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MCQEasyJEE 2025Electric Current & Drift Velocity

JEE Physics 2025 Question with Solution

A motor operating on 100 V100 \, \text{V} draws a current of 1 A1 \, \text{A}. If the efficiency of the motor is 91.6%91.6\%, then the loss of power in units of cal/s is

  • A

    44

  • B

    8.48.4

  • C

    22

  • D

    6.26.2

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Voltage = 100 V100 \, \text{V}, current = 1 A1 \, \text{A}, efficiency = 91.6%=0.91691.6\% = 0.916.

Find: Loss of power in cal/s.

First calculate the input power:

Pinput=VI=100×1=100 WP_{\text{input}} = VI = 100 \times 1 = 100 \, \text{W}

Using efficiency,

η=PoutputPinput\eta = \frac{P_{\text{output}}}{P_{\text{input}}}

so

Poutput=ηPinput=0.916×100=91.6 WP_{\text{output}} = \eta P_{\text{input}} = 0.916 \times 100 = 91.6 \, \text{W}

Hence the power loss is

Ploss=Pinput−Poutput=100−91.6=8.4 WP_{\text{loss}} = P_{\text{input}} - P_{\text{output}} = 100 - 91.6 = 8.4 \, \text{W}

Now convert watts to calories per second using

1 cal=4.184 J1 \, \text{cal} = 4.184 \, \text{J}

and

1 W=1 J/s=14.184 cal/s1 \, \text{W} = 1 \, \text{J/s} = \frac{1}{4.184} \, \text{cal/s}

Therefore,

Ploss=8.4×14.184≈2.0076 cal/sP_{\text{loss}} = 8.4 \times \frac{1}{4.184} \approx 2.0076 \, \text{cal/s}

Rounding to the nearest whole number, the loss of power is 2 cal/s2 \, \text{cal/s}. Therefore, the correct option is C.

Unit Conversion Approach

Given: Input voltage = 100 V100 \, \text{V}, current = 1 A1 \, \text{A}, efficiency = 91.6%91.6\%.

Find: Power lost in cal/s.

  1. Input power:
Pinput=V×I=100×1=100 WP_{\text{input}} = V \times I = 100 \times 1 = 100 \, \text{W}
  1. Output power from efficiency:
η=PoutputPinput×100%\eta = \frac{P_{\text{output}}}{P_{\text{input}}} \times 100\% Poutput=91.6×100100=91.6 WP_{\text{output}} = \frac{91.6 \times 100}{100} = 91.6 \, \text{W}
  1. Power loss:
Ploss=Pinput−Poutput=100−91.6=8.4 WP_{\text{loss}} = P_{\text{input}} - P_{\text{output}} = 100 - 91.6 = 8.4 \, \text{W}
  1. Use the conversion
1 W=0.239 cal/s1 \, \text{W} = 0.239 \, \text{cal/s}

So,

Ploss=8.4×0.239=2.0076 cal/sP_{\text{loss}} = 8.4 \times 0.239 = 2.0076 \, \text{cal/s}

Thus the loss of power is approximately 2 cal/s2 \, \text{cal/s}. The correct option is C.

Common mistakes

  • Using efficiency as the fraction of power lost. Efficiency is PoutputPinput\frac{P_{\text{output}}}{P_{\text{input}}}, not PlossPinput\frac{P_{\text{loss}}}{P_{\text{input}}}. First find output power from efficiency, then subtract from input power.

  • Forgetting to convert from watts to cal/s. The loss is first obtained as 8.4 W8.4 \, \text{W}, but the question asks for cal/s. Use 1 cal=4.184 J1 \, \text{cal} = 4.184 \, \text{J} or equivalently 1 W=0.239 cal/s1 \, \text{W} = 0.239 \, \text{cal/s}.

  • Treating 91.6%91.6\% as 91.691.6 instead of 0.9160.916 in calculations. Percentage must be converted into decimal form before multiplying by input power.

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