Full Syllabus Test Series of Physics
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1. Two plane mirrors are inclined at 70°. A ray incident on one mirror at an angle after reflection falls on the second mirror and is reflected from there parallel to the first mirror. The angle of incidence is:
A. 50°
B. 45°
C. 30°
D. 55°
The correct answer is A. 50°.
Explanation:
When two mirrors are inclined, the total angle covered by the reflection path can be calculated using the mirror angles. Here, the angle of incidence ensures the final reflection aligns parallel to the initial mirror. Thus, an angle of 50° satisfies this condition.
Explanation:
When two mirrors are inclined, the total angle covered by the reflection path can be calculated using the mirror angles. Here, the angle of incidence ensures the final reflection aligns parallel to the initial mirror. Thus, an angle of 50° satisfies this condition.
2. A particle undergoing simple harmonic motion (SHM) along the x-axis has its acceleration defined by \( a = 2 - x \), where \( a \) is in \( \text{m/s}^2 \) and \( x \) is in meters. If the speed of the particle at \( x = 1 \) is zero, the correct statement is:
A. The time period of oscillation is 1 second
B. The amplitude of oscillation is 1 m
C. The speed of the particle at \( x = 1.5 \, \text{m} \) is \( \frac{2}{3} \, \text{m/s} \)
D. The speed of the particle at \( x = 3 \, \text{m} \) is zero
The correct answer is B. The amplitude of oscillation is 1 m.
Explanation:
For SHM, acceleration \( a \) relates to displacement \( x \) with a restoring force. Here, the condition \( a = 2 - x \) implies equilibrium around \( x = 2 \), with amplitude measured from \( x = 1 \) to \( x = 3 \), yielding an amplitude of 1 m.
Explanation:
For SHM, acceleration \( a \) relates to displacement \( x \) with a restoring force. Here, the condition \( a = 2 - x \) implies equilibrium around \( x = 2 \), with amplitude measured from \( x = 1 \) to \( x = 3 \), yielding an amplitude of 1 m.
3. An ideal gas undergoes a cyclic process \( abcda \), shown by a pressure-density (PV) curve. Which of the following statements is correct?
A. Work done by the gas in the process 'bc' is zero.
B. Work done by the gas in the process 'cd' is negative.
C. Internal energy of the gas at point 'a' is greater than at state 'c'.
D. Net work done by the gas in the cycle is negative.
The correct answer is A. Work done by the gas in the process 'bc' is zero.
Explanation:
In a pressure-density (PV) cycle for an ideal gas, specific segments like 'bc' can involve constant volume or pressure changes, affecting the work done. For process 'bc', the gas undergoes no change in volume, resulting in zero work done. The overall cycle work depends on the enclosed area.
Explanation:
In a pressure-density (PV) cycle for an ideal gas, specific segments like 'bc' can involve constant volume or pressure changes, affecting the work done. For process 'bc', the gas undergoes no change in volume, resulting in zero work done. The overall cycle work depends on the enclosed area.
4. A river of width 100 m is flowing with a velocity of 1.5 m/s. A man starts from one end with zero initial velocity relative to the river. He rows with an acceleration of 2 m/s² relative to the river. If the man wants to cross the river in the minimum time, by how much distance (in meters) will he be drifted in the direction of river flow during the crossing?
A. 50 m
B. 15 m
C. 100 m
D. 150 m
The correct answer is B. 15 m.
Solution:
Step 1: To cross the river in minimum time, the man rows perpendicular to the river's flow. The time taken to cross can be found using the formula for motion under constant acceleration:
\[ s = \frac{1}{2} a t^2 \] where \( s = 100 \, \text{m} \) and \( a = 2 \, \text{m/s}^2 \). Solving for \( t \), we get: \[ 100 = \frac{1}{2} \times 2 \times t^2 \Rightarrow t = 10 \, \text{s} \]
Step 2: The drift due to river flow is given by \( d = v_{\text{river}} \times t \), where \( v_{\text{river}} = 1.5 \, \text{m/s} \). So: \[ d = 1.5 \times 10 = 15 \, \text{m} \]
Conclusion:
The drift in the direction of river flow during the crossing is 15 m.
Solution:
Step 1: To cross the river in minimum time, the man rows perpendicular to the river's flow. The time taken to cross can be found using the formula for motion under constant acceleration:
\[ s = \frac{1}{2} a t^2 \] where \( s = 100 \, \text{m} \) and \( a = 2 \, \text{m/s}^2 \). Solving for \( t \), we get: \[ 100 = \frac{1}{2} \times 2 \times t^2 \Rightarrow t = 10 \, \text{s} \]
Step 2: The drift due to river flow is given by \( d = v_{\text{river}} \times t \), where \( v_{\text{river}} = 1.5 \, \text{m/s} \). So: \[ d = 1.5 \times 10 = 15 \, \text{m} \]
Conclusion:
The drift in the direction of river flow during the crossing is 15 m.
5. An object moves in front of a fixed plane mirror. The velocity of the image of the object is
A. Equal in the magnitude and in the direction to that of the object.
B. Equal in the magnitude and opposite in direction to that of the object.
C. Equal in the magnitude and the direction will be either same or opposite to that of the object.
D. Equal in magnitude and makes any angle with that of the object depending on direction of motion of the object.
The correct answer is D. Equal in magnitude and makes any angle with that of the object depending on direction of motion of the object.
Solution:
When object moves normal to the mirror, image velocity will be opposite to it. When object moves parallel to the mirror, image velocity will be in the same direction.
Solution:
When object moves normal to the mirror, image velocity will be opposite to it. When object moves parallel to the mirror, image velocity will be in the same direction.
6. A point object is moving along the principal axis of a concave mirror with uniform velocity towards the pole. Initially, the object is at infinite distance from the pole on the right side of the mirror. Before the object collides with the mirror, the number of times at which the distance between the object and its image is 40 cm is:
A. 1 time
B. 2 times
C. 3 times
D. 4 times
The correct answer is C. 3 times.
Solution:
As the object moves towards the concave mirror, its image will be formed at varying distances due to the mirror's nature. The image will be real and inverted, and as the object approaches the mirror, the distance between the object and its image changes continuously. The distance of the image from the mirror will initially be very large and will decrease as the object approaches the mirror. The distance between the object and its image will be 40 cm at three different positions: 1. When the object is at a position where the image distance is more than 40 cm. 2. When the object is at a position where the image distance is exactly 40 cm. 3. When the object is close to the mirror and the distance decreases to 40 cm again. Hence, the distance between the object and its image will be 40 cm a total of 3 times before the object collides with the mirror.
Conclusion:
The number of times the distance between the object and its image is 40 cm is 3 times.
Solution:
As the object moves towards the concave mirror, its image will be formed at varying distances due to the mirror's nature. The image will be real and inverted, and as the object approaches the mirror, the distance between the object and its image changes continuously. The distance of the image from the mirror will initially be very large and will decrease as the object approaches the mirror. The distance between the object and its image will be 40 cm at three different positions: 1. When the object is at a position where the image distance is more than 40 cm. 2. When the object is at a position where the image distance is exactly 40 cm. 3. When the object is close to the mirror and the distance decreases to 40 cm again. Hence, the distance between the object and its image will be 40 cm a total of 3 times before the object collides with the mirror.
Conclusion:
The number of times the distance between the object and its image is 40 cm is 3 times.
7. If the tension and diameter of a sonometer wire of fundamental frequency \( n \) are doubled and the density is halved, then its fundamental frequency will become:
A. \( 2n \)
B. \( 4n \)
C. \( n \)
D. \( \frac{n}{2} \)
The correct answer is C. \( n \).
Solution:
The fundamental frequency \( f \) of a sonometer wire is given by the formula: \[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \] When the tension \( T \) is doubled and the diameter \( D \) is doubled (which increases the area and thus affects the linear density), while the density is halved, the new linear density becomes: \[ \mu' = 2\mu \] Thus, the new frequency becomes: \[ f' = \frac{1}{2L} \sqrt{\frac{2T}{2\mu}} = \frac{1}{2L} \sqrt{\frac{T}{\mu}} = n \] Therefore, the fundamental frequency will remain \( n \).
Conclusion:
The fundamental frequency will become \( n \).
Solution:
The fundamental frequency \( f \) of a sonometer wire is given by the formula: \[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \] When the tension \( T \) is doubled and the diameter \( D \) is doubled (which increases the area and thus affects the linear density), while the density is halved, the new linear density becomes: \[ \mu' = 2\mu \] Thus, the new frequency becomes: \[ f' = \frac{1}{2L} \sqrt{\frac{2T}{2\mu}} = \frac{1}{2L} \sqrt{\frac{T}{\mu}} = n \] Therefore, the fundamental frequency will remain \( n \).
Conclusion:
The fundamental frequency will become \( n \).
