SSC JE 2014 Electrical question paper with solution(Evening-shift)

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SSC JE electrical paper 2014 Evening shift with explanation

Ques1. A stove element draws 15A when connected to 230 V line. How long does it take to consume one unit of energy?

3.45 h

2.16 h

1.0 h

0.29 h✔

P = E/T

Where E is the energy and T is time

T = E/P

1 Unit of energy is given as 1 Kwhr = 1000 Watt-hour

T = 1000/230 x 15 [since P = VI]

T = 0.29 hour

Ques 2. The R_{eq} for the circuit shown in figure is

14.4 Ω

14.57 Ω✔

15.27 Ω

15.88 Ω

In the above figure 6Ω and 3Ω are parallel therefore

(6 x 3)(6 + 3) = 2Ω

Now 2Ω and 2Ω, as well as 1Ω and 5Ω, are in series therefore

2Ω + 2Ω = 4Ω

5Ω + 1Ω = 6Ω

Ques3. The SI unit of conductivity is

Ohm-m

Ohm/m

mho-m

mho/m✔

Conductivity (or specific conductance) is a measure of its ability to conduct electricity. The SI unit of conductivity is mho per meter.

Ques 4. Calculate the voltage drop across 14.5 Ω Resistance

14.5 V

18 V

29 V✔

30.5 V

In the given figure the resistances are in series, so the same current I will flow across the three resistors.

Requ = 14.5Ω + 25.5Ω + 60Ω = 100Ω

Current I = V/R

= 200/100 = 2A

Voltage drop across the 14.5Ω resistance

V = IR

= 2 x 14.5 = 29 V

Ques 5. For the network shown in the figure, the value of current in 8Ω resistors is

4. 8 A

2.4 A✔

1.5 A

1.2 A

The resistance between the point A and C will be

R_{AB} || (R_{AC} + R_{BC})

AB = 20 x(12 + 8)/(20 + 12 + 8) = 10Ω

I = V/R = 48/10 = 4.8 A

Since AC and BC are parallel with AB, therefore, the current will divide equally into both side

hence the current through the Branch ABC is = I/2

= 4.8/2 = 2.4 A

Ques 6. A piece of oil-soaked paper has been inserted between the plates of a parallel plate capacitor. Then the potential difference between the plates will

Increase

Decrease✔

Remain Unaltered

Become Zero

The paper sheet is the poor conductor of electricity so it does not allow the flow of electric current or electric charges between two parallel plates. However, the paper sheet allows electric field through it. Therefore, the paper sheet placed between the parallel plates acts as the barrier for the electric current. Hence the potential difference between the plates will decrease.

Ques 7. The current drawn by a tungsten filament lamp is measured by an ammeter. The ammeter reading under steady-state condition will be ________ the ammeter reading when the supply is switched on

Same

Less✔

Greater

Double

When a filament lamp is cold(when supply is switched off), it behaves differently than when it’s at its normal operating temperature. The resistance of the filament of the bulb increases with increase in temperature. Hence when it glows its resistances is greater than when it is cold. and in the process the lamp current changes. In the case of a cold incandescent lamp (a filament lamp), the inrush current peaks at approximately ten times the steady-state operating current, but it lasts a relatively short duration, no more than a few cycles.

Therefore the ammeter reading under the steady-state condition is less than then ammeter reading when the supply is switched on.

Ques 8. Tesla is same as

Weber/meter

Weber/(meter)^{2}✔

Greater than

Double

The Tesla (symbol T) is the SI derived unit used to measure magnetic fields. One tesla is equal to one weber per square meter.

Ques 9. The unit of volume of resistivity is

Ohm-m^{3}/m^{2}

Ohm-m^{2}/m✔

Ohm-gram-m/gram

Ohm-m^{4}/m^{3}

Resistivity is the resistance of a unit volume of a material. In the metric system, the unit of length is the meter, and the area is the square meter. Thus, resistivity is measured in units of Ohm – meters squared per meter (Ohm-m^{2}/m), often abbreviated as Ohm-m.

Ques 10. Four resistance 2Ω, 4Ω, 5Ω, 20Ω are connected in parallel. Their combined resistance is

1Ω✔

2Ω

4Ω

5Ω

Since the resistance is connected in parallel therefore

1/ Req = 1/2Ω + 1/4Ω + 1/5Ω + 1/20Ω

=10 + 5 + 4 + 1/20

Req= 1Ω

Note:- The total equivalent resistor connected in parallel is always less than the smallest given resistance. Here in the question, the smallest resistance is 2Ω and in the given option only option 1 i.e 1 Ω is less than the smallest resistance

so in the above case you can directly find the answer without any calculation.

## Leave a Reply

10 Comments on "SSC JE 2014 Electrical question paper with solution(Evening-shift)"

Next pages are nor open

Next pages are not open

Sorry for the inconvenience sai..but all the pages are working fine

Very very good

Sir q17 need full explanation..plz help me to understand the same

15 question value of voltage…. after solving is wrong

corrected @chetan

Explanation is correct but you didn’t put the correct option sir…. it’s B in 98

question 18 is wrng.. check out..

but site really helpful thanku

Answer is right @sanjay Kumar but if you know different approach to solve the problem please share with us