Physics M.C.Q (1 Marks)

1. The charge crossing in a given time interval.

4. Free-electron density.

Explanation:

$\text{v}_\text{f}=\Big(\frac{\text{e}}{\text{qm}}\Big)\tau$

$\text{E}=\frac{\text{i}}{\text{A ne}}$

$\text{j}=\frac{\text{i}}{\text{A}}$

Vd → drift speed

j → current density

i → current

A → cross-section area.

2. Increase.

Explanation:

Power $=\frac{\text{V}^2}{\text{R}},\text{R}\downarrow\text{thanpower}\uparrow$

Because Power $\propto\frac{1}{\text{R}}.$

3. B is correct but A is wrong.

Explanation:

Equivalent emf $\in_0=\frac{\in_1\text{r}_1+\in_2\text{r}_2}{\text{r}_1+\text{r}_2}$

Equivalent resistance $=\text{r}_0=\frac{\text{r}_1\text{r}_2}{\text{r}_1+\text{r}_2}$

$\phi$ The equivalent emf is smaller than either of the two emfs.

$\phi$ The equivalent internal resistance is smaller than either of the two internal resistances.

4. The information is not sufficient to find the relation between P_A and P_B.

Explanation:

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

Resistance is depend on Material, length & Area.

So RA < RB is information is not sufficient to ding.

The relation between $\rho_\text{A}$ and $\rho_\text{B}.$

2. A is correct but B is wrong.

Explanation:

Equivalent emf $=\hat{\text{I}}_1+\hat{\text{I}}_2$

$\text{R}_\text{eq}=\text{r}_1+\text{r}_2$

Equivalent emf $=\hat{\text{I}}_1+\hat{\text{I}}_2\Big\{\hat{\text{I}}_1>\hat{\text{I}}_2\Big\}$

$\text{R}_\text{eq}=\text{r}_1+\text{r}_2$

$\phi$ The equivalent emf is larger than either of the two emfs.

$\phi$ The equivalent internal resistance is smaller than either of the two internal resistances.

1. 1 : 2.

Explanation:

Thermal Energy developed = I²Rt (Because in series, current is same)

$\frac{\text{Thermal Energy developedin "R"}}{\text{Thermal Energy developedin "2R"}}=\frac{\text{I}^2\text{Rt}}{\text{I}^2(2\text{R})\text{t}}=\frac{1}{2}$

4. None of these.

Explanation:

Conductivity $\sigma=\frac{1}{\rho }$ Where r is resistivily.

Product of conductity and resistivity = 1.

1. Increase.

Explanation:

If the number of collisions of the free electrons with the lattice is decreased, then the drift velocity of the electrons increases.

Current i is directly proportional to the drift velocity 'V_d' and is given by the following relation:

$\text{i}=\text{neAV}_\text{d}.$

1. Both A and B are correct.

​​​​​​​Explanation:

$\phi$ Kirchhoff's Junction Law follows from conservation of charge.

$\phi$ Kirchhoff's loop law fallows from conservation nature of electric field.

1. Increases.

Explanation:

Temperature of a cunductor increases that causes resistivity (r) is increases & due conductivity (s) is decrease.

$\therefore\sigma=\frac{1}{\rho}$

$\Rightarrow\text{ratio of}\frac{\text{resistivity}}{\text{conductivity}}=\frac{\rho}{\sigma}=\rho^2$ is increase.

2. Q₁ > Q₂, t₁ < t₂

Explanation:

Condition for charging capacitor-

$\text{Q}=\text{Q}_0\Big(1-\text{e}^\frac{-\text{t}}{\text{Rc}}\Big)$

$\text{Q}=\text{Q}_0\Big(1-\text{e}^\frac{-10\text{m}}{\text{Rc}}\Big)\ ...(1)$

$\text{Q}_1=\text{Q}_2=\text{Q}_0\Big(1-\text{e}^\frac{{-(10\text{m}+10\text{m})}}{\text{Rc}}\Big)$

$\text{Q}_1=\text{Q}_2=\text{Q}_0\Big(1-\text{e}^\frac{{-20\text{m}}}{\text{Rc}}\Big)\ ...(2)$

From eq. (1) and (2) we get-

$\text{Q}_1>\text{Q}_2$

Given

$\text{Q}=\text{Q}_0\Big(1-\text{e}^\frac{-\text{t}}{\text{Rc}}\Big)$

$10\text{mc}=\text{Q}_0\Big(1-\text{e}^{\frac{-\text{t}}{\text{Rc}}}\Big)\ ...(3)$