8. An aeroplane revolves in a circle above the surface of the earth at a fixed height with speed 100 km/hr. What is the magnitude of change in velocity after completing \( \frac{1}{2} \) revolution?
A. 100 km/hr
B. 200 km/hr
C. 150 km/hr
D. 50 km/hr
The correct answer is B. 200 km/hr.
Solution:
The airplane is moving in a circle with a speed of \( 100 \, \text{km/hr} \). After completing \( \frac{1}{2} \) revolution, its initial velocity is directed horizontally to the right, and the final velocity is directed horizontally to the left.
Initial velocity: \( \vec{v}_i = 100 \, \text{km/hr} \, \hat{i} \)
Final velocity: \( \vec{v}_f = -100 \, \text{km/hr} \, \hat{i} \)
The change in velocity is given by: \[ \Delta \vec{v} = \vec{v}_f - \vec{v}_i = -200 \, \text{km/hr} \] Thus, the magnitude of the change in velocity is \( 200 \, \text{km/hr} \).
Conclusion:
The magnitude of change in velocity after completing \( \frac{1}{2} \) revolution is \( 200 \, \text{km/hr} \).
Solution:
The airplane is moving in a circle with a speed of \( 100 \, \text{km/hr} \). After completing \( \frac{1}{2} \) revolution, its initial velocity is directed horizontally to the right, and the final velocity is directed horizontally to the left.
Initial velocity: \( \vec{v}_i = 100 \, \text{km/hr} \, \hat{i} \)
Final velocity: \( \vec{v}_f = -100 \, \text{km/hr} \, \hat{i} \)
The change in velocity is given by: \[ \Delta \vec{v} = \vec{v}_f - \vec{v}_i = -200 \, \text{km/hr} \] Thus, the magnitude of the change in velocity is \( 200 \, \text{km/hr} \).
Conclusion:
The magnitude of change in velocity after completing \( \frac{1}{2} \) revolution is \( 200 \, \text{km/hr} \).
9. An inclined plane is inclined at an angle \( \theta \) with the horizontal. A body of mass \( m \) rests on it. If the coefficient of friction is \( \mu \), what is the minimum force that has to be applied on the body parallel to the inclined plane to make the body just move up the inclined plane?
A. \( mg \sin \theta \)
B. \( mg (\sin \theta - \mu \cos \theta) \)
C. \( mg (\sin \theta + \mu \cos \theta) \)
D. \( mg \cos \theta \)
The correct answer is C. \( mg (\sin \theta + \mu \cos \theta) \).
Solution:
To initiate movement up the inclined plane, the applied force \( F \) must overcome both the gravitational force component down the incline and the frictional force:
\[ F = mg \sin \theta + \mu (mg \cos \theta) \]
Factoring out \( mg \) gives: \[ F = mg (\sin \theta + \mu \cos \theta) \]
Conclusion:
The minimum force required to move the body up the inclined plane is \( mg (\sin \theta + \mu \cos \theta) \).
Solution:
To initiate movement up the inclined plane, the applied force \( F \) must overcome both the gravitational force component down the incline and the frictional force:
\[ F = mg \sin \theta + \mu (mg \cos \theta) \]
Factoring out \( mg \) gives: \[ F = mg (\sin \theta + \mu \cos \theta) \]
Conclusion:
The minimum force required to move the body up the inclined plane is \( mg (\sin \theta + \mu \cos \theta) \).
10. Two bodies are projected with the same velocity, one at an angle of \( 30^\circ \) and the other at an angle of \( 60^\circ \) to the horizontal. What is the ratio of the maximum heights reached by the two bodies?
A. 3:1
B. 1:3
C. 1:1
D. 2:1
The correct answer is B. 1:3.
Solution:
The maximum height \( H \) for a projectile is given by:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
For \( \theta = 30^\circ \): \[ H_1 = \frac{u^2 \sin^2(30^\circ)}{2g} = \frac{u^2}{8g} \]
For \( \theta = 60^\circ \): \[ H_2 = \frac{u^2 \sin^2(60^\circ)}{2g} = \frac{3u^2}{8g} \]
Therefore, the ratio of maximum heights is: \[ \text{Ratio} = \frac{H_1}{H_2} = \frac{1}{3} \]
Conclusion:
The ratio of the maximum heights reached by the two bodies is \( 1:3 \).
