Q.How many significant figures are present in the following?

Showing the 12 most recent of 16 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.If 'B' is magnetic induction, 'e' is charge of electron, 'm' is mass and 'c' is speed of light in vacuum, then the physical quantity having dimensions of $\dfrac{4\pi mc}{Be}$ is (A) Energy (B) Electric potential (C) Length (D) Time

   ›Reveal solutionSolution

   Dimensional analysis of $\dfrac{4\pi mc}{Be}$ cancels mass and time to leave $[L]$, so the quantity is a length — option (C).

   The factor $4\pi$ is dimensionless and can be ignored. We need the dimensions of $B$, which follow from the Lorentz force $F=qvB$, i.e. $[B]=\dfrac{[F]}{[q][v]}$.

   1. Dimensions of each symbol.

   $[m]=[M],\quad [c]=[LT^{-1}],\quad [e]=[IT],\quad [B]=\frac{[MLT^{-2}]}{[IT][LT^{-1}]}=[MT^{-2}I^{-1}].$

   2. Numerator.

   $[mc]=[M][LT^{-1}]=[MLT^{-1}].$

   3. Denominator.

   $[Be]=[MT^{-2}I^{-1}][IT]=[MT^{-1}].$

   4. Divide. …

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2. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.If the distance between Earth and Jupiter is $810 \times 10^6$ km and the angular diameter of Jupiter measured from Earth is $36''$, then the diameter of Jupiter (in km) is nearly (A) $2.4 \times 10^6$ (B) $8.2 \times 10^6$ (C) $4.9 \times 10^5$ (D) $1.4 \times 10^5$

   ›Reveal solutionSolution

   The angular diameter formula $D = \theta \cdot d$ (with $\theta$ in radians) gives Jupiter’s diameter as roughly $1.4 \times 10^5$ km, matching option (D).

   The key idea here is the small-angle approximation in astronomy. When an object is very far away, the tiny angle it subtends at the observer is directly proportional to its actual size — the relationship is simply $D = \theta \cdot d$, where $\theta$ is in radians. This works because for small angles, the arc length (which is the object’s diameter) is nearly equal to the chord length, and the geometry becomes a clean linear proportion.

   The trap most students fall into is forgetting to convert the angular diameter from arcseconds into radians. The given $36''$ is an angle in seconds of arc, not in radians — and the formula only works in radians. Let’s walk through it carefully.

   1. Convert angular diameter to radians. One degree is $3600$ arcseconds ($1^\circ = 3600''$). So $36''$ is

   $\theta = \frac{36}{3600} = 0.01^\circ.$

   Now convert degrees to radians using $\pi$ rad $= 180^\circ$:

   $\theta = 0.01 \times \frac{\pi}{180} = \frac{\pi}{18000} \text{ radians}.$

   Numerically, $\pi \approx 3.1416$, so

   $\theta \approx \frac{3.1416}{18000} \approx 1.7453 \times 10^{-4} \text{ rad}.$

   2. Apply the small-angle formula. The distance $d = 810 \times 10^6$ km. The diameter $D$ is

   $D = \theta \cdot d = (1.7453 \times 10^{-4}) \times (810 \times 10^6).$

   Multiply the numbers:

   $1.7453 \times 810 \approx 1413.7, \quad \text{and} \quad 10^{-4} \times 10^6 = 10^2.$

   So

   $$$$

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3. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.In hydrogen atom, an electron is transferred from an orbit of radius 1.3225 nm to another orbit of radius 0.2116 nm. What is the energy (in J) of emitted radiation? (A) $1.635 \times 10^{-18}$ (B) $3.027 \times 10^{-19}$ (C) $4.087 \times 10^{-19}$ (D) $0.4578 \times 10^{-18}$

   ›Reveal solutionSolution

   The energy of the emitted photon equals the difference in the electron’s total energy between the two orbits. Using the Bohr radius formula $r_n = n^2 a_0$ to find the principal quantum numbers, then $E_n = -13.6\,\text{eV}/n^2$, converting to joules gives $E \approx 4.087 \times 10^{-19}\,\text{J}$, which corresponds to option (C).