$10\mu\text{c}+10\mu\text{c}=\text{Q}_0\Big(1-\text{e}^{\frac{-(\text{t}_1+\text{t}_2)}{\text{Rc}}}\Big)\rightarrow$

$20\mu\text{c}=\text{Q}_0\Big(1-\text{e}^{\frac{-(\text{t}_1+\text{t}_2)}{\text{Rc}}}\Big)\ ...(4)$

From eq. (3) and (4) we get,

$\text{t}_2>\text{t}_1$

1. 5s
2. 50s
3. 500s
4. 500s

Explanation:

$\text{Q}=\text{CE}\Big(1-\text{e}^{\frac{-\text{t}}{\text{Rc}}}\Big)$

$\text{C}-500\times10^{-6}\text{F}$

$\text{R}=10^4\text{W}$

$\text{t}=\text{Rc}=10^4\times500\times10^{-6}=5$

$\text{t}=5\text{sec}$

$\text{Q}=\text{C}\in\Big(1-\text{e}^{-\text{t}}\Big)=\text{C}\in\Big(1-\frac{1}{\text{e}}\Big)=0.63\text{c}\in$

Thus, 63% of the maximum charge is deposited in one time constant.

With the help of the figure we can say that the capacitor in the first 5s is larger than the charge stored in the next any second.

$\Rightarrow\text{at}\ \text{t}=\infty$

$\text{Q}=\text{Q}_0\Big(1-\text{e}^{-\infty}\Big)=\text{Q}_0=\text{C}\in$

$\therefore\text{t}_\infty-\text{t}_5=\text{C}\in-63=37$

after t = 5sec., maximum charge is deposited is only 37%.

1. $2\Omega$

Explanation:

$\text{R}=\frac{\rho}{\text{A}}=50$

resistance of '5' equal parts are same.

$\text{R}'=\frac{\frac{\rho}{5}}{\text{A}}=\frac{50}{5}=10\Omega$

All '5' equal parts connect in parallel,

$\frac{1}{\text{R}_\text{eq}}=\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+\frac{1}{10}=\frac{5}{10}$

$\text{R}_\text{eq}=2\Omega$

2. 2 : 1

Explanation:

Thermal Energy developed $=\frac{\text{v}_2}{\text{R}}\text{t}$ (Because in Parallel, voltage is same)

$\frac{\text{Thermal Energy developedin "R"}}{\text{Thermal Energy developedin "2R"}}=\frac{\frac{\text{v}^2}{\text{R}}\text{t}}{\frac{\text{v}^2}{2\text{R}}\text{t}}=2:1$

2. The currents in  the two discharging circuits at t = 0 are equal but not zero.

4. C₁ loses 50% of its initial charge sooner than C₂ loses 50% of its initial charge.

Explanation:

Charging Þ

Discharging þ

$\text{Q}=\text{C}_1\in\text{e}^{\frac{-\text{t}}{\text{Rc}_1}}$

$\text{Q}=\text{C}_2\in\text{e}^{\frac{-\text{t}}{\text{Rc}_2}}$

$\text{i}_1=\frac{\text{dQ}}{\text{dt}}=\frac{-\text{C}_1\in}{\text{RC}_1}\text{e}^\frac{-\text{t}}{\text{RC}_1}$

$\text{i}_1=\frac{\text{dQ}}{\text{dt}}=\frac{-\in}{\text{R}}\text{e}^\frac{-\text{t}}{\text{RC}_2}$

$\text{i}_1=\frac{-\in}{\text{R}}\text{e}^\frac{-\text{t}}{\text{RC}_1}$

$\text{at t}=0$

$\text{at t}=0$

$\text{i}_1=\frac{-\in}{\text{R}}$

$\text{i}_2=\frac{-\in}{\text{R}}$

3. K₁ > K₂.

Explanation:

vd drift velocity $=\frac{1}{2}\Big(\frac{\text{eE}}{\text{m}}\Big)\text{t}$

$\text{K.E}.=\frac{1}{2}\text{mv}^2_\text{d}=\frac{1}{2}\text{m}\Big(\frac{1}{4}\frac{\text{e}^2\text{E}^2\text{t}^2}{\text{m}^2}\Big)$

$\text{K.E}.=\frac{1}{8}\frac{\text{e}^2\text{E}^2\text{t}^2}{\text{m}}$

$\text{p}=\text{K.E}.\propto\frac{1}{\text{m}}$

Mass of electron < mass of metalions.

K.E. of electron > K.E. of metalions.