1. state the physical quantities whose dimensions are given below:
(a) ( ml^2t^{-1} );
(b) ( m^{-1}l^3t^{-2} );
(c) ( ml^2t^{-4} ).
3 marks

2. state three uses of polar satellites.
3 marks

3. mention three uses of ferromagnetic materials.
3 marks

4. a spiral spring of length 20 cm was compressed to a length of 15 cm by a contact force, f. if the workdone in compressing the spring is 10 j. calculate the magnitude of f.
3 marks

5. a projectile hit its projection plane in 5 seconds after travelling with a speed of ( 80 , ext{ms}^{-1} ). calculate its angle of projection. ( g = 10 , ext{ms}^{-2} )
3 marks

6. fig. 1.0 represents energy bands for three different electrical materials. if a and b represents the valence and conduction bands respectively, identify the materials x, y and z.

image of fig. 1.0: three boxes x, y, z. x has a (valence) and b (conduction) with no gap. y has a and b overlapping (energy gap 0?). z has a and b with an energy gap. fig. 1.0 is below the boxes.
3 marks

7. the expression ( lambda = \frac{h}{p} ) represents the de-broglie’s equation. what does ( lambda ), ( h ), and ( p ) represent?
3 marks

Question

1. state the physical quantities whose dimensions are given below:
(a) ( ml^2t^{-1} );
(b) ( m^{-1}l^3t^{-2} );
(c) ( ml^2t^{-4} ).
3 marks

2. state three uses of polar satellites.
3 marks

3. mention three uses of ferromagnetic materials.
3 marks

4. a spiral spring of length 20 cm was compressed to a length of 15 cm by a contact force, f. if the workdone in compressing the spring is 10 j. calculate the magnitude of f.
3 marks

5. a projectile hit its projection plane in 5 seconds after travelling with a speed of ( 80 , 	ext{ms}^{-1} ). calculate its angle of projection. ( g = 10 , 	ext{ms}^{-2} )
3 marks

6. fig. 1.0 represents energy bands for three different electrical materials. if a and b represents the valence and conduction bands respectively, identify the materials x, y and z.

image of fig. 1.0: three boxes x, y, z. x has a (valence) and b (conduction) with no gap. y has a and b overlapping (energy gap 0?). z has a and b with an energy gap. fig. 1.0 is below the boxes.
3 marks

7. the expression ( lambda = \frac{h}{p} ) represents the de-broglie’s equation. what does ( lambda ), ( h ), and ( p ) represent?
3 marks

Accepted Answer

### Question 4 Solution (Step-by-Step Format)
# Explanation:
## Step1: Recall work on spring formula
The work done on a spring is given by $ W = \frac{1}{2}kx^2 $, and also $ F = kx $ (Hooke's law). We can relate $ W $ and $ F $ by substituting $ k=\frac{F}{x} $ into the work formula.
First, find the compression $ x $: initial length $ L_0 = 20\space cm = 0.2\space m $, final length $ L = 15\space cm = 0.15\space m $, so $ x = L_0 - L = 0.2 - 0.15 = 0.05\space m $.

## Step2: Relate work and force
From $ W=\frac{1}{2}kx^2 $ and $ F = kx $, we can express $ W=\frac{1}{2}Fx $ (since $ kx = F $, so $ k=\frac{F}{x} $, substitute into $ W=\frac{1}{2}kx^2 $ gives $ W=\frac{1}{2}(\frac{F}{x})x^2=\frac{1}{2}Fx $).
Now solve for $ F $: $ F=\frac{2W}{x} $.

## Step3: Substitute values
Given $ W = 10\space J $, $ x = 0.05\space m $, so $ F=\frac{2\times10}{0.05}=\frac{20}{0.05}=400\space N $.

# Answer:
The magnitude of $ F $ is $ 400\space N $.

### Question 5 Solution (Step-by-Step Format)
# Explanation:
## Step1: Recall projectile motion time of flight
For a projectile, time of flight $ T=\frac{2u\sin\theta}{g} $, where $ u $ is initial velocity, $ \theta $ is angle of projection, $ g $ is acceleration due to gravity.