   ---

   Concept and Intuition

   In the Bohr model of the hydrogen atom, the electron orbits the nucleus only in certain allowed circular orbits. Each orbit has a quantized radius and a quantized total energy. When an electron jumps from a higher (larger radius) orbit to a lower (smaller radius) orbit, it emits a photon whose energy equals the difference between the two energy levels. The key is that the radius of the $n$-th orbit is $r_n = n^2 a_0$, where $a_0 = 0.0529\,\text{nm}$ is the Bohr radius. So given the radii, we can find the quantum numbers $n$, then compute the energies.

   ---

   Step-by-step solution

   1. Find the principal quantum numbers from the radii The Bohr radius is $a_0 = 0.0529\,\text{nm}$. For any orbit, $r_n = n^2 a_0$. For the larger radius $r_1 = 1.3225\,\text{nm}$:

   $n_1^2 = \frac{r_1}{a_0} = \frac{1.3225}{0.0529} = 25.0 \quad \Rightarrow \quad n_1 = 5.$

   For the smaller radius $r_2 = 0.2116\,\text{nm}$:

   $n_2^2 = \frac{0.2116}{0.0529} = 4.0 \quad \Rightarrow \quad n_2 = 2.$

   So the electron goes from $n=5$ to $n=2$.

   2. Determine the energy of each orbit In the Bohr model, the total energy of the electron in the $n$-th orbit is

   $E_n = -\frac{13.6\,\text{eV}}{n^2}.$

   Thus:

   $E_5 = -\frac{13.6}{25} = -0.544\,\text{eV}, \qquad E_2 = -\frac{13.6}{4} = -3.4\,\text{eV}.$

   3. Compute the energy difference (photon energy) The emitted photon’s energy is the absolute difference:

   $\Delta E = E_2 - E_5 = (-3.4) - (-0.544) = -2.856\,\text{eV}.$

   The magnitude is $2.856\,\text{eV}$.

   4. Convert electronvolts to joules Use $1\,\text{eV} = 1.602 \times 10^{-19}\,\text{J}$:

   $\Delta E = 2.856 \times 1.602 \times 10^{-19} = 4.575 \times 10^{-19}\,\text{J}.$

   This is approximately $4.58 \times 10^{-19}\,\text{J}$.

   5. Match with the given options

      The options are:

      (A) $1.635 \times 10^{-18}$

      (B) $3.027 \times 10^{-19}$

      (C) $4.087 \times 10^{-19}$

      (D) $0.4578 \times 10^{-18}$ (which is $4.578 \times 10^{-19}$)

      Our computed value $4.575 \times 10^{-19}\,\text{J}$ is closest to option (D) if we read it as $0.4578 \times 10^{-18} = 4.578 \times 10^{-19}$. However, careful: $0.4578 \times 10^{-18} = 4.578 \times 10^{-19}$, which matches our result. But wait — option (C) is $4.087 \times 10^{-19}$, which is slightly smaller. Let’s check precision. …

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4. TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQ
   Q.In Hydrogen atom, an electron jumped from an orbit of radius 2592.1 pm to another orbit of radius 211.6 pm. What is the energy difference (in J) between these two states? (A) $2.18 \times 10^{-18}$ (B) $5 \times 10^{-19}$ (C) $5 \times 10^{-20}$ (D) $5 \times 10^{-18}$

   ›Reveal solutionSolution

   The energy difference is found by first identifying the principal quantum numbers from the given Bohr radii, then using the hydrogen energy formula. The result is $5 \times 10^{-19}$ J, which corresponds to option (B).

   The key idea here is that in the Bohr model of hydrogen, the radius of an orbit is directly tied to the principal quantum number $n$: $r_n = n^2 a_0$, where $a_0 = 52.9$ pm is the Bohr radius. Once you know $n$ for each orbit, you can find the energy of each level using $E_n = -\frac{13.6 \text{ eV}}{n^2}$, and then convert the difference to joules.

   Let’s walk through it step by step.

