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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.

    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.

    Free download ncert class 12 physics chapter 7 exercise solutions PDF for CBSE exam.

    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

    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:

    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.

    View All

    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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    Ethical Hacker

    A career as ethical hacker involves various challenges and provides lucrative opportunities in the digital era where every giant business and startup owns its cyberspace on the world wide web. Individuals in the ethical hacker career path try to find the vulnerabilities in the cyber system to get its authority. If he or she succeeds in it then he or she gets its illegal authority. Individuals in the ethical hacker career path then steal information or delete the file that could affect the business, functioning, or services of the organization.

    3 Jobs Available
    Database Architect

    If you are intrigued by the programming world and are interested in developing communications networks then a career as database architect may be a good option for you. Data architect roles and responsibilities include building design models for data communication networks. Wide Area Networks (WANs), local area networks (LANs), and intranets are included in the database networks. It is expected that database architects will have in-depth knowledge of a company's business to develop a network to fulfil the requirements of the organisation. Stay tuned as we look at the larger picture and give you more information on what is db architecture, why you should pursue database architecture, what to expect from such a degree and what your job opportunities will be after graduation. Here, we will be discussing how to become a data architect. Students can visit NIT Trichy, IIT Kharagpur, JMI New Delhi

    3 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
    Bank Probationary Officer (PO)

    A career as Bank Probationary Officer (PO) is seen as a promising career opportunity and a white-collar career. Each year aspirants take the Bank PO exam. This career provides plenty of career development and opportunities for a successful banking future. If you have more questions about a career as  Bank Probationary Officer (PO), what is probationary officer or how to become a Bank Probationary Officer (PO) then you can read the article and clear all your doubts. 

    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
    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
    Finance Executive

    A career as a Finance Executive requires one to be responsible for monitoring an organisation's income, investments and expenses to create and evaluate financial reports. His or her role involves performing audits, invoices, and budget preparations. He or she manages accounting activities, bank reconciliations, and payable and receivable accounts.  

    3 Jobs Available
    Investment Banker

    An Investment Banking career involves the invention and generation of capital for other organizations, governments, and other entities. Individuals who opt for a career as Investment Bankers are the head of a team dedicated to raising capital by issuing bonds. Investment bankers are termed as the experts who have their fingers on the pulse of the current financial and investing climate. Students can pursue various Investment Banker courses, such as Banking and Insurance, and Economics to opt for an Investment Banking career path.

    3 Jobs Available
    Bank Branch Manager

    Bank Branch Managers work in a specific section of banking related to the invention and generation of capital for other organisations, governments, and other entities. Bank Branch Managers work for the organisations and underwrite new debts and equity securities for all type of companies, aid in the sale of securities, as well as help to facilitate mergers and acquisitions, reorganisations, and broker trades for both institutions and private investors.

    3 Jobs Available
    Treasurer

    Treasury analyst career path is often regarded as certified treasury specialist in some business situations, is a finance expert who specifically manages a company or organisation's long-term and short-term financial targets. Treasurer synonym could be a financial officer, which is one of the reputed positions in the corporate world. In a large company, the corporate treasury jobs hold power over the financial decision-making of the total investment and development strategy of the organisation.

    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
    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
    Architect

    Individuals in the architecture career are the building designers who plan the whole construction keeping the safety and requirements of the people. Individuals in architect career in India provides professional services for new constructions, alterations, renovations and several other activities. Individuals in architectural careers in India visit site locations to visualize their projects and prepare scaled drawings to submit to a client or employer as a design. Individuals in architecture careers also estimate build costs, materials needed, and the projected time frame to complete a build.

    2 Jobs Available
    Landscape Architect

    Having a landscape architecture career, you are involved in site analysis, site inventory, land planning, planting design, grading, stormwater management, suitable design, and construction specification. Frederick Law Olmsted, the designer of Central Park in New York introduced the title “landscape architect”. The Australian Institute of Landscape Architects (AILA) proclaims that "Landscape Architects research, plan, design and advise on the stewardship, conservation and sustainability of development of the environment and spaces, both within and beyond the built environment". Therefore, individuals who opt for a career as a landscape architect are those who are educated and experienced in landscape architecture. Students need to pursue various landscape architecture degrees, such as M.Des, M.Plan to become landscape architects. If you have more questions regarding a career as a landscape architect or how to become a landscape architect then you can read the article to get your doubts cleared. 