## Step2: Substitute values and solve for $ \sin\theta $
Given $ T = 5\space s $, $ u = 80\space m/s $, $ g = 10\space m/s^2 $.
Substitute into $ T=\frac{2u\sin\theta}{g} $: $ 5=\frac{2\times80\times\sin\theta}{10} $.
Simplify: $ 5=\frac{160\sin\theta}{10}=16\sin\theta $.
Then $ \sin\theta=\frac{5}{16} $? Wait, no, wait: $ 5=\frac{2\times80\times\sin\theta}{10} $ → $ 5\times10 = 160\sin\theta $ → $ 50 = 160\sin\theta $ → $ \sin\theta=\frac{50}{160}=\frac{5}{16} $? Wait, that can't be. Wait, no, time of flight formula: $ T = \frac{2u_y}{g} $, where $ u_y = u\sin\theta $. So $ T=\frac{2u\sin\theta}{g} $. So $ 5=\frac{2\times80\times\sin\theta}{10} $ → $ 5\times10 = 160\sin\theta $ → $ 50 = 160\sin\theta $ → $ \sin\theta=\frac{50}{160}=\frac{5}{16}\approx0.3125 $. Then $ \theta=\arcsin(0.3125)\approx18.21^\circ $. Wait, but maybe I made a mistake. Wait, no, let's check again. Wait, $ u = 80\space m/s $, $ T = 5\space s $, $ g = 10\space m/s^2 $. So $ T=\frac{2u\sin\theta}{g} $ → $ \sin\theta=\frac{Tg}{2u}=\frac{5\times10}{2\times80}=\frac{50}{160}=\frac{5}{16}\approx0.3125 $, so $ \theta=\arcsin(0.3125)\approx18.2^\circ $. Alternatively, maybe I messed up the formula. Wait, time of flight is the time to go up and come back, so $ T = \frac{2u\sin\theta}{g} $, that's correct. So solving for $ \theta $:

## Step3: Calculate $ \theta $
$ \sin\theta=\frac{Tg}{2u}=\frac{5\times10}{2\times80}=\frac{50}{160}=\frac{5}{16}\approx0.3125 $
$ \theta=\arcsin(0.3125)\approx18.2^\circ $ (or in exact terms, but usually we can calculate it. Wait, maybe I made a mistake in the problem. Wait, the projectile hits its projection plane in 5 seconds, so time of flight is 5s. So $ T = 5 = \frac{2u\sin\theta}{g} $, so $ \sin\theta=\frac{5\times10}{2\times80}=\frac{50}{160}=\frac{5}{16} $, so $ \theta=\arcsin(\frac{5}{16})\approx18.2^\circ $.

# Answer:
The angle of projection is approximately $ \boldsymbol{18.2^\circ} $ (or using exact fraction, but decimal is more common).

### Question 7 Solution (Answer-Explanation Format)
# Brief Explanations:
In de - Broglie's equation $ \lambda=\frac{h}{p} $:
- $ \lambda $ (lambda) represents the de - Broglie wavelength of a particle. It is the wavelength associated with a moving particle (like an electron, proton, etc.) and is a measure of the wave - like nature of the particle.
- $ h $ is Planck's constant. It has a value of approximately $ 6.626\times10^{-34}\space J\cdot s $ and is a fundamental constant in quantum mechanics that relates the energy of a photon to its frequency ($ E = h
u $) and also appears in the de - Broglie relation to relate the wavelength of a particle to its momentum.
- $ p $ represents the momentum of the particle. For a non - relativistic particle, momentum $ p = mv $, where $ m $ is the mass of the particle and $ v $ is its velocity.

# Answer:
- $ \lambda $: de - Broglie wavelength of the particle.
- $ h $: Planck's constant.
- $ p $: Momentum of the particle.

### Question 1 Solutions (Answer-Explanation Format)
#### Part (a)
# Brief Explanations:
The dimensional formula $ ML^{2}T^{-1} $ corresponds to angular momentum. The formula for angular momentum $ L = r\times p $, where $ r $ (position vector) has dimensions $ L $ and $ p $ (momentum) has dimensions $ MLT^{-1} $. The cross - product of $ r $ and $ p $ gives dimensions $ L\times MLT^{-1}=ML^{2}T^{-1} $. It can also correspond to the moment of momentum.
# Answer:
Physical quantity with dimension $ ML^{2}T^{-1} $ is angular momentum (or moment of momentum).

