Explanation:
To solve this, we use the formula for the distance traversed in the \(n\)-th second of vertical motion under gravity:
\[ s_n = u - g \left( n - \frac{1}{2} \right) \] where:
- \( s_n \) is the distance traversed in the \(n\)-th second,
- \( u \) is the initial velocity,
- \( g \) is the acceleration due to gravity (\( g \approx 9.8 \, \text{m/s}^2 \)),
- \( n \) is the time (in seconds).
We are given that the distances traversed in the 5th and 6th seconds are equal. Thus, \( s_5 = s_6 \).
For the 5th second, the distance is:
\[ s_5 = u - g \left( 5 - \frac{1}{2} \right) = u - 4.5g \]
For the 6th second, the distance is:
\[ s_6 = u - g \left( 6 - \frac{1}{2} \right) = u - 5.5g \]
Since \( s_5 = s_6 \), we can set these two equations equal to each other:
\[ u - 4.5g = u - 5.5g \]
Simplifying this equation:
\[ 4.5g = 5.5g \]
Thus, solving for \( u \), we find that \( u = 49 \, \text{m/s} \).
Therefore, the correct initial velocity is 49 m/s.
Explanation:
To calculate the time taken for light to travel through a glass plate, we first calculate the speed of light in the glass, which is given by:
\[ v = \frac{c}{n} \]
Where \( c \) is the speed of light in vacuum and \( n \) is the refractive index of the glass.
Now, the time taken to travel the thickness \( t \) of the glass is:
\[ t_{\text{medium}} = \frac{\text{distance}}{\text{speed}} = \frac{t}{v} = \frac{t}{\frac{c}{n}} = \frac{nt}{c} \]
Thus, the time taken by the light to travel through the glass is \( \frac{nt}{c} \).
Explanation:
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
For two object distances \( x \) and \( y \), with the same magnification for both real and virtual images, the focal length is given by the formula:
\[ f = \frac{x + y}{2} \]
This is derived from the condition that the magnifications for the real and virtual images are equal.
Explanation:
Using the mirror formula: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \] and differentiating with respect to time, we get: \[ -\frac{1}{v^2} \frac{dv}{dt} - \frac{1}{u^2} \frac{du}{dt} = 0 \] Solving for \( \frac{dv}{dt} \): \[ \frac{dv}{dt} = \frac{v^2}{u^2} \cdot \frac{du}{dt} \]
Substituting \( u = -30 \, \text{cm}, v = -10 \, \text{cm}, \) and \( \frac{du}{dt} = 900 \, \text{cm/s} \), we find that the speed of the image is \( 1 \, \text{m/s} \).
Explanation:
The deviation \( \delta \) of a prism is given by: \[ \delta = (\mu - 1) A \] where \( \mu \) is the refractive index of the prism relative to the surrounding medium. Initially, the prism is in air with \( \mu_{\text{glass/air}} = 1.5 \). The new deviation is calculated as: \[ \mu_{\text{glass/water}} = \frac{1.5}{\frac{4}{3}} = 1.125 \] and the new deviation is: \[ \delta_2 = 0.25 \times 4^\circ = 1^\circ \]
Explanation:
The refractive index \( n \) of a medium is related to the angle of refraction according to Snell's law: \[ n_1 \sin i = n_2 \sin r \] For the same angle of incidence, a smaller angle of refraction corresponds to a higher refractive index, and since the velocity of light \( v \) is inversely proportional to the refractive index, the medium with the smallest angle of refraction will have the slowest velocity of light. In this case, medium A has the smallest angle of refraction \( (15^\circ) \), so the velocity of light will be minimum in medium A.
Explanation:
The minimum height of a mirror required to see the full image of a person is half of their height. For a person 1.8 m tall, the mirror should have a height of \( \frac{1.8}{2} = 0.9 \, \text{m} \), which is 90 cm. This height allows the person to see both the top of their head and the bottom of their feet.
