Physics 5 Marks Questions

$\text{C}_\text{eq}=\big[(3\mu\text{fp}\ 3\mu\text{f})\text{s}(1\mu\text{f p}1\mu\text{f})\big]\text{p}(1\mu\text{f})$

$=\big[(3+3)\mu\text{f s}(2\mu\text{f})\big]\text{p}1\mu\text{f}$

$=\frac{3}{2}+1=\frac{5}{2}\mu\text{f}$

$\text{V}=100\text{V}$

$\text{Q}=\text{CV}=\frac{5}{2}\times100=250\mu\text{c}$

Charge stored across $1\mu\text{f}$ capacitor $=100\mu\text{C}$

C_eq between A and B is $=6\mu\text{f}=\text{C}$

Potential drop across $\text{AB}=\text{V}=\frac{\text{Q}}{\text{C}}=25\text{V}$

Potential drop across BC = 75V.

1. $\text{At}^2=\text{Q}$

$\Rightarrow\text{A}=\frac{\text{Q}}{\text{t}^2}=\frac{\text{A}'\text{T}'}{\text{T}^{-2}}=\text{A}^1\text{T}^{-1}$

2. $\text{Bt}=\text{Q}$

$\Rightarrow\text{B}=\frac{\text{Q}}{\text{t}}=\frac{\text{A}'\text{T}'}{\text{T}}=\text{A}$

3. $\text{C}=[\text{Q}]$

$\Rightarrow\text{C}=\text{A}'\text{T}'$

4. Current $\text{t}=\frac{\text{dQ}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(\text{At}^2+\text{Bt}+\text{C})$

$=2\text{At}+\text{B}=2\times5\times5+3=53\text{A}.$

1. $\text{at t}=0,\text{q}=60\mu\text{c}$

2. $\text{at t}=30\mu\text{s},\text{ q}=\text{Qe}^{\frac{-\text{t}}{\text{RC}}}$

$=60\times10^{-6}\times\text{e}^{-0.3}=44\mu\text{C}$

3. $\text{at t}=120\mu\text{s},\text{ q}=60\times10^{-6}\times\text{e}^{-1.2}=18\mu\text{c}$

4. $\text{at t}=1.0\text{ms, q}=60\times10^{-6}\times\text{e}^{-10}$ $=0.00272=0.003\mu\text{C}.$

1. $\text{I}=\frac{\text{V}}{\text{R}}=\frac{6}{24}=0.25\text{A}$

2. $\text{q}=\text{Q}\Big(1-\text{e}^{\frac{-\text{t}}{\text{RC}}}\Big)$

$=(8\times10^{-6}\times6)\big[1-\text{c}^{-1}\big]$

$=48\times10^{-6}\times0.63=3.024\times10^{-5}$

$\text{V}=\frac{\text{Q}}{\text{C}}=\frac{3.024\times10^{-5}}{8\times10^{-6}}=3.78$

$\text{E}=\text{V}+\text{iR}$

$\Rightarrow6=3.78+\text{i}24$

$\Rightarrow\text{i}=0.09\mathring{\text{A}}$

1. $\text{q}=\text{VC}\Big(1-\text{e}^{\frac{-\text{t}}{\text{CR}}}\Big)$

I = Current $=\frac{\text{dq}}{\text{dt}}=\text{VC}.(-)\text{e}^{\frac{-\text{t}}{\text{RC}}},\frac{-1}{\text{RC}}$

$=\frac{\text{V}}{\text{R}}\text{e}^{\frac{-\text{t}}{\text{RC}}}=\frac{\text{V}}{\text{R}\cdot\text{e}^{\frac{\text{t}}{\text{RC}}}}=\frac{24}{10^6}\times\frac{1}{\text{e}^{\frac{\text{t}}{100}}}$

$=24\times10^{-6}\frac{1}{\text{e}^{\frac{\text{t}}{100}}}$

$\text{t}=10\text{min},600\text{sec}.$

$\text{Q}=24\times10+-4\times\big(1-\text{e}^{-6}\big)=23.99\times10^{-4}$

$\text{I}=\frac{24}{10^6}\times\frac{1}{\text{e}^6}=5.9\times10^{-8}\text{Amp}.$

2. $\text{q}=\text{VC}\Big(1-\text{e}^\frac{-\text{t}}{\text{CR}}\Big)$

1. No.of electrons transferred

$\text{i}=\frac{\text{V}}{\text{R}}=\frac{20}{10^3}=20\times10^{-3}=2\times10^{-2}\text{A}$

