Answer: Option B. -> voltmeter
:
B
The device which measures the electric potential difference between two points is knownas a voltmeter. It is attached in parallel to the electric component through which we need to find the voltage drop.

Answer: Option B. -> 1 Ω
:
B
Two parallel combination of resistors:
When going from A to B, the firsttwo resistors are in parallel to each other and so are the next two. Therefore
$R_{parallel}$ =$\frac{1\times 1}{1+1}$= $\frac{1}{2}\Omega$ for both of them.
Now these equivalent resistors are in series with each other
Therefore, net equivalent resistance is:
$R=0.5+0.5=1\Omega$

Answer: Option C. -> 3
:
C
All electric cells have two terminals, positive and negative terminal. If two cells are connected with opposite polarities as shown, the net potential difference across them is the difference of the individual EMFs of the cells.. Hence, there is no net potential difference in the case the cells are identical. Therefore, current will not flow and the bulb will not glow.

Answer: Option C. ->  H∝I2
:
C
According to Joule's law, theheat produced in a resistor is;
(i) directly proportional to the square of current for a given resistance.
(ii) directly proportional to resistance for a given current.
(iii) directly proportional to the time for which the current flows through the resistor.
That is $H=I^2\times R\times t$
Hence$H\propto I^2$ is the right relation.

Answer: Option B. -> False
:
B
In alloys,the increase in temperature results in an increase of its resistance, but unlike pure metals, the increase is relatively small and irregular.As the temperature increases, the molecules in the atom vibrate with more frequency, they collide with each other more frequently which makes it tough for the movement of free electrons. Thus the resistance increase with increasing temperature in the metal.

Answer: Option A. -> True
:
A
Resistivity depends on temperature. Withthe increasein temperature, the random motion of electrons increases. Asa result, the number of collisions of electrons with the positive ions increases. Hence, the resistivity of all pure metals increases with the rise in temperature and decreases with the decrease in temperature.

Answer: Option A. -> R
:
A
The connection of wires is as shown below.
We have to calculate equivalent resistance across the opposite corners i.e. across Aand Dor Band C. If we take two corners B and C, then we can clearly see that the resistances of AB and AC, and CD and BD are in series.
So the equivalent resistance across CAB is $2R$ ( CA and AB are in series) and CDB is also $2R$ ( CD and BD are in series).
The two equivalent resistances across CAB and CDB are in parallel. Hence, the equivalent resistance across the two corners i.e. B and Cis $\frac{1}{R_{eq}}=\frac{1}{2R}+\frac{1}{2R}$.
On solving we get, $R_{eq}=R$. Similarly, equivalent resistance across Aand Dwill also be$R$.
Hence the equivalent resistance across the opposite corners is $R$.

Answer: Option D. -> 25
:
D
Given the wire iscut into 5 equal parts.
Let the initial resistance of the wire be $R$
where $R=\rho \frac{l}{A}\to R\propto l\left(lengthofthewire\right)$
then, resistance of each one of the five parts $=\left(\frac{R}{5}\right)$
Equivalent resistance (R')of resistorsconnected in parallel is given by$\frac{1}{R^{\prime }}=\frac{1}{R_1}+\frac{1}{R_2}..\frac{1}{R_n}$
Equivalent resistance (R') of five parts connected in parallel is given by:
$\frac{1}{R^{\prime }}=\frac{1}{\left(\frac{R}{5}\right)}+\frac{1}{\left(\frac{R}{5}\right)}+\frac{1}{\left(\frac{R}{5}\right)}+\frac{1}{\left(\frac{R}{5}\right)}+\frac{1}{\left(\frac{R}{5}\right)}{R}^{\prime }=\frac{R}{25}$
Therefore, $\frac{R}{R^{{}^{\prime }}}=25$

Answer: Option B. -> 4:1
:
B
Volume of a solid is the product of length and cross-sectional area. $V=AL$
Since volume is same in both the cases, the length of second wire will be half of the length of 1st wire.
Accordingly, let length of 1st wire = L
area of 1st wire= A = 1$m{m}^2$
So resistance $R=\rho \frac{L}{A}=\rho \frac{L}{1}=\rho L$
Length of the second wire $=\frac{L}{2}$ and
area of 2ndwire$A=2m{m}^2$
So resistance of 2nd wire $R^{\prime }=\rho \frac{\frac{L}{2}}{A}$
$R^{\prime }=\rho \frac{L}{2\times 2}$
$R^{\prime }=\frac{1}{4}\left(\rho L\right)$
$R^{\prime }=\frac{1}{4}\left(R\right)$
$\frac{R}{R^{\prime }}=\frac{4}{1}$
R : R' = 4 : 1
we get the resistance of the 1st wire is 4 times the value of resistance of the 2nd wire.

Answer: Option B. -> electric power
:
B
The physical quantity that determines the rate at which energy is delivered by an electric current is electricpower. The SI unit of electricpower is watt (W). 1 W electrical poweris equivalent to 1 Joule ofelectricalenergy consumed in 1 second time. It is mathematically expressed as $P=\frac{E}{t}$ where $E$ is the electrical energyand $t$ is time.