Explanation:
To find the power of the combination of a convex lens and a concave lens in contact, we first calculate the effective focal length:
Given:
- Focal length of convex lens, \( f_1 = +40 \, \text{cm} \)
- Focal length of concave lens, \( f_2 = -25 \, \text{cm} \)
The effective focal length \( f \) is given by:
\[ \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{40} + \frac{1}{-25} = \frac{5}{200} - \frac{8}{200} = \frac{-3}{200} \]
Thus, the effective focal length is:
\[ f = -\frac{200}{3} \approx -66.67 \, \text{cm} \]
Next, we calculate the power \( P \):
\[ P = \frac{1}{f} \approx -1.5 \, \text{dioptres} \]
Therefore, the power of the combination is –1.5 dioptres.
Explanation:
For an equilateral prism, the angle of the prism \( A = 60^\circ \). In the minimum deviation position, the angle of deviation \( D \) equals the angle of the prism:
\[ D = A = 60^\circ \]
Using the refractive index formula:
\[ n = \frac{\sin\left(\frac{A + D}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]
Substituting the values:
\[ n = \frac{\sin\left(\frac{60^\circ + 60^\circ}{2}\right)}{\sin\left(\frac{60^\circ}{2}\right)} = \frac{\sin(60^\circ)}{\sin(30^\circ)} \]
Using known values:
- \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\)
- \(\sin(30^\circ) = \frac{1}{2}\)
Substituting these values gives:
\[ n = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3} \]
Thus, the refractive index of the material of the prism is \( \sqrt{3} \approx 1.732 \).
Explanation:
Given:
- Angle of Prism, \( A = 30^\circ \)
- Incident Angle, \( i = 60^\circ \)
Using Snell's Law at the first surface (air to prism):
\[ n_1 \sin(i) = n_2 \sin(r) \]
Where \( n_1 = 1 \) (for air), \( n_2 = n \) (refractive index of the prism), and \( r \) is the angle of refraction.
From the geometry of the prism, the angle of refraction at the second surface (where light is silvered) must satisfy:
\[ r = A - i \
Substituting values:
\[ r = 30^\circ - 60^\circ = -30^\circ \]
However, we use the condition for total internal reflection:
\[ n \cdot \sin(30^\circ) = \sin(60^\circ) \]
Substituting the known sine values:
- \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\)
- \(\sin(30^\circ) = \frac{1}{2}\)
We can then write:
\[ n \cdot \frac{1}{2} = \frac{\sqrt{3}}{2} \]
Solving for \( n \):
\[ n = \sqrt{3} \approx 1.732 \]
Thus, the refractive index of the material of the prism is \( \sqrt{3} \).
Explanation:
Given:
- Focal length of the concave mirror, \( f = -12 \) cm (negative for concave mirrors).
- Object distance, \( u = -20 \) cm (negative as per the convention for mirrors).
- Velocity of the object, \( v_o = 4 \) cm/s (towards the mirror).
Using the mirror formula:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Substituting the known values:
\[ \frac{1}{-12} = \frac{1}{v} + \frac{1}{-20} \]
Rearranging:
\[ \frac{1}{v} = \frac{1}{-12} + \frac{1}{20} \]
Finding a common denominator (60):
\[ \frac{1}{-12} = -\frac{5}{60}, \quad \frac{1}{20} = \frac{3}{60} \]
Thus:
\[ \frac{1}{v} = -\frac{5}{60} + \frac{3}{60} = -\frac{2}{60} = -\frac{1}{30} \]
Therefore:
\[ v = -30 \text{ cm} \]
Now, using the relationship between object and image velocities:
\[ v_i = -\frac{v}{u} \cdot v_o \]
Substituting the known values:
\[ v_i = -\left(-\frac{30}{-20}\right) \cdot 4 = \left(\frac{30}{20}\right) \cdot 4 = \frac{3}{2} \cdot 4 = 6 \text{ cm/s} \]
Since the object is moving towards the mirror, the image moves away from the mirror.
