Calculate the length of the wire required to make a 50ohms coil from a wire of resistivity 1uohm metre and diameter 0.2mm. An electron occupies a certain energy level before to ground state. If the Uncertainty in determining the time is 6.3×10⁻³ secs, Calculate the Uncertainty in determining it's energy level at that level. (h=1.055×10⁻³⁴Js) A.) E≥1.60×10⁻³² B.) E≥1.04×10⁻³² C.) E≥1.68×10⁻³² D.) E≥1.82×10⁻³² In gas experiment, pressure is plotted against reciprocal of volume at constant pressure. Calculate the unit of slope of the resulting curve. The force that causes the steering wheel of a car to rotate is... A.) Normal force. B.) Friction C.) Weight. D.) Couple Gauge pressure for a manometer is described as... A.) Difference between atmospheric pressure and absolute pressure. B.) Difference between atmospheric pressure and partial pressure. C.) Product of partial pressure and atmospheric pressure. D.) Sum of absolute pressure and gas pressure. Oxygen gas is heated at constant volume to a temperature of 27°C. If mass is 16g, what is the heat needed by the gas. (Molar Heat capacity=0.85mol-¹K-¹) A.) 1200. B.) 12. C.) 1.2. D.) 120

Physics

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11) Calculate the length of the wire required to make a 50ohms coil from a wire of resistivity 1uohm metre and diameter 0.2mm.

  1. Convert the given values to SI units.

    • Resistance (RR) = 50 ohms50 \text{ ohms}
    • Resistivity (ρ\rho) = 1 uohm metre=1×10−6 ohm⋅m1 \text{ uohm metre} = 1 \times 10^{-6} \text{ ohm} \cdot \text{m}
    • Diameter (dd) = 0.2 mm=0.2×10−3 m0.2 \text{ mm} = 0.2 \times 10^{-3} \text{ m}
    • Radius (rr) = d2=0.2×10−3 m2=0.1×10−3 m\frac{d}{2} = \frac{0.2 \times 10^{-3} \text{ m}}{2} = 0.1 \times 10^{-3} \text{ m}
  2. Calculate the cross-sectional area (AA) of the wire.

    A=πr2A = \pi r^2 A=π(0.1×10−3 m)2A = \pi (0.1 \times 10^{-3} \text{ m})^2 A=π(0.01×10−6 m2)A = \pi (0.01 \times 10^{-6} \text{ m}^2) A=1×10−8π m2A = 1 \times 10^{-8} \pi \text{ m}^2
  3. Use the formula for resistance to find the length (LL).

    The formula for resistance is R=ρLAR = \rho \frac{L}{A}. Rearranging for LL:

    L=R⋅AρL = \frac{R \cdot A}{\rho} L=50 ohms×(1×10−8π m2)1×10−6 ohm⋅mL = \frac{50 \text{ ohms} \times (1 \times 10^{-8} \pi \text{ m}^2)}{1 \times 10^{-6} \text{ ohm} \cdot \text{m}} L=50π×10−81×10−6 mL = \frac{50 \pi \times 10^{-8}}{1 \times 10^{-6}} \text{ m} L=50π×10−2 mL = 50 \pi \times 10^{-2} \text{ m} L=0.50π mL = 0.50 \pi \text{ m}

    Using π≈3.14159\pi \approx 3.14159:

    L≈0.50×3.14159 mL \approx 0.50 \times 3.14159 \text{ m} L≈1.5708 mL \approx 1.5708 \text{ m}

    The length of the wire required is 1.57 m\boxed{\text{1.57 m}}.

    12) An electron occupies a certain energy level before to ground state. If the Uncertainty in determining the time is 6.3×10^-3 secs, Calculate the Uncertainty in determining it's energy level at that level. (h=1.055×10^-34Js)

  4. Identify the given values and the relevant principle.

    • Uncertainty in time (Δt\Delta t) = 6.3×10−3 s6.3 \times 10^{-3} \text{ s}
    • The given value h=1.055×10−34 J⋅sh = 1.055 \times 10^{-34} \text{ J} \cdot \text{s} is the reduced Planck's constant (ℏ\hbar).
    • Heisenberg's Uncertainty Principle for energy and time states: ΔEΔt≥ℏ2\Delta E \Delta t \ge \frac{\hbar}{2} However, in many contexts, especially for multiple-choice questions, a simplified form ΔEΔt≥ℏ\Delta E \Delta t \ge \hbar is used when ℏ\hbar is provided directly. We will use this form to match the options.
  5. Calculate the minimum uncertainty in energy (ΔE\Delta E).

    ΔE≥ℏΔt\Delta E \ge \frac{\hbar}{\Delta t} ΔE≥1.055×10−34 J⋅s6.3×10−3 s\Delta E \ge \frac{1.055 \times 10^{-34} \text{ J} \cdot \text{s}}{6.3 \times 10^{-3} \text{ s}} ΔE≥1.0556.3×10−34−(−3) J\Delta E \ge \frac{1.055}{6.3} \times 10^{-34 - (-3)} \text{ J} ΔE≥0.16746×10−31 J\Delta E \ge 0.16746 \times 10^{-31} \text{ J} ΔE≥1.6746×10−32 J\Delta E \ge 1.6746 \times 10^{-32} \text{ J}

    This value is approximately 1.68×10−32 J1.68 \times 10^{-32} \text{ J}.

