Universe Questions in English

(C) The brightness of a planet as seen from Earth depends on its distance from Earth and its albedo (reflectivity). Venus is the closest planet to Earth and has a very high albedo due to its thick cloud cover,which reflects most of the sunlight falling on it. Therefore,it appears brighter than other planets.

(B) Hubble's law states that the recessional velocity $v$ of a galaxy is directly proportional to its distance $r$ from the observer (Earth).
Mathematically,this is expressed as $v = H_0 r$,where $H_0$ is the Hubble constant.
Therefore,the velocity is proportional to the distance of the galaxy from the Earth.

(A) Radio telescopes detect radio waves from celestial objects.
Radio waves can pass through dust,clouds,and fog,allowing the telescope to operate during the day,at night,and in cloudy weather.
Due to the enormous size of their reflectors,they can detect very faint radio signals.
Therefore,the statement that it cannot work at night is incorrect.

(B) The apparent brightness $(I)$ of a celestial object is inversely proportional to the square of its distance $(r)$ from the observer,given by the relation $I \propto \frac{1}{r^2}$.
Venus is a planet within our solar system,making it significantly closer to the Earth compared to distant stars.
Due to this proximity,the light reflected by Venus reaches the Earth with much higher intensity,making it appear brighter than the stars in the night sky.

(C) The apparent brightness $I$ of a star is given by the formula $I = \frac{L}{r^2}$,where $L$ is the real brightness (luminosity) and $r$ is the distance from the observer.
Since both stars appear to have the same brightness,we have $I_1 = I_2$.
Therefore,$\frac{L_1}{r_1^2} = \frac{L_2}{r_2^2}$.
Given $r_1 = 1$ light year and $r_2 = 10$ light years.
Substituting these values,we get $\frac{L_1}{L_2} = \frac{r_1^2}{r_2^2} = \left( \frac{1}{10} \right)^2 = \frac{1}{100}$.
Thus,the ratio of their real brightness is $1:100$.

(D) The correct answer is $(d)$. $A$ group of stars that appears to form a recognizable pattern or shape in the night sky is known as a constellation. These stars may appear close together from Earth,even if they are at vastly different distances from us.

(B) Modern astronomy,as defined by the International Astronomical Union $(IAU)$,divides the entire celestial sphere into $88$ distinct constellations.
These constellations serve as a standard way to map and identify regions of the sky.

(A) The $Big$ $Bang$ theory is widely accepted as the most satisfactory scientific explanation for the origin of the universe.
It proposes that the universe began as a hot,dense point approximately $13.8$ billion years ago and has been expanding and cooling ever since.
This theory is supported by significant observational evidence,such as the cosmic microwave background radiation and the observed redshift of distant galaxies.

(B) The brightness of a planet as seen from Earth depends on its distance from the Sun, its distance from Earth, and its albedo (reflectivity).
Venus has a very high albedo because it is covered by thick, reflective clouds of sulfuric acid.
Additionally, it comes closer to Earth than any other planet.
Therefore, $Venus$ appears as the brightest planet in the night sky.

(D) The atmospheres of Venus and Mars are primarily composed of carbon dioxide $(CO_2)$.
Specifically, Venus's atmosphere is about $96.5\%$ $CO_2$, and Mars's atmosphere is about $95\%$ $CO_2$.
Therefore, the correct pair is Venus and Mars.

(C) The current scientific consensus,based on measurements of the cosmic microwave background radiation and the expansion rate of the universe (Hubble constant),estimates the age of the universe to be approximately $13.8$ billion years. Among the given options,the range $10-20$ billion years is the most accurate representation of this scientific estimate. Therefore,option $C$ is correct.

(B) The planet $Venus$ is often referred to as the Earth's sister planet. This is because $Venus$ and $Earth$ are very similar in terms of size, mass, density, composition, and gravity. Due to these similarities, $Venus$ is frequently called the Earth's twin or sister planet.

(A) Asteroids are a group of rock pieces moving around the Sun in the region between $Mars$ and $Jupiter$.
They are believed to be the remains of a large planet that exploded due to the gravitational attraction of the $Sun$ and that planet.
Because of their nature and size,they are often referred to as small planets or minor planets.

(D) While $Saturn$ is the most famous for its prominent ring system, $Jupiter$, $Uranus$, and $Neptune$ also possess ring systems. However, in the context of standard textbook questions where only one option is typically expected as the primary answer, $Saturn$ is the correct choice. All four giant planets in our solar system have rings.

(D) The Milky Way is a barred spiral galaxy that contains our Solar System. It is one of the enormous galaxies present in the universe. Therefore,the correct option is $D$.

(B) According to Hubble's law,the recessional velocity $v$ of a galaxy is directly proportional to its distance $r$ from the observer (Earth).
Mathematically,this is expressed as $v = H_0 r$,where $H_0$ is the Hubble constant.
Therefore,the velocity is proportional to the distance of the galaxy from the Earth.
Thus,the correct option is $B$.

(C) The correct answer is $C$.
Although $Mercury$ is the closest planet to the $Sun$, $Venus$ is the hottest planet in the solar system.
This is because $Venus$ has a very thick atmosphere composed primarily of $CO_2$, which traps heat through a runaway greenhouse effect, leading to surface temperatures of approximately $464^{\circ}C$.

(A) The Milky Way is a barred spiral galaxy.
Its diameter or length is estimated to be approximately $100,000$ light years.
Therefore,the correct option is $A$.

(B) Hubble's law states that the recession velocity $(v)$ of a galaxy is directly proportional to its distance $(r)$ from the observer.
Mathematically,it is expressed as $v = H_0 r$,where $H_0$ is the Hubble constant.
Therefore,Hubble's law is related to the speed of galaxies.

(A) The term 'Albedo' refers to the measure of the diffuse reflection of solar radiation out of the total solar radiation received by an astronomical body.
It is essentially the reflecting power of a heavenly body.
Therefore,the correct option is $A$.

(D) According to the pulsating theory of the universe,the universe undergoes cycles of expansion and contraction. The theory suggests that the universe expands from a singularity (Big Bang),reaches a maximum size,and then contracts back to a singularity (Big Crunch). This cycle is estimated to repeat approximately every $80$ billion years. Therefore,the correct option is $D$.

(B) Meteors are small pieces of rock or dust,often remnants of comets,that enter the Earth's atmosphere.
When these particles enter the atmosphere at high speeds,they burn up due to friction with atmospheric gases,creating a streak of light in the sky.
Therefore,the correct description is that they are burnt pieces of comets or asteroids that fall toward the Earth.

(D) The correct answer is $D$.
According to the Doppler effect,when a source of light moves away from an observer,the frequency of the emitted light decreases,which corresponds to an increase in wavelength.
This shift towards longer wavelengths (the red end of the visible spectrum) is known as the red shift.
Since light observed from distant galaxies consistently shows a red shift,it indicates that these galaxies are moving away from us.
This observation provides strong evidence that the universe is expanding.

(B) The albedo (reflection power) is maximum for $Venus$,because it reflects $85\%$ of the incident sunlight. Its albedo value is $0.85$.

(B) According to stellar evolution theory,stars with an initial mass significantly greater than the mass of the Sun $(M)$ undergo a supernova explosion at the end of their life cycle.
After the red giant phase,if the remaining core mass is sufficiently large (typically greater than $3\, M$),the gravitational collapse continues until it forms a black hole.
Therefore,for a star with an original mass greater than $5\, M$,the final remnant is a black hole.

(A) When a comet approaches the Sun,the solar heat causes the substances on the comet,such as ice and frozen gases,to vaporize.
This process is known as sublimation.
The radiation pressure and solar wind then push these vaporized particles away from the Sun,creating the characteristic tail of the comet.
Therefore,the correct option is $A$.

(B) According to the current definition established by the International Astronomical Union $(IAU)$ in $2006$,there are $8$ recognized planets in our solar system: Mercury,Venus,Earth,Mars,Jupiter,Saturn,Uranus,and Neptune. Pluto is classified as a dwarf planet. Therefore,the correct option is $B$.

