Question 2: A parallel plate capacitor has a capacitance of 10 mu F. If the distance between the plates is halved, what will happen to the capacitance?

Here are the solutions to the remaining questions:

Question 2: A parallel plate capacitor has a capacitance of $10 \mu F$. If the distance between the plates is halved, what will happen to the capacitance?

• Step 1: Recall the formula for capacitance. The capacitance ($C$) of a parallel plate capacitor is given by:

  $$C = \frac{\epsilon A}{d}$$

  where $\epsilon$ is the permittivity of the dielectric, $A$ is the area of the plates, and $d$ is the distance between the plates.

• Step 2: Analyze the effect of halving the distance. If the distance $d$ is halved, the new distance $d'$ becomes $\frac{d}{2}$. The new capacitance $C'$ will be:

  $$C' = \frac{\epsilon A}{d'} = \frac{\epsilon A}{(d/2)} = 2 \frac{\epsilon A}{d}$$

  Since $C = \frac{\epsilon A}{d}$, we can write:

  $$C' = 2C$$

  This means the new capacitance will be double the original capacitance.

The final answer is $\boxed{\text{A. It will double}}$.

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Question 3: The graphical representation of pressure law is always a straight line passing through the origin only if the temperature scale is

• Concept: The pressure law (also known as Gay-Lussac's Law or Amontons's Law) states that for a fixed mass of gas at constant volume, the pressure is directly proportional to its absolute temperature.
• Explanation: A graph of pressure versus temperature will be a straight line passing through the origin only when the temperature is measured on an absolute scale, where zero corresponds to absolute zero. The Kelvin scale is the absolute thermodynamic temperature scale.

The final answer is $\boxed{\text{B. Thermodynamic (Kelvin)}}$.

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Question 4: The density of water is $1 \text{ g/cm}^3$, while the density of ice is $0.9 \text{ g/cm}^3$. Calculate the change in volume when $90 \text{ g}$ of ice is completely melted.

• Step 1: Calculate the volume of ice. The formula for volume is $V = \frac{m}{\rho}$. Given mass of ice ($m_{ice}$) = $90 \text{ g}$ Given density of ice ($\rho_{ice}$) = $0.9 \text{ g/cm}^3$

  $$V_{ice} = \frac{90 \text{ g}}{0.9 \text{ g/cm}^3} = 100 \text{ cm}^3$$

• Step 2: Calculate the volume of water. When ice melts, its mass remains the same. So, mass of water ($m_{water}$) = $90 \text{ g}$. Given density of water ($\rho_{water}$) = $1 \text{ g/cm}^3$

  $$V_{water} = \frac{90 \text{ g}}{1 \text{ g/cm}^3} = 90 \text{ cm}^3$$

• **Step 3: