A metallic wire having a resistance of 20 is cut into two eal parts in length. These parts are then connected in parallel. The resistance of this parallel combination is eal to

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Q: 79 (NDA-II/2020)
A metallic wire having a resistance of 20 is cut into two eal parts in length. These parts are then connected in parallel. The resistance of this parallel combination is eal to

question_subject: 

Science

question_exam: 

NDA-II

stats: 

0,6,5,1,4,6,0

When a metallic wire with a resistance of 20 is cut into two equal parts, each part will have a resistance of 10. When these two parts are connected in parallel, the total resistance of the parallel combination can be calculated using the formula 1/Rtotal = 1/R1 + 1/R2, where Rtotal is the total resistance and R1 and R2 are the resistances of the two parts.

Substituting the values, we get:

1/Rtotal = 1/10 + 1/10

Combining the fractions, we have:

1/Rtotal = 2/10

Multiplying both sides by 10, we get:

10/Rtotal = 2

Dividing both sides by 2, we get:

Rtotal = 5

Therefore, the resistance of the parallel combination is equal to 5.

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