(A) The length of a day on a planet is determined by its rotational period (the time it takes to complete one full rotation on its axis).
Among the given options,Venus has the slowest rotation period.
Venus takes approximately $243$ Earth days to complete one rotation on its axis.
In comparison,Earth takes $24$ hours,Mars takes about $24.6$ hours,and Mercury takes about $58.6$ Earth days.
Therefore,Venus has the longest day.

(B) The constellation $Ursa$ $Major$ is a large constellation in the northern sky.
In Indian astronomy,a prominent group of seven stars within $Ursa$ $Major$ is known as $Saptarishi$.

(B) The correct answer is $B$. Small pieces of rock and metal that enter the Earth's atmosphere and burn up due to friction are called $Meteors$ (often referred to as shooting stars). If a piece is large enough to survive the passage through the atmosphere and hits the Earth's surface,it is called a $Meteorite$. Since the question specifies they burn out,the term $Meteors$ is the most accurate.

(B) The observation that galaxies are moving away from each other is known as the expansion of the universe.
This phenomenon is explained by the $Red \ shift$ in the light spectra emitted by these galaxies.
As galaxies move away from an observer,the light waves they emit are stretched to longer wavelengths,shifting them toward the red end of the electromagnetic spectrum.
This is a direct consequence of the $Doppler \ effect$ applied to light,providing evidence for the Big Bang theory.

(A) According to Hubble's Law,the speed of recession $(v)$ of a galaxy is directly proportional to its distance $(r)$ from the observer.
Mathematically,this is expressed as $v = H_0 r$,where $H_0$ is the Hubble constant.
Therefore,the speed of recession is directly proportional to the distance.

(C) The Great Bear,also known as Ursa Major,is a constellation. $A$ constellation is a group of stars that forms a recognizable pattern in the night sky.

(B) The correct answer is $B$.
Although $Pluto$ is currently classified as a dwarf planet,in the context of traditional astronomical questions,it is considered the coldest due to its extreme distance from the $Sun$.
Planets closer to the $Sun$ receive more solar radiation,while those farther away receive significantly less,leading to much lower surface temperatures.

(A) Hubble's law states that the recessional velocity $(v)$ of a galaxy is directly proportional to its distance $(r)$ from the observer.
Mathematically,this is expressed as $v = H_0 r$,where $H_0$ is the Hubble constant.
Since the redshift $(Z)$ is directly proportional to the recessional velocity $(v)$ for non-relativistic speeds,we have $Z \propto v$.
Substituting the relationship,we get $Z \propto r$.
Therefore,the redshift of a receding galaxy is directly proportional to its distance from Earth.

(D) The relationship between the brightness ratio and the magnitude difference is given by the formula: $\frac{I_1}{I_2} = (100)^{(m_2 - m_1)/5}$.
Given the magnitude of the bright star $m_1 = -5$ and the reference star $m_2 = 0$.
The difference in magnitude is $m_2 - m_1 = 0 - (-5) = 5$.
Substituting these values into the formula:
$\frac{I_1}{I_2} = 100^{(5/5)} = 100^1 = 100$.
Therefore,the bright star is $100$ times brighter than the reference star.

(B) The relationship between the apparent magnitude $(m)$ and the luminosity $(l)$ of a star is given by the formula: $m_2 - m_1 = -2.5 \log_{10} \left( \frac{l_2}{l_1} \right)$.
Given that the star quadruples its light output,we have $\frac{l_2}{l_1} = 4$.
Substituting this value into the formula:
$m_2 - m_1 = -2.5 \log_{10}(4)$.
Since $\log_{10}(4) = 2 \log_{10}(2) \approx 2 \times 0.3010 = 0.6020$.
Therefore,$m_2 - m_1 = -2.5 \times 0.6020 = -1.505 \approx -1.5$.
Thus,the magnitude changes by $-1.5$.