$\text{q}=\text{it}=2\times10^{-2}\times1=2\times10^{-2}\text{C}$

No. of electrons transferred $=\frac{2\times10^{-2}}{1.6\times10^{-19}}=\frac{2\times10^{-17}}{1.6}=1.25\times10^{17}.$

2. Current density of wire

​​​​​​​​​​​​​​​​​​​​​$=\frac{\text{i}}{\text{A}}=\frac{2\times10^{-2}}{\pi\times10^{-8}}=\frac{2}{3.14}\times10^6$

$=0.6369\times10^6=6.37\times10^5\text{A}/\text{m}^2.$

$\text{G}=25\Omega,\text{lg}=1\text{ma},\text{l}=2\text{A},\text{S}=?$

Potential across A B is same

$25\times10^{-3}=(2-10^{-3})\text{S}$

$\Rightarrow\text{S}=\frac{25\times10^{-3}}{2-10^{-3}}=\frac{25\times10^{-3}}{1.999}$

$=12.5\times10^{-3}=1.25\times10^{-2}.$

1. $\text{R}_\text{eff}=\frac{180}{3}=60\Omega$

$\text{i}=\frac{60}{60}=1\text{A}$

2. $\text{R}_\text{eff}=\frac{180}{2}=90\Omega$

$\text{i}=\frac{60}{90}=0.67\text{A}$

3. $\text{R}_\text{eff}=180\Omega\Rightarrow\text{i}=\frac{60}{180}=0.33$

$\text{R}_\text{eff}=(1150+50)\Omega=1200\Omega$

$\text{i}=\Big(\frac{12}{1200}\Big)\text{A}=0.01\text{A}.$

(The resistor of $50\Omega$ can tolerate)

Let R be the resistance of sheet used.

The potential across both the resistors is same.

$0.01\times50=1.99\times\text{R}$

$\Rightarrow\text{R}=\frac{0.01\times50}{1.99}=\frac{50}{199}=0.251\Omega$

1. Current through (1) $4\Omega$ resistor = 0
2. Current through (2) and (3)

net E = 4V - 2V = 2V

(2) and (3) are in series,

$\text{R}_\text{eff}=4+6=10\Omega$

$\text{i}=\frac{2}{10}=0.2\text{A}$

Current through (2) and (3) are 0.2A.

1. When switch is open, no current passes through the ammeter. In the upper part of the circuit the Voltmenter has $\infty$ resistance. Thus current in it is 0.

$\therefore$ Voltmeter read the emf. (There is not Pot. Drop across the resistor).

2. When switch is closed current passes through the circuit and if its value of i.

The voltmeter reads

$\in-\text{ir}=1.45$

$\Rightarrow1.52-\text{ir}=1.45$

$\Rightarrow\text{ir}=0.07$

$\Rightarrow1\text{r}=0.07\Rightarrow\text{r}=0.07\Omega.$

If the wire is connected to the potentiometer wire so that $\frac{\text{R}_\text{AD}}{\text{R}_\text{DB}}=\frac{8}{12},$ then according to wheat stone’s bridge no current will flow through galvanometer.

$\frac{\text{R}_\text{AB}}{\text{R}_\text{DB}}=\frac{\text{L}_\text{AB}}{\text{L}_\text{B}}=\frac{8}{12}=\frac{2}{3}$ (Acc. To principle of potentiometer).

$\text{I}_\text{AB}+\text{I}_\text{DB}=40\text{cm}$

$\Rightarrow\text{I}_\text{DB}\frac{2}{3}+\text{I}_\text{DB}=40\text{cm}$

$\Rightarrow\Big(\frac{2}{3}+1\Big)\text{I}_\text{DB}=40\text{cm}$

$\Rightarrow\frac{5}{3}\text{I}_\text{DB}=40$

$\Rightarrow\text{L}_\text{DB}=\frac{40\times3}{5}=24\text{cm}.$

$\text{I}_\text{AB}=(40-24)\text{cm}=16\text{cm}.$

1. Total emf = n₁E

in 1 row

Total emf in all news = n₁E

Total resistance in one row = n₁r

Total resistance in all rows $=\frac{\text{n}_1\text{r}}{\text{n}_2}$

Net resistance $=\frac{\text{n}_1\text{r}}{\text{n}_2}+\text{R}$

Current $=\frac{\text{n}_1\text{E}}{\frac{\text{n}_1}{\text{n}_2\text{r}}+\text{R}}=\frac{\text{n}_1\text{n}_2\text{E}}{\text{n}_1\text{r}+\text{n}_2\text{R}}$

2. $\text{I}=\frac{\text{n}_1\text{n}_2\text{E}}{\text{n}_1\text{r}+\text{n}_2\text{R}}$

for I = max,

n₁r + n₂R = min

$\Rightarrow\Big(\sqrt{\text{n}_1\text{r}}-\sqrt{\text{n}_2\text{R}}\Big)^2+2\sqrt{\text{n}_1\text{rn}_2\text{R}}=\text{min}$

it is min, when

$\sqrt{\text{n}_1\text{r}}=\sqrt{\text{n}_2\text{R}}$

$\Rightarrow\text{n}_1\text{r}=\text{n}_2\text{R}$

I is max when n₁r = n₂R.

