If \( R \) is the radius of the earth and the acceleration due to gravity on the surface of the earth is \( g = \pi^2 \, \text{m/s}^2 \), then the length of the second's pendulum at a height \( h = 2R \) from the surface of the earth will be:
- \( \frac{2}{9} \, \text{m} \)
- \( \frac{1}{9} \, \text{m} \)
- \( \frac{4}{9} \, \text{m} \)
- \( \frac{8}{9} \, \text{m} \)
The Correct Option is B
Approach Solution - 1
The time period \(T\) of a simple pendulum is related to the acceleration due to gravity \(g\) by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
where \(L\) is the length of the pendulum and \(g\) is the acceleration due to gravity.
Step 1: Gravitational acceleration at height \(h\)
At a height \(h = 2R\) from the surface of the Earth, the acceleration due to gravity \(g'\) is given by:
\[ g' = g \left( \frac{R}{R + h} \right)^2 \]
Substituting \(h = 2R\):
\[ g' = g \left( \frac{R}{3R} \right)^2 = \frac{g}{9} \]
Therefore, the new value of gravitational acceleration at height \(h = 2R\) is \(\frac{g}{9}\).
Step 2: Time period of the pendulum
The time period \(T\) of a pendulum is related to the length \(L\) and the acceleration due to gravity \(g\) by:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
At height \(h = 2R\), the new time period \(T'\) will be:
\[ T' = 2\pi \sqrt{\frac{L'}{g'}} \]
Since the time period remains 2 seconds, we equate the time periods:
\[ 2 = 2\pi \sqrt{\frac{L'}{g/9}} \]
Squaring both sides:
\[ 1 = \pi^2 \frac{L'}{g} \]
Solving for \(L'\):
\[ L' = \frac{g}{\pi^2} \]
Substitute \(g = \pi^2 \, \text{m/s}^2\) into the equation:
\[ L' = \frac{\pi^2}{9\pi^2} = \frac{1}{9} \, \text{m} \]
Thus, the length of the second's pendulum at a height \(h = 2R\) is \(\frac{1}{9} \, \text{m}\).
Approach Solution -2
We are given: Earth's radius \(R\), gravity at surface \(g = \pi^2 \ \text{m/s}^2\). We must find the length of a seconds pendulum at height \(h = 2R\) above the surface. A seconds pendulum has a period \(T = 2 \ \text{s}\).
Concept Used:
The period of a simple pendulum is \(T = 2\pi \sqrt{\frac{L}{g'}}\), where \(g'\) is the acceleration due to gravity at that location. For a seconds pendulum, \(T = 2 \ \text{s}\).
Gravity varies with height as \(g' = g \left( \frac{R}{R + h} \right)^2\).
Step-by-Step Solution:
Step 1: Find length \(L_0\) at Earth's surface.
At surface: \(T = 2\pi \sqrt{\frac{L_0}{g}} = 2 \ \text{s}\). Given \(g = \pi^2\):
\[ 2\pi \sqrt{\frac{L_0}{\pi^2}} = 2 \] \[ 2\pi \cdot \frac{\sqrt{L_0}}{\pi} = 2 \] \[ 2\sqrt{L_0} = 2 \] \[ \sqrt{L_0} = 1 \] \[ L_0 = 1 \ \text{m} \]
Step 2: Find \(g'\) at height \(h = 2R\).
At height \(h\), \(g' = g \left( \frac{R}{R + h} \right)^2 = g \left( \frac{R}{R + 2R} \right)^2 = g \left( \frac{R}{3R} \right)^2 = \frac{g}{9}\).
Given \(g = \pi^2\), so \(g' = \frac{\pi^2}{9}\).
Step 3: Find length \(L\) at height \(h\) for seconds pendulum.
At height \(h\), \(T = 2\pi \sqrt{\frac{L}{g'}} = 2\):
\[ 2\pi \sqrt{\frac{L}{\pi^2/9}} = 2 \] \[ 2\pi \cdot \frac{\sqrt{L}}{\pi/3} = 2 \] \[ 2\pi \cdot \frac{3\sqrt{L}}{\pi} = 2 \] \[ 6\sqrt{L} = 2 \] \[ \sqrt{L} = \frac{1}{3} \] \[ L = \frac{1}{9} \ \text{m} \]
Hence, the length of the second's pendulum at height \(2R\) is \(\frac{1}{9} \ \text{m}\).
Top JEE Main Physics Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
- $I _{ CM }$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc $I_{A B}$ is it's moment of inertia about an axis $AB$ perpendicular to plane and parallel to axis $CM$ at a distance $\frac{2}{3} R$ from center Where $R$ is the radius of the dise The ratio of $I _{ AB }$ and $I _{ CM }$ is $x: 9$ The value of $x$ is______
- A nucleus disintegrates into two smaller parts, which have their velocities in the ratio $3: 2$ The ratio of their nuclear sizes will be $\left(\frac{x}{3}\right)^{\frac{1}{3}}$ The value of ' $x$ ' is:-
Top JEE Main Gravitation Questions
- A planet takes 200 days to complete one revolution around the Sun. If the distance of the planet from Sun is reduced to one fourth of the original distance, how many days will it take to complete one revolution ?
- Four identical particles of mass \( m \) are kept at the four corners of a square. If the gravitational force exerted on one of the masses by the other masses is \[ \left( \frac{2 \sqrt{2} + 1}{32} \right) \frac{G m^2}{L^2}, \] the length of the sides of the square is:
- Assertion: The angular velocity of the moon revolving around the earth is more than that of the earth revolving around the Sun.
Reason: The time taken by the moon to revolve around the earth is less than the time taken by the earth to revolve around the sun. - A spherical body of mass 100 g is dropped from a height of 10 m from the ground. After hitting the ground, the body rebounds to a height of 5 m. The impulse of force imparted by the ground to the body is given by: (given \( g = 9.8 \, \text{m/s}^2 \))
- Four equal masses m are kept at corners of a square of side a. If net gravitational force on a mass is given by \(\bigg(\frac{2\sqrt2+1}{32}\bigg) \frac{GM^2}{L^2}\) Find the value of a in the terms of \(L\).
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.