3 Marks Question

Question 13 Marks

Is inversion temperature always double of the neutral temperature? Does the unit of temperature have an effect in deciding this question?

Answer

Inversion temperature is equal to 2 × neutral temperature + temperature of cold junction.
In case cold junction is at zero temperature the condition is satisfied.
yes, In case C this condition does not stand but in case ok this condition justifies.

View full question & answer→

Question 23 Marks

The specification on a heater coil is $250V, 500W.$ Calculate the resistance of the coil. What will be the resistance of a coil of $1000W$ to operate at the same voltage?

Answer

The power consumed by a coil of resistance $R$ when connected across a supply $v$ is $\text{P}=\frac{\text{v}^2}{\text{R}}$

The resistance of the heater coil is, therefore $\text{R}=\frac{\text{v}^2}{\text{P}}=\frac{(250)^2}{500}=125\Omega$

If, $\text{P}=1000\text{w}$

then, $\text{R}=\frac{\text{v}^2}{\text{P}}=\frac{(250)^2}{1000}=62.5\Omega$

View full question & answer→

Question 33 Marks

Do all the thermocouples have a neutral temperature?

Answer

As neutral temperature is The temperature of the hot junction of a thermocouple at which the electromotive force of the thermocouple attains its maximum value, when the cold junction is maintained at a constant temperature of 0°C.
So, answer is yes.

View full question & answer→

Question 43 Marks

An electric bulb, when connected across a power supply of $220V,$ consumes a power of $60W.$ If the supply drops to $180V,$ what will be the power consumed? If the supply is suddenly increased to $240V,$ what will be the power consumed?

Answer

$\text{E}=220\text{v}$
$\text{P}=60\text{w}$
$\text{R}=\frac{\text{V}^2}{\text{P}}=\frac{220\times220}{60}$
$=\frac{220\times11}{3}\Omega$

$\text{E}=180\text{v}$

$\text{P}=\frac{\text{V}^2}{\text{R}}$
$=\frac{180\times180\times3}{220\times11}=40.16\approx40\text{w}$

$\text{E}=240\text{v}$

$\text{P}=\frac{\text{V}^2}{\text{R}}$
$=\frac{240\times240\times3}{220\times11}=71.4\approx71\text{w}$

View full question & answer→

Question 53 Marks

A plate of area $10cm^2$ is to be electroplated with copper (density $9000kg/m^3$) to a thickness of 10 micrometres on both sides, using a cell of 12V. Calculate the energy spent by the cell in the process of deposition. If this energy is used to heat 100g of water, calculate the rise in the temperature of the water. ECE of copper = $3 \times 10^{-7}kg C^{-1}$ and specific heat capacity of water = $4200Jkg^{-1}$.

Answer

Surface area of the plate, $A=10 \mathrm{~cm}^2=10 \times 10^{-4} \mathrm{~m}^2$ Thickness of copper deposited, $\mathrm{t}=10 \mu \mathrm{~m}=10^{-5} \mathrm{~m}$
Density of copper $=9000 \mathrm{~kg} / \mathrm{m}^3$ Volume of copper deposited, $\mathrm{V}=\mathrm{A}(2 \mathrm{t}) \mathrm{V}=10 \times 10^{-4} \times 2 \times 10 \times 10^{-6}=2 \times 10^2 \times$ $10^{-10}=2 \times 10^{-8} \mathrm{~m}^3$ Mass of copper deposited, $\mathrm{m}=$ Volume $\times$ Density $=2 \times 10^{-8} \times 9000 \Rightarrow \mathrm{~m}=18 \times 10^{-5} \mathrm{~kg}$ Using the formula, $\mathrm{m}=\mathrm{ZQ}$. We get, $18 \times 10^{-5} \mathrm{~kg}=3 \times 10^{-7} \times \mathrm{Q} \Rightarrow \mathrm{Q}=6 \times 10^2 \mathrm{C}$
Energy spent by the cell = Work done by the cell
$\Rightarrow \mathrm{W}=\mathrm{VQ}$
$=2 \times 6 \times 10^2$
$=72 \times 10^2=7.2 \mathrm{~kJ}$
Let $\Delta \theta$ be the rise in temperature of water. When this energy is used to heat 100 g of water,
We have,
$72 \times 10^3$
$=100 \times 10^{-3} \times 4200 \times \Delta \theta$
$\Rightarrow \Delta \theta=17 \mathrm{~K}$

View full question & answer→

Question 63 Marks

A coil of resistance $100\Omega$ is connected across a battery of emf 6.0V. Assume that the heat developed in the coil is used to raise its temperature. If the heat capacity of the coil is 4.0J/K, how long will it take to raise the temperature of the coil by 15°C?

Answer

$\text{R}=100\Omega,$$\text{E}=6\text{v}$
Heat capacity of the coil $=4\text{J/k}$
$\Delta\text{T}=15^\circ$
Heat liberate $\Rightarrow\frac{\text{E}^2}{\text{Rt}}=4\text{J/K}\times15$
$\Rightarrow\frac{6\times6}{100}\times\text{t}=60$
$\Rightarrow\text{t}=166.67\text{sec}=2.8\text{min}$

View full question & answer→

Physics 3 Marks Question

Take a timed test