   1. Find the principal quantum numbers from the radii. The Bohr radius is $a_0 = 52.9$ pm. For an orbit of radius $r$, we have $r = n^2 a_0$, so $n = \sqrt{r / a_0}$. For the larger radius $r_1 = 2592.1$ pm:

   $n_1 = \sqrt{\frac{2592.1}{52.9}} = \sqrt{49} = 7$

   For the smaller radius $r_2 = 211.6$ pm:

   $n_2 = \sqrt{\frac{211.6}{52.9}} = \sqrt{4} = 2$

   So the electron jumped from $n=7$ to $n=2$.

   2. Recall the energy of a hydrogen level. The energy of the $n$-th orbit in hydrogen is:

   $E_n = -\frac{13.6 \text{ eV}}{n^2}$

   This is a standard result from the Bohr model. The negative sign means the electron is bound.

   3. Compute the energy difference in eV. The energy difference $\Delta E = E_{\text{final}} - E_{\text{initial}} = E_2 - E_7$.

   $E_2 = -\frac{13.6}{4} = -3.4 \text{ eV}$

   $E_7 = -\frac{13.6}{49} \approx -0.2776 \text{ eV}$

   So:

   $\Delta E = (-3.4) - (-0.2776) = -3.1224 \text{ eV}$ …

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5. TG EAPCET 2024Set ap-2024-05-07-AN1 markMCQ
   Q.What is the energy (in J) required to transfer the electron from n = 1 to n = 2 state in Li$^{2+}$ (K = constant = $2.18 \times 10^{-18}$ J) (A) $\dfrac{4K}{27}$ (B) $9K$ (C) $8K$ (D) $\dfrac{27K}{4}$

   ›Reveal solutionSolution

   For hydrogen-like ions, the energy of a level is $E_n = -K \frac{Z^2}{n^2}$, where $K = 2.18 \times 10^{-18}$ J. The transition energy from $n=1$ to $n=2$ in Li$^{2+}$ ($Z=3$) is $\frac{27K}{4}$ J, matching option (D).

   The key idea here is that Li$^{2+}$ is a hydrogen-like ion — it has only one electron, so the Bohr model applies directly. The constant $K$ given is the ground-state energy of hydrogen ($-2.18 \times 10^{-18}$ J for $n=1$, $Z=1$). For any hydrogen-like ion, the energy scales with $Z^2$, the square of the atomic number.

   For lithium, $Z = 3$. So the energy levels are:

   $E_n = -K \frac{Z^2}{n^2} = -K \frac{9}{n^2}$

   The energy required to move the electron from $n=1$ to $n=2$ is the difference:

   $\Delta E = E_2 - E_1$

   Let’s work it through step by step.

   1. Write the energies for $n=1$ and $n=2$

      For $n=1$: $E_1 = -K \frac{9}{1^2} = -9K$

      For $n=2$: $E_2 = -K \frac{9}{2^2} = -K \frac{9}{4}$

   2. Find the difference

   $\Delta E = E_2 - E_1 = \left(- \frac{9K}{4}\right) - (-9K) = -\frac{9K}{4} + 9K$

   3. Combine the terms Write $9K$ as $\frac{36K}{4}$: $\Delta E = \frac{36K}{4} - \frac{9K}{4} = \frac{27K}{4}$ …

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6. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
   Q.The ratio of the radii of a planet and the earth is 1:2, the ratio of their mean densities is 4:1. If the acceleration due to gravity on the surface of the earth is $9.8\ \mathrm{ms}^{-2}$, then the acceleration due to gravity on the surface of the planet is (A) $4.9\ \mathrm{ms}^{-2}$ (B) $8.9\ \mathrm{ms}^{-2}$ (C) $29.4\ \mathrm{ms}^{-2}$ (D) $19.6\ \mathrm{ms}^{-2}$

   ›Reveal solutionSolution

   The acceleration due to gravity on a planet depends on its radius and density. Using $g = \frac{4}{3}\pi G \rho R$, the planet’s gravity is $19.6\ \mathrm{ms}^{-2}$, which is option (D).