    The correct option is (C).

    • C) E≥1.68×10^-32

    13) In gas experiment, pressure is plotted against reciprocal of volume at constant pressure. Calculate the unit of slope of the resulting curve.

  6. Identify the relationship between pressure and volume at constant temperature.

    According to Boyle's Law, for a fixed amount of gas at constant temperature, the pressure (PP) is inversely proportional to the volume (VV).

    P∝1VP \propto \frac{1}{V}

    This can be written as:

    P=k(1V)P = k \left(\frac{1}{V}\right)

    where kk is a constant.

  7. Determine the slope of the curve.

    If pressure (PP) is plotted on the y-axis and reciprocal of volume (1/V1/V) is plotted on the x-axis, the equation is in the form y=mxy = mx, where mm is the slope.
    Thus, the slope (mm) is equal to the constant kk. From Boyle's Law, PV=kPV = k. Therefore, the slope is PVPV.

  8. Calculate the unit of the slope.

    The unit of pressure (PP) is Pascal (Pa\text{Pa}) or Newton per square meter (N/m2\text{N/m}^2).
    The unit of volume (VV) is cubic meter (m3\text{m}^3).
    The unit of the slope is the unit of P×VP \times V:

    Unit of slope=Unit of P×Unit of V\text{Unit of slope} = \text{Unit of } P \times \text{Unit of } V Unit of slope=Pa⋅m3\text{Unit of slope} = \text{Pa} \cdot \text{m}^3 Unit of slope=(Nm2)⋅m3\text{Unit of slope} = \left(\frac{\text{N}}{\text{m}^2}\right) \cdot \text{m}^3 Unit of slope=N⋅m\text{Unit of slope} = \text{N} \cdot \text{m}

    A Newton-meter (N⋅m\text{N} \cdot \text{m}) is the unit of work or energy, which is the Joule (J\text{J}).

    The unit of the slope of the resulting curve is Joule (J)\boxed{\text{Joule (J)}}.

    14) The force that causes the steering wheel of a car to rotate is...

    The rotation of a steering wheel is caused by a couple. A couple consists of two forces of equal magnitude, acting in opposite directions, and separated by a perpendicular distance. This arrangement produces a pure rotational effect (torque) without causing any translational motion.

    The correct option is (D).

    • D) Couple

    15) Gauge pressure for a manometer is described as...

    Gauge pressure is the pressure measured relative to the ambient atmospheric pressure. It is defined as the difference between the absolute pressure of a fluid and the local atmospheric pressure.

    Pgauge=Pabsolute−PatmosphericP_{\text{gauge}} = P_{\text{absolute}} - P_{\text{atmospheric}}

    Let's analyze the given options:

    • A) Difference between atmospheric pressure and absolute pressure. This option describes Patmospheric−PabsoluteP_{\text{atmospheric}} - P_{\text{absolute}}, which is the negative of the standard gauge pressure. However, it is the only option that correctly identifies gauge pressure as a difference between atmospheric and absolute pressure. The other options are fundamentally incorrect descriptions of gauge pressure.

    The correct option is (A).

    • A) Difference between atmospheric pressure and absolute pressure.

    **16) Oxygen gas is heated at constant volume to a temperature of 27°C. If mass is 16g, what is the heat needed by the gas. (Molar Heat capacity=0.85mol-¹K-¹) **

  9. Identify the given values and make necessary assumptions.

    • Gas: Oxygen (O2\text{O}_2)
    • Process: Heated at constant volume.
    • Mass (mm) = 16 g16 \text{ g}
    • Molar heat capacity at constant volume (CvC_v) = 0.85 J⋅mol−1K−10.85 \text{ J} \cdot \text{mol}^{-1}\text{K}^{-1}
    • Temperature: "to a temperature of 27∘C27^\circ\text{C}". Since no initial temperature is given, we assume this refers to the change in temperature, ΔT=27 K\Delta T = 27 \text{ K}. (A change of 27∘C27^\circ\text{C} is equivalent to a change of 27 K27 \text{ K}).
  10. Calculate the number of moles (nn) of oxygen gas.

    • Molar mass of O2=2×16.00 g/mol=32.00 g/mol\text{O}_2 = 2 \times 16.00 \text{ g/mol} = 32.00 \text{ g/mol}
    n=massmolar mass=16 g32 g/mol=0.5 moln = \frac{\text{mass}}{\text{molar mass}} = \frac{16 \text{ g}}{32 \text{ g/mol}} = 0.5 \text{ mol}
  11. Calculate the heat needed (QQ) using the formula for heat transfer at constant volume.

    Q=nCvΔTQ = n C_v \Delta T Q=0.5 mol×0.85 J⋅mol−1K−1×27 KQ = 0.5 \text{ mol} \times 0.85 \text{ J} \cdot \text{mol}^{-1}\text{K}^{-1} \times 27 \text{ K} Q=0.5×0.85×27 JQ = 0.5 \times 0.85 \times 27 \text{ J} Q=0.425×27 JQ = 0.425 \times 27 \text{ J} Q=11.475 JQ = 11.475 \text{ J}
  12. Compare the calculated value to the given options.

    The calculated value 11.475 J11.475 \text{ J} is closest to 12 J12 \text{ J}.

    The correct option is (B).

    • B) 12
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