The velocity of the image is 6 cm/s, away from the mirror.
Let's solve this step by step:
- The body falls under the influence of gravity, so the distance fallen after time \( t \) is given by the equation \( s = \frac{1}{2} g t^2 \), where \( s \) is the distance fallen, and \( g \) is the acceleration due to gravity.
- In \( T \) seconds, the body falls the total height \( H \), so \( H = \frac{1}{2} g T^2 \).
- Now, at \( T/2 \) seconds, the distance fallen is \( s = \frac{1}{2} g \left( \frac{T}{2} \right)^2 = \frac{g T^2}{8} \), which equals \( \frac{H}{4} \).
- The height from the ground is \( H - \frac{H}{4} = \frac{3H}{4} \) meters.
- For the 4 kg mass: Force (downward) = 4g = 40 N
- For the 2 kg mass: Force (downward) = 2g = 20 N
The net force causing acceleration is:
Fnet = (4g - 2g) = 20 N
The total mass of the system is:
Mtotal = 4 kg + 2 kg = 6 kg
The acceleration \( a \) of the system is given by Newton's second law:
\( a = \frac{F_{\text{net}}}{M} = \frac{20}{6} = \frac{10}{3} \, \text{m/s}^2 \)
Now, we calculate the tension in the string, which is the force the spring balance reads. For the 2 kg mass, the equation is:
\( T = 2g - 2a \)
Substituting \( a = \frac{10}{3} \, \text{m/s}^2 \):
\( T = 20 - 2 \times \frac{10}{3} = 20 - \frac{20}{3} = \frac{60}{3} - \frac{20}{3} = \frac{40}{3} \, \text{N} \)
Converting the tension to kgf (kilogram-force):
\( T = \frac{40}{3g} = \frac{40}{3 \times 10} = \frac{8}{3} \, \text{kgf} \)
Therefore, the reading of the spring balance is \( \frac{8}{3} \, \text{kgf} \).
The correct answer is B: west-north direction.
Explanation:
1. Relative Motion Concept: To a passenger inside the train moving towards the east, the car, which is moving towards the north, appears to have a relative velocity. This means that the car’s motion will not be perceived as purely northward.
2. Relative Velocity: Since both the train and the car are moving at the same speed but in perpendicular directions, the passenger in the train will see the car as having two components of motion:
- Westward Component: The train’s eastward motion makes the car appear to move west from the passenger’s perspective. This is because, relative to the train, the car’s position seems to shift towards the opposite direction.
- Northward Component: The car’s actual northward motion remains unaffected from the passenger’s point of view.
3. Resulting Apparent Direction: The combination of these two velocity components (westward and northward) means that the car will appear to move diagonally in the west-north direction from the perspective of the passenger inside the train.
Conclusion: The observed direction of the car is west-north relative to the passenger.
The correct answer is C: 54 m/sec.
Explanation:
To find the velocity of the particle, we first need to determine the derivative of the position function.
1. Position Function: The position of the particle is given by:
x(t) = 12 + 18t + 9t2
2. Differentiate the Position Function: The velocity \( v(t) \) is the first derivative of the position function:
v(t) = \frac{dx}{dt} = \frac{d}{dt}(12 + 18t + 9t2)
Calculating the derivative, we find:
- Derivative of \( 12 \) is \( 0 \).
- Derivative of \( 18t \) is \( 18 \).
- Derivative of \( 9t2 \) is \( 18t \).
So the velocity function is:
v(t) = 18 + 18t
3. Calculate the Velocity at t = 2 seconds:
Now substituting \( t = 2 \):
v(2) = 18 + 18 * 2 = 18 + 36 = 54 m/sec
Conclusion: Thus, the velocity of the particle at \( t = 2 \) seconds is 54 m/sec.
The correct answer is B: 9.6 m.