Solution:
The maximum height \( H \) for a projectile is given by:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
For \( \theta = 30^\circ \): \[ H_1 = \frac{u^2 \sin^2(30^\circ)}{2g} = \frac{u^2}{8g} \]
For \( \theta = 60^\circ \): \[ H_2 = \frac{u^2 \sin^2(60^\circ)}{2g} = \frac{3u^2}{8g} \]
Therefore, the ratio of maximum heights is: \[ \text{Ratio} = \frac{H_1}{H_2} = \frac{1}{3} \]
Conclusion:
The ratio of the maximum heights reached by the two bodies is \( 1:3 \).
11. A body covers a distance of \( L \) m along a curved path of a quarter circle. What is the ratio of distance to displacement?
A. \( \frac{1}{\sqrt{2}} \)
B. \( \frac{\pi}{2\sqrt{2}} \)
C. \( \frac{L}{2} \)
D. \( \frac{L}{\sqrt{2}} \)
The correct answer is B. \( \frac{\pi}{2\sqrt{2}} \).
Solution:
For a body traveling along a quarter circle, the distance covered is \( L \) and the displacement \( d \) is \( r\sqrt{2} \). The ratio of distance to displacement is:
\[ \text{Ratio} = \frac{L}{d} = \frac{L}{r\sqrt{2}} \]
Since \( L = \frac{\pi r}{2} \), substituting gives:
\[ \text{Ratio} = \frac{\frac{\pi r}{2}}{r\sqrt{2}} = \frac{\pi}{2\sqrt{2}} \]
Conclusion:
The ratio of distance to displacement is \( \frac{\pi}{2\sqrt{2}} \).
Solution:
For a body traveling along a quarter circle, the distance covered is \( L \) and the displacement \( d \) is \( r\sqrt{2} \). The ratio of distance to displacement is:
\[ \text{Ratio} = \frac{L}{d} = \frac{L}{r\sqrt{2}} \]
Since \( L = \frac{\pi r}{2} \), substituting gives:
\[ \text{Ratio} = \frac{\frac{\pi r}{2}}{r\sqrt{2}} = \frac{\pi}{2\sqrt{2}} \]
Conclusion:
The ratio of distance to displacement is \( \frac{\pi}{2\sqrt{2}} \).
12. A ball of mass 'm', moving with uniform speed, collides elastically with another stationary ball. The incident ball will lose maximum kinetic energy when the mass of the stationary ball is:
A. \( \frac{m}{2} \)
B. \( m \)
C. \( 2m \)
D. \( 3m \)
The correct answer is B. \( m \).
Solution:
For maximum kinetic energy loss during an elastic collision, the mass of the stationary ball should equal the mass of the incident ball. In this case, both kinetic energy and momentum are conserved. The kinetic energy lost by the incident ball is maximized when:
\[ \text{When } M = m \]
Conclusion:
The incident ball will lose maximum kinetic energy when the mass of the stationary ball is equal to the mass of the incident ball.
Solution:
For maximum kinetic energy loss during an elastic collision, the mass of the stationary ball should equal the mass of the incident ball. In this case, both kinetic energy and momentum are conserved. The kinetic energy lost by the incident ball is maximized when:
\[ \text{When } M = m \]
Conclusion:
The incident ball will lose maximum kinetic energy when the mass of the stationary ball is equal to the mass of the incident ball.
13. A man of mass 60 kg standing on a platform executing S.H.M. in the vertical plane. The displacement from the mean position varies as \( y = 0.5 \sin(2 \pi f t) \). The value of \( f \), for which the man will feel weightlessness at the highest point is:
A. 0.5 Hz
B. 1.0 Hz
C. 1.41 Hz
D. 2.0 Hz
The correct answer is C. 1.41 Hz.
Solution:
To determine the frequency \( f \) at which the man feels weightlessness at the highest point of his motion on the platform, we first need to understand the dynamics of the forces acting on him at that moment.
Step 1: **Understanding Weightlessness**
A person feels weightless when the net force acting on them becomes zero. This occurs at the highest point of their motion in S.H.M. At this point, the gravitational force \( mg \) acts downwards, while the acceleration of the platform (upwards) provides the necessary centripetal force to keep the person moving in a circular path.
