2 Marks Questions - Page 2 - Physics STD 12 Science Questions - Vidyadip

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2 Marks Questions

Question 512 Marks

If a constant potential difference is applied across a bulb, the current slightly decreases as time passes and then becomes constant. Explain.

Answer

As current passes through wire of bulb it heats up and thus the resistance of wire (element) increase as its metal so current decreases as resistance increases after element is heated well the resistance stops increasing thus current stops decreases.

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Question 522 Marks

A non-ideal battery is connected to a resistor. Is work done by the battery equal to the thermal energy developed in the resistor? Will your answer change if the battery is ideal?

Answer

Yes, the answer will change if the battery is ideal. An ideal battery has no internal resistance. Hence, the work done by an ideal battery will be equal to the thermal energy developed in the resistor, assuming that the resistance of the wires used for connection is negligible.

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Question 532 Marks

Power P is to be delivered to a device via transmission cables having resistance R_C. If V is the voltage across R and I the current through it, find the power wasted and how can it be reduced.

Answer

Power wasted P_C = I²R_C where, R_C is the resistance of the connecting wires.
$\Rightarrow\ \text{P}_\text{C}=\Big(\frac{\text{P}^2}{\text{V}^2}\Big)\text{R}_\text{C}$ [as, P = VI]
In order to reduce P_C, power should be transmitted at high voltage.

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Question 542 Marks

The electric current existing in a discharge tube is $2.0\mu\text{A}.$ How much charge is transferred across a cross-section of the tube in 5 minutes?

Answer

$\text{i}'=2\mu\text{A},\text{t}=5\text{min}=5\times60\text{sec}$
$\text{q}=\text{i t}$
$=2\times10^{-6}\times5\times60$
$=10\times60\times10^{-6}\text{c}=6\times10^{-4}\text{c}$

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Question 552 Marks

The potential difference across the terminals of a battery of emf 12V and internal resistance $2\Omega$ drops to 10V when it is connected to a silver voltameter. Find the silver deposited at the cathode in half an hour. Atomic weight of silver is 107.9g/mole.

Answer

For potential drop, t = 30min = 180sec
$\text{V}_\text{i}=\text{V}_\text{f}+\text{iR}$
$\Rightarrow12=10+2\text{i}\Rightarrow\text{i}=1\text{Amp}$
$\text{m}=\text{Zit}=\frac{107.9}{96500}\times1\times30\times60$
$=2.01\text{g}\approx2\text{g}$

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Question 562 Marks

Why is Wheatstone bridge (Metre Bridge) method not suitable for measurement of very low and very high resistances?

Answer

Because, in order to ensure sensitivity of the bridge, all other resistances used should either have low value or very high value. This also requires a galvanometer of very low resistance or very high resistance and introduces error in the results.

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Question 572 Marks

Two cells of emf 10V and 2V and internal resistance $10\Omega$ and $5\Omega$ respectively, are connected in parallel as shown. Find the effective voltage across R.

Answer

The effective voltage across R is geven by $\varepsilon_\text{eq}=\frac{\Big(\frac{\varepsilon_1}{\text{r}_1}+\frac{\varepsilon_2}{\text{r}_2}\Big)} {\Big(\frac{2}{\text{r}_1}+\frac{1}{\text{r}_2}\Big)}=\frac{\Big(\frac{10}{10}-\frac25\Big)}{\Big(\frac{1}{10}+\frac15\Big)}$
$\Rightarrow\varepsilon_\text{eq}=2\text{V}$

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Question 582 Marks

A bulb is made using two filaments. A switch selects whether the filaments are used individually or in parallel. When used with a 15V battery, the bulb can be operated at 5W, 10W or 15W. What should be the resistances of the filaments?

Answer

The various resistances of the bulbs $=\frac{\text{V}^2}{\text{P}}$
Resistances are $\frac{(15)^2}{10},\frac{(15)^2}{10},\frac{(15)^2}{15},=45,22.5,15.$
Since two resistances when used in parallel have resistances less than both.
The resistances are 45 and 22.5.

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Question 592 Marks

An electric current of 2.0A passes through a wire of resistance $25\Omega$. How much heat will be developed in 1minute?

Answer

$\text{i}=2\text{A},$
$\text{r}=25\Omega,$
$\text{t}=1\text{min}=60\text{sec}$
Heat developed $=\text{i}^2\text{RT}=2\times2\times25\times60=6000\text{J}$

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Question 602 Marks

An electroplating unit plates 3.0g of silver on a brass plate in 3.0 minutes. Find the current used by the unit. The electrochemical equivalent of silver is 1.12 × 10^-⁶kg C^-¹.

Answer

t = 3min = 180sec

w = 2g

E.C.E = 1.12 × 10^-6kg/c

$\Rightarrow3\times10^{-3}=1.12\times10^{-6}\times\text{i}\times180$

$\Rightarrow\text{i}=\frac{3\times10^{-3}}{1.12\times10^{-6}\times180}$

$=\frac{1}{6.72}\times10^2\approx15\text{Amp}$

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Question 612 Marks

What is the effect of heating of a conductor on the drift velocity of free electrons?

Answer

$\upsilon_\text{d}=\frac{\text{eE}}{\text{m}}\tau$
By heating a conductor, the collisions of electrons occur more frequently; so relaxation time decreases and hence drift velocity decreases.

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Question 622 Marks

In a Wheatstone’s bridge experiment, a student by mistake, connects key (K) in place of galvanometer and galvanometer (G) in place of key (K). What will be the test for the balance of the bridge?

Answer

When the bridge is balanced, there will be no current in key, therefore a constant current will flow in the galvanometer. In balanced position, there will be a constant deflection in galvanometer and hence no change in its deflection on pressing the key.

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Question 632 Marks

The temperatures of the junctions of a bismuth-silver thermocouple are maintained at 0^°C and 0.001^°C. Find the thermo-emf (Seebeck emf) developed. For bismuth-silver, a = -46 × 10^-⁶V^°C-¹ and b = -0.48 × 10^-⁶V^°C^-².

Answer

$\theta=0.001^\circ\text{C}$
$\text{a}=-46\times10^{-6}\text{v/deg},$
$\text{b}=-0.48\times10^{-6}\text{v/deg}^2$
$\text{Emf}=\text{a}_{\text{BlAg}}\theta+\Big(\frac{1}{2}\Big)\text{b}_{\text{BlAg}}\theta^2$
$=-46\times10^{-6}\times0.001-\Big(\frac{1}{2}\Big)\times0.48\times10^{-6}(0.001)^2$
$=-46\times10^{-9}-0.24\times10^{-12}$
$=-46.00024\times10^{-9}$
$=-4.6\times10^{-8}\text{V}$

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Question 642 Marks

If the radius of a copper wire is doubled, will its specific resistance increase, decrease or remain same?

Answer

The specific resistance of a wire depends only on the material (at a given temperature).
Therefore by changing the radius, the specific resistance of copper remains unchanged.

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Question 652 Marks

The amount of electric charge passing through a cross-section of wire in time t is Q (t) = At² + Bt + C
Where A, B and C are constants having values 5, 4 and 1 respectively.
Calculate the value of electric current at t = 4s.

Answer

We have, $\text{Q}(\text{t})=\text{At}^2+\text{Bt}+\text{C}$
$\frac{\text{dQ}}{\text{dt}}=2\text{At}+\text{B}$
$=2\times5\times4+4(\text{at t}=4\text{s})$
$\text{I}=44\text{A}$

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Question 662 Marks

Define the term resistivity and write the SI unit.

Answer

The resistivity of a material of a conductor is defined as the resistance offered by a conductor of length 1m and area of cross-section 1m². Its SI unit is ohm × metre ($\Omega\text{m}$).

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Question 672 Marks

Two voltameters, one having a solution of silver salt and the other of a trivalent-metal salt, are connected in series and a current of 2A is maintained for 1.50 hours. It is found that 1.00g of the trivalent-metal is deposited.

What is the atomic weight of the trivalent metal?
How much silver is deposited during this period? Atomic weight of silver is 107.9 g/mole.

Answer

$\text{w}_1=\text{Zit}$
$\Rightarrow1=\frac{\text{mm}}{3\times96500}\times2\times2.5\times3600$
$\Rightarrow\text{mm}=\frac{3\times96500}{2\times1.5\times3600}=26.8\text{g/mole}$
$\frac{\text{E}_1}{\text{E}_2}=\frac{\text{W}_1}{\text{W}_2}$
$\Rightarrow\frac{107.9}{\Big(\frac{\text{mm}}{3}\Big)}=\frac{\text{W}_1}{1}$
$\Rightarrow\text{W}_1=\frac{107.9\times3}{26.8}=12.1\text{gm}$

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Question 682 Marks

A capacitor of capacitance C is given a charge Q. At t = 0, it is connected to an ideal battery of emf $\in$ through a resistance R. Find the charge on the capacitor at time t.

Answer

The capacitor is given a charge Q. It will discharge and the capacitor will be charged up when connected with battery.
Net charge at time $\text{t}=\text{Qe}^\frac{-\text{t}}{\text{RC}}+\text{Q}\Big(1-\text{e}^\frac{-\text{t}}{\text{RC}}\Big).$

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Question 692 Marks

An electric bulb marked 220V, 100W will get fused if it is made to consume 150W or more. What voltage fluctuation will the bulb withstand?

Answer

$\text{V}=220\text{v}$
$\text{P}=100\text{w}$
$\text{R}=\frac{\text{V}^2}{\text{P}}$
$=\frac{220\times220}{100}=484\Omega$
$\text{P}=150\text{w}$
$\text{V}=\sqrt{\text{PR}}=\sqrt{150\times22\times22}$
$=22\sqrt{150}=269.4\approx270\text{v}$

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Question 702 Marks

Two heating coils, one of fine wire and other of thick wire, made of the same material and of the same length are connected one by one to a source of electricity. Which coil will produce heat at a greater rate?

Answer

$\text{Q}\propto\frac{1}{\text{R}}$ and $\text{R}=\frac{\rho\text{l}}{\pi\text{r}^2}$
$\therefore\text{Q}\propto\frac{\pi\text{r}^2}{\rho\text{l}}$
Clearly, thick wire will produce heat at a greater rate.

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Question 712 Marks

Find the time required to liberate 1.0 litre of hydrogen at STP in an electrolytic cell by a current of 5.0A.

Answer

$\frac{\text{H}_2}{22.4\text{L}}\rightarrow2\text{g}$
$1\text{L}\rightarrow\frac{2}{22.4}$
$\text{m}=\text{Zit}$
$\frac{2}{22.4}=\frac{1}{96500}\times5\times\text{T}$
$\Rightarrow\text{T}=\frac{2}{22.4}\times\frac{96500}{5}$
$=1732.21\text{sec}\approx28.7\text{min}\approx29\text{min}$

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Question 722 Marks

By evaluating $\int\text{i}^2\text{Rdt},$ show that when a capacitor is charged by connecting it to a battery through a resistor, the energy dissipated as heat equals the energy stored in the capacitor.

Answer

$\int\text{i}^2\text{Rdt}=\int\text{i}_0^2\text{Re}^{\frac{-2\text{t}}{\text{RC}}}\text{dt}=\text{i}_0^2\text{R}\int\text{e}^{\frac{-2\text{t}}{\text{RC}}}\text{dt}$
$=\text{i}^2_0\text{R}\Big(\frac{-\text{RC}}{2}\Big)\text{e}^{\frac{-2\text{t}}{\text{RC}}}=\frac{1}{2}\text{Ci}_0^2\text{R}^2\text{e}^{\frac{-2\text{t}}{\text{RC}}}=\frac{1}{2}\text{CV}^2$ (Proved).

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Question 732 Marks

Two unequal resistances R₁ and R₂ are connected across two identical batteries of emf $\varepsilon$ and internal resistance r. Can the thermal energies developed in R₁ and R₂be equal in a given time. If yes, what will be the condition?

Answer

As, Q = i²Rt
EMF is identical thus change in R would result into different thermal energies at given time A all the other components are same but R₁ and R₂ are different.

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Question 742 Marks

Why do we prefer a potentiometer to measure the emf of a cell rather than a voltmeter?

Answer

A voltmeter has a finite resistance and draws current from a cell, therefore voltmeter measures terminal potential difference rather than emf, while a potentiometer at balance condition, does not draw any current from the cell; so the cell remains in open circuit. Hence potentiometer reads the actual value of emf.

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Question 752 Marks

An immersion heater rated 1000W, 220V is used to heat 0.01m³ of water. Assuming that the power is supplied at 220V and 60% of the power supplied is used to heat the water, how long will it take to increase the temperature of the water from 15^°C to 40^°C?

Answer

$\text{P}=1000$
$\text{V}=220\text{v}$
$\text{R}\frac{\text{V}^2}{\text{P}}=\frac{48400}{1000}=48.4\Omega$
Mass of water $=\frac{1}{100}\times1000=10\text{kg}$
Heat required to raise the temp. of given amount of water $=\text{ms}\Delta\text{t}$
$=10\times4200\times25=1050000$
Now heat liberated is only 60%.
So, $\frac{\text{V}^2}{\text{R}}\times\text{T}\times60\%=1050000$
$\Rightarrow\frac{(220)^2}{48.4}\times\frac{60}{100}\times\text{T}=1050000$
$\Rightarrow\text{T}=\frac{10500}{6}\times\frac{1}{60}\text{nub}=29.16\text{min}$

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Question 762 Marks

A proton beam is going from east to west. Is there an electric current? If yes, in what direction?

Answer

A proton beam means positive charge moving in west direction that means current flowing from east to west. As electric current is in direction of positive charge flowing or opposite to negative charge flowing.

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Question 772 Marks

Suppose you have three resistors of $20\Omega,50\Omega$ and $100\Omega.$ What minimum and maximum resistance can you obtain from these resistors?

Answer

$\text{R}_{\text{max}}=(20+50+100)\Omega=170\Omega$
$\text{R}_{\text{min}}=\frac{1}{\Big(\frac{1}{20}+\frac{1}{50}+\frac{1}{100}\Big)}=\frac{100}{8}=12.5\Omega.$

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Question 782 Marks

Current is allowed to flow in a metallic wire at a constant potential difference. When the wire becomes hot, cold water is poured on half of its portion. By doing so, its other half portion becomes still more hot. Explain its reason.

Answer

When cold water is poured on half the wire, the resistance of this portion decreases, and hence the current $\text{I}=\frac{\text{V}}{\text{R}}$ in whole wire increases and so the other half portion becomes still more hot.

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Question 792 Marks

Two 120V light bulbs, one of 25W and the other of 200W were connected in series across a 240V line. One bulb burnt out almost instantaneously. Which one was burnt and why?

Answer

Resistance of bulb $\text{R}=\frac{\text{V}^2}{\text{P}}\propto\frac{1}{\text{P}};$ so 25W bulb has higher resistance. In series current remains the same; so pd across 25W bulb will be more than that across 200W bulb; so 25W bulb was burnt out immediately.

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Question 802 Marks

A wire has a length of 2.0m and a resistance of $5.0\Omega.$ Find the electric field existing inside the wire if it carries a current of 10A.

Answer

$\text{l}=2\text{m},\text{R}=5\Omega,\text{i}=10\text{A},\text{E}=?$
$\text{V}=\text{iR}=10\times5=50\text{V}$
$\text{E}=\frac{\text{V}}{\text{l}}=\frac{50}{2}=25\text{V}/\text{m}$

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Question 812 Marks

Can the potential difference across a battery be greater than its emf?

Answer

In general potential difference can be equal or less than the emf.
E.m.f can never exceed the potential difference.
If the emf of a battery is E Volt, the potential difference across a battery is given by,
V = E - I r where I is the current in the circuit and r is the internal resistance.

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Question 822 Marks

In an electrolyte, the positive ions move from left to right and the negative ions from right to left. Is there a net current? If yes, in what direction?

Answer

As positive current moves from left to right and negative current moves from right to left in both cases as explained in previous question the current would be flowing from left to right.

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Question 832 Marks

On increasing the current drawn from a cell, the net potential difference across its terminals is lowered, why?

Answer

The terminal potential difference V = E - Ir.
Clearly if I is increased, the terminal potential difference falls.

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Question 842 Marks

The potential difference between the terminals of a 6.0V battery is 7.2V when it is being charged by a current of 2.0A. What is the internal resistance of the battery?

Answer

$\text{V}=\in+\text{ir}$

$\Rightarrow7.2=6+2\times\text{r}$

$\Rightarrow1.2=2\text{r}\Rightarrow\text{r}=0.6\Omega.$

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Question 852 Marks

Figure shows an electrolyte of AgCl through which a current is passed. It is observed that 2.68g of silver is deposited in 10 minutes on the cathode. Find the heat developed in the $20\Omega$ resistor during this period. Atomic weight of silver is 107.9g/mole.

Answer

$\text{W}=\text{Zit}$

$2.68=\frac{107.9}{96500}\times\text{i}\times10\times60$

$\Rightarrow\text{l}=\frac{2.68\times965}{107.9\times6}=3.99\approx4\text{Amp}$

Heat developed in the $20\Omega$ resister$=(4)^2\times20\times10\times60=192000\text{J}=192\text{KJ}$

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Question 862 Marks

A voltmeter consists of a $25\Omega$ coil connected in series with a $575\Omega$ resistor. The coil takes 10mA for full scale deflection. What maximum potential difference can be measured by this voltmeter?

Answer

Full deflection current $=10\text{mA}=(10\times10^{-3})\text{A}$

$\text{R}_\text{eff}=(575+25)\Omega=600\Omega$

$\text{V}=\text{R}_\text{eff}\times\text{i}=600\times10\times10^{-3}=6\text{V}.$

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Question 872 Marks

Consider the situation shown in figure. The switch is closed at t = 0 when the capacitors are uncharged. Find the charge on the capacitor C₁ as a function of time t.

Answer

$\text{C}_\text{eff}=\frac{\text{C}_1\text{C}_2}{\text{C}_1+\text{C}_2}$

$\text{Q}=\text{C}_\text{eff}\text{E}\Big(1-\text{e}^{\frac{-\text{t}}{\text{RC}}}\Big)=\frac{\text{C}_1\text{C}_2}{\text{C}_1+\text{C}_2}\text{E}\Big(1-\text{e}^\frac{-\text{t}}{\text{RC}}\Big)$

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Question 882 Marks

Consider the circuit shown in figure. Find the current through the $10\Omega$ resistor when the switch S is (a) Open (b) Closed.

Answer

When S is open,

$\text{R}_\text{eq}=(10+20)\Omega=30\Omega$

i = When S is closed,

$\text{R}_\text{eq}=10\Omega$

$\text{i}=\Big(\frac{3}{10}\Big)\Omega=0.3\Omega.$

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Question 892 Marks

Find the equivalent resistance of the network shown in figure. between the points a and b.

Answer

Eq. Resistance $=\frac{\text{r}}{3}$

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Question 902 Marks

Find the charge on each of the capacitors 0.20ms after the switch S is closed in the figure.

Answer

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

$=25(2+2)\times10^{-6}\bigg(1-\text{e}^\frac{-0.2\times10^{-3}}{25\times4\times10^{-6}}\bigg)$

$=24\times10^{-6}(1-\text{e}^2)=20.75$

Charge on each capacitor $=\frac{20.75}{2}=10.3$

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Question 912 Marks

The internal resistance of an accumulator battery of emf 6V is $10\Omega$ when it is fully discharged. As the battery gets charged up, its internal resistance decreases to $1\Omega.$
The battery in its completely discharged state is connected to a charger that maintains a constant potential difference of 9V. Find the current through the battery (a) just after the connections are made and (b) after a long time when it is completely charged.

Answer

Net emf while charging

$9-6=3\text{V}$

Current $=\frac{3}{10}=0.3\text{A}$

When completely charged

Internal resistance $'\text{r}'=1\Omega$

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

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Question 922 Marks

What should be the value of R in the figure. for which the current in it is zero?

Answer

For an value of R, the current in the branch is 0.

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Question 932 Marks

Consider a circuit containing an ideal battery connected to a resistor. Do "work done by the battery" and "the thermal energy developed" represent two names of the same physical quantity?

Answer

No as the work done by the battery is work done by the battery like ouput. and the thermal energy developed is due to lesser efficiency of battery. In case of ideal battery there would be no thermal energy developed.

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2 Marks Questions - Page 2 - Physics STD 12 Science Questions - Vidyadip