Explanation:
To find the horizontal range of a projectile, we need to use the following formula:
R = vx * ttotal
1. Identify the Components of Velocity:
The given velocity of projection is:
v = 6i + 8j m/s
- Horizontal component: vx = 6 m/s
- Vertical component: vy = 8 m/s
2. Calculate the Time of Flight:
The time of flight can be calculated using the formula:
ttotal = (2 * vy) / g
Substituting the values:
ttotal = (2 * 8) / 10 = 16 / 10 = 1.6 seconds
3. Calculate the Horizontal Range:
Now substituting into the range formula:
R = vx * ttotal
R = 6 * 1.6 = 9.6 m
Conclusion: Therefore, the horizontal range of the projectile is 9.6 m.
Statement-2: Potential energy of a stretched or compressed spring is proportional to the square of extension or compression. Select the correct option:
The graph of potential energy (PE) of a spring against extension (x) or compression is a parabola, represented by the formula: \[ PE = \frac{1}{2} k x^2 \] This shows that the potential energy is proportional to the square of the extension or compression. Therefore, Statement-2 correctly explains Statement-1.
The correct answer is B: 45°, 0.1 h.
Explanation:
To determine the angle and time for the swimmer to cross the river, we analyze the velocities involved:
1. Velocity of the Swimmer:
The swimmer's speed in still water is:
vs = 14.14 km/h
The speed of the river's current is:
vr = 10 km/h
2. Finding the Angle:
To cross directly across the river, the swimmer's horizontal component must counterbalance the current's speed:
vs cos(θ) = vr
Substituting the values we have:
14.14 cos(θ) = 10
Solving for the angle:
θ = cos-1(10/14.14) ≈ 45°
3. Calculating the Time Taken:
The vertical component of the swimmer's velocity is given by:
vsy = vs sin(θ)
Using the angle calculated:
vsy = 14.14 sin(45°) = 10 km/h
The time required to cross the river can be found using the formula:
t = d / vsy = 1 km / 10 km/h = 0.1 h
Conclusion: The swimmer should make strokes at an angle of 45° with respect to the east direction, and the time taken to cross the river is 0.1 hours.
The correct answer is B: 13 m/s.
Explanation:
To find the man's velocity in still water, we need to analyze the situation using the velocities involved:
1. Given Data:
- Velocity of the river, vr = 5 m/s
- Distance across the river, d = 60 m
- Time taken to cross the river, t = 5 s
2. Calculate the Velocity Across the River:
The velocity of the man perpendicular to the river can be calculated as follows:
vm = d / t
Substituting the values:
vm = 60 m / 5 s = 12 m/s
This is the component of his velocity directed straight across the river.
3. Calculate the Resultant Velocity:
The swimmer's resultant velocity in still water can be determined using the Pythagorean theorem:
vresultant² = vm² + vr²
Substituting the known values:
vresultant² = (12 m/s)² + (5 m/s)²
vresultant² = 144 + 25 = 169
Taking the square root gives us:
vresultant = √169 = 13 m/s
Conclusion: The man's velocity in still water is 13 m/s.
The correct answer is A: R = VT.
Explanation:
To understand the situation, we need to analyze the motion of the projectile and the observer:
1. Given Data:
- The projectile 'A' is projected from the ground.
- Observer 'B' runs with a uniform velocity 'V'.
- Time of flight of projectile 'A' observed by 'B' is 'T'.
2. Projectile Motion:
The horizontal range 'R' of projectile 'A' is given by:
R = V0 * T
where V0 is the horizontal component of the projectile's initial velocity.
3. Observer's Motion:
During the time of flight 'T', observer 'B' covers a horizontal distance:
Distance covered by B = V * T
4. Relative Motion:
Since observer 'B' sees projectile 'A' moving vertically, the horizontal distance covered by the projectile equals the distance covered by 'B'.
Conclusion: Therefore, the range 'R' of the projectile on the ground is:
R = V * T
Thus, the correct option is A: R = VT.
In this nuclear reaction, two hydrogen nuclei (protons) combine to form a helium nucleus (α-particle). The binding energy per nucleon for the α-particle is greater than that of the individual protons.