    2 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
    Carpenter

    Carpenters are typically construction workers. They stay involved in performing many types of construction activities. It includes cutting, fitting and assembling wood.  Carpenters may help in building constructions, bridges, big ships and boats. Here, in the article, we will discuss carpenter career path, carpenter salary, how to become a carpenter, carpenter job outlook.

    2 Jobs Available
    Welder

    An individual who opts for a career as a welder is a professional tradesman who is skilled in creating a fusion between two metal pieces to join it together with the use of a manual or fully automatic welding machine in their welder career path. It is joined by intense heat and gas released between the metal pieces through the welding machine to permanently fix it. 

    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
    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

    A veterinary doctor is a medical professional with a degree in veterinary science. The veterinary science qualification is the minimum requirement to become a veterinary doctor. There are numerous veterinary science courses offered by various institutes. He or she is employed at zoos to ensure they are provided with good health facilities and medical care to improve their life expectancy.

    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
    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
    Surgical Technologist

    When it comes to an operation theatre, there are several tasks that are to be carried out before as well as after the operation or surgery has taken place. Such tasks are not possible without surgical tech and surgical tech tools. A single surgeon cannot do it all alone. It’s like for a footballer he needs his team’s support to score a goal the same goes for a surgeon. It is here, when a surgical technologist comes into the picture. It is the job of a surgical technologist to prepare the operation theatre with all the required equipment before the surgery. Not only that, once an operation is done it is the job of the surgical technologist to clean all the equipment. One has to fulfil the minimum requirements of surgical tech qualifications. 

    Also Read: Career as Nurse

    3 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
    Dentist

    Those who wish to make a dentist career in India must know that dental training opens up a universe of expert chances. Notwithstanding private practice, the present dental school graduates can pick other dental profession alternatives, remembering working in medical clinic crisis rooms, leading propelled lab examinations, teaching future dental specialists, or in any event, venturing to the far corners of the planet with International health and relief organizations.

    2 Jobs Available
    Health Inspector

    Individuals following a career as health inspectors have to face resistance and lack of cooperation while working on the sites. The health inspector's job description includes taking precautionary measures while inspecting to save themself from any external injury and the need to cover their mouth to avoid toxic substances. A health inspector does the desk job as well as the fieldwork. Health inspector jobs require one to travel long hours to inspect a particular place.

    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
    Talent Agent

    The career as a Talent Agent is filled with responsibilities. A Talent Agent is someone who is involved in the pre-production process of the film. It is a very busy job for a Talent Agent but as and when an individual gains experience and progresses in the career he or she can have people assisting him or her in work. Depending on one’s responsibilities, number of clients and experience he or she may also have to lead a team and work with juniors under him or her in a talent agency. In order to know more about the job of a talent agent continue reading the article.

    If you want to know more about talent agent meaning, how to become a Talent Agent, or Talent Agent job description then continue reading this article.

    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
    Producer

    An individual who is pursuing a career as a producer is responsible for managing the business aspects of production. They are involved in each aspect of production from its inception to deception. Famous movie producers review the script, recommend changes and visualise the story. 

    They are responsible for overseeing the finance involved in the project and distributing the film for broadcasting on various platforms. A career as a producer is quite fulfilling as well as exhaustive in terms of playing different roles in order for a production to be successful. Famous movie producers are responsible for hiring creative and technical personnel on contract basis.

    2 Jobs Available
    Fashion Blogger

    Fashion bloggers use multiple social media platforms to recommend or share ideas related to fashion. A fashion blogger is a person who writes about fashion, publishes pictures of outfits, jewellery, accessories. Fashion blogger works as a model, journalist, and a stylist in the fashion industry. In current fashion times, these bloggers have crossed into becoming a star in fashion magazines, commercials, or campaigns. 