#### Part (b)
# Brief Explanations:
The dimensional formula $ M^{1}L^{3}T^{-2} $ corresponds to gravitational constant $ G $. From Newton's law of gravitation $ F = G\frac{m_1m_2}{r^{2}} $, so $ G=\frac{F r^{2}}{m_1m_2} $. The dimensions of force $ F $ are $ MLT^{-2} $, $ r^{2} $ is $ L^{2} $, $ m_1m_2 $ is $ M^{2} $. So dimensions of $ G $ are $ \frac{MLT^{-2}\times L^{2}}{M^{2}}=M^{-1}L^{3}T^{-2} $? Wait, wait, the given dimension is $ M^{1}L^{3}T^{-2} $. Wait, maybe it's the dimension of gravitational potential energy per unit mass? No, wait, let's re - check. Wait, the formula for gravitational potential $ V = -\frac{GM}{r} $, dimensions of $ V $ are $ \frac{ML^{3}T^{-2}\times M}{L}=M^{2}L^{2}T^{-2} $? No, I think I made a mistake. Wait, the given dimension is $ M^{1}L^{3}T^{-2} $. Another quantity: the dimension of torque is $ ML^{2}T^{-2} $, no. Wait, the dimension of gravitational constant $ G $ is $ M^{-1}L^{3}T^{-2} $, but the given is $ M^{1}L^{3}T^{-2} $. Wait, maybe it's the dimension of the product of mass and gravitational constant times something? No, perhaps the question has a typo, but assuming it's $ M^{-1}L^{3}T^{-2} $, it's $ G $. But if we take the given as $ M^{1}L^{3}T^{-2} $, maybe it's the dimension of a quantity like $ \frac{Force\times r^{3}}{m} $, but more likely, the intended quantity is gravitational constant (maybe a typo in the exponent of $ M $). So we'll go with gravitational constant (correcting the $ M $ exponent to - 1, but as per the given $ M^{1}L^{3}T^{-2} $, maybe it's a different quantity, but the most probable with $ L^{3}T^{-2} $ and $ M $ is related to gravitation.
# Answer:
Physical quantity with dimension $ M^{1}L^{3}T^{-2} $ (assuming a possible typo, and the correct $ M^{-1}L^{3}T^{-2} $) is gravitational constant $ G $.

#### Part (c)
# Brief Explanations:
The dimensional formula $ ML^{2}T^{-4} $ corresponds to power of a lens? No, power of lens is $ L^{-1} $. Wait, the dimension of intensity of radiation (like light) or the dimension of a quantity related to energy per unit area per unit time? Wait, the formula for intensity $ I=\frac{Power}{Area} $, power has dimensions $ ML^{2}T^{-3} $, area has $ L^{2} $, so $ I $ has $ MT^{-3} $. No. Wait, the dimension of a quantity like the product of mass, length squared, and inverse time to the fourth. Wait, the dimension of force is $ MLT^{-2} $, pressure is $ ML^{-1}T^{-2} $, energy is $ ML^{2}T^{-2} $, power is $ ML^{2}T^{-3} $. Wait, the given is $ ML^{2}T^{-4} $. Ah, the dimension of a quantity like the rate of change of power? No, more likely, the dimension of a quantity related to the second - derivative of energy with respect to time? No, let's think of the dimension of a quantity like $ \frac{Energy}{Time^{2}} $, energy is $ ML^{2}T^{-2} $, so $ \frac{ML^{2}T^{-2}}{T^{2}}=ML^{2}T^{-4} $. But a more common quantity: the dimension of the coefficient of thermal conductivity? No, thermal conductivity has dimensions $ MLT^{-3}\Theta^{-1} $. Wait, the dimension of a quantity like the power per unit time? No, power is $ ML^{2}T^{-3} $. Wait, maybe it's the dimension of a quantity related to the force - time - length product? No. Wait, the correct quantity with dimension $ ML^{2}T^{-4} $ is the dimension of a quantity like the “stiffness” in a different context, but more likely, it's the dimension of the quantity related to the second - order time derivative of angular momentum? No. Wait, the dimension of a quantity like $ \frac{Torque}{Time^{2}} $, torque is $ ML^{2}T^{-2} $, so $ \frac{ML^{2}T^{-2}}{T^{2}}=ML^{2}T^{-4} $. But the most probable physical quantity with dimension $ ML^{2}T^{-4} $ is the dimension of a quantity related to the rate of change of power, but actually, the correct physical quantity is the **dimension of a quantity like the “moment of a force - time” related quantity**? No, I think the intended quantity is the **dimension of a quantity related to the power - time derivative**, but more accurately, the physical quantity with dimension $ ML^{2}T^{-4} $ is the **dimension of a quantity like the “force - length - time^{-2}”**? Wait, no, let's re - calculate. The dimension of energy is $ ML^{2}T^{-2} $, so if we take the second derivative of energy with respect to time, $ \frac{d^{2}E}{dt^{2}} $, its dimension is $ \frac{ML^{2}T^{-2}}{T^{2}}=ML^{2}T^{-4} $. But a more common physical quantity: the dimension of the **intensity of a wave**? No, intensity is $ MT^{-3} $. Wait, maybe the question has a typo, but the most probable physical quantity with dimension $ ML^{2}T^{-4} $ is related to the **rate of change of power** (power is $ ML^{2}T^{-3} $, so rate of change of power is $ ML^{2}T^{-4} $).
# Answer:
Physical quantity with dimension $ ML^{2}T^{-4} $ is the rate of change of power (or a quantity with similar dimensional derivation).

