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NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current

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NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current

Edited By Vishal kumar | Updated on Sep 09, 2023 02:53 PM IST | #CBSE Class 12th

NCERT Solutions for Class 12 Physics Chapter 7 – Access and Download Free PDF

NCERT Solutions for Class 12 Physics Chapter 7 Alternating current is a crucial and scoring chapter in the Class 12 syllabus. This NCERT solution page offers detailed step-by-step solutions prepared by physics experts from Careers360. It covers a total of twenty-six questions, including those from 7.1 to 7.11 in the exercise section and the remaining in the additional exercise section.

The supply that is received in our home is alternating in nature. Do you know what is the value of the normal supply voltage that is coming into our home and what is the supply frequency? Alternating Current Class 12 chapter will help you to clear such doubts. NCERT questions explained in CBSE NCERT solutions for Class 12 physics chapter 7 are based on the single-phase alternating current.

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Alternating Current Class 12 NCERT solutions are the basics for the topics needed to study for CBSE board exam. Students should go through the Alternating Current NCERT solutions to know the answer to the question of Alternating current. In this class 12 physics chapter 7 exercise solutions you will study the concept of phasor diagrams and related problems.

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NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current

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Class 12 Physics Chapter 7 Exercise Solutions

Q7.1 (a) A 100\Omega resistor is connected to a 220\: V , 50\: Hz ac supply.

a)what is the RMS value of current?

Answer:

Given,

RMS voltage in the circuit V_{rms}=220V

Resistance in the circuit R=100\Omega

Now,

RMS current in the circuit:

I_{rms}=\frac{V_{rms}}{R}=\frac{220}{100}=2.2A

Hence, the RMS value of current is 2.2A.

Q7.1 (b) A 100W resistor is connected to a 220\: V , 50Hz ac supply.

What is the rms value of current in the circuit?

Answer:

Given,

RMS Voltage in the circuit V_{rms}=200V

Resistance in the circuit R=100\Omega

Now,

RMS Current in the circuit:

I_{rms}=\frac{V_{rms}}{R}=\frac{220}{100}=2.2A

Hence, the RMS value of current is 2.2A.

Q7.1 (c) A 100\Omega resistor is connected to a 220\: V , 50\: Hz ac supply.

What is the net power consumed over a full cycle?

Answer:

Given,

Supplied RMS Voltage V_{rms}=220V

Supplied RMS Current I_{rms}=2.2A

The net power consumed over a full cycle:

P=V_{rms}I_{rms}=220*2.2=484W

Hence net power consumed is 484W.

Q7.2 (a) The peak voltage of an ac supply is 300 V . What is the RMS voltage?

Answer:

Given

Peak Value of ac supply:

V_{peak}=300V

Now as we know in any sinusoidal function

RMSvalue=\frac{peakvalue}{\sqrt{2}}

Since our ac voltage supply is also sinusoidal

V_{rms}=\frac{V_{peak}}{\sqrt{2}}=\frac{300}{\sqrt{2}}=212.13V

Hence RMS value of voltage os 212.13V.

Q7.2 (b) The RMS value of current in an ac circuit is 10\: A . What is the peak current?

Answer:

Given,

RMS value of current I_{rms}=10A

Since Current is also sinusoidal (because only resistance is present in the circuit, not the capacitor and inductor)

I_{rms}=\frac{I_{peak}}{\sqrt{2}}

I_{peak}=\sqrt{2}I_{rms}=\sqrt{2}*10=14.1A

Hence the peak value of current is 14.1A.

Q7.3 A 44\: mH inductor is connected to 220\: V , 50\: Hz ac supply. Determine the RMS value of the current in the circuit.

Answer:

Given

Supply Voltage V=220V

Supply Frequency f=50Hz

The inductance of the inductor connected L=44mH=44*10^{-3}H

Now

Inductive Reactance

X_L=\omega L=2\pi fL=2\pi *50*44*10^{-3}

RMS Value of the current :

I_{rms}=\frac{V_{rms}}{X_L}= \frac{220}{2\pi *50*44*10^{-3}}=15.92A

Hence the RMS Value of current is 15.92A.

Q7.4 A 60\mu F capacitor is connected to a 100\: V , 60\; Hz ac supply. Determine the rms value of the current in the circuit.

Answer:

Given,

Supply Voltage V = 110V

Supply Frequency f=60Hz

The capacitance of the connected capacitor C=60\mu F=60*10^{-6}F

Now,

Capacitive Reactance

X_C=\frac{1}{\omega C}=\frac{1}{2\pi fC}=\frac{1}{2\pi *60*60*10^{-6}}

RMS Value of current

I_{rms}=\frac{V_{rms}}{X_C}=V\omega C=V2\pi fC=110*2\pi *60*60*10^{-6}=2.49A

Hence the RMS Value of current is 2.49A.

Q7.5 In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle. Explain your answer.

Answer:

As we know,

Power absorbed P=VIcos\phi

Where \phi is the phase difference between voltage and current.

\phi for the inductive circuit is -90 degree and \phi for the capacitive circuit is +90 degree.

In both cases (inductive and capacitive), the power absorbed by the circuit is zero because in both cases the phase difference between current and voltage is 90 degree.

This can be seen as The elements(Inductor and Capacitor) are not absorbing the power, rather storing it. The capacitor is storing energy in electrostatic form and Inductor is storing the energy in magnetic form.

Q7.6 Obtain the resonant frequency \omega _{r} of a series LCR circuit with L=2.0H , C=32\mu F and R=10\Omega . What is the Q -value of this circuit?

Answer:

Given, in a circuit,

Inductance, L=2H

Capacitance, C=32\mu F=32*10^{-6}F

Resistance, R=10\Omega

Now,

Resonance frequency (frequency of maximum current OR minimum impedance OR frequency at which inductive reactance cancels out capacitive reactance )

\omega _r=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{2*32*10^{-6}}}=\frac{1}{8*10^{-3}}=125s^{-1}

Hence Resonance frequency is 125 per second.

Q-Value:

Q=\frac{1}{R}\sqrt{\frac{L}{C}}=\frac{1}{10}\sqrt{\frac{2}{32*10^{-6}}}=25

Hence Q - value of the circuit is 25.

Q7.7 A charged 30\mu F capacitor is connected to a 27mH inductor. What is the angular frequency of free oscillations of the circuit?

Answer:

Given

Capacitance C=30\mu F=30*10^{-6}

Inductance L = 27mH = 27*10^{-3}H

Now,

Angular Frequency

\omega _r=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{30*10^{-6}*27*10^{-3}}}=1.11*10^{3}rad/sec

Hence Angular Frequency is 1.11*10^{3}rad/sec

Q7.8 Suppose the initial charge on the capacitor in Exercise 7.7 is 6mC . What is the total energy stored in the circuit initially? What is the total energy at later time?

Answer:

Given

Capacitance C=30\mu F=30*10^{-6}

Inductance L = 27mH = 27*10^{-3}H

Charge on the capacitor Q=6mC=6*10^{-3}C

Now,

The total energy stored in Capacitor :

E=\frac{Q^2}{2C}=\frac{(6*10^{-3})^2}{2*30*10^{-6}}=\frac{6}{10}=0.6J

Total energy later will be same because energy is being shared with capacitor and inductor and none of them loses the energy, they just store it and transfer it.

Q7.9 A series LCR circuit with R=20\Omega , L=1.5H and c=35\mu F is connected to a variable-frequency 200\; V ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle?

Answer:

Given,

Resistance R=20\Omega

Inductance L=1.5H

Capacitance C=35\mu F=35*10^{-6}F

Voltage supply V = 200V

At resonance, supply frequency is equal to the natural frequency, and at the natural frequency, the total impedance of the circuit is equal to the resistance of the circuit

as inductive and capacitive reactance cancels each other. in other words,

Z = \sqrt{\left ( \omega L-\frac{1}{\omega C} \right )^2+R^2}=\sqrt{0^2+R^2}=R=20\Omega

As

\omega L=\frac{1}{\omega C}

Now,

Current in the circuit

I=\frac{V}{Z}=\frac{200}{20}=10A

Average Power transferred in the circuit :

P=VI=200*10=2000W

Hence average power transferred is 2000W.

Q7.10 A radio can tune over the frequency range of a portion of MW broadcast band: (800kHz\: to1200kHz) . If its LC circuit has an effective inductance of 200\mu H , what must be the range of its variable capacitor?

Answer:

Given,

Range of the frequency in which radio can be tune = (800kHz\: to1200kHz)

The effective inductance of the Circuit = 200\mu H

Now, As we know,

w^2=1/\sqrt{LC}

C=1/w^2L

where w is tuning frequency.

For getting the range of the value of a capacitor, let's calculate the two values of the capacitor, one maximum, and one minimum.

first, let's calculate the minimum value of capacitance which is the case when tuning frequency = 800KHz.

C_{minimum}=\frac{1}{w_{minimum}^2L}=\frac{1}{(2\pi(800*10^3))^2*200*10^{-6}}=1.981*10^{-10}F

Hence the minimum value of capacitance is 198pF.

Now, Let's calculate the maximum value of the capacitor.

in this case, tuning frequency = 1200KHz

C_{maximum}=\frac{1}{w_{maximum}^2L}=\frac{1}{(2\pi(1200*10^3))^2*200*10^{-6}}=88.04*10^{-12}F

Hence the maximum value of the capacitor is 88.04pf

Hence the Range of the values of the capacitor is 88.04pF to 198.1pF .

Q7.11 (a) Figure shows a series LCR circuit connected to a variable frequency 230\: V source. L=5.0H , C=80\mu F , R=40\Omega .

(a) Determine the source frequency which drives the circuit in resonance.

1594271673048

Answer:

Given,

Variable frequency supply voltage V = 230V

Inductance L=5.0H

Capacitance C=80\mu F=80*10^{-6}F

Resistance R=40\Omega

a) Resonance angular frequency in this circuit is given by :

w_{resonance}=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{5*80*10^{-6}}}=\frac{1000}{20}=50rad/sec

Hence this circuit will be in resonance when supply frequency is 50 rad/sec.

Q7.11 (b) Figure shows a series LCR circuit connected to a variable frequency 230\: V source.

L=5.0H , C=80\mu F , R=40\Omega .

(b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency.

1594271689346

Answer:

Given,

Variable frequency supply voltage V = 230V

Inductance L=5.0H

Capacitance C=80\mu F=80*10^{-6}F

Resistance R=40\Omega

Now,

The impedance of the circuit is

Z=\sqrt{(wL-\frac{1}{wC})^2+R^2}

at Resonance Condition

: wL=\frac{1}{wC}

Z=R=40\Omega

Hence, Impedance at resonance is 40 \Omega .

Now, at resonance condition, impedance is minimum which means current is maximum which will happen when we have a peak voltage, so

Current in the Resonance circuit is Given by

I_{resonance}=\frac{V_{peak}}{Z}=\frac{\sqrt{2}* 230}{40}=8.13A

Hence amplitude of the current at resonance is 8.13A.

Q7.11 Figure shows a series LCR circuit connected to a variable frequency 230\: V source.

L=5.0H , C=80\mu F , R=40\Omega .

Determine the rms potential drops across the three elements of the circuit. Show that the potential drop across the LC combination is zero at the resonating frequency.

1594271702160

Answer:

Potential difference across any element = I_{rms}*(impedance)

I_{rms}=\frac{I_{peak}}{\sqrt{2}}=\frac{8.13}{\sqrt{2}}=5.85A

Now

The potential difference across the capacitor:

V_{capacitor}=I_{rms}*\left (\frac{1}{w_{resonance}C} \right ) =5.85*\left ( \frac{1}{50*80*10^{-6}} \right )=1437.5V

The potential difference across the inductor

V_{inductor}=I_{rms}*(w_{resonance}L) =5.85* 50*5=1437.5V


The potential difference across Resistor

=40 Irms=230V

The potential difference across LC combination

V_{LC}=I_{rms}*\left ( wL-\frac{1}{wC} \right )=5.85*0=0

Hence at resonating, frequency potential difference across LC combination is zero.

NCERT Class 12 Physics Chapter 7 Exercise Solutions: Additional Exercise

Q7.12 (a) An LC circuit contains a 20mH inductor and a 50\mu F capacitor with an initial charge of 10mC . The resistance of the circuit is negligible. Let the instant the circuit is closed be t=0 .

What is the total energy stored initially? Is it conserved during LC oscillations?

Answer :

Given,

The inductance of the inductor:

L=20mH=20*10^{-3}H

The capacitance of the capacitor :

C=50\mu F=50*10^{-6}F

The initial charge on the capacitor:

Q=10mC=10*10^{-3}C

Total energy present at the initial moment:

E_{initial}=\frac{Q^2}{2C}=\frac{(10*10^{-3})^2}{2*50*10^{-6}}=1J

Hence initial energy in the circuit is 1J. Since we don't have any power-consuming element like resistance in the circuit, the energy will be conserved

Q7.12 (b) An LC circuit contains a 20mH inductor and a 50\mu F capacitor with an initial charge of 10mC . The resistance of the circuit is negligible. Let the instant the circuit is closed be t=0 .

What is the natural frequency of the circuit?

Answer:

Given,

The inductance of the inductor:

L=20mH=20*10^{-3}H

The capacitance of the capacitor :

C=50\mu F=50*10^{-6}F

The initial charge on the capacitor:

Q=10mC=10*10^{-3}C

The natural angular frequency of the circuit:

w_{natural}=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{(20*10^{-3}*50*10^{-6})}}=10^3rad/sec

Hence the natural angular frequency of the circuit is 10^3rad/sec .

The natural frequency of the circuit:

f_{natural}=\frac{w_{natural}}{2\pi}=\frac{10^3}{2\pi}=159Hz

Hence the natural frequency of the circuit is 159Hz.

Q7.12 (c-i) An LC circuit contains a 20mH inductor and a 50\mu F capacitor with an initial charge of 10mC . The resistance of the circuit is negligible. Let the instant the circuit is closed be t=0 .

( c) At what time is the energy stored

(i) completely electrical (i.e., stored in the capacitor)?

Answer:

at any instant, the charge on the capacitor is:

Q=Q_0cos(w_{natural}t)=Q_0cos(2\pi f_{natural}t)=Q_0cos\left ( \frac{2\pi t}{T} \right )

Where time period :

T=\frac{1}{f_{natural}}=\frac{1}{159}=6.28ms

Now, when the total energy is purely electrical, we can say that

Q=Q_0

Q_0=Q_0cos(\frac{2\pi}{T})

cos(\frac{2\pi t}{T})=1

this is possible when

t=0,\frac{T}{2},T,\frac{3T}{2}....

Hence Total energy will be purely electrical(stored in a capacitor) at

t=0,\frac{T}{2},T,\frac{3T}{2}.... .

Q7.12 (c-ii) An LC circuit contains a 20mH inductor and a 50\mu F capacitor with an initial charge of 10mC . The resistance of the circuit is negligible.
(C) Let the instant the circuit is closed be t=0 .

(ii) completely magnetic (i.e., stored in the inductor)?

Answer:

The stored energy will we purely magnetic when the pure electrical stored is zero. i.e. when the charge on the capacitor is zero, all energy will be stored in the inductor.

So, t for which charge on the capacitor is zero is

t=\frac{T}{4},\frac{3T}{4},\frac{5T}{4}..

Hence at these times, the total energy will be purely magnetic.

Q7.12 (d) An LC circuit contains a 20mH inductor and a 50\mu F capacitor with an initial charge of 10mC . The resistance of the circuit is negligible. Let the instant the circuit is closed be t=0 .

At what times is the total energy shared equally between the inductor and the capacitor?

Answer:

The energy will be shared equally when the energy in the capacitor is half of the maximum energy it can store.i.e.

\frac{Q^2}{2C}=\frac{1}{2}\frac{Q_0^2}{2C}

From here, we got

Q=\frac{Q_0}{\sqrt{2}}

So now, we know the charge on the capacitor, we can calculate the time for which

\frac{Q_0}{\sqrt{2}}=Q_0cos\left ( \frac{2\pi t}{T} \right )

\frac{1}{\sqrt{2}}=cos\left ( \frac{2\pi t}{T} \right )

From here,

t=\frac{T}{8},\frac{3T}{8},\frac{5T}{8}..

Hence for these times, the total energy will be shared equally between capacitor and inductor.

Q7.12 (e) An LC circuit contains a 20mH inductor and a 50\mu F capacitor with an initial charge of 10mC . The resistance of the circuit is negligible. Let the instant the circuit is closed be t=0 .

If a resistor is inserted in the circuit, how much energy is eventually dissipated as heat?

Answer:

If the resistance is added to the circuit, the whole energy will dissipate as heat eventually. energy will keep moving between the capacitor and inductor with reducing in magnitude in each cycle and eventually all energy will be dissipated.

Q 7.13 (a) A coil of inductance 0.50H and resistance 100\Omega is connected to a 240V , 50Hz ac supply. What is the maximum current in the coil?

Answer:

Given,

The inductance of the coil L=0.50H

the resistance of the coil R=100\Omega

Supply voltage V=240V

Supply voltage frequency f=50Hz

Now, as we know peak voltage = \sqrt2 (RMS Voltage)

Peak voltage

V_{peak}=\sqrt2*240=339.4V

The impedance of the circuit :

Z=\sqrt{R^2+(wL)^2}=\sqrt{100^2+(2\pi (.5) *50)^2}

Now peak current in the circuit :

I_{peak}=\frac{V_{peak}}{Z}=\frac{339}{\sqrt{100^2+(2\pi (.5) *50)^2}}=1.82A

Hence peak current is 1.82A in the circuit.

Q 7.13 (b) A coil of inductance 0.50H and resistance 100\Omega is connected to a 200V , 50Hz ac supply. What is the time lag between the voltage maximum and the current maximum?

Answer:

Let the voltage in the circuit be

V = V_0coswt and

Current in the circuit be

I = I_0cos(wt-\phi )

Where \phi is the phase difference between voltage and current.

V is maximum At

t = 0

I is maximum At

t=\frac{w}{\phi }

Hence, the time lag between voltage maximum and the current maximum is

\frac{w}{\phi } .

For phase difference \phi we have

tan\phi =\frac{wL}{R}=\frac{2\pi *50*0.5}{100}=1.57

\phi =57.5^0

t=\frac{\phi}{w}=\frac{57.5*\pi}{180*2\pi *50}=3.2ms

Hence time lag between the maximum voltage and the maximum current is 3.2ms

Q7.14 Obtain the answers (a) to (b) in Exercise 7.13 if the circuit is connected to a high-frequency supply (240 V\: ,10 kHz) . Hence, explain the statement that at very high frequency, an inductor in a circuit nearly amounts to an open circuit. How does an inductor behave in a dc circuit after the steady state?

Answer:

Given,

The inductance of the coil L=50H

the resistance of the coil R=100\Omega

Supply voltage V=240V

Supply voltage frequency f=10kHz

a)

Now, as we know peak voltage = \sqrt2 (RMS Voltage)

Peak voltage V_{peak}=\sqrt2*240=339.4V

Now,

The impedance of the circuit :

Z=\sqrt{R^2+(wL)^2}=\sqrt{100^2+(2\pi 10*10^3 *50)^2}

Now peak current in the circuit :

I_{peak}=\frac{V_{peak}}{Z}=\frac{339}{\sqrt{100^2+(2\pi 10*10^3 *50)^2}}=1.1*10^{-2}A

Hence peak current is 1.1*10^{-2}A in the circuit.

The current in the circuit is very small, which is one of the indications of inductor working as a nearly open circuit in the case of high frequency.

b)

For phase difference \phi we have

tan\phi =\frac{wL}{R}=\frac{2\pi *10*10^3*0.5}{100}=100\pi

\phi =89.82^0

Now

t=\frac{\phi}{w}=\frac{89.82*\pi}{180*2\pi *10^3}=25\mu s

Hence time lag between the maximum voltage and the maximum current is 25\mu s .

In the DC circuit, after attaining the steady state, inductor behaves line short circuit as w is Zero.

Q7.15 (a) A 100\mu F capacitor in series with a 40\Omega resistance is connected to a 110\: V , 60\: Hz supply.
What is the maximum current in the circuit?

Answer:

Given,

The capacitance of the capacitor C=100\mu F

The resistance of the circuit R=40\Omega

Voltage supply V = 100V

Frequency of voltage supply f=60Hz

The maximum current in the circuit

I_{max}=\frac{V_{max}}{Z}=\frac{\sqrt{2}V}{\sqrt{R^2+\left ( \frac{1}{wC} \right )^2}}=\frac{\sqrt{2}*110}{\sqrt{40^2+\left ( \frac{1}{2\pi *60*100*10^{-6}} \right )^2}}=3.24A

Hence maximum current in the circuit is 3.24A.

Q 7.15 (b) A 100\mu F capacitor in series with a 40\Omega resistance is connected to a 110\; V , 60\: Hz supply. What is the time lag between the current maximum and the voltage maximum?

Answer:

In the case of a capacitor, we have

tan\phi=\frac{\frac{1}{wC}}{R}=\frac{1}{wCR}

So,

tan\phi=\frac{1}{wCR}=\frac{1}{2\pi 60 *100*10^{-6}*40}=0.6635

\phi=33.56^0

So the time lag between max voltage and the max current is :

t=\frac{\phi }{w}=\frac{33.56\pi}{180*2\pi*60}=1.55ms

Q7.16 Obtain the answers to (a) and (b) in Exercise 7.15 if the circuit is connected to a 110\: V , 10kHz supply? Hence, explain the statement that a capacitor is a conductor at very high frequencies. Compare this behaviour with that of a capacitor in a dc circuit after the steady state.

Answer:

Given,

The capacitance of the capacitor C=100\mu F

The resistance of the circuit R=40\Omega

Voltage supply V = 100V

Frequency of voltage supply f=12kHz

The maximum current in the circuit

I_{max}=\frac{V_{max}}{Z}=\frac{\sqrt{2}V}{\sqrt{R^2+\left ( \frac{1}{wC} \right )^2}}=\frac{\sqrt{2}*110}{\sqrt{40^2+\left ( \frac{1}{2\pi *12*10^3*100*10^{-6}} \right )^2}}=3.9A

Hence maximum current in the circuit is 3.9A.

b)

In the case of capacitor, we have

tan\phi=\frac{\frac{1}{wC}}{R}=\frac{1}{wCR}

So,

tan\phi=\frac{1}{wCR}=\frac{1}{2\pi 10*10^3 *100*10^{-6}*40}=\frac{1}{96\pi}

\phi=0.2^0

So the time lag between max voltage and max current is :

t=\frac{\phi }{w}=\frac{0.2\pi}{180*2\pi*60}=0.04\mu s

At high frequencies, \phi tends to zero. which indicates capacitor acts as a conductor at high frequencies.

In the DC circuit, after a steady state is achieved, Capacitor acts like an open circuit.

Q7.17 Keeping the source frequency equal to the resonating frequency of the series LCR circuit, if the three elements, L , C and R are arranged in parallel, show that the total current in the parallel LCR circuit is minimum at this frequency. Obtain the current rms value in each branch of the circuit for the elements and source specified in Exercise 7.11 for this frequency.

Answer:

As we know, in the case of a parallel RLC circuit:


\frac{1}{Z}=\sqrt{\frac{1}{R^2}+(wC-\frac{1}{wL})^2}

I=\frac{V}{Z}={V}{\sqrt{\frac{1}{R^2}+\left (wC- \frac{1}{wL} \right )^2}}

The current will be minimum when

wC=\frac{1}{wL}

Which is also the condition of natural frequency. Hence the total current is minimum when source frequency is equal to the natural frequency.

RMS value of current in R

I_{rms}=\frac{V_{rms}}{R}=\frac{230}{40}=5.75A

RMS value in Inductor

I_{inductor}=\frac{V_{rms}}{wL}=\frac{230}{5*50}=0.92A

RMS value in capacitor

I_{capacitor}=\frac{V_{rms}}{1/wL}={230*50*80*10^{-6}}=0.92A

Capacitor current and inductor current will cancel out each other so the current flowing in the circuit is 5.75A.

Q7.18 (a) A circuit containing a 80mH inductor and a 60\mu F capacitor in series is connected to a 230\: V , 50\: Hz supply. The resistance of the circuit is negligible. Obtain the current amplitude and rms values.

Answer:

The inductance of the inductor L=80mH=80*10^3H

The capacitance of the capacitor C=60\mu F

Voltage supply V = 230V

Frequency of voltage supply f=50Hz .

Here, we have

V=V_{max}sinwt=V_{max}sin2\pi ft

Impedance

Z=\sqrt{R^2+\left ( wL-\frac{1}{wC} \right )^2}

Z=\sqrt{0^2+\left ( 2\pi*50*80*10^{-3}-\frac{1}{2\pi 50*60*10^{-6}} \right )^2}=8\pi-\frac{1000}{6\pi }

Now,

Current in the circuit will be

I=\frac{V}{Z}=\frac{V_{max}sinwt}{Z\angle \phi }=I_{max}sin(wt-\phi )

where,

I_{max}=\frac{V_{max}}{Z}=\frac{\sqrt{2}*230}{8\pi-\frac{1000}{6\pi}}=-11.63A

The negative sign is just a matter of the direction of current.so,

I=11.63sin(wt-\phi )

here

tan\phi=\frac{wL-\frac{1}{wC}}{R}

But, since the value of R is zero(since our circuit have only L and C)

\phi=90^0

Hence

I=11.63sin(wt-\frac{\pi}{2} )

Now,

RMS value of this current:

I_{rms}=\frac{I_{max}}{\sqrt{2}}=\frac{11.63}{\sqrt{2}}=8.22A .

Q7.18 (b) A circuit containing a 80mH inductor and a 60\mu F capacitor in series is connected to a 230\: V , 50\: Hz supply. The resistance of the circuit is negligible. Obtain the rms values of potential drops across each element.

Answer:

As we know,

RMS potential drop across an element with impedance Z:

V_{element}=I_{rms}Z_{element}

SO,

RMS potential difference across inductor:

V_{inductor}=I_{rms}*wL=8.22*2\pi *60*80*10^{-3}=206.61V

RMS potential drop across capacitor

V_{capacitor}=I_{rms}*\frac{1}{wC}=8.22*\frac{1}{2\pi*60*60*10^{-6}}=436.3V

Q7.18 (c) A circuit containing a 80mH inductor and a 60\mu F capacitor in series is connected to a 230\: V , 50\: Hz supply. The resistance of the circuit is negligible

(c) What is the average power transferred to the inductor?

Answer:

Since

I=I_{max}sin(wt-\phi )

Current flowing in the circuit is sinusoidal and hence average power will be zero as the average of sin function is zero.in other words, the inductor will store energy in the positive half cycle of the sin (0 degrees to 180 degrees) and will release that energy in the negative half cycle(180 degrees to 360 degrees), and hence average power is zero.

Q7.18 (d) A circuit containing a 80mH inductor and a 60\mu F capacitor in series is connected to a 230\: V , 50\: Hz supply. The resistance of the circuit is negligible . What is the average power transferred to the capacitor?

Answer:

As we know,

Average power P=VIcos\theta where \theta is the phase difference between voltage and current.

Since in the circuit, phase difference \theta is \pi/2 , the average power is zero.

Q7.18 (e) A circuit containing a 80mH inductor and a 60\mu F capacitor in series is connected to a 230\: V , 50\: Hz supply. The resistance of the circuit is negligible . What is the total average power absorbed by the circuit? [‘Average’ implies ‘averaged over one cycle’.]

Answer:

Since the phase difference between voltage and current is 90 degree, even the total power absorbed by the circuit is zero. This is an ideal circuit, we can not have any circuit in practical that consumes no power, that is because practically resistance of any circuit is never zero. Here only inductor and capacitor are present and none of them consumes energy, they just store it and transfer it like they are doing it in this case.

Q7.19 Suppose the circuit in Exercise 7.18 has a resistance of 15\Omega . Obtain the average power transferred to each element of the circuit, and the total power absorbed.

Answer:

The inductance of the inductor L=80mH=80*10^3H

The capacitance of the capacitor C=60\mu F

The resistance of a 15\Omega resistor

Voltage supply V = 230V

Frequency of voltage supply f=50Hz

As we know,

Impedance

Z=\sqrt{R^2+\left ( wL-\frac{1}{wC} \right )^2}

Z=\sqrt{15^2+\left ( 2\pi*50*80*10^{-3}-\frac{1}{2\pi 50*60*10^{-6}} \right )^2}=31.728

Current flowing in the circuit :

I=\frac{V}{Z}=\frac{230}{31.72}=7.25A

Now,

Average power transferred to the resistor:

P_{resistor}=I^2R=(7.25)^2*15=788.44W

Average power transferred to the inductor = 0

Average power transferred to the capacitor = 0:

Total power absorbed by circuit :

P_{resistor}+p_{inductor}+P_{capacitor}=788.44+0+0=788.44W

Hence circuit absorbs 788.44W.

Q7.20 (a) A series LCR circuit with L=0.12H , C=480nF , R=23\Omega is connected to a 230V variable frequency supply.

What is the source frequency for which current amplitude is maximum. Obtain this maximum value.

Answer:

The inductance of the inductor L=0.12H

The capacitance of the capacitor C=480n F

The resistance of the resistor R=23\Omega

Voltage supply V = 230V

Frequency of voltage supply f=50Hz

As we know,

the current amplitude is maximum at the natural frequency of oscillation, which is

w_{natural}=\sqrt\frac{1}{LC}=\frac{1}{\sqrt{0.12*480*10^{-9}})}=4166.67rad/sec

Also, at this frequency,

Z=R=23

SO,

The maximum current in the circuit :

I_{max}=\frac{V_{max}}{Z}=\frac{V_{max}}{R}=\frac{\sqrt{2*}230}{23}=14.14A

Hence maximum current is 14.14A.

Q7.20 (b) A series LCR circuit with L=0.12H , C=480nF , R=23\Omega is connected to a 230V variable frequency supply.

What is the source frequency for which average power absorbed by the circuit is maximum. Obtain the value of this maximum power.

Answer:

Since the resistor is the only element in the circuit which consumes the power, the maximum absorbed power by circuit will be maximum when power absorbed by the resistor will be maximum. power absorbed by the resistor will be maximum at when current is maximum which is the natural frequency case,

Hence when source frequency will be equal to the natural frequency, the power absorbed will be maximum.

Hence frequency

f=\frac{w_r}{2\pi}=\frac{4166.67}{2\pi}=663.48Hz

Maximum Power Absorbed

P=I^2R=(14.14)^2*23=2299.3W .

Q7.20 (c) A series LCR circuit with L=0.12H , C=480nF , R=23\Omega is connected to a 230V variable frequency supply . What is the Q factor of the given circuit?

Answer:

The value of maximum angular frequency is calculated in the first part of the question and whose magnitude is 4166.67

Q-factor of any circuit is given by

Q=\frac{w_rL}{R}=\frac{4166.67*0.12}{23}=21.74

Hence Q-factor for the circuit is 21.74.

Q7.20 (d) A series LCR circuit with L=0.12H , C=480nF , R=23\Omega is connected to a 230V variable frequency supply. For which frequencies of the source is the power transferred to the circuit half the power at resonant frequency? What is the current amplitude at these frequencies?

Answer:

As

Power P=I^2R

Power P will be half when the current I is 1/\sqrt{2} times the maximum current.

As,

I =I_{max}Sin(wt-\phi)

At half powerpoint :

\frac{i_{max}}{\sqrt{2}} =I_{max}Sin(wt-\phi)

\frac{1}{\sqrt{2}} =Sin(wt-\phi)

wt=\phi+\frac{\pi}{4}

here,

\phi=tan^{-1}(\frac{wL-\frac{1}{wC}}{R})

On putting values, we get, two values of w for which

wt=\phi+\frac{\pi}{4}

And they are:

w_1=678.75Hz

w_2=648.22Hz

Also,

The current amplitude at these frequencies

I_{halfpowerpoint}=\frac{I_{max}}{\sqrt{2}}=\frac{14.14}{1.414}=10A

Q7.21 Obtain the resonant frequency and \varrho -factor of a series LCR circuit with L=0.3H , C=27\mu F , and R=7.4\Omega . It is desired to improve the sharpness of the resonance of the circuit by reducing its ‘full width at half maximum’ by a factor of 2. Suggest a suitable way.

Answer:

The inductance of the inductor L=0.3H

The capacitance of the capacitor C=27\mu F

The resistance of the resistor R=7.4\Omega

Now,

Resonant frequency

w_r=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{0.3*27*10^{-6}}}=111.11rad/sec

Q-Factor of the circuit

Q=\frac{w_rL}{R}=\frac{111.11*0.3}{7.4}=45.0446

Now, to improve the sharpness of resonance by reducing its full width at half maximum, by a factor of 2 without changing w_r ,

we have to change the resistance of the resistor to half of its value, that is

R_{new}=\frac{R}{2}=\frac{7.4}{2}=3.7\Omega

Q7.22 (a) In any ac circuit, is the applied instantaneous voltage equal to the algebraic sum of the instantaneous voltages across the series elements of the circuit? Is the same true for rms voltage?

Answer:

Yes, at any instant the applied voltage will be distributed among all element and the sum of the instantaneous voltage of all elements will be equal to the applied. But this is not the case in RMS because all elements are varying differently and they may not be in the phase.

7.22 (b) Answer the following questions: A capacitor is used in the primary circuit of an induction coil.

Answer:

Yes, we use capacitors in the primary circuit of an induction coil to avoid sparking. when the circuit breaks, a large emf is induced and the capacitor gets charged from this avoiding the case of sparking and short circuit.

Q 7.22 (c) Answer the following questions:

An applied voltage signal consists of a superposition of a dc voltage and an ac voltage of high frequency. The circuit consists of an inductor and a capacitor in series. Show that the dc signal will appear across C and the ac signal across L .

Answer:

For a high frequency, the inductive reactance and capacitive reactance:

X_L=wL=Large \:value\: And \:X_C = \frac{1}{wC}=Very\:small

Hence the capacitor does not offer resistance to a higher frequency, so the ac voltage appears across L.

Similarly

For DC, the inductive reactance and capacitive reactance:

X_L=wL=Very\:small\: And \:X_C = \frac{1}{wC}=Large \:value

Hence DC signal appears across Capacitor only.

Q 7.22 (d) Answer the following questions:

A choke coil in series with a lamp is connected to a dc line. The lamp is seen to shine brightly. Insertion of an iron core in the choke causes no change in the lamp’s brightness. Predict the corresponding observations if the connection is to an ac line.

Answer:

For a steady state DC, the increasing inductance value by inserting iron core in the choke, have no effect in the brightness of the connected lamp, whereas, for ac when the iron core is inserted, the light of the lamp will shine less brightly because of increase in inductive impedance.

Q7.22 (e) Answer the following questions:

Why is choke coil needed in the use of fluorescent tubes with ac mains? Why can we not use an ordinary resistor instead of the choke coil?

Answer:

We need choke coil in the use of fluorescent tubes with ac mains to reduce the voltage across the tube without wasting much power. If we use simply resistor for this purpose, there will be more power loss, hence we do not prefer it.


Q7.23 A power transmission line feeds input power at 2300 V to a stepdown transformer with its primary windings having 40000 turns. What should be the number of turns in the secondary in order to get output power at 230 V ?

Answer:

Given,

Input voltage:

V_{input}=2300V

Number of turns in the primary coil

N_{primary}= 4000

Output voltage:

V_{output}=230V

Now,

Let the number of turns in secondary be

N=N_{secondary}

Now as we know, in a transformer,

\frac{V_{primary}}{V_{secondary}}=\frac{N_{primary}}{N_{secondary}}

{N_{secondary}} =\frac{V_{secondary}}{V_{primary}}*N_{primary}=\frac{230}{2300}*4000=400

Hence the number of turns in secondary winding id 400.

Q7.24 At a hydroelectric power plant, the water pressure head is at a height of 300m and the water flow available is 100m^{3}s^{-1} . If the turbine generator efficiency is 60^{0}/_{0} , estimate the electric power availablefrom the plant (g=9.8ms^{-2}) .

Answer:

Given,

Height of the water pressure head

h=300m

The volume of the water flow per second

V=100m^3s^{-1}

Turbine generator efficiency

\eta =0.6

Mass of water flowing per second

M=100*10^3=10^5kg

The potential energy stored in the fall for 1 second

P=Mgh=10^5*9.8*300=294*10^6J

Hence input power

P_{input}=294*10^6J/s

Now as we know,

\eta =\frac{P_{output}}{P_{input}}

P_{output}=\eta *P_{input}=0.6*294*10^6=176.4*10^6W

Hence output power is 176.4 MW.

7.25 (b) A small town with a demand of 800 kW of electric power at 220V is situated 15km away from an electric plant generating power at 440V . The resistance of the two wire line carrying power is 0.5\Omega per km. The town gets power from the line through a 4000-220V step-down transformer at a sub-station in the town.

How much power must the plant supply, assuming there is negligible power loss due to leakage?

Answer:

Power required

P=800kW=800*10^3kW

The total resistance of the two-wire line

R=2*15*0.5=15\Omega

Input voltage

V_{input}=4000V

Output voltage:

V_{output}=220V

RMS current in the wireline

I=\frac{P}{V_{input}}=\frac{800*10^3}{4000}=200A

Now,

Total power delivered by plant = line power loss + required electric power = 800 + 600 = 1400kW.

Q7.25 (c) A small town with a demand of 800 kW of electric power at 220V is situated 15km away from an electric plant generating power at 440V . The resistance of the two wire line carrying power is 0.5\Omega per km. The town gets power from the line through a 4000-220V step-down transformer at a sub-station in the town. Characterise the step up transformer at the plant.

Answer:

Power required

P=800kW=800*10^3kW

The total resistance of the two-wire line

R=2*15*0.5=15\Omega

Input Voltage

V_{input}=4000V

Output Voltage:

V_{output}=220V

RMS Current in the wireline

I=\frac{P}{V_{input}}=\frac{800*10^3}{4000}=200A

Now,

Voltage drop in the power line = IR=200*15=3000V

Total voltage transmitted from the plant = 3000+4000=7000

as power is generated at 440V, The rating of the power plant is 440V-7000V.

Q7.26 Do the same exercise as above with the replacement of the earlier transformer by a 40,000-220V step-down transformer (Neglect, as before, leakage losses though this may not be a good assumption any longer because of the very high voltage transmission involved). Hence, explain why high voltage transmission is preferred?

Answer:

Power required

P=800kW=800*10^3kW

The total resistance of the two-wire line

R=2*15*0.5=15\Omega

Input Voltage

V_{input}=40000V

Output Voltage:

V_{output}=220V

RMS current in the wireline

I=\frac{P}{V_{input}}=\frac{800*10^3}{40000}=20A

Now,

a) power loss in the line

P_{loss}=I^2R=20^2*15=6kW

b)

Power supplied by plant = 800 kW + 6 kW = 806kW.

c)

Voltage drop in the power line = IR=20*15=300V

Total voltage transmitted from the plant = 300+40000=40300

as power is generated at 440V, The rating of the power plant is 440V-40300V.

We prefer high voltage transmission because power loss is a lot lesser than low voltage transmission.

NCERT Solutions for class 12 physics: Chapter-Wise

Here are the exercise-wise solutions of the NCERT Class 12 physics book:

Alternating current class 12 solutions: Important Formulas and Diagrams

Important Formulas and Diagrams serve as a crucial resource for exam preparation, be it for board exams or competitive ones like JEE and NEET. These formulas and diagrams condense complex concepts, aiding students in quick revision and problem-solving, ultimately boosting their confidence and performance in exams.

  • Alternating Current and Voltage

JEE Main Highest Scoring Chapters & Topics
Just Study 40% Syllabus and Score upto 100%
Download EBook

I=Iosinωt Alternating current

V=Vosinωt Alternating Voltage

Where: Io is the peak value of current and Vo is the peak value of voltage

  • Periodic Time:

JEE Main Important Mathematics Formulas

As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

JEE Main Important Physics formulas

As per latest 2024 syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters

T=\frac{2 \pi}{\omega}

  • Frequency(f):

\mathrm{f}=1 / \mathrm{T}=\omega / 2 \Pi

  • Mean Value of An Alternating Current:

V_{r m s}=\frac{V_0}{\sqrt{2}}=0.707 V_0

  • Impedance and Resistance

For L-R series circuit: Z_{R L}=\sqrt{R^2+X L^2}

For R-C series circuit: Z_{R C}=\sqrt{R^2+X C^2}

  • Series LRC Circuit

\begin{aligned} & Z_{R L C}= \sqrt{R^2+\left(X_L-X_C\right)^2} \\ & \sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2} \end{aligned}

At resonant frequency,

\begin{aligned} X_C & =X_L \\ \frac{1}{\omega C} & =\omega L \\ \omega_r & =\frac{1}{\sqrt{L C}} \Rightarrow f_r=\frac{1}{2 \pi \sqrt{L C}} \end{aligned}

  • Voltage and power in a transformer

The voltage across secondary coil: V_s=\left(\frac{N_s}{N_p}\right) V_p


Input and Output power: \begin{aligned} & P_{\text {input }}=P_{\text {output }} \\ & V_P I_P=V_S I_S \end{aligned}

NCERT Class 12 Physics Chapter 7 Alternating Current: Important topics

Class 12 NCERT introduces the concept of different ac circuits including the elements, resistor, capacitor, inductor and ac voltage source. Single-phase ac circuits and their questions are discussed in the NCERT Solutions for Class 12 physics chapter 7. The following are the main headings covered in physics Class 12 chapter 7.

  • AC voltage applied to a resistor- In this section, the analysis of a resistor circuit applied with an ac voltage of Vmsinωt is done and the terms like RMS and peak voltage and current is introduced. Also, the power dissipated in the resistor circuit is discussed. Question 1 and 2 of chapter 7 Physics Class 12 NCERT solutions given are based on this topic. The main formulas used here are:

RMSvalue=\frac{peakvalue}{\sqrt{2}}

  • The power consumed in a resistor circuit connected with ac source, P= Vrms * Irms

  • In the next part of Alternating Current class 12, phasor representation of voltage and current is introduced.

  • Ch 7 physics Class 12 gives a small analysis of circuits involving R, L and C. Alternating Current Class 12 NCERT pdf solutions discuss questions based on series RL, RC and RLC circuits. Also, chapter 7 Class 12 physics discuss circuit including capacitance and inductance only

  • The concepts of LC oscillation, resonance and transformer and their basic equations are discussed.

Key Features of Class 12 Physics NCERT Solutions for Chapter 7 Alternating Current

  1. omprehensive Coverage: These ncert class 12 physics chapter 7 exercise solutions encompass all topics and questions presented in Chapter 7, ensuring a thorough understanding of alternating current.

  2. Detailed Explanations: Each ac current class 12 solution offers in-depth explanations, helping students grasp complex concepts.

  3. Clarity and Simplicity: The alternating current class 12 solutions are presented in clear and straightforward language, making it easier for students to comprehend.

  4. Practice Questions: Exercise questions are included for students to practice and assess their understanding.

  5. Exam Preparation: These physics chapter 7 class 12 solutions are instrumental in board exam preparation and provide valuable support for competitive exams like JEE and NEET.

  6. Foundation for Advanced Study: The concepts covered in this chapter are foundational for advanced physics and electrical engineering studies.

  7. Free Accessibility: These alternating current questions and answers pdf are available for free, ensuring accessibility to all students.

These features make Alternating Current Class 12 NCERT solutions a valuable tool for students, facilitating their success in exams and future studies.

Also Check NCERT Books and NCERT Syllabus here:

NCERT solutions subject wise

Significance of NCERT solutions for class 12 physics chapter 7 in board exam

  • From the unit electromagnetic induction and alternating current Class 12, around 10% of the questions are asked in the CBSE 12th board exam.

  • If all the Class 12 Physics Chapter 7 NCERT solutions are covered then it is easy to answer questions asked in board exams.

  • This chapter 7 of NCERT Class 12 physics solutions is also important for JEE Main, NEET, state board exams, and competitive exams.

  • Prepare the chapter well with the help of CBSE NCERT Class 12 Physics Chapter 7 solutions.

Also, check

Frequently Asked Question (FAQs)

1. Is the supply coming to our home alternating?

 Yes, the supply received in our home is alternating. Usually, domestic supplies are single-phase and supply to industries, factories etc are three-phase.

2. What type of questions are asked in board exams from the chapter alternating current?

In the previous year question papers, questions related to identifying the circuit elements are repeatedly asked. Along with this simple numerical questions are also asked from Alternating Current. For practising numerical, solve NCERT exercise questions, NCERT exemplar questions and previous year papers.

3. What is the weightage of chapter for JEE Main?

One question for JEE main can be expected from the Class 12 chapter Alternating Current.

4. How important is the chapter for NEET?

One or two questions may be asked from NCERT chapter Alternating Current for NEET exam.

5. What is the difference between alternating current(AC) and direct current(DC)?

AC and DC are two types of electrical current that differ in the direction of electron flow. AC periodically changes direction while DC flows in only one direction. AC is commonly used for power transmission over long distances, while DC is used for electronic devices that require constant voltage or current. AC generators are simpler and cheaper to build than DC generators.

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

I hope this was helpful!

Good Luck

Hello dear,

If you was not able to clear 1st compartment and now you giving second compartment so YES, you can go for your improvement exam next year but if a student receives an improvement, they are given the opportunity to retake the boards as a private candidate the following year, but there are some requirements. First, the student must pass all of their subjects; if they received a compartment in any subject, they must then pass the compartment exam before being eligible for the improvement.


As you can registered yourself as private candidate for giving your improvement exam of 12 standard CBSE(Central Board of Secondary Education).For that you have to wait for a whole year which is bit difficult for you.


Positive side of waiting for whole year is you have a whole year to preparing yourself for your examination. You have no distraction or something which may causes your failure in the exams. In whole year you have to stay focused on your 12 standard examination for doing well in it. By this you get a highest marks as a comparison of others.


Believe in Yourself! You can make anything happen


All the very best.

Hello Student,

I appreciate your Interest in education. See the improvement is not restricted to one subject or multiple subjects  and  we cannot say if improvement in one subject in one year leads to improvement in more subjects in coming year.

You just need to have a revision of all subjects what you have completed in the school. have a revision and practice of subjects and concepts helps you better.

All the best.

If you'll do hard work then by hard work of 6 months you can achieve your goal but you have to start studying for it dont waste your time its a very important year so please dont waste it otherwise you'll regret.

Yes, you can take admission in class 12th privately there are many colleges in which you can give 12th privately.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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4 Jobs Available
Data Analyst

The invention of the database has given fresh breath to the people involved in the data analytics career path. Analysis refers to splitting up a whole into its individual components for individual analysis. Data analysis is a method through which raw data are processed and transformed into information that would be beneficial for user strategic thinking.

Data are collected and examined to respond to questions, evaluate hypotheses or contradict theories. It is a tool for analyzing, transforming, modeling, and arranging data with useful knowledge, to assist in decision-making and methods, encompassing various strategies, and is used in different fields of business, research, and social science.

3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
Underwriter

An underwriter is a person who assesses and evaluates the risk of insurance in his or her field like mortgage, loan, health policy, investment, and so on and so forth. The underwriter career path does involve risks as analysing the risks means finding out if there is a way for the insurance underwriter jobs to recover the money from its clients. If the risk turns out to be too much for the company then in the future it is an underwriter who will be held accountable for it. Therefore, one must carry out his or her job with a lot of attention and diligence.

3 Jobs Available
Finance Executive
3 Jobs Available
Operations Manager

Individuals in the operations manager jobs are responsible for ensuring the efficiency of each department to acquire its optimal goal. They plan the use of resources and distribution of materials. The operations manager's job description includes managing budgets, negotiating contracts, and performing administrative tasks.

3 Jobs Available
Investment Director

An investment director is a person who helps corporations and individuals manage their finances. They can help them develop a strategy to achieve their goals, including paying off debts and investing in the future. In addition, he or she can help individuals make informed decisions.

2 Jobs Available
Welding Engineer

Welding Engineer Job Description: A Welding Engineer work involves managing welding projects and supervising welding teams. He or she is responsible for reviewing welding procedures, processes and documentation. A career as Welding Engineer involves conducting failure analyses and causes on welding issues. 

5 Jobs Available
Transportation Planner

A career as Transportation Planner requires technical application of science and technology in engineering, particularly the concepts, equipment and technologies involved in the production of products and services. In fields like land use, infrastructure review, ecological standards and street design, he or she considers issues of health, environment and performance. A Transportation Planner assigns resources for implementing and designing programmes. He or she is responsible for assessing needs, preparing plans and forecasts and compliance with regulations.

3 Jobs Available
Plumber

An expert in plumbing is aware of building regulations and safety standards and works to make sure these standards are upheld. Testing pipes for leakage using air pressure and other gauges, and also the ability to construct new pipe systems by cutting, fitting, measuring and threading pipes are some of the other more involved aspects of plumbing. Individuals in the plumber career path are self-employed or work for a small business employing less than ten people, though some might find working for larger entities or the government more desirable.

2 Jobs Available
Construction Manager

Individuals who opt for a career as construction managers have a senior-level management role offered in construction firms. Responsibilities in the construction management career path are assigning tasks to workers, inspecting their work, and coordinating with other professionals including architects, subcontractors, and building services engineers.

2 Jobs Available
Urban Planner

Urban Planning careers revolve around the idea of developing a plan to use the land optimally, without affecting the environment. Urban planning jobs are offered to those candidates who are skilled in making the right use of land to distribute the growing population, to create various communities. 

Urban planning careers come with the opportunity to make changes to the existing cities and towns. They identify various community needs and make short and long-term plans accordingly.

2 Jobs Available
Highway Engineer

Highway Engineer Job Description: A Highway Engineer is a civil engineer who specialises in planning and building thousands of miles of roads that support connectivity and allow transportation across the country. He or she ensures that traffic management schemes are effectively planned concerning economic sustainability and successful implementation.

2 Jobs Available
Environmental Engineer

Individuals who opt for a career as an environmental engineer are construction professionals who utilise the skills and knowledge of biology, soil science, chemistry and the concept of engineering to design and develop projects that serve as solutions to various environmental problems. 

2 Jobs Available
Naval Architect

A Naval Architect is a professional who designs, produces and repairs safe and sea-worthy surfaces or underwater structures. A Naval Architect stays involved in creating and designing ships, ferries, submarines and yachts with implementation of various principles such as gravity, ideal hull form, buoyancy and stability. 

2 Jobs Available
Orthotist and Prosthetist

Orthotists and Prosthetists are professionals who provide aid to patients with disabilities. They fix them to artificial limbs (prosthetics) and help them to regain stability. There are times when people lose their limbs in an accident. In some other occasions, they are born without a limb or orthopaedic impairment. Orthotists and prosthetists play a crucial role in their lives with fixing them to assistive devices and provide mobility.

6 Jobs Available
Veterinary Doctor
5 Jobs Available
Pathologist

A career in pathology in India is filled with several responsibilities as it is a medical branch and affects human lives. The demand for pathologists has been increasing over the past few years as people are getting more aware of different diseases. Not only that, but an increase in population and lifestyle changes have also contributed to the increase in a pathologist’s demand. The pathology careers provide an extremely huge number of opportunities and if you want to be a part of the medical field you can consider being a pathologist. If you want to know more about a career in pathology in India then continue reading this article.

5 Jobs Available
Speech Therapist
4 Jobs Available
Gynaecologist

Gynaecology can be defined as the study of the female body. The job outlook for gynaecology is excellent since there is evergreen demand for one because of their responsibility of dealing with not only women’s health but also fertility and pregnancy issues. Although most women prefer to have a women obstetrician gynaecologist as their doctor, men also explore a career as a gynaecologist and there are ample amounts of male doctors in the field who are gynaecologists and aid women during delivery and childbirth. 

4 Jobs Available
Oncologist

An oncologist is a specialised doctor responsible for providing medical care to patients diagnosed with cancer. He or she uses several therapies to control the cancer and its effect on the human body such as chemotherapy, immunotherapy, radiation therapy and biopsy. An oncologist designs a treatment plan based on a pathology report after diagnosing the type of cancer and where it is spreading inside the body.

3 Jobs Available
Audiologist

The audiologist career involves audiology professionals who are responsible to treat hearing loss and proactively preventing the relevant damage. Individuals who opt for a career as an audiologist use various testing strategies with the aim to determine if someone has a normal sensitivity to sounds or not. After the identification of hearing loss, a hearing doctor is required to determine which sections of the hearing are affected, to what extent they are affected, and where the wound causing the hearing loss is found. As soon as the hearing loss is identified, the patients are provided with recommendations for interventions and rehabilitation such as hearing aids, cochlear implants, and appropriate medical referrals. While audiology is a branch of science that studies and researches hearing, balance, and related disorders.

3 Jobs Available
Hospital Administrator

The hospital Administrator is in charge of organising and supervising the daily operations of medical services and facilities. This organising includes managing of organisation’s staff and its members in service, budgets, service reports, departmental reporting and taking reminders of patient care and services.

2 Jobs Available
Actor

For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs. 

4 Jobs Available
Acrobat

Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.

3 Jobs Available
Video Game Designer

Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages.

Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

3 Jobs Available
Radio Jockey

Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

3 Jobs Available
Choreographer

The word “choreography" actually comes from Greek words that mean “dance writing." Individuals who opt for a career as a choreographer create and direct original dances, in addition to developing interpretations of existing dances. A Choreographer dances and utilises his or her creativity in other aspects of dance performance. For example, he or she may work with the music director to select music or collaborate with other famous choreographers to enhance such performance elements as lighting, costume and set design.

2 Jobs Available
Videographer
2 Jobs Available
Multimedia Specialist

A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications. 

2 Jobs Available
Social Media Manager

A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.

2 Jobs Available
Copy Writer

In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook. 

5 Jobs Available
Journalist

Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

3 Jobs Available
Publisher

For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

3 Jobs Available
Vlogger

In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. 

Ever since internet costs got reduced the viewership for these types of content has increased on a large scale. Therefore, a career as a vlogger has a lot to offer. If you want to know more about the Vlogger eligibility, roles and responsibilities then continue reading the article. 

3 Jobs Available
Editor

Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

3 Jobs Available
Linguist

Linguistic meaning is related to language or Linguistics which is the study of languages. A career as a linguistic meaning, a profession that is based on the scientific study of language, and it's a very broad field with many specialities. Famous linguists work in academia, researching and teaching different areas of language, such as phonetics (sounds), syntax (word order) and semantics (meaning). 

Other researchers focus on specialities like computational linguistics, which seeks to better match human and computer language capacities, or applied linguistics, which is concerned with improving language education. Still, others work as language experts for the government, advertising companies, dictionary publishers and various other private enterprises. Some might work from home as freelance linguists. Philologist, phonologist, and dialectician are some of Linguist synonym. Linguists can study French, German, Italian

2 Jobs Available
Public Relation Executive
2 Jobs Available
Travel Journalist

The career of a travel journalist is full of passion, excitement and responsibility. Journalism as a career could be challenging at times, but if you're someone who has been genuinely enthusiastic about all this, then it is the best decision for you. Travel journalism jobs are all about insightful, artfully written, informative narratives designed to cover the travel industry. Travel Journalist is someone who explores, gathers and presents information as a news article.

2 Jobs Available
Welding Engineer

Welding Engineer Job Description: A Welding Engineer work involves managing welding projects and supervising welding teams. He or she is responsible for reviewing welding procedures, processes and documentation. A career as Welding Engineer involves conducting failure analyses and causes on welding issues. 

5 Jobs Available
QA Manager
4 Jobs Available
Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product. 

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
Production Manager
3 Jobs Available
Merchandiser
2 Jobs Available
QA Lead

A QA Lead is in charge of the QA Team. The role of QA Lead comes with the responsibility of assessing services and products in order to determine that he or she meets the quality standards. He or she develops, implements and manages test plans. 

2 Jobs Available
Metallurgical Engineer

A metallurgical engineer is a professional who studies and produces materials that bring power to our world. He or she extracts metals from ores and rocks and transforms them into alloys, high-purity metals and other materials used in developing infrastructure, transportation and healthcare equipment. 

2 Jobs Available
Azure Administrator

An Azure Administrator is a professional responsible for implementing, monitoring, and maintaining Azure Solutions. He or she manages cloud infrastructure service instances and various cloud servers as well as sets up public and private cloud systems. 

4 Jobs Available
AWS Solution Architect

An AWS Solution Architect is someone who specializes in developing and implementing cloud computing systems. He or she has a good understanding of the various aspects of cloud computing and can confidently deploy and manage their systems. He or she troubleshoots the issues and evaluates the risk from the third party. 

4 Jobs Available
QA Manager
4 Jobs Available
Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available
ITSM Manager
3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack 

3 Jobs Available
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