The condition for weightlessness is given by: \[ mg = m a \] Here, \( a \) is the acceleration of the platform. Step 2: **Acceleration in S.H.M.**
The acceleration \( a \) in simple harmonic motion can be expressed as: \[ a = -\omega^2 y \] where \( \omega \) is the angular frequency and is related to the frequency \( f \) by: \[ \omega = 2 \pi f \] At the highest point of motion, the displacement \( y \) is equal to the amplitude \( A \), which is given as 0.5 m. Thus, we have: \[ y = A = 0.5 \, \text{m} \] Step 3: **Setting Up the Equation**
The acceleration at the highest point can be written as: \[ a = -\omega^2 A = - (2 \pi f)^2 (0.5) \] For weightlessness, we equate the gravitational force to the centripetal force: \[ mg = m(-\omega^2 A) \] Canceling the mass \( m \) from both sides yields: \[ g = -\omega^2 A \] Substituting \( \omega = 2 \pi f \) into this equation gives: \[ g = -(2 \pi f)^2 A \] Step 4: **Solving for Frequency \( f \)**
Rearranging the equation, we find: \[ g = (2 \pi f)^2 (0.5) \] Thus, we can solve for \( f \): \[ f^2 = \frac{g}{\pi^2 \cdot 0.5} \] Taking the square root: \[ f = \sqrt{\frac{g}{\pi^2 \cdot 0.5}} \] Step 5: **Substituting Values**
Using \( g \approx 9.81 \, \text{m/s}^2 \): \[ f = \sqrt{\frac{9.81}{\pi^2 \cdot 0.5}} = \sqrt{\frac{9.81 \cdot 2}{\pi^2}} = \sqrt{\frac{19.62}{\pi^2}} \] Calculating \( f \): \[ f = \frac{\sqrt{19.62}}{\pi} \approx \frac{4.43}{3.14} \approx 1.41 \, \text{Hz} \] Conclusion:
The frequency \( f \) at which the man will feel weightlessness at the highest point is approximately \( \boxed{1.41 \, \text{Hz}} \).
Solution:
To determine the frequency \( f \) at which the man feels weightlessness at the highest point of his motion on the platform, we first need to understand the dynamics of the forces acting on him at that moment.
Step 1: **Understanding Weightlessness**
A person feels weightless when the net force acting on them becomes zero. This occurs at the highest point of their motion in S.H.M. At this point, the gravitational force \( mg \) acts downwards, while the acceleration of the platform (upwards) provides the necessary centripetal force to keep the person moving in a circular path.
The condition for weightlessness is given by: \[ mg = m a \] Here, \( a \) is the acceleration of the platform. Step 2: **Acceleration in S.H.M.**
The acceleration \( a \) in simple harmonic motion can be expressed as: \[ a = -\omega^2 y \] where \( \omega \) is the angular frequency and is related to the frequency \( f \) by: \[ \omega = 2 \pi f \] At the highest point of motion, the displacement \( y \) is equal to the amplitude \( A \), which is given as 0.5 m. Thus, we have: \[ y = A = 0.5 \, \text{m} \] Step 3: **Setting Up the Equation**
The acceleration at the highest point can be written as: \[ a = -\omega^2 A = - (2 \pi f)^2 (0.5) \] For weightlessness, we equate the gravitational force to the centripetal force: \[ mg = m(-\omega^2 A) \] Canceling the mass \( m \) from both sides yields: \[ g = -\omega^2 A \] Substituting \( \omega = 2 \pi f \) into this equation gives: \[ g = -(2 \pi f)^2 A \] Step 4: **Solving for Frequency \( f \)**
Rearranging the equation, we find: \[ g = (2 \pi f)^2 (0.5) \] Thus, we can solve for \( f \): \[ f^2 = \frac{g}{\pi^2 \cdot 0.5} \] Taking the square root: \[ f = \sqrt{\frac{g}{\pi^2 \cdot 0.5}} \] Step 5: **Substituting Values**
Using \( g \approx 9.81 \, \text{m/s}^2 \): \[ f = \sqrt{\frac{9.81}{\pi^2 \cdot 0.5}} = \sqrt{\frac{9.81 \cdot 2}{\pi^2}} = \sqrt{\frac{19.62}{\pi^2}} \] Calculating \( f \): \[ f = \frac{\sqrt{19.62}}{\pi} \approx \frac{4.43}{3.14} \approx 1.41 \, \text{Hz} \] Conclusion:
The frequency \( f \) at which the man will feel weightlessness at the highest point is approximately \( \boxed{1.41 \, \text{Hz}} \).