    2 Jobs Available
    Photographer

    Photography is considered both a science and an art, an artistic means of expression in which the camera replaces the pen. In a career as a photographer, an individual is hired to capture the moments of public and private events, such as press conferences or weddings, or may also work inside a studio, where people go to get their picture clicked. Photography is divided into many streams each generating numerous career opportunities in photography. With the boom in advertising, media, and the fashion industry, photography has emerged as a lucrative and thrilling career option for many Indian youths.

    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 cost got reduced the viewership for these types of content has increased on a large scale. Therefore, the career as vlogger has a lot to offer. If you want to know more about the career as vlogger, how to become a vlogger, so on and so forth then continue reading the article. Students can visit Jamia Millia Islamia, Asian College of Journalism, Indian Institute of Mass Communication to pursue journalism degrees.

    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
    Fashion Journalist

    Fashion journalism involves performing research and writing about the most recent fashion trends. Journalists obtain this knowledge by collaborating with stylists, conducting interviews with fashion designers, and attending fashion shows, photoshoots, and conferences. A fashion Journalist  job is to write copy for trade and advertisement journals, fashion magazines, newspapers, and online fashion forums about style and fashion.

    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
    Corporate Executive

    Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.

    2 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
    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
    Production Manager

    Production Manager Job Description: A Production Manager is responsible for ensuring smooth running of manufacturing processes in an efficient manner. He or she plans and organises production schedules. The role of Production Manager involves estimation, negotiation on budget and timescales with the clients and managers. 

    Resource Links for Online MBA 

    3 Jobs Available
    QA Manager

    Quality Assurance Manager Job Description: A QA Manager is an administrative professional responsible for overseeing the activity of the QA department and staff. It involves developing, implementing and maintaining a system that is qualified and reliable for testing to meet specifications of products of organisations as well as development processes. 

    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
    Reliability Engineer

    Are you searching for a Reliability Engineer job description? A Reliability Engineer is responsible for ensuring long lasting and high quality products. He or she ensures that materials, manufacturing equipment, components and processes are error free. A Reliability Engineer role comes with the responsibility of minimising risks and effectiveness of processes and equipment. 

    2 Jobs Available
    Safety Manager

    A Safety Manager is a professional responsible for employee’s safety at work. He or she plans, implements and oversees the company’s employee safety. A Safety Manager ensures compliance and adherence to Occupational Health and Safety (OHS) guidelines.

    2 Jobs Available
    Corporate Executive

    Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.

    2 Jobs Available
    ITSM Manager

    ITSM Manager is a professional responsible for heading the ITSM (Information Technology Service Management) or (Information Technology Infrastructure Library) processes. He or she ensures that operation management provides appropriate resource levels for problem resolutions. The ITSM Manager oversees the level of prioritisation for the problems, critical incidents, planned as well as proactive tasks. 

    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
    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
    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
    IT Consultant

    An IT Consultant is a professional who is also known as a technology consultant. He or she is required to provide consultation to industrial and commercial clients to resolve business and IT problems and acquire optimum growth. An IT consultant can find work by signing up with an IT consultancy firm, or he or she can work on their own as independent contractors and select the projects he or she wants to work on.

    2 Jobs Available
    Data Architect

    A Data Architect role involves formulating the organisational data strategy. It involves data quality, flow of data within the organisation and security of data. The vision of Data Architect provides support to convert business requirements into technical requirements. 

    2 Jobs Available
    Security Engineer

    The Security Engineer is responsible for overseeing and controlling the various aspects of a company's computer security. Individuals in the security engineer jobs include developing and implementing policies and procedures to protect the data and information of the organisation. In this article, we will discuss the security engineer job description, security engineer skills, security engineer course, and security engineer career path.

    2 Jobs Available
    UX Architect

    A UX Architect is someone who influences the design processes and its outcomes. He or she possesses a solid understanding of user research, information architecture, interaction design and content strategy. 

     

    2 Jobs Available
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