   The key idea is that surface gravity $g$ is proportional to both the planet’s radius and its mean density. Instead of memorizing a formula for mass, we can combine the definition $g = \frac{GM}{R^2}$ with $M = \rho \cdot \frac{4}{3}\pi R^3$ to get a direct proportionality: $g \propto \rho R$. This lets us compare planets without needing numerical values for $G$ or the actual masses.

   1. Write the relation for surface gravity. The acceleration due to gravity on the surface of a spherical body is

   $g = \frac{GM}{R^2},$

   where $M$ is the mass and $R$ is the radius.

   2. Express mass in terms of density and radius. For a sphere, $M = \rho \cdot \frac{4}{3}\pi R^3$. Substituting:

   $g = \frac{G \cdot \frac{4}{3}\pi \rho R^3}{R^2} = \frac{4}{3}\pi G \rho R.$

   So $g$ is directly proportional to the product $\rho R$.

   3. Set up the ratio for the planet and Earth. Let subscripts $p$ and $e$ denote the planet and Earth. Then

   $\frac{g_p}{g_e} = \frac{\rho_p R_p}{\rho_e R_e}.$

   We are given:

   $\frac{R_p}{R_e} = \frac{1}{2}, \quad \frac{\rho_p}{\rho_e} = \frac{4}{1}.$

   Therefore: …

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7. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
   Q.If the radius of first orbit of hydrogen like ion is $1.763 \times 10^{-2}$ nm, the energy associated with that orbit (in J) is (A) $+1.962 \times 10^{-17}$ (B) $-1.962 \times 10^{-17}$ (C) $-0.872 \times 10^{-17}$ (D) $-2.18 \times 10^{-18}$

   ›Reveal solutionSolution

   The key idea is that the radius of the first orbit of a hydrogen-like ion scales as $r_n = \frac{n^2 a_0}{Z}$, and the energy scales as $E_n = -\frac{Z^2 \cdot 13.6\ \text{eV}}{n^2}$. Using the given radius, we find $Z = 3$, then compute the energy in joules, obtaining $-1.962 \times 10^{-17}\ \text{J}$, which corresponds to option (B).

   ---

   Concept & Intuition

   For hydrogen-like ions (one electron around a nucleus of charge $Ze$), both the radius and energy of the $n$-th Bohr orbit are determined by the atomic number $Z$. The first orbit ($n=1$) radius is $r_1 = \frac{a_0}{Z}$, where $a_0 = 0.529\ \text{Å} = 0.0529\ \text{nm}$ is the Bohr radius. The energy of that orbit is $E_1 = -Z^2 \cdot 13.6\ \text{eV}$.

   If we are given the actual radius, we can solve for $Z$, then plug into the energy formula and convert to joules. The negative sign is crucial: bound electrons have negative energy.

   ---

   Step-by-step solution

   1. Recall the Bohr radius formula for hydrogen-like ions The radius of the $n$-th orbit is

   $r_n = \frac{n^2 a_0}{Z}$

   where $a_0 = 0.0529\ \text{nm}$ (Bohr radius). For the first orbit, $n=1$, so

   $r_1 = \frac{a_0}{Z}.$

   2. Use the given radius to find $Z$ Given $r_1 = 1.763 \times 10^{-2}\ \text{nm}$.

   $\frac{a_0}{Z} = 1.763 \times 10^{-2} \ \text{nm}$

   $Z = \frac{a_0}{1.763 \times 10^{-2}} = \frac{0.0529}{0.01763} \approx 3.00.$

   So the ion is $\text{Li}^{2+}$ (lithium with one electron).