14. A particle of mass 'm' executes SHM according to the equation \( \frac{d^2 x}{dt^2} + kx = 0 \). Its time period will be:
A. \( \frac{2\pi}{k} \)
B. \( \frac{2\pi}{\sqrt{k}} \)
C. \( \frac{2\pi \sqrt{m}}{k} \)
D. \( 2\pi \sqrt{\frac{m}{k}} \)
The correct answer is B. \( \frac{2\pi}{\sqrt{k}} \).
Solution:
The given equation for the particle in simple harmonic motion (SHM) is: \[ \frac{d^2 x}{dt^2} + kx = 0 \] This is a second-order differential equation that describes simple harmonic motion. We can identify the parameters that characterize the motion from this equation.
The standard form of the equation for SHM is: \[ \frac{d^2 x}{dt^2} + \omega^2 x = 0 \] where \( \omega \) is the angular frequency of the motion. By comparing this with our given equation, we can see that: \[ \omega^2 = k \] Therefore, we have: \[ \omega = \sqrt{k} \] The time period \( T \) of a particle in SHM is related to the angular frequency \( \omega \) by the formula: \[ T = \frac{2\pi}{\omega} \] Substituting for \( \omega \): \[ T = \frac{2\pi}{\sqrt{k}} \] Conclusion:
Thus, the time period of the particle executing SHM according to the given equation is: \[ T = \frac{2\pi}{\sqrt{k}} \] Therefore, the correct answer is \( \boxed{\frac{2\pi}{\sqrt{k}}} \).
Solution:
The given equation for the particle in simple harmonic motion (SHM) is: \[ \frac{d^2 x}{dt^2} + kx = 0 \] This is a second-order differential equation that describes simple harmonic motion. We can identify the parameters that characterize the motion from this equation.
The standard form of the equation for SHM is: \[ \frac{d^2 x}{dt^2} + \omega^2 x = 0 \] where \( \omega \) is the angular frequency of the motion. By comparing this with our given equation, we can see that: \[ \omega^2 = k \] Therefore, we have: \[ \omega = \sqrt{k} \] The time period \( T \) of a particle in SHM is related to the angular frequency \( \omega \) by the formula: \[ T = \frac{2\pi}{\omega} \] Substituting for \( \omega \): \[ T = \frac{2\pi}{\sqrt{k}} \] Conclusion:
Thus, the time period of the particle executing SHM according to the given equation is: \[ T = \frac{2\pi}{\sqrt{k}} \] Therefore, the correct answer is \( \boxed{\frac{2\pi}{\sqrt{k}}} \).
15. Assuming Earth to be a perfectly spherical body, the time period of a simple pendulum is maximum:
A. At the equator
B. At the poles
C. At mid-latitudes
D. At any latitude
The correct answer is A. At the equator.
Solution:
The time period \( T \) of a simple pendulum is given by the formula: \[ T = 2\pi \sqrt{\frac{L}{g}} \] where: - \( L \) is the length of the pendulum, - \( g \) is the acceleration due to gravity. The acceleration due to gravity \( g \) varies with latitude due to the Earth's rotation and its shape (oblate spheroid). It is given by the formula: \[ g = g_0 \cdot \cos(\phi) - \omega^2 R \cdot \sin^2(\phi) \] where: - \( g_0 \) is the gravitational acceleration at the poles, - \( \omega \) is the angular velocity of the Earth, - \( R \) is the radius of the Earth, - \( \phi \) is the latitude. To maximize the time period \( T \), we need to minimize the value of \( g \). Since \( g \) is maximum at the poles (where \( \phi = 0^\circ \)) and decreases towards the equator (where \( \phi = 90^\circ \)), the time period is maximized when \( g \) is at its minimum. Therefore, the time period is maximum at the equator where the effective value of gravity is lower due to the centrifugal effect caused by Earth's rotation. Conclusion:
Hence, assuming Earth to be a perfectly spherical body, the time period of a simple pendulum is maximum at the equator.
Solution:
The time period \( T \) of a simple pendulum is given by the formula: \[ T = 2\pi \sqrt{\frac{L}{g}} \] where: - \( L \) is the length of the pendulum, - \( g \) is the acceleration due to gravity. The acceleration due to gravity \( g \) varies with latitude due to the Earth's rotation and its shape (oblate spheroid). It is given by the formula: \[ g = g_0 \cdot \cos(\phi) - \omega^2 R \cdot \sin^2(\phi) \] where: - \( g_0 \) is the gravitational acceleration at the poles, - \( \omega \) is the angular velocity of the Earth, - \( R \) is the radius of the Earth, - \( \phi \) is the latitude. To maximize the time period \( T \), we need to minimize the value of \( g \). Since \( g \) is maximum at the poles (where \( \phi = 0^\circ \)) and decreases towards the equator (where \( \phi = 90^\circ \)), the time period is maximized when \( g \) is at its minimum. Therefore, the time period is maximum at the equator where the effective value of gravity is lower due to the centrifugal effect caused by Earth's rotation. Conclusion:
Hence, assuming Earth to be a perfectly spherical body, the time period of a simple pendulum is maximum at the equator.