(A) According to the Doppler effect,the recession speed $v$ of the galaxy is given by $v = c \frac{\Delta \lambda}{\lambda}$.
Given $\lambda' = 1.1\lambda$,the change in wavelength is $\Delta \lambda = \lambda' - \lambda = 0.1\lambda$.
Thus,$v = c \times \frac{0.1\lambda}{\lambda} = 0.1c$.
Using Hubble's Law,$v = Hr$,where $H = 19.3 \text{ mm/s/ly} = 19.3 \times 10^{-3} \text{ km/s/ly} = 19.3 \times 10^{-6} \text{ km/s/ly}$ (assuming standard units for $H \approx 19.3 \text{ km/s per million ly}$ or similar constant).
Given $c = 3 \times 10^8 \text{ m/s} = 3 \times 10^5 \text{ km/s}$.
$r = \frac{v}{H} = \frac{0.1 \times 3 \times 10^5 \text{ km/s}}{19.3 \times 10^{-6} \text{ km/s/ly}} \approx 1.55 \times 10^9 \text{ ly} \approx 1.6 \times 10^9 \text{ ly}$.

(A) The relationship between the brightness ratio of two celestial objects and their magnitudes is given by the formula: $\frac{I_1}{I_2} = 100^{(m_2 - m_1)/5}$.
Here,for Venus,the magnitude $m_1 = -4$.
For the dimmest visible star,the magnitude $m_2 = 6$.
Substituting these values into the formula:
$\frac{I_1}{I_2} = 100^{(6 - (-4))/5} = 100^{10/5} = 100^2$.
Calculating the result: $100^2 = 10,000$.
Therefore,Venus is $10,000$ times brighter than the dimmest visible star.

(A) Hubble's Law is given by $v = H \times r$,where $v$ is the velocity,$H$ is the Hubble constant,and $r$ is the distance.
$H = \frac{v}{r} = \frac{8600 \ km \ s^{-1}}{430 \times 10^6 \ ly} = 2 \times 10^{-5} \ km \ s^{-1} \ ly^{-1}$.
The age of the universe $t_0$ is given by $t_0 = \frac{1}{H}$.
To calculate the age in years,we convert the distance $r$ from light-years to kilometers: $1 \ ly \approx 9.46 \times 10^{12} \ km$.
$r = 430 \times 10^6 \times 9.46 \times 10^{12} \ km = 4.0678 \times 10^{21} \ km$.
$t_0 = \frac{r}{v} = \frac{4.0678 \times 10^{21} \ km}{8600 \ km \ s^{-1}} \approx 4.73 \times 10^{17} \ s$.
Converting seconds to years: $t_0 = \frac{4.73 \times 10^{17}}{3600 \times 24 \times 365.25} \approx 1.49 \times 10^{10} \ \text{years}$.

(A) Let $S$ be the total energy emitted by the Sun per second and $r$ be the average distance between the Earth and the Sun. The intensity of solar radiation at the Earth's distance is $I = \frac{S}{4\pi r^2}$.
The Earth intercepts this radiation over its cross-sectional area,which is $\pi R^2$,where $R$ is the radius of the Earth.
Energy intercepted by Earth per second $= I \times \pi R^2 = \frac{S}{4\pi r^2} \times \pi R^2 = \frac{S R^2}{4 r^2}$.
The percentage of the Sun's total energy reaching the Earth is given by:
$\text{Percentage} = \left( \frac{\text{Energy intercepted}}{\text{Total energy emitted}} \right) \times 100 = \left( \frac{S R^2 / 4 r^2}{S} \right) \times 100 = \frac{R^2}{4 r^2} \times 100$.
Substituting the values $R \approx 6.4 \times 10^6 \ m$ and $r \approx 1.5 \times 10^{11} \ m$:
$\text{Percentage} = \frac{(6.4 \times 10^6)^2}{4 \times (1.5 \times 10^{11})^2} \times 100 \approx \frac{40.96 \times 10^{12}}{9 \times 10^{22}} \times 100 \approx 4.5 \times 10^{-10} \times 100 \approx 4.5 \times 10^{-8} \%$.
Given the standard approximation in physics problems of this type,the order of magnitude is $10^{-7} \%$.

(A) The solar constant is defined as the amount of solar energy incident per unit area per unit time on a surface perpendicular to the sun's rays at the average distance of the Earth from the Sun.
The value of the solar constant is approximately $1.36 \ kW/m^2$ to $1.4 \ kW/m^2$.
Therefore,the correct option is $A$.