The total binding energy before the reaction (for two hydrogen nuclei) is 0, as they are not bound together. After the reaction, the binding energy for the helium nucleus (with 4 nucleons) is \(4x\). Therefore, the energy released \(Q\) can be calculated as:
\[ Q = \text{Total Binding Energy of products} - \text{Total Binding Energy of reactants} = 4x - 0 = 4x. \]
Hence, since we express this in terms of binding energies, the energy released is given by \(4(x - x)\).
Statement-2: The unit of instantaneous power is watt-hour.
Select the correct option:
Statement-1 is true as instantaneous power is calculated using the dot product of velocity and force. Statement-2 is false because the unit of instantaneous power is watt (W), not watt-hour, which is a unit of energy.
The correct answer is B: 1/2 hr.
Explanation:
To determine the time it takes for car B to catch up with car A, we need to analyze the motion of both cars:
1. Given Data:
- Speed of car A: 60 km/h
- Speed of car B: 70 km/h
- Initial distance between the cars: 2.5 km
- Deceleration of car B: -20 km/h²
2. Initial Relative Speed:
The initial speed difference between car B and car A is:
V_{relative} = V_B - V_A = 10 km/h
3. Distance Covered:
The distance covered by car A in time t is:
d_A = V_A * t = 60t
The distance covered by car B can be calculated using:
d_B = V_B * t + \frac{1}{2} a t^2 = 70t - 10t^2
4. Catch-Up Condition:
For car B to catch up with car A:
70t - 10t^2 = 60t + 2.5
5. Rearranging the Equation:
After rearranging, we get:
10t^2 - 10t + 2.5 = 0
6. Solving Using the Quadratic Formula:
Using the quadratic formula:
t = \frac{-b ± √(b² - 4ac)}{2a} = \frac{4 ± 0}{8} = 0.5 hr
Thus, car B will catch up with car A in 0.5 hours or 30 minutes.
The correct answer is A: H = 0.4u²/g.
Explanation:
When a stone is thrown vertically upward from the top of a tower, it first moves up and then comes down to the ground. We can use the equations of motion to analyze this situation.
1. Given Data:
- Initial speed of the stone: u
- Final speed when reaching the ground: v = 3u
- Acceleration due to gravity: g \approx 10 \, \text{m/s}²
- Height of the tower: H
2. Equation of Motion:
We can use the equation:
v² = u² + 2gH
Substituting the values into the equation:
(3u)² = u² + 2gH
3. Expanding the Equation:
We get:
9u² = u² + 2gH
4. Rearranging:
Subtracting u² from both sides:
9u² - u² = 2gH
Which simplifies to:
8u² = 2gH
5. Solving for Height:
Dividing both sides by 2g gives:
H = \frac{8u²}{2g} = \frac{4u²}{g}
When using g \approx 10 \, \text{m/s}², we find:
H = 0.4u²
This shows the height of the tower in terms of the initial speed u and gravity g.
To find the point of collision of the two balls, we analyze their motions:
- **Ball A** is thrown upwards with an initial speed of 20 m/s. Its height at any time \( t \) can be calculated using the equation:
\( h_A = 20t - 5t^2 \)
- **Ball B** is thrown downwards from a height of 40 m with the same initial speed of 20 m/s. Its height at any time \( t \) is given by:
\( h_B = 40 - 20t - 5t^2 \)
To find when they collide, we set the heights equal:
\( 20t - 5t^2 = 40 - 20t - 5t^2 \)
This simplifies to:
\( 40t = 40 \)
Therefore, \( t = 1 \, \text{s} \).
Now substituting \( t = 1 \) into the equation for \( h_A \):
\( h_A = 20(1) - 5(1)^2 = 15 \, \text{m} \)
Thus, the balls collide at a height of **15 m above the ground**.
Statement-2: On increasing temperature, collision of electrons becomes more frequent and number of free electrons in the metallic conductor decreases.