### Question 2 Solution (Answer-Explanation Format)
# Brief Explanations:
1. **Weather Forecasting**: Polar satellites orbit at a low altitude (around 800 - 1000 km) and have a polar orbit (pass over the Earth's poles). They can monitor weather patterns, cloud cover, and atmospheric conditions over the entire Earth's surface, including polar regions, which is crucial for accurate weather prediction.
2. **Remote Sensing of Earth's Resources**: They are used to study and map Earth's resources such as forests, water bodies, mineral deposits, and agricultural land. The polar orbit allows them to cover the entire Earth's surface over a period of time, providing comprehensive data for resource management.
3. **Military and Strategic Surveillance**: Due to their ability to cover the entire Earth, including polar regions, they can be used for military surveillance, monitoring of enemy activities, and strategic reconnaissance. They can also be used for monitoring border areas and coastal regions.

# Answer:
Three uses of polar satellites are:
1. Weather forecasting (monitoring weather patterns and atmospheric conditions).
2. Remote sensing of Earth's resources (mapping forests, water bodies, mineral deposits, etc.).
3. Military and strategic surveillance (monitoring enemy activities and border areas).

### Question 3 Solution (Answer-Explanation Format)
# Brief Explanations:
1. **Electromagnets**: Ferromagnetic materials like iron, nickel, and their alloys are used in the cores of electromagnets. When a current passes through the coil around the ferromagnetic core, it gets magnetized strongly, producing a strong magnetic field. Electromagnets are used in devices like cranes (to lift heavy magnetic materials), electric bells, and relays.
2. **Transformers**: The cores of transformers are made of ferromagnetic materials (usually silicon - steel laminations). These materials have high magnetic permeability, which helps in confining the magnetic flux within the core, thereby increasing the efficiency of the transformer in transferring electrical energy between primary and secondary coils.
3. **Permanent Magnets**: Some ferromagnetic materials can be magnetized permanently to form permanent magnets. These are used in devices like speakers, microphones, magnetic compasses (in some cases), and in magnetic storage devices like hard disk drives (the read/write heads or the magnetic media can use ferromagnetic materials).

# Answer:
Three uses of ferromagnetic materials are:
1. Used in the cores of electromagnets (e.g., in cranes, electric bells).
2. Used in transformer cores (to enhance magnetic flux and efficiency).
3. Used to make permanent magnets (e.g., in speakers, hard disk drives).

### Question 6 Solution (Answer-Explanation Format)
# Brief Explanations:
- **Material X**: In material X, the valence band (A) and conduction band (B) overlap. This is the characteristic of a conductor. In conductors, there is no energy gap between the valence and conduction bands, so electrons can easily move from the valence band to the conduction band, allowing electrical conduction.
- **Material Y**: In material Y, the valence band (A) and conduction band (B) are very close to each other (the energy gap is zero or almost zero). This is the characteristic of a semiconductor at high temperatures or an intrinsic semiconductor at room temperature? No, wait, in a semiconductor, there is a small energy gap, but in material Y, the bands are overlapping? Wait, no, looking at the figure, for Y, the bands A and B are overlapping (no energy gap). Wait, no, the figure shows for X: A and B overlap; for Y: A and B are very close (maybe a small energy gap, but in the figure, it's shown as overlapping? Wait,

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QUESTION IMAGE

Question

Explanation:

Question 4 Solution (Step-by-Step Format)

Step1: Recall work on spring formula

Step2: Relate work and force

Step3: Substitute values

Step1: Recall projectile motion time of flight

Step2: Substitute values and solve for \( \sin\theta \)

Step3: Calculate \( \theta \)

Answer:

Question 5 Solution (Step-by-Step Format)