   3. Energy of the first orbit for a hydrogen-like ion The energy in electronvolts is

   $E_1 = -Z^2 \cdot 13.6\ \text{eV}.$

   With $Z=3$:

   $E_1 = -9 \times 13.6\ \text{eV} = -122.4\ \text{eV}.$

   4. Convert energy from eV to joules Use $1\ \text{eV} = 1.602 \times 10^{-19}\ \text{J}$.

   $E_1 = -122.4 \times 1.602 \times 10^{-19} \ \text{J}$

   $= - (122.4 \times 1.602) \times 10^{-19} \ \text{J}$

   $$$$

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8. TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
   Q.The energy equivalent to a mass of 1 kg is (A) $9 \times 10^{13}$ J (B) $9 \times 10^{9}$ J (C) $9 \times 10^{16}$ J (D) $9 \times 10^{6}$ J

   ›Reveal solutionSolution

   The energy equivalent of mass is given by Einstein’s equation $E = mc^2$. For $m = 1\ \text{kg}$ and $c = 3 \times 10^8\ \text{m/s}$, the result is $9 \times 10^{16}\ \text{J}$, which corresponds to option (C).

   The core idea here is mass–energy equivalence, one of the most profound results in physics. It tells us that mass is a form of energy, and the conversion factor is the square of the speed of light — a huge number. That’s why even a tiny amount of mass can release an enormous amount of energy (as in nuclear reactions). For 1 kg, the energy is staggeringly large, far beyond everyday experience.

   1. Recall the famous formula

      Einstein’s relation is $E = mc^2$, where:

      • $E$ = energy (in joules),
      • $m$ = mass (in kilograms),
      • $c$ = speed of light in vacuum ($3 \times 10^8\ \text{m/s}$).

   2. Plug in the given mass

      Here $m = 1\ \text{kg}$, so:

   $E = 1 \times (3 \times 10^8)^2$

   3. Square the speed of light

   $(3 \times 10^8)^2 = 9 \times 10^{16}$

   (Because $3^2 = 9$ and $(10^8)^2 = 10^{16}$.)

   4. State the result …

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9. TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
   Q.The energy of second orbit of hydrogen atom is $-5.45 \times 10^{-19}$ J. What is the energy of first orbit of Li$^{2+}$ ion (in J)? (A) $-1.962 \times 10^{-18}$ (B) $-1.962 \times 10^{-17}$ (C) $-3.924 \times 10^{-17}$ (D) $-3.924 \times 10^{-18}$

   ›Reveal solutionSolution

   The energy of a hydrogen-like atom scales as $E_n = -13.6\,\frac{Z^2}{n^2}$ eV. Using the given energy for H (Z=1, n=2) to find the constant, then applying it to Li²⁺ (Z=3, n=1) gives $-1.962 \times 10^{-17}$ J, which is option (B).

   The key idea is that for any hydrogen-like ion (one electron around a nucleus of charge $Ze$), the energy levels follow the same simple formula:

   $E_n = -13.6\,\frac{Z^2}{n^2}\ \text{eV}$

   or in joules:

   $E_n = -2.18 \times 10^{-18}\,\frac{Z^2}{n^2}\ \text{J}$

   The problem gives us a specific data point for hydrogen (Z=1, n=2) so we can verify the constant, then scale it for Li²⁺ (Z=3, n=1). This avoids memorising the constant — we derive it from the given number.

   1. Identify the pattern For hydrogen-like atoms, the energy of the $n$-th orbit is

   $E_n = -k \frac{Z^2}{n^2}$

   where $k$ is a constant (the Rydberg energy in joules). For hydrogen (Z=1), the ground state (n=1) energy is $-2.18 \times 10^{-18}$ J. The problem gives the n=2 energy for H as $-5.45 \times 10^{-19}$ J. Let’s check consistency:

   $E_2(\text{H}) = -k \frac{1^2}{2^2} = -\frac{k}{4}$

   So $-\frac{k}{4} = -5.45 \times 10^{-19}$ → $k = 4 \times 5.45 \times 10^{-19} = 2.18 \times 10^{-18}$ J. Perfect — that matches the known Rydberg constant.

