of energy$\frac{1}{2} m v_1^2-\frac{G M_S m}{r_1}=\frac{1}{2} m v_2^2-\frac{G M_S m}{r_2}$.
Angular momentumis conserved, that is $m v_1 r_1=m v_2 r_2$
or $v_2=v_1 \frac{r_1}{r_2} \Rightarrow \frac{1}{2} m v_1^2-\frac{G M_S m}{r_1}=\frac{1}{2} m\left(\frac{v_1 r_1}{r_2}\right)^2-\frac{G M_S m}{r_2}$
or $v_1=\sqrt{\frac{2 G M_S r_2}{r_1\left(r_1+r_2\right)}} \Rightarrow L=m v_1 r_1=\sqrt{\frac{2 G M_S m^2 r_1 r_2}{r_1+r_2}}$
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MCQ 631 Mark
Consider a binary star system consisting of two stars of masses $M_1$ and $M_2$ separated by a distance of $30 AU$ with a period of revolution equal to $30$ years. If one of the two stars is $5$ times farther from the centre of mass than the other. The masses of the two stars in terms of solar masses are
  • A
    $5,15$
  • ✓
    $25,5$
  • C
    $25,10$
  • D
    $7,25$
Answer
Correct option: B.
$25,5$
$ M_1+M_2=\frac{4 \pi^2}{G} \cdot \frac{r^3}{T^2}$
If $T$ is measured in years, $r$ in A.U. and masses in Solar masses then $G=4 \pi^2$.
$\therefore M_1+M_2=\frac{r^3}{T^2}=\frac{(30)^3}{(30)^2}=30$
Now $r_1+r_2=30 \Rightarrow r_1+5 r_1=60$
$\Rightarrow r_1=5$ and $r_2=25$ Again $M_1 r_1=M_2 r_2 \Rightarrow \frac{M_1}{M_2}=5$
After solving (i) and (ii) we get $M_1=25$ and $M_2=5$
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MCQ 641 Mark
A galaxy is observed to be moving with a velocity of $8600\ km - sec$. If it is at a distance of $430$ million light year from us, Hubble constant and corresponding age of the universe are respectively
  • ✓
    $2 \times 10^{-5} \frac{ kms ^{-1}}{ ly }, 1.49 \times 10^{10}$ year
  • B
    $2 \times 10^{-6} \frac{ kms ^{-1}}{ ly }, 1.58 \times 10^3$ year
  • C
    $10^6 \frac{ kms ^{-1}}{l y}, 1.49 \times 10^{10}$ year
  • D
    None of these
Answer
Correct option: A.
$2 \times 10^{-5} \frac{ kms ^{-1}}{ ly }, 1.49 \times 10^{10}$ year
$H=\frac{v}{r}=\frac{8600}{430 \times 10^6}=2 \times 10^{-5} \frac{\mathrm{kms}^{-1}}{\mathrm{ly}}$
Age of the universe, $t_0=\frac{1}{H}=\frac{r}{v}$
Taking $r=430 \times 10^{\circ} \mathrm{l} y=430 \times 10^{\circ} \times 9.46 \times 10^{\circ} \mathrm{km}$
$ \Rightarrow t_0=\frac{430 \times 10^6 \times 9.46 \times 10^{12}}{8600} \mathrm{sec}$
$=\frac{430 \times 10^6 \times 9.46 \times 10^{12}}{8600 \times 3600 \times 24 \times 365}=1.49 \times 10^{10} \text { year }$
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MCQ 651 Mark
Assuming that the dimmest visible star to the naked eye has a magnitude of about $6.$ Brightness of planet Venus (magnitude $=-4$ ) w.r.t. this star is
  • ✓
    $10,000$ times brighter
  • B
    $2000$ times brighter
  • C
    $15000$ times brighter
  • D
    $4000$ times brighter
Answer
Correct option: A.
$10,000$ times brighter
Here, for Venus $m_1=-4$, for star $m_2=6$ using
$\frac{l_1}{l_2}=100^{\left(m_2-m_1\right) / 5}=100^{[6-(-4)] / 5}=100^2=10,000 \text {. }$
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MCQ 661 Mark
A particular emission line, detected in the light from a galaxy, has a wavelength $\lambda^{\prime}=1.1 \lambda$, where $\lambda$ is the proper wavelength of the line. The galaxy distance from us
  • ✓
    $1.6 \times 10^9 l y$
  • B
    $0.97 \times 10^9 l y$
  • C
    $2.4 \times 10^9 l y$
  • D
    $1.62 \times 10^{11} l y$
Answer
Correct option: A.
$1.6 \times 10^9 l y$
From Hubble's law $v=H r$ where $H=$ Hubble's constant $=19.3$ $\mathrm{mm} / \mathrm{sec}-\mathrm{ly}$ and $r=$ Distance of Galaxy from us.
According to Doppler's effect speed of Galaxy $v=\frac{c \Delta \lambda}{\lambda}$
$\Rightarrow r=\frac{c \Delta \lambda}{H \lambda}=\frac{c \times 0.1 \lambda}{H \lambda}=\frac{0.1 \times 3 \times 10^8}{19.3 \times 3 \times 10^{-3}}=1.6 \times 10^9 l y$
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MCQ 671 Mark
There are certain types of stars called visible stars which undergo periodic change in their light output. If such a star quadruple it's light output, how much does it's magnitude change
  • A
    $ -1.25$
  • ✓
    $-1.5$
  • C
    $-1.75$
  • D
    $-2$
Answer
Correct option: B.
$-1.5$
$\frac{l_2}{l_1} =4 \Rightarrow m_2-m_1=-2.5 \log \left(\frac{l_2}{l_1}\right)=-2.5 \log 4$
$ =-2.5 \times 0.6021=-1.5 .$
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MCQ - Page 2 - JEE physics STD 12 Science Questions - Vidyadip