Science M.C.Q-[Phy-1M]

4. One-fourth.

Explanation:

$\text{R}=\rho\frac{\text{I}}{\text{A}}$

When the diameter is doubled, d = 2d

Radius, r' = 2r

Area of cross section, $\text{A}'=\pi\text{r}'^2=\pi(2\text{r})=4\pi\text{r}^2=4\text{A}$

The area of cross-section will increase by four times.

Then the new resistance, $\text{R}'=\frac{\rho\text{I}}{\text{A}'}$

$\text{R}'=\frac{\rho\text{I}}{4\text{A}}$

$\text{R}'=\frac{\text{R}}{4}$

Thus, the resistance will get reduced by four times.

2. Half.

Explanation:

We know that the resistance of a conductor is given by:

$\text{R}=\rho\frac{\text{I}}{\text{A}}$

where $\rho=\text{resistivity}$

l = length of the conductor

A = area of the cross-section of the conductor

Let the new resistance be R' when the area of cross-section of the conductor is doubled

$\text{R}'=\rho\frac{\text{I}}{2\text{A}}$

2. 3

Explanation:

The number of coulombs passing through the resistor is the current passing through it.

$\text{Current}=\frac{\text{Voltage}}{\text{Resistance}}$

$\text{I}=\frac{\text{V}}{\text{R}}$

$\text{I}=\frac{12}{4}$

$\text{I}={3\text{A}}$

Thus, when a $4 Ω$ resistor is connected across the terminals of a 12V battery, the number of coulombs passing through the resistor per second will be 3.

4. A semiconductor.

Explanation:

The resistivity of a certain material is $0.6 Ω\ \text{m}. $The material is most likely to be a semiconductor because it has moderate resistivity.

4. $30 Ω$

Explanation:

Resistance, $\text{R}=\frac{\rho\text{I}}{\text{A}}$

Length, I = 300m

Cross section area, A = 1.0mm² = 10^-6 m²

Resistivity, $\rho=1.0\times10^{-7}\Omega\ \text{m}$

Resistance, $\text{R}=\frac{10^{-7}\times300}{10^{-6}}$

$\text{R}=30\Omega$

3. 2A.

Explanation:

If the p.d. across a $3\Omega$ resistor is 6V, the current flowing in the resistor will be 2A as current (I) is given by the equation, $\text{I}=\frac{\text{V} (\text{Voltage})}{\text{R}(\text{Resistance})}.$

or $\text{I}=\frac{\text{V}}{\text{R}}$

$\text{I}=6\ \frac{\text{V}}{3}\ Ω$

$\text{I} = 2\text{A.}$

4. $7\frac{1}{2}\Omega$

Explanation:

The resistors of 6I and 2I are connected in parallel.

$\therefore\frac{1}{\text{R}}=\frac{1}{\text{R}_1}+\frac{1}{\text{R}_2}$ Here,

$\text{R}_1=6\Omega$

$\text{R}_2=2\Omega\ \frac{1}{\ \text{R}}=\frac{1}{6}+\frac{1}{2}\frac{1}{\text{R}}=\frac{4}{6}\ \text{R}=\frac{6}{4}$

This arrangement is further connected in series with the 6I resistor.

$\therefore$ Net resistance $=\frac{6}{4}+6=7\frac{1}{2}\Omega$

1. $9\Omega$

Explanation:

The two resistors of 6I are connected in parallel with each other. So, their net resistance 3I is connected in series with a resistance of 6I. So, the net resistance of the complete arrangement is $9 Ω.$

4. Double.

Explanation:

As we know from Ohm’s law:

Voltage = Current × Resistance

V = IR

If the voltage is constant, the resistance of the circuit is halved. That is, it becomes $\frac{\text{R}}{2}$

Current, $\text{I}=\frac{\text{V}}{\text{R}}$

$\text{I}=\frac{\text{V}}{\text{R}}$

$\text{I}=\frac{\frac{\text{V}}{\text{R}}}{2}=2\text{I}$

Thus by keeping the p.d. constant, the resistance of a circuit is halved and the current is doubled.

3. 1V

Explanation:

The resistors of $1 Ω, 2 Ω$ and $3 Ω$ are connected in series. Therefore, the net resistance,

$\text{R} = \text{R}_1 + \text{R}_2 +\text{R}_3$

$\text{R}=1\Omega+2\Omega+3\Omega=6\Omega$

Current in the circuit will be,

$\text{I}=\frac{\text{V}}{\text{R}}$

or $\text{I}=\frac{2}{6}=\frac{1}{3}\text{A}$

Current $=\frac{1}{3}\text{A}$

Therefore, the voltage across the 3I resistor,

V = IR

or $\text{V}=\frac{1}{3}\times3=1\text{V}$

3. $20\Omega$

Explanation:

When $25 Ω$ and $15 Ω$ are connected in series, then:

Total resistance, R = 25 + 15 = 40I

This 40I is connected in parallel with the 40I resistor.

Therefore, the net resistance $=\frac{40}{2}=20\text{I}$

3. $60 Ω$

Explanation:

If the resistance of a certain copper wire is $1 Ω,$ the resistance of a similar nichrome wire will be about $60 Ω$ because the resistivity of nichrome is 60 times the resistivity of copper.

4. One-fourth.

Explanation:

Since $\text{P}=\frac{\text{V}^2}{\text{R}}, $ when the potential difference is halved, the power becomes one-fourth.

3. Four times.

Explanation:

Resistance of the wire is given by:

$\text{R}=\rho\frac{\text{I}}{\text{A}}$

When the diameter is halved:

$\text{d}'=\frac{\text{d }}{2}$

Radius:

$\text{r}'=\frac{\text{r}}{2}$

Area of cross-section:

$\text{A}'=\pi\text{r}'^2=\pi\Big(\frac{\text{r}}{2}\Big)^2=\frac{\pi\text{r}}{4}=\frac{\text{A}}{4}$

Area of cross-section will get reduced by four times.

Then the new resistance:

$\text{R}'=\frac{\rho\text{I}}{\text{A}}$

$\text{R}'=\frac{4\rho\text{I}}{\text{A}}$

$\text{R}'=4\text{R}$

Thus, the resistance will increase by four times.

4. 50C.

Explanation:

Here, current, I = 5A

Time, t = 10s

So, using the relation, $\text{I} =\frac{\text{Q}}{\text{t}}$

Q = I × t

Charge, Q = 5 × 10 = 50C

Thus, a charge of 50C is passed in 10s.

3. 0.5A.

Explanation:

Here, charge, Q = 300C

Time, t = 10 minutes = (10 × 60)s = 600s

Then, current, $\text{I}=\frac{\text{Q}}{\text{t}}$

$=\frac{300}{600}=0.5\text{A}.$

4. 25

Explanation:

If the resistance wire is cut into five pieces, the resistance of each wire is $\frac{\text{R}}{5}.$ If we connected the pieces in parallel, we will get the net resistance as $\frac{\text{R}}{25}$ Therefore, the ratio will be 25.

3. 293

Explanation:

El Total power provided in the house, P = VI = 100A × 220V = 22000W

This is the maximum power that can be drawn in the house.

Total power drawn by x bulbs = 75 x W

Maximum power x bulbs can draw = 22000W.

$\Rightarrow75\text{x}=22000$

$\Rightarrow \text{x}=\frac{22000}{75}\approx293$