   2. Apply to Li²⁺ Lithium ion Li²⁺ has Z=3 (nucleus with 3 protons) and only one electron. We want the first orbit (n=1). Using the same constant $k$:

   $E_1(\text{Li}^{2+}) = -k \frac{Z^2}{n^2} = -2.18 \times 10^{-18} \times \frac{3^2}{1^2}$

   $= -2.18 \times 10^{-18} \times 9 = -1.962 \times 10^{-17}\ \text{J}$

   3. Match to options …

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10. TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
    Q.A cube of edge length 1 cm is divided into smaller cubes of uniform size of length 1 mm. Assuming that no voids are present, the ratio of total surface area of all the cubes of 1 nm edge length to the surface area of the initial cube is (A) $10^9$ (B) $10^7$ (C) $10^6$ (D) $10^5$

    ›Reveal solutionSolution

    When a larger cube is divided into many smaller cubes, the total volume remains constant, but the total surface area increases significantly. The ratio of the total surface area of the smaller cubes to the surface area of the initial cube is simply the ratio of their respective edge lengths. For a 1 cm cube divided into 1 nm cubes, this ratio is $\boxed{10^7}$.

    Concept and Intuition

    When a large object is broken down into many smaller pieces, its total volume remains the same (assuming no material is lost and no voids are created). However, the total surface area changes dramatically. Imagine cutting a block of cheese: each cut creates new surfaces that were previously internal. The sum of the surface areas of all the smaller pieces will always be greater than the surface area of the original block.

    In this problem, a large cube is divided into many smaller cubes.

    1. Volume Conservation: The total volume of all the smaller cubes must equal the volume of the original large cube. This principle allows us to determine how many smaller cubes are formed.
    2. Surface Area Calculation: The surface area of a cube with edge length $x$ is given by $6x^2$ (since a cube has 6 identical square faces).
    3. Ratio of Surface Areas: We need to find the ratio of the total surface area of all the small cubes to the surface area of the initial large cube. This means we'll calculate the surface area of one small cube, multiply it by the number of small cubes, and then divide by the surface area of the large cube.

    Important

    The statement "A cube of edge length 1 cm is divided into smaller cubes of uniform size of length 1 mm" describes a scenario. However, the question then asks for "the ratio of total surface area of all the cubes of 1 nm edge length to the surface area of the initial cube". This means we should use the 1 cm and 1 nm lengths for the ratio calculation, and the 1 mm length mentioned initially is a distractor.

    Step-by-Step Solution

    1. Identify Edge Lengths and Convert Units:

       We are given the edge length of the initial cube and the edge length of the smaller cubes for which we need to calculate the total surface area. It's crucial to work with consistent units. Let's convert everything to meters.

       • Edge length of the initial cube, $L = 1 \text{ cm} = 1 \times 10^{-2} \text{ m}$.
       • Edge length of the smaller cubes, $l = 1 \text{ nm} = 1 \times 10^{-9} \text{ m}$.

    2. Calculate the Surface Area of the Initial Cube:

       The surface area of a cube with edge length $L$ is $6L^2$.

    $SA_{initial} = 6L^2$

    3. Determine the Number of Smaller Cubes: When the initial cube is divided into smaller cubes, the total volume remains constant.
       • Volume of the initial cube: $V_{initial} = L^3$.
       • Volume of one smaller cube: $v_{small} = l^3$. The number of smaller cubes, $N$, is the total volume divided by the volume of one small cube:

    $N = \frac{V_{initial}}{v_{small}} = \frac{L^3}{l^3} = \left(\frac{L}{l}\right)^3$

    4. Calculate the Total Surface Area of All the Smaller Cubes: The surface area of one smaller cube is $sa_{small} = 6l^2$. …

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11. TG EAPCET 2022Set ap-2022-07-31-AN1 markMCQ
    Q.The number of revolutions made by an electron in the 1$^{\text{st}}$ orbit of hydrogen atom per minute is approximately (A) $3.9 \times 10^{15}$ (B) $3.9 \times 10^{16}$ (C) $3.9 \times 10^{17}$ (D) $3.9 \times 10^{14}$

    ›Reveal solutionSolution

    The number of revolutions made by an electron in the 1st orbit of a hydrogen atom per minute is found by calculating its orbital frequency (speed divided by circumference) and then converting from revolutions per second to revolutions per minute. The approximate value is $\boxed{3.9 \times 10^{17}}$.