(D) The statement “the universe is expanding” refers to the observation that the space between galaxies is stretching over time. It does not mean the universe is expanding into a pre-existing space,nor does it imply an infinite universe becoming 'more infinite'. Therefore,the correct interpretation is that the metric of space itself is expanding,causing galaxies to move away from each other. Since none of the provided options $A, B,$ or $C$ accurately describe this physical phenomenon,the correct choice is $D$.

(A) The galaxy in which we live is a spiral galaxy.
Our galaxy,the Milky Way,is classified as a spiral galaxy.

(D) The solar radiation received per unit area (intensity $I$) follows the inverse square law: $I \propto \frac{1}{r^2}$.
Let $r_1$ be the distance of the Sun from the Earth,$r_1 = 1.5 \times 10^{11} \text{ m}$.
Let $r_2$ be the distance of Alpha Centauri,$r_2 = 4 \text{ light-years} = 4 \times 9.46 \times 10^{15} \text{ m}$.
The ratio of the intensity received at the new distance $(I_2)$ to the original intensity $(I_1)$ is given by:
$\frac{I_2}{I_1} = \frac{r_1^2}{r_2^2} = \left( \frac{1.5 \times 10^{11}}{4 \times 9.46 \times 10^{15}} \right)^2$.
Calculating the value:
$\frac{I_2}{I_1} = \left( \frac{1.5}{37.84} \times 10^{-4} \right)^2 \approx (0.0396 \times 10^{-4})^2 \approx (3.96 \times 10^{-6})^2 \approx 1.57 \times 10^{-11}$.
Thus,the factor by which the radiation decreases is approximately $1.5 \times 10^{-11}$.

(B) Hubble's Law is given by the formula $v = H \times r$,where $v$ is the recession velocity,$H$ is the Hubble constant,and $r$ is the distance.
Given:
Velocity $v = 6.48 \times 10^6 m s^{-1} = 6480 km s^{-1}$.
Distance $r = 430$ million light years.
We need to find $H$ in units of $km s^{-1}$ per million light years.
$H = \frac{v}{r} = \frac{6480 km s^{-1}}{430 \text{ million light years}}$.
$H \approx 15.07 km s^{-1} \text{ per million light year}$.
Rounding to the nearest integer,we get $15 km s^{-1}$ per million light year.
Therefore,the correct option is $B$.

(C) The relationship between the magnitudes $(m)$ and the brightness $(I)$ of two stars is given by the formula: $m_B - m_A = -2.5 \log_{10} \left( \frac{I_B}{I_A} \right)$.
Given $m_A = 2.5$ and $m_B = -5$.
Substituting these values into the formula: $-5 - 2.5 = -2.5 \log_{10} \left( \frac{I_B}{I_A} \right)$.
$-7.5 = -2.5 \log_{10} \left( \frac{I_B}{I_A} \right)$.
Dividing both sides by $-2.5$: $3 = \log_{10} \left( \frac{I_B}{I_A} \right)$.
Therefore,the brightness ratio is $\frac{I_B}{I_A} = 10^3$.

(C) The classification of stars based on their surface temperature is known as the Harvard spectral classification system.
In this system,stars are categorized into types $O, B, A, F, G, K,$ and $M$.
$O$ type stars are the hottest,with surface temperatures exceeding $30,000 \ K$.
$M$ type stars are the coolest,with surface temperatures below $3,700 \ K$.
Therefore,the correct answer is $O$ type.

(C) Venus is a planet,not a star. It appears brighter than most stars in the night sky primarily because it is much closer to the Earth than any star (excluding the Sun).
Additionally,Venus has a high albedo,meaning it reflects a large portion of the sunlight that hits its thick cloud cover,making it appear very bright to an observer on Earth.

(B) The evolution of a star depends on its initial mass. Stars with an initial mass roughly between $1.4M$ (the Chandrasekhar limit) and about $3M$ (the Tolman-Oppenheimer-Volkoff limit) typically end their lives as neutron stars after a supernova explosion. Among the given options,the range $2M$ to $4M$ is the most appropriate representation for the progenitor mass of a neutron star. Therefore,option $B$ is correct.