16. A uniform disk of mass 300 kg is rotating freely about a vertical axis through its center with constant angular velocity. A boy of mass 30 kg starts from the center and moves along a radius to the edge of the disk. The angular velocity of the disk now is:
A. Less than initial angular velocity
B. Equal to the initial angular velocity
C. Greater than initial angular velocity
D. Cannot be determined
The correct answer is A. Less than initial angular velocity.
Solution:
This problem involves the conservation of angular momentum. The key principle here is that in a closed system (no external torques acting), the total angular momentum before and after the event must remain constant.
Step 1: **Define Initial Conditions**
Let: - \( M = 300 \, \text{kg} \) (mass of the disk), - \( m = 30 \, \text{kg} \) (mass of the boy), - \( \omega_i \) = Initial angular velocity of the disk. The initial angular momentum \( L_i \) of the system (disk + boy) when the boy is at the center is given by: \[ L_i = I_d \cdot \omega_i \] where \( I_d \) is the moment of inertia of the disk about its center. For a uniform disk: \[ I_d = \frac{1}{2} M R^2 \] (where \( R \) is the radius of the disk, which we do not need to know for this ratio). The boy at the center contributes no angular momentum, so: \[ L_{i, \text{total}} = I_d \cdot \omega_i + 0 = I_d \cdot \omega_i \] Step 2: **Define Final Conditions**
When the boy moves to the edge of the disk, the moment of inertia of the system changes. The new moment of inertia \( I_f \) will be: \[ I_f = I_d + mR^2 = \frac{1}{2} M R^2 + mR^2 = \left( \frac{1}{2} M + m \right) R^2 \] The final angular momentum \( L_f \) is given by: \[ L_f = I_f \cdot \omega_f \] Step 3: **Apply Conservation of Angular Momentum**
According to the conservation of angular momentum: \[ L_i = L_f \] Thus, \[ I_d \cdot \omega_i = \left( \frac{1}{2} M + m \right) R^2 \cdot \omega_f \] Rearranging gives: \[ \omega_f = \frac{I_d \cdot \omega_i}{\left( \frac{1}{2} M + m \right) R^2} \] Since \( \frac{1}{2} M + m > \frac{1}{2} M \), we have: \[ \omega_f < \omega_i \] Conclusion:
Therefore, the angular velocity of the disk after the boy moves to the edge is less than the initial angular velocity. Thus, the correct answer is: \[ \boxed{\text{Less than initial angular velocity}} \]
Solution:
This problem involves the conservation of angular momentum. The key principle here is that in a closed system (no external torques acting), the total angular momentum before and after the event must remain constant.
Step 1: **Define Initial Conditions**
Let: - \( M = 300 \, \text{kg} \) (mass of the disk), - \( m = 30 \, \text{kg} \) (mass of the boy), - \( \omega_i \) = Initial angular velocity of the disk. The initial angular momentum \( L_i \) of the system (disk + boy) when the boy is at the center is given by: \[ L_i = I_d \cdot \omega_i \] where \( I_d \) is the moment of inertia of the disk about its center. For a uniform disk: \[ I_d = \frac{1}{2} M R^2 \] (where \( R \) is the radius of the disk, which we do not need to know for this ratio). The boy at the center contributes no angular momentum, so: \[ L_{i, \text{total}} = I_d \cdot \omega_i + 0 = I_d \cdot \omega_i \] Step 2: **Define Final Conditions**
When the boy moves to the edge of the disk, the moment of inertia of the system changes. The new moment of inertia \( I_f \) will be: \[ I_f = I_d + mR^2 = \frac{1}{2} M R^2 + mR^2 = \left( \frac{1}{2} M + m \right) R^2 \] The final angular momentum \( L_f \) is given by: \[ L_f = I_f \cdot \omega_f \] Step 3: **Apply Conservation of Angular Momentum**
According to the conservation of angular momentum: \[ L_i = L_f \] Thus, \[ I_d \cdot \omega_i = \left( \frac{1}{2} M + m \right) R^2 \cdot \omega_f \] Rearranging gives: \[ \omega_f = \frac{I_d \cdot \omega_i}{\left( \frac{1}{2} M + m \right) R^2} \] Since \( \frac{1}{2} M + m > \frac{1}{2} M \), we have: \[ \omega_f < \omega_i \] Conclusion:
Therefore, the angular velocity of the disk after the boy moves to the edge is less than the initial angular velocity. Thus, the correct answer is: \[ \boxed{\text{Less than initial angular velocity}} \]
17. A man is holding an umbrella at an angle of 30° with the vertical, with the lower end towards himself. If his horizontal velocity is 10 m/s, what is the appropriate angle to protect him from rain?