- **Statement-1** is true because as temperature increases, the conductivity of metallic conductors decreases due to increased lattice vibrations causing more frequent collisions among electrons.
- **Statement-2** is also true regarding increased collisions; however, it incorrectly implies that the number of free electrons decreases. In metals, while the number of free electrons remains relatively constant, their mobility decreases with increased temperature due to these collisions.
Thus, while both statements are true, Statement-2 does not accurately explain Statement-1.
To find the initial velocity of the body thrown horizontally:
- **Vertical Motion:** Using the formula for free fall:
\( h = \frac{1}{2} g t^2 \)
Substituting values gives us:
\( 5 = \frac{1}{2} \cdot 10 \cdot t^2 \) ⟹ \( t = 1 \, \text{s} \)
- **Horizontal Motion:** The horizontal distance (R) is given by:
\( R = u \cdot t \)
Substituting values gives:
\( 10 = u \cdot 1 \) ⟹ \( u = 10 \, \text{m/s} \)
Therefore, the initial velocity of the body is **10 m/s**.
To find the focal lengths of the lenses used in the telescope:
- The formula for magnifying power \( M \) is given by:
\( M = \frac{f_o}{f_e} \)
Given \( M = 8 \), we have:
\( f_o = 8 f_e \)
- The distance between the lenses is:
\( f_o + f_e = 45 \, \text{cm} \)
Substituting gives us:
\( 8f_e + f_e = 45 \) ⟹ \( 9f_e = 45 \) ⟹ \( f_e = 5 \, \text{cm} \)
- Now substituting \( f_e \) back to find \( f_o \):
\( f_o = 8 \times 5 = 40 \, \text{cm} \)
Therefore, the focal lengths of the lenses are **40 cm** and **5 cm**.
To find the angle at which the ray emerges into air:
- The **critical angle** \( C \) is calculated as:
\( \sin C = \frac{\mu_a}{\mu_g} = \frac{1}{\frac{3}{2}} = \frac{2}{3} \) ⟹ \( C \approx 41.81° \)
- When the ray moves from glass to water, using Snell's Law, we find the angle of incidence at the water-air interface.
The calculations yield an angle of incidence in water \( i_w \) of approximately **48.59°**.
- Applying Snell's Law at the water-air interface gives:
\( \mu_w \sin(i_w) = \mu_a \sin(r) \)
Solving yields \( \sin(r) = 1 \), meaning the angle \( r \) at which the ray emerges into air is **90°**.
To find the angle of the prism, we use the formula:
\( A = \frac{D}{\mu - 1} \)
Where \( D \) is the minimum deviation and \( \mu \) is the refractive index.
Substituting the given values:
\( A = \frac{4.8}{1.62 - 1} = \frac{4.8}{0.62} \approx 7.74° \)
Thus, the angle of the prism is **7.74°**.
According to Gauss's law, the electric flux through a closed surface is proportional to the net charge enclosed within that surface. Since each dipole consists of equal and opposite charges, the total charge from eight dipoles is zero. Hence, the total electric flux out of the cube is:
\( \Phi_E = \frac{Q_{\text{enc}}}{\varepsilon_0} = \frac{0}{\varepsilon_0} = 0 \).
To find the power of the combination of lenses, we use the formula:
\( P = P_1 + P_2 \)
Where \( P_1 \) is the power of the convex lens and \( P_2 \) is the power of the concave lens.
The focal lengths are:
Convex Lens: \( f_1 = 25 \, \text{cm} = 0.25 \, \text{m} \) → \( P_1 = \frac{1}{0.25} = 4 \, \text{D} \)
Concave Lens: \( f_2 = -25 \, \text{cm} = -0.25 \, \text{m} \) → \( P_2 = \frac{1}{-0.25} = -4 \, \text{D} \)
Combining the powers gives:
\( P = 4 \, \text{D} + (-4 \, \text{D}) = 0 \, \text{D} \)
Thus, the total power of the combination is **Zero**.
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