1. Potential difference between A and B is 6V.

B is at 0 potential.

Thus potential of A point is 6V.

The potential difference between Ac is 4V.

$\text{V}_\text{A}-\text{V}_\text{C}=0.4$

$\text{V}_\text{C}=\text{V}_\text{A}-4=6-4=2\text{V}$

2. The potential at $\text{D}=2\text{V},\text{V}_\text{AD}=4\text{V};\text{V}_\text{BD}=\text{OV}$

Current through the resisters R₁ and R₂ are equal

Thus, $\frac{4}{\text{R}_1}=\frac{2}{\text{R}_2}$

$\Rightarrow\frac{\text{R}_1}{\text{R}_2}=2$

$\Rightarrow\frac{\text{I}_1}{\text{I}_2}=2$ (Acc. to the law of potentiometer)

$\text{I}_1+\text{I}_2=100\text{cm}$

$\Rightarrow\text{I}_1+\frac{\text{I}_1}{2}=100\text{cm}\Rightarrow\frac{3\text{I}_1}{2}=100\text{cm}$

$\Rightarrow\text{I}_1=\frac{200}{3}\text{cm}=66.67\text{cm}.$

$\text{AD}=66.67\text{cm}$

3. When the points C and D are connected by a wire current flowing through it is 0 since the points are equipotential.

4. Potential at A = 6 v

Potential at C = 6 - 7.5 = -1.5V

The potential at B = 0 and towards A potential increases.

Thus -ve potential point does not come within the wire.

1. In circuit ABFGA,

$\text{i}_150+2\text{i}+\text{i}-4.3=0$

$\Rightarrow50\text{i}_1+3\text{i}=4.3\ ...(1)$

In circuit BEDCB,

$50\text{i}-200\text{i}+200\text{i}_1=0$

$\Rightarrow250\text{i}_1-200\text{i}=0$

$\Rightarrow50\text{i}_1-40\text{i}=0\ ...(2)$

From (1) and (2)

$43\text{i}=4.3\Rightarrow\text{i}=0.1$

$5\text{i}_1=4\times\text{i}=4\times0.1$

$\Rightarrow\text{i}_1=\frac{4\times0.1}{5}=0.08\text{A}$

Ammeter reads a current $=\text{i}=0.1\text{A}$

Voltmeter reads a potential difference equal to $\text{i}_1\times50=0.08\times50=4\text{V}.$

2. In circuit ABEFA,

$50\text{i}_1+2\text{i}_1+1\text{i}-4.3=0$

$\Rightarrow52\text{i}_1+\text{i}=4.3$

$\Rightarrow200\times52\text{i}_1+200\text{i}=4.3\times200\ ...(1)$

In circuit BCDEB,

$(\text{i}-\text{i}_1)200-\text{i}_12-\text{i}_150=0$

$\Rightarrow200\text{i}-200\text{i}_1-2\text{i}_1-50\text{i}_1=0$

$\Rightarrow200\text{i}-252\text{i}_1=0\ ...(2)$

From (1) and (2)

$\text{i}_1(10652)=4.3\times2\times100$

$\Rightarrow\text{i}_1=\frac{4.3\times2\times100}{10652}=0.08$

$\text{i}=4.3-52\times0.08=0.14$

Reading of the ammeter = 0.08a

Reading of the voltmeter $=(\text{i}-\text{i}_1)200=(0.14-0.08)\times200=12\text{V}.$

1. $0.1\text{i}_1+1\text{i}_1-6+1\text{i}_1-6=0$

$\Rightarrow0.1\text{i}_1+1\text{i}_1+1\text{i}_1=12$

$\Rightarrow\text{i}_1=\frac{12}{2.1}$

$\text{ABCDA}$

$\Rightarrow0.1\text{i}_2+1\text{i}-6=0$

$\Rightarrow0.1\text{i}_2+1\text{i}$

$\text{ADEFA}$

$\Rightarrow\text{i}-6+6-(\text{i}_2-\text{i})1=0$

$\Rightarrow\text{i}-\text{i}_2+\text{i}=0$

$\Rightarrow2\text{i}-\text{i}_2=0\Rightarrow-2\text{i}\pm0.2\text{i}=0$

$\Rightarrow\text{i}_2=0.$

2. $1\text{i}_1+1\text{i}_1-6+1\text{i}_1=0$

$\Rightarrow3\text{i}_1=12\Rightarrow\text{i}_1=4$

$\text{DCFED}$

$\Rightarrow\text{i}_2+\text{i}-6=0\Rightarrow\text{i}_2+\text{i}=6$

$\text{ABCDA}$

$\text{i}_2+(\text{i}_2-\text{i})-6=0$

$\Rightarrow\text{i}_2+\text{i}_2-\text{i}=6\Rightarrow2\text{i}_2-\text{i}=6$

$\Rightarrow-2\text{i}_2\pm2\text{i}=6\Rightarrow\text{i}=-2\text{i}_2+\text{i}=6$

$\Rightarrow\text{i}_2-2=6\Rightarrow\text{i}_2=8$

$\frac{\text{i}_1}{\text{i}_2}=\frac{4}{8}=\frac{1}{2}.$

3. $10\text{i}_1+1\text{i}_1-6+1\text{i}_1-6=0$

$\Rightarrow12\text{i}_1=12\Rightarrow\text{i}_1=1$

$10\text{i}_2-\text{i}_1-6=0$

$\Rightarrow10\text{i}_2-\text{i}_1=6$

$\Rightarrow10\text{i}_2+(\text{i}_2-\text{i})1-6=0$

$\Rightarrow11\text{i}_2=6$

$\Rightarrow-\text{i}_2=0.$

1. $\text{R}_\text{eff}=\frac{100\times400}{500}+200=280$

$\text{i}=\frac{84}{280}=0.3$

$100\text{i}=(0.3-\text{i})400$

$\Rightarrow\text{i}=1.2-4\text{i}$

$\Rightarrow5\text{i}=1.2\Rightarrow\text{i}=0.24$

Voltage measured by the voltmeter $=\frac{0.24\times100}{24\text{V}}$

2. If voltmeter is not connected

$\text{R}_\text{eff}=(200+100)=300\Omega$

$\text{i}=\frac{84}{300}=0.28\text{A}$

Voltage across $100\Omega=(0.28\times100)=28\text{V}.$

$\text{i}_\text{min}=\frac{5.5\times3}{110}=0.15$

$\text{i}_\text{min}=\frac{5.5\times3}{20}=\frac{16.5}{20}=0.825.$

1. $\text{C} = 12.0 \mu\text{F} = 12 \times 10^{-6}$

$\text{emf} = 6.00 \text{V, R} = 1\Omega$

$\text{t} = 12 \mu\text{c}, \text{ i} = \text{i}_0 \text{e}^{\frac{-\text{t}}{\text{RC}}}$

$=\frac{\text{CV}}{\text{T}}\times\text{e}^\frac{-\text{t}}{\text{RC}}=\frac{12\times10^{-6}\times6}{12\times10^{-6}}\times\text{e}^{-1}$

$=2.207=2.1\text{A}$

2. Power delivered by battery

We known, $\text{V}=\text{V}_0\text{e}^{\frac{-\text{t}}{\text{RC}}}$ (where V and V₀ are potential VI)

$\text{VI}=\text{V}_0\text{I}\text{e}^{\frac{-\text{t}}{\text{RC}}}$

$\Rightarrow\text{VI}=\text{V}_0\text{I}\times\text{e}^{-1}=6\times6\times\text{e}^{-1}=13.24\text{W}$

3. $\text{U}=\frac{\text{CV}^2}{\text{T}}\Big(\text{e}^{\frac{-\text{t}}{\text{RC}}}\Big)^2$ $\Big[\frac{\text{CV}^2}{\text{T}}$ = energy drawing per unit time$\Big]$

$=\frac{12\times10^{-6}\times36}{12\times10^{-6}}\times\big(\text{e}^{-1}\big)^2=4.872.$

1. Applying Kirchoff’s law,

$10\text{i}-6+5\text{i}-12=0$

$\Rightarrow10\text{i}+5\text{i}=18$

$\Rightarrow15\text{i}=18$

$\Rightarrow\text{i}=\frac{18}{15}=\frac{6}{5}=1.2\text{A}.$

2. Potential drop across $5\Omega$ resistor,

$\text{i}5=1.2\times5\text{V}=6\text{V}$

3. Potential drop across $10\Omega$ resistor

$\text{i}10=1.2\times10\text{V}=12\text{V}$

4. $10\text{i}-6+5\text{i}-12=0$

$\Rightarrow10\text{i}+5\text{i}=18$

$15\text{i}=18$

$\Rightarrow\text{i}=\frac{18}{15}=\frac{6}{5}=1.2\text{A}.$

Potential drop across $5\Omega$ resistor = 6V

Potential drop across $10\Omega$ resistor = 12V.

1. Let the equation resistance of the combination be R.

$\Big(\frac{2\text{R}}{\text{R}+2}\Big)+1=\text{R}$

$\Rightarrow\frac{2\text{R}+\text{R}+2}{\text{R}+2}=\text{R}\Rightarrow3\text{R}+2=\text{R}^2+2\text{R}$

$\Rightarrow\text{R}^2-\text{R}-2=0$

$\Rightarrow\text{R}=\frac{+1\pm\sqrt{1+4\times1\times2}}{2.1}$

$=\frac{1\pm\sqrt{9}}{2}=\frac{1\pm3}{2}=2\Omega.$

2. Total current sent by battery $=\frac{6}{\text{R}_\text{eff}}=\frac{6}{2}=3$

Potential between A and B

$3.1+2\text{i}=6$

$\Rightarrow3+2\text{i}=6\Rightarrow2\text{i}=3$

$\Rightarrow\text{i}=1.5\text{a}$

1. $\text{R}_\text{eff}=\frac{\Big(\frac{2\text{r}}{2}\Big)\times\text{r}}{\Big(\frac{2\text{r}}{2}\Big)+\text{r}}$

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

2. At 0 current coming to the junction is current going from BO = Current going along OE.

Current on CO = Current on OD

Thus it can be assumed that current coming in OC goes in OB.

Thus the figure becomes

$\Bigg[\text{r}+\Big(\frac{2\text{r}\times\text{r}}{3\text{r}}\Big)+\text{r}\Bigg]=2\text{r}+\frac{2\text{r}}{3}=\frac{8\text{r}}{3}$

$\text{R}_\text{eff}=\frac{\Big(\frac{8\text{r}}{6}\Big)\times2\text{r}}{\Big(\frac{8\text{r}}{6}\Big)+2\text{r}}=\frac{\frac{8\text{r}^2}{3}}{\frac{20\text{r}}{6}}=\frac{8\text{r}^2}{3}\times\frac{6}{20}=\frac{8\text{r}}{10}=4\text{r}.$

1. Potential difference between terminals of ‘a’ is 10V.

i through $\text{a}=\frac{10}{10}=1\text{A}$

Potential different between terminals of b is 10 - 10 = 0V

i through $\text{b}=\frac{0}{10}=0\text{A}.$

2. Potential difference across ‘a’ is 10V

i through $\text{a}=\frac{10}{10}=1\text{A}$

Potential different between terminals of b is 10 - 10 = 0V

i through $\text{b}=\frac{0}{10}=0\text{A}.$

1. $\text{R}_\text{eff}=\frac{\frac{15\times5}{6}\times\frac{15}{6}}{\frac{15\times5}{6}+\frac{15}{6}}=\frac{\frac{15\times5\times15}{6\times6}}{\frac{75+15}{6}}$

$=\frac{15\times5\times15}{6\times90}=\frac{25}{12}=2.08\Omega$

2. Across AC,

$\text{R}_\text{eff}=\frac{\frac{15\times4}{6}\times\frac{15\times2}{6}}{\frac{15\times4}{6}+\frac{15\times2}{6}}=\frac{\frac{15\times4\times15\times2}{6\times6}}{\frac{60+30}{6}}$

$=\frac{15\times4\times15\times2}{6\times90}=\frac{10}{3}=3.33\Omega$

3. Across AD,

$\text{R}_\text{eff}=\frac{\frac{15\times3}{6}\times\frac{15\times3}{6}}{\frac{15\times3}{6}+\frac{15\times3}{6}}=\frac{\frac{15\times3\times15\times3}{6\times6}}{\frac{60+30}{6}}$

$=\frac{15\times3\times15\times3}{6\times90}=\frac{15}{4}=3.75\Omega.$

$\text{i}_1\times20=\text{i}_2\times10$

$\Rightarrow\frac{\text{i}_1}{\text{i}_2}=\frac{10}{20}=\frac{1}{2}$

i₁ = 4mA, i₂ = 8mA

Current in $20\text{K}\Omega$ resistor = 4mA

Current in $10\text{K}\Omega$ resistor = 8mA

Current in $100\text{K}\Omega$ resistor = 12mA

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

$=5\text{K}\Omega\times12\text{mA}+10\text{K}\Omega\times8\text{mA}\\+100\text{K}\Omega\times12\text{mA}$

$=60+80+1200=1340\text{ volts}.$