A. Greater than 30°
B. Equal to 30°
C. Less than 30°
D. Cannot be determined
The correct answer is A. Greater than 30°.
Solution:
To determine the appropriate angle at which the man should hold the umbrella to effectively protect himself from rain, we must analyze the relative motion of the rain with respect to the man.
Step 1: **Understanding the Rain's Motion**
Rain generally falls vertically downward under the influence of gravity. However, when the man is moving horizontally with a velocity of \( v_m = 10 \, \text{m/s} \), the effective velocity of the rain (as perceived by the man) is the vector sum of the vertical component of the rain's velocity and the horizontal component due to the man's movement.
Let's denote: - \( v_r \) = Vertical velocity of the rain (which we consider to be downward). - \( v_m = 10 \, \text{m/s} \) (horizontal velocity of the man). The effective rain's velocity (\( \vec{v}_{\text{effective}} \)) as perceived by the man can be represented as a right triangle where: - One leg is the vertical component (falling rain) and - The other leg is the horizontal component (man's velocity). Step 2: **Calculating the Angle of the Umbrella**
The angle \( \theta \) at which the umbrella should be held to protect against the rain is determined by the resultant vector of the rain's motion relative to the man. To ensure maximum protection, the umbrella should be aligned along the resultant vector. We can apply the tangent function: \[ \tan(\theta) = \frac{v_m}{v_r} \] Therefore, the angle \( \theta \) can be calculated as: \[ \theta = \tan^{-1}\left(\frac{v_m}{v_r}\right) \] Since the man is moving horizontally, this angle \( \theta \) will be greater than the angle at which the umbrella is currently held (30°). To adequately protect himself from the rain, the angle of the umbrella must adjust to be steeper (greater than 30°). Step 3: **Conclusion**
As the horizontal velocity increases, the angle of the umbrella must also increase to maintain coverage from the rain effectively. Therefore, the correct answer is: \[ \boxed{\text{Greater than 30°}} \]
Solution:
To determine the appropriate angle at which the man should hold the umbrella to effectively protect himself from rain, we must analyze the relative motion of the rain with respect to the man.
Step 1: **Understanding the Rain's Motion**
Rain generally falls vertically downward under the influence of gravity. However, when the man is moving horizontally with a velocity of \( v_m = 10 \, \text{m/s} \), the effective velocity of the rain (as perceived by the man) is the vector sum of the vertical component of the rain's velocity and the horizontal component due to the man's movement.
Let's denote: - \( v_r \) = Vertical velocity of the rain (which we consider to be downward). - \( v_m = 10 \, \text{m/s} \) (horizontal velocity of the man). The effective rain's velocity (\( \vec{v}_{\text{effective}} \)) as perceived by the man can be represented as a right triangle where: - One leg is the vertical component (falling rain) and - The other leg is the horizontal component (man's velocity). Step 2: **Calculating the Angle of the Umbrella**
The angle \( \theta \) at which the umbrella should be held to protect against the rain is determined by the resultant vector of the rain's motion relative to the man. To ensure maximum protection, the umbrella should be aligned along the resultant vector. We can apply the tangent function: \[ \tan(\theta) = \frac{v_m}{v_r} \] Therefore, the angle \( \theta \) can be calculated as: \[ \theta = \tan^{-1}\left(\frac{v_m}{v_r}\right) \] Since the man is moving horizontally, this angle \( \theta \) will be greater than the angle at which the umbrella is currently held (30°). To adequately protect himself from the rain, the angle of the umbrella must adjust to be steeper (greater than 30°). Step 3: **Conclusion**
As the horizontal velocity increases, the angle of the umbrella must also increase to maintain coverage from the rain effectively. Therefore, the correct answer is: \[ \boxed{\text{Greater than 30°}} \]
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