    The electron in a hydrogen atom's orbit can be thought of as moving in a circular path around the nucleus. The number of revolutions it makes per unit time is its frequency of revolution. To find this frequency, we need two pieces of information: the speed of the electron in that orbit and the circumference of the orbit. The Bohr model provides formulas for both the orbital speed and the orbital radius for a given energy level (principal quantum number $n$).

    Here's how to calculate the number of revolutions:

    1. Recall the formulas for orbital speed and radius in the Bohr model: For a hydrogen atom ($Z=1$), the speed of an electron in the $n$-th orbit, $v_n$, and the radius of the $n$-th orbit, $r_n$, are given by:

    $v_n = \frac{v_0}{n}$

    $r_n = n^2 a_0$

    where $v_0$ is the speed of the electron in the first Bohr orbit (approximately $2.188 \times 10^6 \text{ m/s}$) and $a_0$ is the Bohr radius (approximately $0.529 \times 10^{-10} \text{ m}$). These are fundamental constants derived from the basic physical constants.
    

    2. Determine the speed and radius for the 1st orbit ($n=1$):

    For the 1st orbit, $n=1$.

    The speed of the electron is:

    $v_1 = \frac{v_0}{1} = v_0 = 2.188 \times 10^6 \text{ m/s}$

    The radius of the orbit is:
    

    $r_1 = (1)^2 a_0 = a_0 = 0.529 \times 10^{-10} \text{ m}$

    3. Calculate the frequency of revolution (revolutions per second): The frequency $f$ is the number of revolutions per second. It is given by the ratio of the electron's speed to the circumference of its orbit ($2\pi r$).

    $f = \frac{v_1}{2\pi r_1}$

    Substitute the values for $v_1$ and $r_1$: …
    

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12. TG EAPCET 2022Set eng-2022-07-18-FN1 markMCQ
    Q.Three particles, each of mass $M$, situated at the vertices of an equilateral triangle of side length '$l$'. The only forces acting on the particles are their mutual gravitational forces. It is desired that each particle moves in a circle while maintaining the original separation '$l$'. The initial speed that should be given to each particle is (A) $\sqrt{\dfrac{2GM}{l}}$ (B) $\sqrt{\dfrac{GM}{2l}}$ (C) $\sqrt{\dfrac{GM}{l}}$ (D) $\sqrt{\dfrac{3GM}{l}}$

    ›Reveal solutionSolution

    The three masses rotate as a rigid equilateral triangle about their common centre of mass; the required centripetal force is provided by the net gravitational attraction from the other two masses, leading to an orbital speed of $\sqrt{GM/l}$.

    Concept & Intuition

    When three equal masses are placed at the vertices of an equilateral triangle and released with the right initial velocities, they can orbit their common centre of mass in a circle while keeping the triangle’s shape. The key is that each mass feels the gravitational pull from the other two, and the vector sum of those pulls points directly toward the centre of the triangle. That net force must equal the centripetal force needed for circular motion at the given radius.

    Step-by-step reasoning

    1. Geometry of the system The three masses form an equilateral triangle of side $l$. Their centre of mass (CM) is at the triangle’s centroid. For an equilateral triangle, the distance from any vertex to the centroid is

    $R = \frac{l}{\sqrt{3}}.$

    This is the radius of the circle each mass will travel around the CM.

    2. Gravitational force on one mass Consider one mass, say at vertex A. The other two masses (at B and C) each exert a gravitational force of magnitude

    $F_{\text{pair}} = \frac{G M^2}{l^2}.$

    The directions of these forces are along the sides AB and AC. Because the triangle is equilateral, the angle between these two force vectors is $60^\circ$.

    3. Net gravitational force toward the centre The vector sum of the two equal forces at $60^\circ$ has magnitude

    $F_{\text{net}} = 2 \cdot \frac{G M^2}{l^2} \cdot \cos(30^\circ) = 2 \cdot \frac{G M^2}{l^2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}\, G M^2}{l^2}.$

    By symmetry, this net force points directly toward the centroid (the centre of the circle).

    4. Centripetal force requirement …

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